Class 12

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Students can Download Commerce Chapter 11 Employee Selection Process Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Commerce Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Samacheer Kalvi 12th Commerce Employee Selection Process Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The recruitment and Selection Process aimed at right kind of people.
(a) At right people
(b) At right time
(c) To do right things
(d) All of the above
Answer:
(d) All of the above

Convert CGPA into Percentage out of 4 … You must divide the percentage by 25 to get the percentage

Question 2.
The poor quality of selection will mean extra cost on _________ and supervision.
(a) Training
(b) Recruitment
(c) work quality
(d) None of these
Answer:
(a) Training

Question 3.
_________ refers to the process of identifying and attracting job seekers so as to build a pool of qualified job applicants.
(a) Selection
(b) Training
(c) Recruitment
(d) Induction
Answer:
(c) Recruitment

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 4.
Selection is usually considered as a _________ process.
(a) Positive
(b) Negative
(c) Natural
(d) None of these
Answer:
(b) Negative

Question 5.
Which of the following test is used to measure the various characteristics of the candidate?
(a) physical Test
(b) Psychological Test
(c) attitude Test
(d) Proficiency tests
Answer:
(b) Psychological Test

Question 6.
Wfifich of the following orders is followed in a typical selection process? .
(a) application form test and or interview, reference check and physical examination
(b) Application form test and or interview, reference check, and physical examination
(c) Reference check, application form, test and interview and physical examination
(d) physical examination test and on interview application term and reference check.
Answer:
(b) Application form test and or interview, reference check, and physical examination

Question 7.
The purpose of an application blank is to gather information about the
(a) Company
(b) Candidate
(c) Questionnaire or Interview Schedule
(d) Competitors
Answer:
(b) Candidate

Question 8.
Identify the test that acts as an instrument to discover the inherent ability of a candidate.
(a) Aptitude Test
(b) Attitude Test
(c) Proficiency Test
(d) Physical Test
Answer:
(a) Aptitude Test

Question 9.
The process of eliminating unsuitable candidate is called _________
(a) Selection
(b) Recruitment
(c) Interview
(d) Induction
Answer:
(a) Selection

Question 10.
Scrutiny of application process is the _________
(a) Last step in Selection process
(b) First step in Selection process
(c) Third step in Selection Process Selection process
(d) None of the above
Answer:
(b) First step in Selection process

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 11.
Scrutiny of application process is the
(a) Locating candidates
(b) Determinining the suitable of the candidates
(c) preparing employees for training
(d) None of the above
Answer:
(b) Determinining the suitable of the candidates

Question 12.
The process of placing the right man on the right job is called _________
(a) Training
(b) Placement
(c) Promotion
(d) Transfer
Answer:
(b) Placement

Question 13.
Probation/Trial period signifies _________
(a) one year to two years
(b) One year to three years
(c) Two years to four years
(d) None of the above
Answer:
(a) one year to two years

Question 14.
Job first man next is one of the principles of _________
(a) Test
(b) Interview
(c) Training
(d) placement
Answer:
(d) placement

II. Very Short Answer Questions

Question 1.
What is selection?
Answer:
Selection is the process of choosing the most suitable person for the vacant position in the organization.

Question 2.
What is an interview?
Answer:
According to Scott and others “an interview is a purpose full exchange of ideas, the answering of questions and communication between two or more persons.”

Question 3.
What is intelligence test?
Answer:
Intelligence tests are one of the psychological tests, that is designed to measure a variety of mental ability, individual capacity of a candidate.

Question 4.
What do you mean by test?
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualification to fit into various positions in the organization.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 5.
What do you understand about bio data?
Answer:
Most of the Public sector undertakings Recruitment boards supply applications forms for selection of jobs. In this form, the candidate fill the details about family background, educational qualifications, experience, co-curricular activities.

Question 6.
What do you mean by placement?
Answer:
Placement is a process of assigning a specific job to each and every candidate selected. It includes initial assignment of new employees and promotion, transfer or demotion of present employees.

III. Short Answer Questions

Question 1.
What is stress interview?
This type of interview is conducted to test the temperament and emotional balance of the candidate interviewed. Interviewer deliberately creates stressful situation and tries to assess the suitability of the candidate by observing his reaction and response to the stressful situations.

Question 2.
What is structured interview?
Answer:
Under this method, a series of questions to be asked by the interviewer are pre-prepared by the interviewer and only these questions are asked in the interview.

Question 3.
Name the types of selection test?
Answer:
Selection tests are of two types: Ability Tests and Personality Tests. Ability tests can further be divided into: aptitude test, achievement test, intelligence test, and judgement test. Personality tests can further be divided into: interest test, personality inventory test, projective test or thematic appreciation test, and attitude test.

Question 4.
What do you mean by achievement test?
Answer:
This test measures a candidate’s capacity to achieve in a particular field. In other words this test measures a candidate’s level of skill in certain areas, accomplishment and knowledge in a particular subject. It is also called proficiency test.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 5.
Why do you think the medical examinations of a candidate is necessary?
Answer:
The last technique used in selection process is medical examination. This is the most important step in the selection because a person of poor health cannot work competently and any investment on him may go waste, if he/she is unable to discharge duties efficiently on medical grounds.

Question 6.
What is aptitude test?
Answer:
Aptitude test is a test to measure suitability of the candidates for the post/role. It actually measures whether the candidate possess a set of skills required to perform a given job. It helps in predicting the ability and future performance of the candidate.

Question 7.
How is panel interview conducted?
Answer:
Where a group of people interview the candidate, it is called panel interview. Usually panel comprises chair person, subject expert, psychological experts, representatives of minorities/ underprivileged groups, nominees of higher bodies and so on. All panel members ask different types of questions on general areas of specialization of the candidate.

Question 8.
List out the various selection interviews.
Answer:
Interview represents a face to face interaction between the interviewer and interviewee

  1. Preliminary Interview
  2. Structured Interview
  3. Unstructured Interview
  4. In-depth Interview
  5. Panel Interview
  6. Stress Interview
  7. Telephone Interview
  8. Online Interview
  9. Group interview
  10. Video Conference Interview

Question 9.
List out the significance of placement.
Answer:
The significance of the placement is as follows:

  1. It improves employee morale.
  2. It helps in reducing employee turnover.
  3. It helps in reducing conflict rates or accidents.
  4. It avoids misfit between the candidates and the job.
  5. It helps the candidate to work as per the predetermined objectives of the organization.

IV. Long Answer Questions

Question 1.
Briefly explain the various types of tests.
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualifications.

A. Ability Test: Ability test may be divided into:

  • Aptitude Test: Aptitude test is a test to measure suitability of the candidates for the post.
  • Achievement Test: This test measures a candidate’s capacity to achieve in a particular field.
  • Intelligence Test: Intelligence test is designed to measure a variety of mental ability, individual capacity of a candidate.
  • Judgment Test: This test is conducted to test the presence of mind and reasoning capacity of the candidates.

B. Personality Test: It refers to the test conducted to find out the non-intellectual traits of a candidate. It can be further divided into:

  • Interest Test: Interest test measures a candidate’s extent of interest in a particular area.
  • Projective Test: This test measures the candidate’s values, personality of the candidate.
  • Attitude Test: measures candidate’s tendencies towards the people, situation, action and related things.

Question 2.
Explain the important methods of interview.
Interview means a face to face interaction between the interviewer and interviewee. Interview may be of various types:-

  1. Preliminary Interview: It is conducted to know the general suitability of the candidates who have applied for the job.
  2. Structured Interview: In this method, a series of questions is to be asked by the interviewer. The questions may be pre-prepared.
  3. In depth Interview: This interview is conducted to test the level of knowledge of the interviewee in a particular field.
  4. Panel Interview: Where a group of people interview the candidate. The panel usually comprises the chair person, subject expert, psychological experts and so on.
  5. Stress Interview: This type of interview is conducted to test the temperament and emotional balance of the candidate.
  6. Online Interview: Due to tremendous growth in information and communication technology, interviews are conducted by means of internet via Skype, Google duo, Whatsapp.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 3.
Explain the principles of placement.
Answer:
The following are the principles of placement:

  1. Job First, Man Next: Man should be placed on the job according to the requirements of the job.
  2. Job Offer: The job should be offered to the man based on his qualification..
  3. Terms and conditions: The employee should be informed about the terms and conditiqns of the organisation.
  4. Aware about the Penalties: The employee should also be made aware of the penalties if he / she commits a mistake.
  5. Loyalty and Co-operation: When placing a person in a new job, an effort should be made to develop a sense of loyalty and co-operation in him.

Samacheer Kalvi 12th Commerce Employee Selection Process Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
The types of Aptitude Test are
(i) Numerical Reasoning Test
(ii) Attitude Test
(iii) Vocabulary Test
(iv) Interest Test
(a) (i) and (ii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(b) (i) and (iii)

Question 2.
Which one of the following is not correctly matched?

(a) In-depth interview Skype
(b) Online interview Group of people interview the candidate
(c) Video conferencing interview Level of knowledge
(d) Telephone interview Face to face interview

Answer:
(d) Telephone interview – Face to face interview

II. Very Short Answer Questions

Question 1.
What is meant by personality test?
Answer:
Personality test refers to the test conducted to find out the non-intellectual traits of a candidate namely temperament, emotional response, capability and stability.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 2.
Write a note on attitude test.
Answer:
Attitude test measures candidate’s tendencies towards the people, situation, action and related things. For example: morale study, values study, etc.

III. Short Answer Questions

Question 1.
What do you mean by online interview?
Answer:
Due to tremendous growth in information and communication technology, these days interviews are conducted by means of internet via Skype. We chat, Google duo, Viber, Whatsapp or Video chat applications. This enables the interviewers to conduct interview with the candidates living in faraway places.

Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

Students can Download Physics Chapter 6 Optics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

Samacheer Kalvi 12th Physics Optics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions 

Question 1.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 2.
A rod of length 10 cm lies along the principal axis of a concav e mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, (AIPMT Main 2012)
(a) 2.5 cm
(b) 5cm
(c) 10 cm
(d) 15cm
Answer:
(b) 5cm
Hint:
By mirror formula, image distance of A
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) ; \(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
\(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) + \(\frac { 1 }{ (-30) }\) = \(\frac { 1 }{ (-10) }\)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-1
∴ vA = – 15 cm
Image distance of C, vc = – 20 cm
The length of image = |vA – vc|
= |-15 + 20| = 5 cm

The full form of ktm is Kraftfahrzeuge Trunkenpolz Mattighofen. It translates into English to mean “motor Vehicle”.

Question 3.
An object is placed in front of a convex mirror of focal length of/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (IEE Main 2009)]
(a) 2ƒ and c
(b) c and ∞
(c) ƒ and O
(d) None of these
Answer:
(d) None of these
Hint:
There is no maximum minimum object distance for convex mirror to form real and inverted image.

SamacheerKalvi.Guru

Question 4.
For light incident from air onto a slab of refractive index 2. Maximum possible angle of refraction is,
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) 30°
Hint:
From Snell’s law, µ = \(\frac { sin i }{ sin r }\)
Now consider an angle of incident is 90°
\(\frac { sin 90° }{ 2 }\)
sin r = sin-1 (0.5)
r = 30°

Question 5.
If the velocity and wavelength of light in air is Va and λa and that in water is Va and λw, then the refractive index of water is,
(a) \(\frac { { V }_{ w } }{ { V }_{ a } } \)
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)
(c) \(\frac { { λ }_{ w } }{ { λ }_{ a } } \)
(d) \(\frac { { V }_{ a }{ \lambda } }{ { V }_{ w }{ \lambda }_{ w } } \)
Answer:
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)

Question 6.
Stars twinkle due to
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction

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Question 7.
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index,
(a) less than one
(b) less than that of glass
(c) greater than that of glass
(d) equal to that of glass
Answer:
(d) equal to that of glass
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ } \) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Biconvex lens dipped in a liquid, acts as a plane sheet of glass, ƒ = ∞; \(\frac { 1 }{ ƒ } \) = 0
\(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) -1 = 0;  n2 = n1

Question 8.
The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be,
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm
Answer:
(b) 10 cm
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ }\) = (n – 1) \(\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
R1 = ∞; R2 = -6
∴ ƒ = \(\frac { R }{ \left( n – 1 \right) } \) (R= 10 cm; n= 1.5
ƒ = \(\frac { 10}{ \left( 1.5 – 1 \right) } \) =20 cm

Question 9.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint:
Let d1 = 5 cm and d2 = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d1n + d2 n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 10.
A ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45°. The ray can undergo total internal reflection for the following n,
(a) n= 1.25
(b) n = 1.33
(c) n= 1.4
(d) n= 1.5
Answer:
(d) n= 1.5
Hint:
For total internal reflection
sin i > sin c            where, i – angle of incidence
But, sin c = 1/n      c – critical angle
sin i > 1/n
n > \(\frac { 1 }{ sin i }\)
n > \(\frac { 1 }{ sin 45 }\) ; n > √2 ; n > 1.414

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Question 11.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Answer:
(d) violet
Hint:
Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.

Question 12.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1 m
(b) 5 m
(c) 3 m
(d) 6m
Answer:
(b) 5 m
Hint:
Resolution limit sin θ = \(\frac { Y }{ d }\) = \(\frac { 1.22c }{ d }\)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-2

Question 13.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
(a) 2D
(b) \(\frac { D }{ 2 }\)
(c) √2D
(d) \(\frac { D }{ √2 }\)
Answer:
(a) 2D
Hint:
Young’s double -slite experiment is
β = \(\frac { λD }{ d}\) ; β’ = \(\frac { λD’ }{ d’}\) ; d’ = 2d
Same fringe space, β = β’
⇒ \(\frac { λD }{ d}\) = \(\frac { λD’ }{ d’}\) ; D’ = 2D

Question 14.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
(a) 5I and I
(b) 5I and 3I
(c) 9I and I
(d) 9I and 3I
Answer:
(c) 9I and I
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-3

Question 15.
When light is incident on a soap film of thickness 5 x 10-5
cm, the wavelength of light reflected maximum in the visible region is 5320 A. Refractive index of the film will be,
(a) 1.22
(b) 1.33
(c) 1.51
(cl) 1.83
Answer:
(b) 1.33
Hint.
The condition for constructive interference, (for reflection)
2µ tcos r = (2n +1) \(\frac { λ }{ 2 }\) [∴ cos r = 1]
µ = \(\frac { \left( 2n+1 \right) \lambda }{ 4t } \)
For visible region, n = 2
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-4

Question 16.
First diffraction minimum due to a single slit of width 1.0 x 10-5 cm is at 30°. Then wavelength of light used is,
(a) 400 Å
(b) 500 Å
(c) 600 Å
Answer:
(b) 500 Å
Hint.
For diffraction minima, d sin θ = nλ
λ = \(\frac { dsin\theta }{ n } =\frac { 1\times { 10 }^{ -5 }\times { 10 }^{ -2 }\times sin30° }{ 1 } \) = 0.5 x 10-7
λ = 500 Å

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Question 17.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) \(\frac { 3 }{ 2 }\)
(c) \(\sqrt { \frac { 3 }{ 2 } } \)
(d) 2
Answer:
(a) √3
Hint.
Angle of refraction r = 60° ; Angle of incident i = 30°
sin i =n x sin r
n = \(\frac {sin 30°}{ sin 60°} \) = √3

Question 18.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-5
(a) get shifted downwards
(b) get shifted upwards
(c) will remain the same
(d) data insufficient to conclude
Answer:
(b) get shifted upwards

Question 19.
Light transmitted by Nicol prism is,
(a) partiallypolarised
(b) unpolarised
(c) plane polarised
(d) elliptically polarised
Answer:
(c) plane polarised

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Question 20.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation

Samacheer Kalvi 12th Physics Optics Short Answer Questions

Question 1.
State the laws of reflection.
Answer:
(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(b) The angle of incidence i is equal to the angle of reflection r. i = r

Question 2.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, d= 180 – (i + r).
As, i = r in reflection, we can write angle of deviation ‘ in reflection at plane surface as, d = 180 – 2i

Question 3.
Give the characteristics of image formed by a plane mirror.
Answer:

  1. The image formed by a plane mirror is virtual, erect, and laterally inverted.
  2. The size of the image is equal to the size of the object.
  3. The image distance far behind the mirror is equal to the object distance in front of it.
  4.  If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as, n = \(\left( \frac { 360 }{ \theta } -1 \right) \)

Question 4.
Derive the relation between ƒ and R for a spherical mirror.
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection. If MP is the perpendicular from M on the principal axis, then from the geometry, The angles ∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-6
\(\frac { 1 }{ f }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
tan i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
As the angles are small, tan i ≈ i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
Simplifying further, 2 = \(\frac { PM }{ PC }\) and tan = \(\frac { PM }{ PF }\) ;2PF = PC
PF is focal length/and PC is the radius of curvature R.
2 ƒ= R (or) ƒ = \(\frac { R }{ 2 }\)
ƒ = \(\frac { R }{ 2 }\) is the relation between ƒ and R.

Question 5.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

  1. The Incident light is taken from left to right (i.e. object on the left of mirror).
  2. All the distances are measured from the pole of the mirror (pole is taken as origin).
  3. The distances measured to the right of pole along the principal axis are taken as positive.
  4. The distances measured to the left of pole along the principal axis are taken as negative.
  5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
  6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

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Question 6.
What is optical path? Obtain the equation for optical path of a medium of thickness d and refractive index n.
Answer:
Optical path of a medium is defined as the distance d’ light travels in vacuum in the same time it travels a distance d in the medium.
Let us consider a medium of refractive index n and thickness d. Light travels with a speed v through the medium in a time t. Then we can write,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-7
v = \(\frac { d }{ t }\); rewritten as, t = \(\frac { d }{ v }\)
In the same time, light can cover a greater distance d’
in vacuum as it travels with greater speed c in vacuum.
Then we have,
c = v = \(\frac { d’ }{ t }\); rewritten as, t = \(\frac { d’ }{ c }\)
As the time taken in both the cases is the same, we can equate the time t as,
\(\frac { d’ }{ c }\) = \(\frac { d }{ v }\)
rewritten for the optical path d’ as d’ = \(\frac { c }{ v }\)d
As, \(\frac { c }{ v }\) = n ; The optical path d’ is, d’ = nd
As n is always greater than 1, the optical path d’ of the medium is always greater than d.

Question 7.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of refractive index of the second medium n2 to that of the refractive index of the first medium n1.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \).

Question 8.
What is angle of deviation due to refraction?
Answer:
The angle between the incident and deviated light is called angle of deviation. When light travels from rarer to denser medium it deviates towards normal. The angle of deviation in this case is, d = i – r

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Question 9.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 10.
What is relative refractive index?
Answer:
Snell’s law, the term \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \) is called relative refractive index of second medium with respect to the first medium which is denoted as (n21). n21 = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)

Question 11.
Obtain the equation for apparent depth.
Answer:
Light from the object O at the bottom of the tank passes from denser medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is n1 and rarer medium is n2. Here, n1 > n2. The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines NN’ and OD are parallel. Thus angle ∠DIB is also r. The angles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell’s law in product form for this refraction is,
n1 sin i = n2 sin r
As the angles i and r are small, we can approximate, sin i ≈ tan i;
n1 tan i = n2 tan i
In triangles ∆DOB and ∆DIB,
tan(i) = \(\frac { DB }{ DO }\) and tan(r) = \(\frac { DB }{ DI }\)
n1 = \(\frac { DB }{ DO }\) n2 = \(\frac { DB }{ DI }\)
DB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d’.
Rearranging the above equation for the apparent depth d’,
d’ = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)d
As the rarer medium is air and its refractive index n2 can be taken as 1, (n2 = 1). And the refractive index n1 of denser medium could then be taken as n, (n1 = n).
In that case, the equation for apparent depth becomes,
d = \(\frac { d }{ n }\)

Question 12.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 13.
What is critical angle and total internal reflection?
Answer:
The angle of incidence in the denser medium for which the refracted ray graces the boundary is called critical angle ic.
The entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.

Question 14.
Obtain the equation for critical angle.
Answer:
Snell’s law in the product form, equation for critical angle incidence becomes,
n1 sini ic = n2 sin 90°
n1 sini ic = n2 (∵ sin 90° = 1)
sini ic = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)
Here, n1 > n2
If the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n2 = 1) and (n1= n).
sini ic = \(\frac { 1 }{ n }\) (or) ic = sin-1 \(\left( \frac { 1 }{ n } \right) \)

Question 15.
Explain the reason for glittering of diamond.
Answer:
Diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which is about only 1.5. The critical angle of diamond is about 24.4°. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4° to 90° inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamond.

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Question 16.
What are mirage and looming?
Answer:
Mirage: Mirage takes place in hot regions. The light from distant objects appears to be reflected from ground. For mirage to form refractive index goes on increasing as we go up. Looming: Looming takes place in cold regions. The light from distant objects appears to be flying. For looming to form refractive index goes on decreasing.

Question 17.
Write a short notes on the prisms making use of total internal reflection.
Answer:
Prisms can be designed to reflect light by 90° or by 180° by making use of total internal reflection. The critical angle ic for the material of the prism must be less than 45°. This is true for both crown glass and flint glass. Prisms are also used to invert images without changing their size.

Question 18.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell’s window.

Question 19.
Write a note on optical fibre.
Answer:
Transmitting signals through optical fibres is possible due to the phenomenon of total internal reflection. Optical fibres consists of inner part called core and outer part called cladding (or) sleeving. The refractive index of the material of the core must be higher than that of the cladding for total internal reflection to happen. Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. Hence, it undergoes repeated total internal reflections along the length of the fibre without undergoing any refraction.

Question 20.
Explain the working of an endoscope.
Answer:
An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient’s body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.

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Question 21.
What are primary focus and secondary focus of convex lens?
Answer:
Primary focus: The primary focus F1 is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
Secondary focus: The secondary focus F2 is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.

Question 22.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).
(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 23.
Arrive at lens equation from lens maker’s formula.
Answer:
From refraction through a double convex lens, the relation between the object distance u, image distance v1 and radius of curvature R1 as
\(\frac { { \mu }_{ 2 } }{ { v }_{ 1 } } -\frac { { \mu }_{ 1 } }{ u } =\frac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ { R }_{ 1 } } \) …… (1)
The relation between the object distance image distance v1 and radius of curvature R2 can be
\(\frac { { \mu }_{ 1 } }{ { v } } -\frac { { { \mu }_{ 2 } } }{ v_{ 1 } } =\frac { { \mu }_{ 1 }-{ \mu }_{ 2 } }{ { R }_{ 2 } } \) …… (2)
Adding equation (1) and (2)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-8
If the object is placed at infinity (µ = ∞), the image will be formed at the focus. i.e.v = ƒ
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-9
This is len’s maker’s formula. When the lens is placed in air µ1 = 1 and µ2 = µ.
Equation (4) becomes,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-10
From equation (3) and (4), we have \(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)
=\(\frac { 1 }{ ƒ }\)
This is the len’s equation.

Question 24.
Obtain the equation for lateral magnification for thin lens.
Answer:
Lateral magnification in terms of u and ƒ.
The thin lens formula is
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\)
Multiplying on both sides by ‘u’
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-11
In terms of v and ƒ multiplying by v, we get
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-12
Hence, lateral magnification for thin lens,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-13

Question 25.
What is power of a lens?
Answer:
The power of a lens P is defined as the reciprocal of its focal length.
p = \(\frac { 1 }{ ƒ }\)
The unit of power is diopter D.

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Question 26.
Derive the equation for effective focal length for lenses in contact.
Answer:
Consider a two thin lenses in contact. In the absence of second lens L2, the first lens L1 will form a real image I’. Using thin lens formula.
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v’ }\)–\(\frac { 1 }{ u }\) ….. (1)
The image I’ acts as a virtual object (u = v’) for the second lens L2 which finally forms its real image I at distance v. Thus
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ v’ }\) ….. (2)
Adding equation (1) and (2) we get,
\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) ….. (3)
For the combination of thin lenses in contact, if ‘f’ is the equivalent focal length, then
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\) ….. (4)
From equations (3) and (4), the effective focal length for lenses in contact.
\(\frac { 1 }{ ƒ }\)=\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)

Question 27.
What is angle of minimum deviation?
Answer:
The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation.

Question 28.
What is dispersion?
Answer:
The phenomenon of spliting of white light into its component colours on passing through a refracting medium is called dispersion.

Question 29.
How are rainbows formed?
Answer:
Rainbow is formed by dispersion of sunlight into its constituent colours by raindrops which disperse sunlight by refraction and deviate the colours by total internal reflection.

Question 30.
What is Rayleigh’s scattering?
Answer:
The scattering of light by particles in a medium without a change in wavelength is called as Rayleigh’s scattering.

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Question 31.
Why does sky appear blue?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \) So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appears blue.

Question 32.
What is the reason for reddish appearance of sky during sunset and sunrise?
Answer:
During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red.

Question 33.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 34.
What are the salient features of corpuscular theory of light?
Answer:

  • According this theory, light is emitted as tiny, massless (negligibly small mass) and perfectly elastic particles called corpuscles.
  • As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time.
  • On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index.
  • The energy of light is the kinetic energy of these corpuscles. When these corpuscles impinge on the retina of the eye, the vision is produced.
  • The different size of the corpuscles is the reason for different colours of light.
  • When the corpuscles approach a surface between two media, they are either attracted or repelled.
  • The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium.

Question 35.
What is wave theory of light?
Answer:
Light is a disturbance from a source that travels as longitudinal mechanical waves through the either medium that was presumed to pervade all space as mechanical wave requires medium for its propagation. The wave theory could successfully explain phenomena of reflection, refraction, interference and diffraction of light.

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Question 36.
What is electromagnetic wave theory of light?
Answer:
Electromagnetic wave theory:
Maxwell (1864) proved that light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. He could also show that no medium is necessary for the propagation of electromagnetic waves. All the phenomenon of light could be successfully explained by this theory.

Question 37.
Write a short note on quantum theory of light.
Answer:
Albert Einstein (1905), endorsing the views of Max Plank (1900), was able to explain photoelectric effect in which light interacts with matter as photons to eject the electrons. A photon is a discrete packet of energy. Each photon has energy E of,
E = hv
Where, h is Plank’s constant (h = 6.625 x 10-34J s) and v is frequency of electromagnetic wave. As light has both wave as well as particle nature it is said to have dual nature. Thus, it is concluded that light propagates as a wave and interacts with matter as a particle.

Question 38.
What is a wave front?
Answer:
A wavefront is the locus of points which are in the same state or phase of vibration.

Question 39.
What is Huygens’ principle?
Answer:
According to Huygens principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

Question 40.
What is interference of light?
Answer:
The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.

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Question 41.
What is phase of a wave?
Answer:
Phase is a particular point in time on the cycle of a waveform, measured as an angle in degrees.

Question 42.
Obtain the relation between phase difference and path difference.
Answer:
Phase difference (Φ):
It is the difference expressed in degrees or radians between two waves having same frequency and referenced to same point in time.
Path difference (δ):
It is the difference between the lengths of two paths of the two different having same frequency and travelling at same velocity. δ =\(\frac { \lambda }{ 2\pi } \) Φ

Question 43.
What are coherent sources?
Answer:
Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.

Question 44.
What is intensity division?
Answer:
Intensity’ or amplitude division: If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams. They will be either in-phase or at constant phase difference.

Question 45.
How does wavefront division provide coherent sources?
Answer:
Wavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.

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Question 46.
How do source and images behave as coherent sources?
Answer:
Source and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel’s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd’s mirror uses a source and its virtual image as two coherent sources.

Question 47.
What is bandwidth of interference pattern?
Answer:
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

Question 48.
What is diffraction?
Answer:
Diffraction is bending of waves around sharp edges into the geometrically shadowed region.

Question 49.
Differentiate between Fresnel and Fraunhofer diffraction.
Answer:

S.No. Fresnel diffraction Fraunhofer diffraction
1. Spherical or cylindrical wavefront undergoes diffraction Plane wavefront undergoes diffraction
2. Light wave is from a source at finite distance Light wave is from a source at infinity
3. For laboratory conditions, convex lenses need not be used In laboratory conditions, convex lenses are to be used
4. difficult to observe and analyse Easy to observe and analyse

Question 50.
Discuss the special cases on first minimum in Fraunhofer diffraction.
Let us consider the condition for first minimum with (n = 1). a sin θ = λ
The first minimum has an angular spread of, sin θ = \(\frac { \lambda }{ a } \). Special cases to discuss on the condition.
1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1.
2. When a ≥ λ, the diffraction is possible.

  • For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°.
  • For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.

3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac { \lambda }{ a } \) = \(\frac { \lambda }{ { 2\lambda } } \) =\(\frac { 1 }{ 2 }\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

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Question 51.
What is Fresnel’s distance? Obtain the equation for Fresnel’s distance.
Answer:
Fresnel’s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
Fresnel’s distance z as, z = \(\frac { { a }^{ 2 } }{ 2\lambda } \).

Question 52.
Mention the differences between interference and diffraction.
Answer:

S.No. Interference Diffraction
1. Superposition of two waves Bending of waves around edges
2. Superposition of waves from two coherent sources. Superposition wavefronts emitted from various points of the same wavefront.
3. Equally spaced fringes. Unequally spaced fringes
4. Intensity of all the bright fringes is almost same Intensity falls rapidly for higher orders
5. Large number of fringes are obtained Less number of fringes are obtained

Question 53.
What is a diffraction grating?
Answer:
A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.

Question 54.
What are resolution and resolving power?
Answer:
Optical resolution describes the ability of an imaging system to resolve detail in the object that is being imaged. Resolving power is the ability of an optical instrument to resolve or separate the image of two nearby point objects so that they can be distinctly seen. It is equal to the reciprocal of the limit of resolution of the optical instrument.

Question 55.
What is Rayleigh’s criterion?
Answer:
The images of two point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other.

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Question 56.
What is polarisation?
Answer:
The phenomenon of restricting the vibrations of light (electric or magnetic field vector) to a particular direction perpendicular to the direction of wave propagation motion is called polarization of light.

Question 57.
Differentiate between polarised and unpolarised light.
Answer:

S.No. Polarised light Unpolarised light
1. Consists of waves having their electric field vibrations in a single plane normal to the direction of ray. Consists of waves having their electric field vibrations equally distributed in all directions normal to the direction of ray.
2. Asymmetrical about the ray direction Symmetrical about the ray direction
3. It is obtained from unpolarised light with the help of polarisers Produced by conventional light sources.

Question 58.
Discuss polarisation by selective absorption.
Answer:
Selective absorption is the property of a material which transmits w’aves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism.

Question 59.
What are polariser and analyser?
Answer:
Polariser:
The Polaroid which plane polarises the unpolarised light passing through it is called a polariser.
Analyser:
The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Question 60.
What are plane polarised, unpolarised and partially polarised light?
Answer:
Plane polarised:
If the vibrations of a wave are present in only one direction in a plane perpendicular to the direction of propagation of the wave is said to be polarised or plane polarised light.

Unpolarised:
A transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.

Partially polarised light:
If the intensity of light varies between maximum and minimum for every’ rotation of 90° of the analyser, the light is said to be partially polarised light.

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Question 61.
State and obtain Malus’ law.
Answer:
When a beam of plane polarised light of intensity I0 is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle 0 between the transmission axis of polariser and analyser. This is known as Malus’ law. I = I0 cos2θ

Question 62.
List the uses of polaroids.
Answer:
Uses of polaroids:

  1. Polaroids are used in goggles and cameras to avoid glare of light.
  2. Polaroids are useful in three dimensional motion pictures i.e., in holography.
  3. Polaroids are used to improve contrast in old oil paintings.
  4. Polaroids are used in optical stress analysis.
  5. Polaroids are used as window glasses to control the intensity of incoming light.

Question 63.
State Brewster’s law.
Answer:
The law states that the tangent of the polarising angle for a transparent medium is equal to its refractive index, tan i = n. This relation is known as Brewster’s law.

Question 64.
What is angle of polarisation and obtain the equation for angle of polarisation.
Answer:
The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster’s angle. It is denoted by ip
ip = 90° – Rp

Question 65.
Discuss about pile of plates.
Answer:
The phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of 33.7° (90° – 56.3°) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3° which is the polarising angle for glass.

The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser.

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Question 66.
What is double refraction?
Answer:
When a ray of unpolarised light is incident on a calcite crystal, two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.

Question 67.
Mention the types of optically active crystals with example.
Answer:
Types of optically active crystals:
Uniaxial crystals:
Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.

Biaxial crystals:
Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.

Question 68.
Discuss about Nicol prism.
Answer:
Nicol prism is an optical device incorporated in optical instruments both for producing and analysing plane polarised light. The construction of a Nicol prism is based on the phenomenon of Double refraction. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.

It is cut into two halves along the diagonal so that their face angles are 72° and 108°. The two halves are joined i together by a layer of Canada balsam, a transparent cement.

Question 69.
How is polarisation of light obtained by scattering of light?
Answer:
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth’s atmosphere. The electric field of light interact with the electrons present in the air molecules.

Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.

Question 70.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  1. Magnification in near point focusing m = 1 + \(\frac { D }{ f }\)
  2. Magnification in normal focusing (angular magnification), m = \(\frac { D }{ f }\)

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Question 71.
What are near point and normal focusing?
Answer:

  • Near point focusing:
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing:
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Question 72.
Why is oil immersed objective preferred in a microscope?
Answer:
It is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.

Question 73.
What are the advantages and disadvantages of using a reflecting telescope?
Answer:
Advantages:

  • The main advantage is reflector telescope can escape from chromatic aberration because wavelength does not effect reflection.
  • The primary mirror is very stable because it is located at the back of the telescope and can be support in the back.
  • More cost effective than refractor of similar size.
  • Easier to make a high quality mirror than lens because mirror need to only concern with one side of the curvature.

Disadvantages:

  • Optical misalignment can occur quite easily.
  • Require frequent cleaning because the inside is expose to the atmosphere.
  • Secondary mirror can cause diffraction of original incoming light rays causing the “Christmas star effect” where a bright object have spikes.

Question 74.
What is the use of an erecting lens in a terrestrial telescope?
Answer:
A terrestrial telescope has an additional erecting lens to make the final image erect.

Question 75.
What is the use of collimator?
Answer:
The collimator is an arrangement to produce a parallel beam of light.

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Question 76.
What are the uses of spectrometer?
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.

Question 77.
What is myopia? What is its remedy?
Answer:
Myopia (or) short sightedness:
It is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.

Remedy (correction):
A myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

Question 78.
What is hypermetropia? What is its remedy?
Answer:
Hypermetropia (or) Long sightedness: It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly.
Remedy (correction): A hypermetropic eye is corrected by using a convex lens of suitable focal length.

Question 79.
What is presbyopia?
Answer:
This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).

Question 80.
What is astigmatism?
Answer:
Astigmatism is the defect arising due to different curvatures along different planes in the eye lens. Astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.

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Samacheer Kalvi 12th Physics Optics Long Answer Questions

Question 1.
Derive the mirror equation and the equation for lateral magnification.
Answer:
The mirror equation:
1. The mirror equation establishes a relation among object distance u, image distance v and focal length/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C’.
2. Let us consider three paraxial rays from point B on the object.
3. The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PBThe third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
4. The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-14
As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B’PA’. The triangles ∆BPA and ∆B’PA’ are similar. Thus, from the rule of similar triangles,
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\) …….. (1)
The other set of similar triangles are, ADPF and A BA.’ F. (PD is almost a straight vertical line)
\(\frac { A’B’ }{ PD }\) = \(\frac { AF’ }{ PF }\)
As, the distances PD = AB the above equation becomes,
\(\frac { A’B’ }{ AB }\) = \(\frac { AF’ }{ PF }\) ……… (2)
From equations (1) and (2) we can write,
\(\frac { PA’ }{ PA }\) = \(\frac { AF’ }{ PF }\)
As, A’ F = PA’ – PF, the above equation becomes,
\(\frac {PA’ }{ PA }\) = \(\frac { PA’-PF’P }{ PF }\) ……….. (3)
We can apply the sign conventions for the various distances in the above equation.
PA = – u, PA’ = -v, PF = -f
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\frac { -v }{ -u }\) = \(\frac { -v\left( -f \right) }{ -f } \)
On further simplification,
\(\frac { -v }{ -u }\) = \(\frac { v-f }{ f }\); \(\frac { v }{ u }\)=\(\frac { v }{ f }\)=1
Dividing either side with v,
\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\)=\(\frac { 1 }{ v }\)
After rearranging,
\(\frac { 1 }{ v }\)+\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\) ……… (4)
The above equation (4) is called mirror equation.

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q1
m=\(\frac { h’ }{ h }\) ……..(5)
Applying proper sign conventions for equation (1),
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\)
A’B’ = -h, AB = h, PA’ = -v, PA = -u
\(\frac { -h’ }{ h }\)=\(\frac { -v }{ -u }\)
On simplifying we get,
m=\(\frac { h’ }{ h }\)=-\(\frac { v }{ u }\) ………. (6)
Using mirror equation, we can further write the magnification as,
m=\(\frac { h’ }{ h }\)–\(\frac { f-v }{ f }\)=\(\frac { f }{ f-u }\) ………… (7)

Question 2.
Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus:
The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism. The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed w’heel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working:
The angular speed of rotation of the toothed wheel was increased from zero to a value to until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light:
The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
v = \(\frac { 2d }{ t }\) c ….. (1)
The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed co of the toothed wheel.
The angular speed ω of the toothed wheel when the light disappeared for the first time is,
ω = \(\frac { θ }{ t }\) ……. (2)
Here, 0 is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t.
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Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q2
θ = \(\frac { 2π }{ 2N }\) = \(\frac { π }{ N }\)
Substituting for 0 in the equation (2) for ,
ω = \(\frac { \pi /n }{ 2 } \) = \(\frac { π }{ Nt }\)
Rewriting the above equation for t,
t = \(\frac { π }{ Nω }\) ……. (3)
Substituting t from equation (3) in equation (1), v = \(\frac { 2d }{ \pi /N\omega } \)
After rearranging,
v = \(\frac { 2dNω }{ π }\) …….. (4)
Fizeau had some difficulty to visually estimate the minimum intensity of the light when blocked by the adjacent tooth, and his value for speed of light was very close to the actual value.

SamacheerKalvi.Guru

Question 3.
Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n2 sin i = n2 sin r for the refraction happening at the point B on the boundary between the two media is,
n1 sin ic = n2 sin 90° ……. (1)
n1 sin ic= n2 ∵ sin 90° = 1
sin ic = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (2)
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From the right angle triangle ∆ABC,
sin ic = \(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) ……. (3)
Equating the above two equation (3) and equation (2).
\(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
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If the rarer medium outside is air, then, n2 = 1, and we can take n1 = n
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Question 4.
Derive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:
Answer:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n1, cladding n2 and the outer medium n3. Assume the light is incident at an angle called acceptance angle i at the outer medium and core boundary at A.
The Snell’s law in the product form, equation for this refraction at the point A.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-20
n3 sin ia = n1 sin ra …(1)
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle ic. Snell’s law in the product form, equation for the refraction at point B is,
n1 sin ic = n2 sin 90° …(2)
n1 sin ic= n2 ∵ sin 90° =1
∴sin ic= \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …(3)
From the right angle triangle ∆ABC,
ic = 90°-ra
Now, equation (3) becomes, sin (90° – ra) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Using trigonometry’, cos ra = ra = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (4)
sin ra = \(\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \)
Substituting for cos ra
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If outer medium is air, then n3 = 1. The acceptance angle ia becomes,
ra=sin-1\(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (9)
Light can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n3 sin ia) is called numerical aperture NA of the optical fibre.
NA =n3 sin ia \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (10)
If outer medium is air, then n3 = 1. The numerical aperture NA becomes,
NA = sin ia = \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (11)

Question 5.
Obtain the equation for lateral displacement of light passing through a glass slab.
Answer:
Consider a glass slab of thickness t and refractive index n is kept in air medium. The path of the light is ABCD and the refractions occur at two points B and C in the glass slab. The angles of incidence i and refraction r are measured with respect to the normal N1and N2 at the two points B and C respectively. The lateral displacement L is the perpendicular distance CE drawn between the path of light and the undeviated path of light at point C. In the right angle triangle ∆BCE,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-22
sin(i – r) = \(\frac { L }{ BC }\); BC= \(\frac { L }{ sin(i-r) } \) …… (1)
In the right angle triangle ∆BCF,
cos(r) = \(\frac { t }{ BC }\) ; BC= \(\frac { t }{ cos(r) } \) …… (2)
Equating equations (1) and (2)
\(\frac { L }{ sin(i-r) } \) = \(\frac { t }{ cos(r) } \)
After rearranging,
L = t\(\left( \frac { sin(i-r) }{ cos(r) } \right) \)
Lateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.

SamacheerKalvi.Guru

Question 6.
Derive the equation for refraction at single spherical surface.
Answer:
Equation for refraction at single spherical surface:
Let us consider two transparent media having refractive indices n1and n2 are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n1. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.
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Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n2 > n1, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
Snell’s law in product form for the refraction at the point N could be written as,
n1 sin i = n2 sin r …(1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n1 i = n2 r …(2)
Let the angles,
∠NOP = α, ∠NCP = β, ∠NIP = γ
tan α = \(\frac { PN }{ PO }\);tan β = \(\frac { PN }{ PC }\);tan γ = \(\frac { PN }{ PI }\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac { PN }{ PO }\); β = \(\frac { PN }{ PC }\); γ = \(\frac { PN }{ PI }\) …….. (3)
For the triangle, ∆ONC,
I = α + β …….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ ……… (5)
Substituting for I and r from equations (4) and (5) in the equation (2).
n1 (α + β) = n2 (β – γ)
Rearranging,
n1 α + n2γ = (n2 – n1
Substituting for α, β and γ from equation (3)
n1 = (\(\frac { PN }{ PO }\)) + n2 = (\(\frac { PN }{ PI }\)) = (n2-n1) (\(\frac { PN }{ PC }\))
Further simplifying by cancelling PN,
\(\frac { { n }_{ 1 } }{ PO } \) + \(\frac { { n }_{ 2 } }{ PI } \) = \(\frac { { n }_{ 2 }-{ n }_{ 1 } }{ PC } \)
Following sign conventions, PO = -u, PI = +v and PC = +R in equation (6),
\(\frac { { n }_{ 2 } }{ -v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
After rearranging, finally we get,
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \) ……… (7)
Equation (7) gives the relation among the object distance n, image distance v, refractive indices of the two media (n1 and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n1 = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac { n }{ v}\)–\(\frac { 1 }{ u}\) = \(\frac { \left( { n }-1 \right) }{ R } \) ………(8)

Question 7.
Obtain lens maker’s formula and mention its significance.
Answer:
Lens maker’s formula and lens equation:
Let us consider a thin lens made up of a medium of refractive index n2, is placed in a medium of refractive index n1. Let R1 and R2 be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
For the refracting surface (1), the light goes from n1ton2.
\(\frac { { n }_{ 2 } }{ v’ } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R }_{ 1 } \) …….. (1)
For the refracting surface (2), the light goes from medium n2ton1.
\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 2 } }{ v’ } \) = \(\frac { \left( { n }_{ 1 }-{ n }_{ 2 } \right) }{ R }_{ 2 } \) …….. (2)
Adding the above two equation (1) and (2)
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\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = (n2 – n1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Furhter simplifying and rearranging,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}-n_{1}}{n_{1}}\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ……….. (3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = ∞, v= f. Then the equation becomes.
\(\frac { 1 }{ f }\)–\(\frac { 1 }{ ∞ }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ f }\)=\(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………… (4)
If the refractive index of the lens is and it is placed in air, then n2 = n and n1 = 1 equation (4) becomes,
\(\frac { 1 }{ f }\) = (n-1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………. (5)
The above equation is called the lens maker’s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) =\(\frac { 1 }{ f }\) …… (6)
This equation is known as lens equation which relates the object distance u and image distance v with the focal length f of the lens. This formula holds good for a any type of lens.

Question 8.
Derive the equation for thin lens and obtain its magnification.
Answer:
Lateral magnification in thin lens:
Let us consider an object 00′ of height h1 placed on the principal axis with its height perpendicular to the principal axis. The ray Op passing through the pole of the lens goes undeviated. The inverted real image II’ formed has a height h2.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-25
The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
m = \(\frac { II’ }{ OO’ }\) …….. (1)
From the two similar triangles ∆ POO’ and ∆ PII’, we can write,
\(\frac { H’ }{ OO’ }\) = \(\frac { PI }{ PO }\) ……..(2)
Applying sign convention,
\(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
Substituting this in the equation (1) for magnification,
m = \(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
After rearranging,
m = \(\frac{h_{2}}{h_{1}}\) = \(\frac { v }{ u}\) …….. (3)
The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one. We can also have the equations for magnification by combining the lens equation with the formula for magnification
as,
m = \(\frac{h_{2}}{h_{1}}=\frac{f}{f+u}\) (or) m = \(\frac{h_{2}}{h_{1}}=\frac{f-v}{f}\) ……… (4)

SamacheerKalvi.Guru

Question 9.
Derive the equation for effective focal length for lenses in out of contact.
Answer:
Consider a two lenses of focal length f1 and f2 arranged coaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produce deviations δ1 and δ2 respectively and The net deviation δ is,
δ = δ1 + δ2 ……. (1)
From Angle of deviation in lens equation, δ = \(\frac { h }{ f }\)
δ1 = \(\frac{h_{1}}{f_{1}}\); δ2 = \(\frac{h_{2}}{f_{2}}\) δ = \(\frac{h_{1}}{f}\) ….. (2)
The equation (1) becomes,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-26
\(\frac{h_{1}}{f}\) = \(\frac{h_{1}}{f_{1}}\) + \(\frac{h_{2}}{f_{2}}\) ……. (3)
From the geometry,
h2 – h1 = P2C = CG
h2 – h1 = BG tan δ1 ≈ BG δ1
h2 – h1 = d\(\frac{h_{1}}{f_{1}}\)
h2 = h1 + d\(\frac{h_{1}}{f_{1}}\) …….. (4)
Substituting the above equation in Equation (3)
\(\frac{h_{1}}{f}=\frac{h_{1}}{f_{1}}+\frac{h_{1}}{f_{2}}+\frac{h_{1} d}{f_{1} f_{2}}\)
On further simplification,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}}\) …….. (5)
The above equation could be used to find the equivalent focal length.

Question 10.
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
Answer:
Angie of deviation produced by prism:
Let light ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face AB are i1 and r1. The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is r2 and i2 respectively. RS is the ray emerging from p the second face. Angle i2 is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
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The two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M.
The deviation d1 at the surface AB is,
angle ∠RQM = d1 = i1 – r1 …(1)
The deviation d2 at the surface AC is,
angle ∠QRM = d2 = i2 – r2 …(2)
Total angle of deviation d produced is,
d = d1 + d2 …(3)
Substituting for d1 and d2,
d=(i1 – r1) + (i2 – r2)
After rearranging,
d = (i1 – r1) + (i2 – r2) …(4)
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
∠A + ∠QNR = 180° …(5)
From the triangle ∆QNR,
r1+ r2 ∠QNR = 180° …(6)
Comparing these two equations (5) and (6) we get,
r1+ r2 = A …(7)
Substituting this in equation (4) for angle of deviation,
d= i1+ i2 -A …(8)
Thus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
Refractive index of the material of the prism:
At minimum deviation,
i1 = i2 = i and r1 = r2 = r
Now, the equation (8) becomes,
D = i1 + i2-A (or) i = \(\frac { \left( A+D \right) }{ 2 } \)
The equation (7) becomes,
r1 + r2 = A ⇒ 2r = A (or) r = \(\frac { A }{ 2 }\)
Substituting i and r in Snell’s law,
n = \(\frac { sin i }{ sin r }\)
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The above equation is used to find the refractive index of the material of the prism.

Question 11.
What is dispersion? Obtain the equation for dispersive power of a medium.
Answer:
Dispersion. Dispersion is splitting of white light into its constituent colours.
Dispersive Power:
Consider a beam of white light passes through a prism; It gets dispersed into its constituent colours. Let δV, δR are the angles of deviation for violet and red light. Let nV and nR are the refractive indices for the violet and red light respectively.
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The refractive index of the material of a prism is given by the equation
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10°, the prism is said to be a small angle prism. When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a small angle prism and 5 the angle of deviation then the prism formula becomes.
n = \(\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) ……… (1)
For small angles of A and δm,
\(\sin \left(\frac{\mathrm{A}+\delta}{2}\right) \approx\left(\frac{\mathrm{A}+\delta}{2}\right)\) ……. (2)
\(\sin \left(\frac{\mathrm{A}}{2}\right) \approx\left(\frac{\mathrm{A}}{2}\right)\) …… (3)
∴ n = \(\frac{\left(\frac{A+\delta}{2}\right)}{\left(\frac{A}{2}\right)}=\frac{A+\delta}{A}=1+\frac{\delta}{A}\)
Further simplifying, \(\frac { δ }{ A }\) = n – 1
δ = (n – 1) A ……. (4)
When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For Violet light, δV = (nV – 1)A …(5)
For Red light, δR = (nR – 1) …(6)
As, angle of deviation for violet colour δV is greater than angle of deviation for red colour δR, the refractive index for violet colour nV is greater than the refractive index for red colour nR. Subtracting δV from δR we get,
δV – δR = (nV – nR)A ….. (7)
The term (δV – δR) is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.

  1. Angle of the prism
  2. Nature of the material of the prism.

If we take 8 is the angle of deviation for any middly ray (green or yellow) and n the corresponding refractive index. Then,
8 = (n – 1) A … (8)
Dispersive power (ω) is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour. Dispersive power (ω),
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Substituting for (δVR)and (δ),
ω = \(\frac{\left(n_{\mathrm{V}}-n_{\mathrm{R}}\right)}{(n-1)}\) ……. (10)
Dispersive power is a dimensionless quality. It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

SamacheerKalvi.Guru

Question 12.
Prove laws of reflection using Huygens’ principle.
Answer:
Proof for laws of reflection using Huygens’ Principle:
Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY. The incident wavefront is AB and the reflected wavefront is A’B’ in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L’, M’ respectively.

By the time point A of the incident wavefront touches the reflecting surface, the point B is yet to travel a distance BB’ to touch the reflecting surface a B’. When the point B falls on the reflecting surface at B’ , the point A would have reached A’. This is applicable to all the points on the wavefront.

Thus, the reflected wavefront A’B’ emanates as a plane wavefront. The two normals N and N’ are considered at the points where the rays L and M fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection. Hence, the time taken for the ray to travel from B to B’ is the same as the time taken for the ray to travel from A to A’. Thus, the distance BB’ is equal to the distance AA’; (AA’= BB’).
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(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence,
∠i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of reflection,
∠r = ∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A
For the two right angle triangles, A ABB’ and A B’A’A, the right angles, ∠B and ∠A’ are equal, (∠B and ∠A’ = 90°); the two sides, AA’ and BB’ are equal, (AA’ = BB’); the side AB’ is the common. Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB’ and ∠A’B’A must also be equal. i = r
Hence, the laws of reflection are proved.

Question 13.
Prove laws of refraction using Huygens’ principle.
Answer:
Let us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface. The incident wavefront AB is in rarer medium (1) and the refracted wavefront A’B’ is in denser medium (2). These wavefronts are perpendicular to the incident rays L, M and refracted rays L’, M’ respectively. By the time the point A of the incident wavefront touches the refracting surface, the point B is yet to travel a distance BB’ to touch the refracting surface at B’
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When the point B falls on the refracting surface at B’, the point A’ would have reached A in the other medium. This is applicable to all the points on the wavefront. Thus, the refracted wavefront A’B’ emanates as a plane wavefront.
The two normals N and N’ are considered at the points where the rays L and M fall on the refracting surface. As refraction happens from rarer medium (1) to denser medium (2), the speed of light is v1 and v2 before and after refraction and v1 is greater than v2 (v1 > v2). But, the time taken t for the ray to travel from B to B’ is the same as the time taken for the ray to
travel from A to A’.
t = \(\frac { BB’ }{ { v }_{ 1 } } \) = \(\frac { AA’ }{ { v }_{ 2 } } \) (or) \(\frac { BB’ }{ AA’ }\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \)
(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence,
i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of refraction,
r =∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A’
For the two right angle triangles ∆ ABB’ and ∆ B’A’A,
\(\frac { sin i }{ sin r }\) = \(\frac{B B^{\prime} / A B^{\prime}}{A A^{\prime} / A B^{\prime}}\) =\(\frac { BB’ }{ AA’ }\) =\(\frac{v_{1}}{v_{2}}=\frac{c / v_{2}}{c / v_{1}}\)
Here, c is speed of light in vacuum. The ratio c/v is the constant, called refractive index of the medium. The refractive index of medium (1) is, c/v1 = n1 and that of medium (2) is, c/v2 = n2.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
In product form,
n1 sin i = n2 sin r
Hence, the laws of refraction are proved.

Question 14.
Obtain the equation for resultant intensify’ due to interference of light.
Answer:
Let us consider two light waves from the two sources S1 and S2 meeting at a point P. The wave from S1 at an instant t at P is,
y1 = a1 sin ωt … (1)
The wave form S2 at an instant t at P is,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-31
y2 = a2 sin (ωt + Φ) … (2)
The two waves have different amplitudes a1 and a2, same angular frequency ω, and a phase difference of Φ between them. The resultant displacement will be given by,
y = y1 + y2 = a1 sin ωt + a1 sin2 (ωt + Φ) … (3)
The simplification of the above equation by using trigonometric identities gives the equation,
y = A sin (ωt + θ) … (4)
Where, A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\) …. (5)
θ = tan-1 \(\frac{a_{2} \sin \phi}{a_{1}+a_{2} \cos \phi}\) …. (6)
The resultant amplitude is maximum,
Amax = \(\sqrt{\left(a_{1}+a_{2}\right)^{2}}\) ; when Φ = 0, ±2π, ±4π …., …… (7)
The resultant amplitude is minimum,
Amax = \(\sqrt{\left(a_{1}-a_{2}\right)^{2}}\) ; when Φ = ±π, ±3,π ±5π …., …… (8)
The intensity of light is proportional to square of amplitude,
I ∝ A2 ….. (9)
Now, equation (5) becomes,
I ∝ I1+I 2+2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) cos Φ …… (10)
In equation (10) if the phase difference, Φ = 0, ±2π, ±4π …., it corresponds to the condition for maximum intensity of light called as constructive interference. The resultant maximum intensity is,
Imax ∝ ( a1 + a2)2 ∝ I1 + I2 +2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) …… (11)
In equation (10) if the phase difference, Φ = ±π, ±3,π ±5π …., it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is,
Imax ∝( a1 – a2)2 ∝ I1 + I2 -2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) ……. (12)

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Question 15.
Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
I Experimental setup:

1. Wavefronts from S1 and S2 spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter s from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
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2. Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from S1 and S2 travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

3. The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

II Equation for path difference:

1. Let d be the distance between the double slits S1 and S2 which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S1 and S2 is C and the midpoint of the screen O is equidistant from S1 and S2. P is any point at a distance y from O.
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2. The waves from S1 and S2 meet at P either inphase or out-of-phase depending upon the path difference between the two waves.
3. The path difference δ between the light waves from S1 and S2 to the point P is,
δ = S2 P – S1
4. A perpendicular is dropped from the point S1 to the line S2 P at M to find the path difference more precisely.
δ = S2 P – MP = S2 M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S2 S1M are equal. ∠OCP = ∠S2 S1 M = θ
In right angle triangle ∆S1 S2 M, the path difference, S2 M = d sin θ
δ = d sin θ
If the angle 0 is small, sin θ ≈ tan θ ≈ θ. From the right angle triangle ∆OCP, tan θ = \(\frac { y }{ D }\)
The path difference, δ = \(\frac { dy }{ D }\)

Question 16.
Obtain the equation for bandwidth in Young’s double slit experiment.
Answer:
Condition for bright fringe (or) maxima
The condition for the constructive interference or the point P to be have a bright fringe is,
Path difference, δ = nλ
where, n = 0, 1, 2,. . .
∴ \(\frac { dy }{ D }\) = nλ
y = n \(\frac { λD }{ d }\) (or) yn = n \(\frac { λD }{ d }\)
This is the condition for the point P to be a bright fringe. The distance is the distance of the nth bright fringe from the point O.
Condition for dark fringe (or) minima
The condition for the destructive interference or the point P to be have a dark fringe is,
Path difference, δ = (2n – 1) \(\frac { λ }{ 2 }\)
where, n = 1, 2, 3 …
∴ \(\frac { dy }{ D }\) = (2n – 1) \(\frac { λ }{ 2 }\)
y = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\) (or) yn = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\)
Equation for bandwidth
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.
The distance between (n +1)th and nth consecutive bright fringes from O is given by,
β = y(n +1) – yn = \(\left((n+1) \frac{\lambda \mathrm{D}}{d}\right)-\left(n \frac{\lambda \mathrm{D}}{d}\right)\)
β = \(\frac { λD }{ d }\)
Similarly, the distance between (n +1)th and nth consecutive dark fringes from O is given by,
β = y(n+1) – yn = \(\left(\frac{(2(n+1)-1)}{2} \frac{\lambda D}{d}\right)-\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right)\)
β = \(\frac { λD }{ d }\)
The above equation show that the bright and dark fringes are of same width equally spaced on either side of central bright fringe.

Question 17.
Obtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
Answer:
Interference in thin films:
Let us consider a thin film of transparent material of refractive index p (not to confuse with order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted.

The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back in to the film. Reflected as well as refracted waves are sent by the film as multiple reflections take place inside the film. The interference is produced by both the reflected and transmitted light.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-34
For transmitted light:
The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occured. The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD.
If we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is traversed in a medium of refractive index µ, the optical path difference is, δ = 2µd.
The condition for constructive interference in transmitted ray is,
2µd = nλ
Similarly, the condition for destructive interference in transmitted ray is,
2µd = (2n – 1) \(\frac { λ }{ 2 }\)
For reflected light:
It is experimentally and theoretically proved that a wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of π. Hence, an additional path difference of \(\frac { λ }{ 2 }\) should be considered.

Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film. The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, δ = 2µd.
The condition for constructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = nλ (or) 2µd = (2n – 1) \(\frac { λ }{ 2 }\)
The additional path difference \(\frac { λ }{ 2 }\) is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = (2n + 1) \(\frac { λ }{ 2 }\) (or) 2µd = nλ

Question 18.
Discuss diffraction at single slit and obtain the condition for nth minimum.
Answer:
Diffraction at single slit:
Let a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O. We would like to find the intensity at any point P on the screen. The lines joining P to the different points on the slit can be treated as parallel lines, making an angle 9 with the normal CO.

All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction. We need to give the condition for the point P to be of various minima.

The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at P with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-35
Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
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The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum. The path difference 8 between waves from these corresponding points
is, δ = \(\frac { a }{ 2 }\) sin θ
The condition for P to be first minimum, \(\frac { a }{ 2 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = λ (first minimum) ….. (1)
Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
The path difference δ between waves from these corresponding points δ = \(\frac { a }{ 4 }\) sin θ.
The condition for P to be first minimum, \(\frac { a }{ 4 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 2λ (second minimum) …(2)
Condition for P to be third order minimum:
The same way the slit is divided into six equal parts to explain the condition for P to be third
minimum is, \(\frac { a }{ 6 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 3λ (third minimum) …(3)
Condition for P to be nth order minimum:
Dividing the slit into 2n number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for nth
order minimum is, \(\frac { a }{ 2n }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = nλ (nth minimum)

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Question 19.
Discuss the diffraction at a grating and obtain the condition for the mth maximum.
Answer:
A plane transmission grating is represented by AB. Let a plane wavefront of monochromatic light with wavelength λ be incident normally on the grating. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.
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A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let us consider a point P at an angle 0 with the normal drawn from the center of the grating to the screen. The path difference 5 between the diffracted waves from one pair of corresponding points is,
δ = (a + b) sin θ
This path difference is the same for any pair of corresponding points. The point P will be bright, when
δ = m λ where m = 0, 1 , 2, 3

Combining the above two equations, we get,
(a + b) sin θ = m λ
Here, m is called order of diffraction.

Condition for zero order maximum, m = 0
For (a + b) sin θ = 0, the position, θ = 0. sin θ = 0 and m = 0. This is called zero order diffraction or central maximum.

Condition for first order maximum, m = 1
If (a + b) sin θ1 = λ, the diffracted light meet at an angle θ1 to the incident direction and the first order maximum is obtained.

Condition for second order maximum, m = 2
Similarly, (a + b) sin θ2 = 2λ, forms the second order maximum at the angular position θ2.

Condition for higher order maximum
On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.
If we take, N = \(\frac { 1 }{ a+b }\)

Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itself. Now, the equation becomes,
\(\frac { 1 }{ N }\) sin θ = mλ, (or) sin θ = Nmλ

Question 20.
Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.
Answer:
Experiment to determine the wavelength of monochromatic light:
The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. Initially all the preliminary adjustments of the spectrometer are made. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined. The telescope is brought in line with collimator to view the image of the slit. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.

The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece. The reading of the position of the telescope is noted. Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between two positions gives 2θ. Half of its value gives θ, the diffraction angle for first order maximum. The wavelength of light is calculated from the equation,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 21.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Answer:
Determination of wavelength of different colours:
When white light is used, the diffraction pattern consists of a white central maximum and on both sides continuous coloured diffraction patterns are formed. The central maximum is white as all the colours meet here constructively with no path difference.

As θ increases, the path difference, (a + b) sin θ, passes through condition for maxima of diffraction of different orders for all colours from violet to red. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 22.
Obtain the equation for resolving power of optical instrument.
Answer:
Resolution:
The effect of diffraction has an adverse impact in the image formation by the optical instruments such as microscope and telescope. For a single rectangular slit, the half angle θ subtended by the spread of central maximum (or position of first minimum) is given by the ‘ relation,
a sin θ = λ … (1)
Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that becomes fainter while moving away from the center. These are known as Airy’s discs. The circle of central maximum has the half angular spread given by the equation,
a sin θ = 1.22λ … (2)
Here, the numerical value 1.22 comes for central maximum formed by circular apertures. This involves higher level mathematics which is avoided in this discussion.
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For small angles, sin θ ≈ θ
a θ = 1.22λ … (3)
Rewriting further, θ = \(\frac { 1.22λ }{ a}\) and \(\frac { { r }_{ 0 } }{ f } \) = \(\frac { 1.22λ }{ a}\)
r0 = \(\frac { 1.22λf }{ a}\) …. (4)

When two point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.

According to Rayleigh’s criterion, for two point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh’s criterion is said to be limit of resolution.

According to Rayleigh’s criterion the two point sources are said to be just resolved when the distance between the two maxima is at least r0. The angular resolution has a unit in radian (rad) and it is given by the equation.
θ = \(\frac { 1.22λf}{ a}\)
It shows that the first order diffraction angle must be as small as possible for greater resolution. This further shows that for better resolution, the wavelength of light used must be as small as possible and the size of the aperture of the instrument used must be as large as possible. The equation (4) is used to calculate spacial resolution. The inverse of resolution is called resolving power. This implies, smaller the resolution, greater is the resolving power of the instrument. The ability of an optical instrument to • separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.

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Question 23.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
Simple microscope:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  • Near point focusing :
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing :
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Magnification in near point focusing:
Object distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
m = \(\frac { v}{ u}\)
With the help of lens equation, \(\frac { 1}{ v}\) – \(\frac { 1}{ u}\) \(\frac { 1}{ f}\) the magnification can further be writen as,
m = 1- \(\frac { v}{ f}\)
Substituting for v with sign convention, v = -Derivem
m = 1 + \(\frac { D}{ f}\)
This is the magnification for near point focusing
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-39
Magnification in normal focusing (angular magnification):
We will now find the magnification for the image formed at infinity. If we take the ratio of height of image to height of object \(\left(m=\frac{h^{\prime}}{h}\right)\) to find the magnification, we will not get a practical relation as the image will also be of infinite size when the image is formed at infinity. Hence, we can practically use the angular magnification. The angular magnification is defined as the ratio of angle 0j subtended by the image with aided eye to the angle 90 subtended by the object with unaided eye.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-40
m = \(\frac{\theta_{i}}{\theta_{0}}\)
For unaided eye, tan θ0 ≈ θ0 = \(\frac { h }{ D }\)
For aided eye, tan θi ≈ θi = \(\frac { h }{ f }\)
The angular magnification is, m = \(\frac{\theta_{i}}{\theta_{0}}\) = \(\frac { h/f }{ h/d }\)
m = \(\frac { d }{ f }\)
This is the magnification for normal focusing. The magnification for normal focusing is one less than that for near point focusing.

Question 24.
Explain about compound microscope and obtain the equation for magnification. Compound microscope:
Answer:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.
Magnification of compound microscope :
From the ray diagram, the linear magnification due to the objective is,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-41
m0 = \(\frac { h’ }{ f}\)
From the figure, tan β = \(\frac { h }{ { f }_{ 0 } } \) =\(\frac { h’ }{ L’}\),then
\(\frac { h’ }{ h}\) = \(\frac { L }{ { f }_{ 0 } } \); m0 =\(\frac { L }{ { f }_{ 0 } } \)
Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length
L of the microscope as f0 and fe are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification me of the eyepiece is,
me = 1 + \(\frac { D }{ { f }_{ e } } \)
The total magnification m in near point focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)
If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is,
me = \(\frac { D }{ { f }_{ e } } \)
The total magnification m in normal focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 25.
Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope:
The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance dmin.
Smaller the value of dmin better will be the resolving power of the microscope.
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The radius of central maxima is, r0 = \(\frac { 1.22λv }{ a }\)
In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin, then the magnification m is, m = \(\frac{r_{o}}{d_{\min }}\)
dmin = \(\frac{r_{o}}{m}\) =\(\frac{1.22 \lambda v}{a m}\) = \(\frac{1.22 \lambda v}{a(v / u)}\) = \(\frac{1.22 \lambda u}{a}\) [∴m = v/u] [∴ [u ≈ ƒ]
On the object side, 2tan β ≈ 2sin β = \(\frac { a }{ f }\)
dmin = \(\frac{1.22 \lambda}{2 sin β }\) ∴ [a = ƒ2 sinβ]
To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n.
dmin = \(\frac{1.22 \lambda}{2n sin β }\)
Such an objective is called oil immersed objective. The term n sin p is called numerical aperture NA.
dmin = \(\frac{1.22 \lambda}{2(NA) }\)

Question 26.
Discuss about astronomical telescope.
Answer:
Astronomical telescope:
An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by astronomical telescope will be inverted. It has an objective of long focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image.
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Magnification of astronomical telescope:
The magnification m is the ratio of the angle β subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye
m = \(\frac { β }{ α }\)
From the diagram, m = \(\frac{h / f_{e}}{h / f_{0}}\)
m = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \)
The length of the telescope is approximately, L = f0 + fe

Question 27.
Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.
Adjustments of the spectrometer:
The following adjustments must be made before doing the experiment using spectrometer.

(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.

(b) Adjustment of the telescope:
The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.

(c) Adjustment of the collimator:
The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.

(d) Levelling the prism table:
The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

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Question 28.
Explain the experimental determination of material of the prism using spectrometer. Determination of refractive index of material of the prism:
Answer:
The preliminary adjustments of the telescope, collimator and the prism table of the spectrometer are made. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum deviation.

(i) Angle of the prism (A):
The prism is placed on the prism table with its refracting edge facing the collimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the collimator fall on the two faces AB and AC. The telescope is rotated to the position T1 until the image of the slit formed by the reflection at the face AB is made to coincide with the vertical cross wire of the telescope.
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The readings of the verniers are noted. The telescope is then rotated to the position T2 where the image of the slit formed by the reflection at the face AC coincides with the vertical cross wire. The readings are again noted.
The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.

(ii) Angle of minimum deviation (D):
The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.
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The readings of the verniers are noted. Now, the prism is removed and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula,
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.

Samacheer Kalvi 12th Physics Optics Conceptual Questions

Question 1.
Why are dish antennas curved?
Answer:
Dish antenna is curved so as it can receive parallel signal rays coming from same direction. These parallel signal rays reflect from parabolic dish, and gathered at main antenna part. This increases directivity of antenna, and gives sufficient amplitude signal.

Question 2.
What type of lens is formed by a bubble inside water?
Answer:
Air bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.

Question 3.
It is possible for two lenses to produce zero power?
Answer:
Yes. It is possible for two lenses to produce zero power. Both the surfaces of lenses are equally curved, i.e. R1 = R2 and hence
Power (P) = (µ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=0\)

Question 4.
Why does sky look blue and clouds look white?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appear blue. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 5.
Why is yellow light preferred to during fog?
Answer:
Yellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). so yellow colour is least scattered and produces sufficient illumination.

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Question 6.
Two independent monochromatic sources cannot act as coherent sources, why?
Answer:
Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms. When they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Question 7.
Does diffraction take place at the Young’s double slit?
Answer:
Both diffraction and interference in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs.

Question 8.
Is there any difference between colored light obtained from prism and colours of soap bubble?
Answer:
Yes. there is a difference between colored light obtained from the prism is the phenomenon of‘dispersion of light’ and colored light obtained from the soap bubble is the phenomenon of ‘interference of light’.

Question 9.
A small disc is placed in the path of the light from distance source. Will the center of the shadow be bright or dark?
Answer:
When a tiny circular small disc is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the disc because, wave diffracted from the edge of the circular disc interface constructively at the centre of the shadow, which produces bright spot.

Question 10.
When a wave undergoes reflection at a denser medium, what happens to its phase?
Answer:
When a wave undergoes a reflection at a denser medium then it’s crest reflected as trough and vice versa. So, its phase changes at 180°.

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Samacheer Kalvi 12th Physics Optics Numerical Problems

Question 1.
An object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.
Solution:
ƒ = – 20 cm
v = – 4u
According to lens formula
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\frac { 1 }{ (-4u) }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\left[-\frac{1}{4}+1\right]\)
= \(\frac { 1 }{ u }\) \(\frac { 3 }{ 4 }\)
u = \(\frac { 3 × 20 }{ 4 }\) = -15 cm.

Question 2.
A compound microscope has a magnification of 30. The focal length of eye piece is 5 cm. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Solution:
Magnification of compound microscope, M = 30
Focal length, f = 5 cm
Least distance of distinct vision, D = 25 cm
Now, M = M0 x Me
= M0 x \(\left[1+\frac{\mathrm{D}}{f_{e}}\right]\)
30 = M0 x \(\left[1+\frac{25}{5}\right]\)
M0 = \(\frac { 30 }{ 6 }\) = 5

Question 3.
An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
Solution:
m = + 3 and m = – 3; f = – 20 cm
When, m = 3
m = \(\frac { v }{ u }\) = 3
v = – 3 u
From mirror equation,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ -3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ u }\) \(\left[-\frac{1}{3}+1\right]\)
\(\frac { 1 }{ -20 }\) = \(\frac { 2 }{ 3u }\)
u = –\(\frac { 40 }{ 3 }\) cm
When, m = – 3 ⇒ v = 3u
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ 3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 4 }{ 3u }\)
u = \(\frac { 4×20 }{ 3 }\)
u = –\(\frac { 80 }{ 3 }\) cm

Question 4.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Solution:
Actual depth of the bulb in water
d1 = 80 cm = 0.8 m
Refractive index of water, p = 1.33
The given situation is shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-46
Where, i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle AT
R= \(\frac { AC }{ 2 }\) = AO = OB
Using snell’s law, the relation for the refractive index of water is sin
µ = \(\frac { sin r }{ sin i }\)
i = sin-1 (0.75) = 48.75°
Using the given figure, we have the relation
tan i = \(\frac { OC }{ BC }\) = \(\frac { R }{ { d }_{ 1 } } \)
R = tan i x d1 = tan 48.75° x 0.8
R = 0.91 m
Area of the surface of water = πR²
= 3.14 x (0.91)² = 2.61 m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

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Question 5.
A thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
Solution:
According to Lens maker’s formula, the focal length for a convex lens placed in air can be obtained as
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-47
\(\frac { 0.5 }{ 5 }\) = \(\frac { 1.5 }{ n }\) -1
0.9 = \(\frac { 1.5 }{ n }\)
n = \(\frac { 1.5 }{ 0.9 }\) = \(\frac { 5 }{ 3 }\) -1

Question 6.
If the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
Solution:
Let us fix the position of object and place the screen to get the enlarged image first. Also, let us fix the position of screen where we get the enlarged image.
Let D be the distance between object and screen. Let us mark the position of lens dv Then let us move the lens away from the object to get a diminished image. Let this position of lens be d2. Let d be the distance between the lens position d1 and d2. Let V be the distance b/w image and lens. Let ‘u’ be the distance between object and lens.
From mirror equation,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-48
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
Let us replace v by substituting v = D – u
\(\frac { 1 }{ D-u }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
we get the equation, u² -Du + fD = 0
the quadratic equation for above equation,
u = \(\frac{\mathrm{D} \pm \sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
When D = 4f, we get only position of lens to get image.
This corresponds to placing the object at 2f and getting the image at 2f on the otherside. Hence, for displacement method we need D > 4 f. When this condition is satisfied we get
u1 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) ; corresponding v1 – D – u2 = \(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) after changing the location
u1 =\(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4f \mathrm{D}}}{2}\) ; corresponding v2 – D – u2 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
now the displacement d = v1 – u1 = u1 – v1 = \(\sqrt{D^{2}-4 f D}\)
Hence we get focal length, ƒ = \(\frac{D^{2}-d^{2}}{4 D}\)

Question 7.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?
Solution:
For first minimum (n = 1) on either side of central maximum.
sin θ = \(\frac { λ }{ a }\)
Where a is width of the slit Since 0 is very small, (sin θ ≈ θ)
θ = \(\frac { λ }{ a }\) …… (1)
sin θ ≈ θ = \(\frac { x }{ 2D }\) …… (2)
Where, x is distance of first dark fringe from central maximum.
D is distance between slit and screen.
From equation (1) and (2)
\(\frac { x }{ 2D }\) = \(\frac { λ }{ a }\)
x = \(\frac { 2Dλ }{ a }\) = \(\frac { 2\times 2\times 600\times { 10 }^{ -9 } }{ 1\times { 10 }^{ -3 } } \)
= 2400 x 10-6 = 2.4 x 10-3m
x = 2.4 mm

Question 8.
In Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0 = 750 nm and λ = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
Solution:
Now from the question we can infer that
D = 2; d = 2
n1 λ1 = n2 λ2
\(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 900 }{ 750 }\) \(\frac { 1 }{ 2 }\)
Thus, we have
\(\frac{n_{1}}{n_{1}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 6 }{ 5 }\)
5th and 6th fringes will coincide respectively. The minimum distance is given as
Xmin = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\) = \(\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}\)
= 4500 x 10-6 = 4.5 x 10-3m
Xmin = 4.5 mm

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Question 9.
In Young’s double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, then what is the number of fringes seen?
Solution:
From young’s double slit experiment,
λ1 = 5893 Å ; λ2 = 4359 Å
\(\frac{n_{1} \lambda_{1} D}{d}\) = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\)
The above condition is total extent of fringes is constant for both wavelengths.
\(\frac{62 \times 5893 \times 10^{-10} \times D}{d}\) = \(\frac{n_{2} \times 4359 \times 10^{-10} \times D}{d}\)
n2 = \(\frac{62 \times 5893}{4359}\) = \(\frac{365366}{4359}\) = 83.8
n2 = 84

Question 10.
A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.
Solution:
Magnifying Power, m = 100
Focal length of the objective,f0 = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
v0e= 6.5 cm ….. (1)
The magnifying power for normal adjustment is given by
m = \(\left(\frac{v_{o}}{u_{o}}\right) \times \frac{\mathrm{D}}{f_{e}}\)
= – \(\left[1-\frac{v_{o}}{f_{\mathrm{o}}}\right] \frac{\mathrm{D}}{f_{e}}\)
100 = – \(\left[1-\frac{v_{o}}{0.5}\right] \times \frac{25}{f_{e}}\)
2v0 -4ƒe= 1
On solving equations (1) and (2), we get
v0 = 4.5 cm and ƒe = 2 cm
Thus, the focal length of the eyepiece is 2 cm.

Samacheer Kalvi 12th Physics Optics Additional Questions

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions

Question 1.
When a ray of light enters a glass slab from air
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint:
Wavelength, λ = \(\frac { Velocity }{ Frequency }\) = \(\frac { u }{ v }\)
When light travels from air to glass, frequency v remains unchanged, velocity u decreases and hence wavelength X also decreases.

Question 2.
A source emits sound of frequency 600 Hz inside water. The frequency heard in air (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s) will be
(a) 300 Hz
(b) 120 Hz
(c) 600 Hz
(d) 6000 Hz
Answer:
(c) 600 Hz
Hint:
Frequency does not change when sound travels from one medium to another
∴ Frequency of sound in air = Frequency of sound in water = 600 Hz

Question 3.
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be
(a) 30° for both the colours
(b) greater for the violet colour
(c) greater for the violet colour
(d) equal but not 30° for both the colours
Answer:
(a) 30° for both the colours
Hint:
For any prism, r1 = r2 = A
In the position of minimum deviation for any wavelength,
r1 = r2 = \(\frac { A }{ 2 }\) = \(\frac { 60° }{ θ }\) = 30°

Question 4.
To get three images of a single object, one should have two plane mirrors at an angle of
(a) 60°
(b) 90°
(c) 120°
(d) 30°
Answer:
(b) 90°
Hint:
The number of images formed,
n = \(\frac { 360° }{ θ }\) – 1 or 3 = \(\frac { 360° }{ θ }\) -1 or θ = 90°

Question 5.
Which of the following is used in optical fibres?
(a) Total internal reflection
(b) Diffraction
(c) Refraction
(d) Scattering
Answer:
(a) Total internal reflection
Hint:
The working of optical fibres is based on total internal reflection.

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Question 6.
Two lenses of power – 15 D and +15D are in contact with each other. The focal length of the combination is
(a) + 10 cm
(b) -20 cm
(c) – 10 cm
(d) + 20cm
Answer:
(c) – 10 cm
Hint P = P1 + P2 = -15 + 5= -10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ -10 }\) m = -10 cm

Question 7.
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let δ1 and δ2 be angles of minimum deviation for red and blue light respectively in a prism of this glass, then
(a) δ1, can be less than or greater than δ2 depending upon the values of δ1 and δ2
(b) δ1 > δ2
(c) δ1 < δ2
(d) δ1 = δ2
Answer:
(c) δ1 < δ2
Hint:
δ1 = (μR – 1)A, δ2 = (μB – 1)A
As, μR < μB ∴ δ1 < δ2

Question 8.
Time image formed by an objective of a compound microscope is
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
Answer:
(c) real and enlarged
Hint
The image formed by the objective of a compound microscope is real and enlarged.

Question 9.
An astronomical telescope has a large aperture to,
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
Answer:
(b) have high resolution

Question 10.
Two plane mirrors are inclined to each other at an angle of 60°. A point object is placed in between them. The total number of images produced by both the mirror is
(a) 2
(b) 4
(c) 5
( d) 6
Answer:
(c) 5
Hint:
Number of images formed, θ = \(\frac { 360° }{ θ }\) – 1 = \(\frac { 360 }{ 60 }\) = 1 = 5.

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Question 11.
A boy 1.5 m tall with his eye level at 1.38 m stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is
(a) 0.75 m
(b) 0.06 m
(c) 0.69 m
(d) 0.12 m
Answer:
(a) 0.75 m
Hint
Minimum length of mirror required \(\frac { 1 }{ 2 }\) x Height of boy = \(\frac { 1 }{ 2 }\) x 1.5 = 0.75 m.

Question 12.
A pencil of light rays falls on a plane mirror and forms a real image, so the incident rays are
(a) parallel
(b) diverging
(c) converging
(d) statement is false
Answer:
(c) converging
Hint:
When converging rays fall on a plane mirror, they get reflected to a point d in front of the mirror forming a real image.

Question 13.
For a real object, which of the following can produce a real image?
(a) plane mirror
(b) concave lens
(c) convex lens
(d) concave mirror
Answer:
(d) concave mirror
Hint:
Only concave mirror produces real image provided the object is not placed between its focus and pole.

Question 14.
Which mirror is to be used to obtain a parallel beam of light from a small lamp?
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of the above
Answer:
(c) Concave mirror
Hint:
When the small lamp is placed at the focus of the concave mirror, the reflected light is a parallel beam.

Question 15.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Answer:
(b) Frequency
Only the frequency of the electromagnetic wave remains unchanged.

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Question 16.
If wavelength of light in air is 2400 x 10-10 m, then what will be the wavelength of light in glass (μ = 1.5)?
(a) 1600 Å
(b) 7200 Å
(c) 1080 Å
(d) None of these
Answer:
(a) 1600 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λλg = \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac{2400 \times 10^{-10} \mathrm{m}}{1.5}\) = 1600 Å

Question 17.
Why is refractive index in a transparent medium greater than one?
(a) Because the speed of light in vacuum is always less than speed in a transparent medium.
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
(c) Frequency of wave changes when it Gasses medium.
(d) None of the above.
Answer:
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-49
As c > v, μ > 1

Question 18.
The wavelength of sodium light in air is 5890 Å. The velocity of light in air is 3 x 108 ms-1. The wavelength of light in a glass of refractive index 1.6 would be close to
(a) 5890 Å
(b) 3681 Å
(c) 9424 Å
(d) 15078 Å
Answer:
(b) 3681 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λg= \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac { 5890 Å }{ 1.6 }\) = 3681 Å

Question 19.
A glass slab (μ = 1.5) of thickness 6 cm is placed over a paper. Qhat is the shift in the letters?
(a) 4 cm
(b) 2 cm
(c) 1 cm
(d) None of these
Answer:
(b) 2 cm
Hint:
Normal shift, x = \(t\left(1-\frac{1}{\mu}\right)=6\left(1-\frac{1}{1.5}\right)\) cm = 2cm

Question 20.
Light traveling from a transparent medium to air undergoes total internal reflection at an angle of incident of 45°. Then refractive index of the medium may be
(a) 1.5
(b) 1.3
(c) 1.1
(d) \(\frac { 1 }{ √2 }\)
Answer:
(a) 1.5
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac{1}{\sin 45^{\circ}}\) = √2 = 1.414 ≈ 1.5.

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Question 21.
A point source of light is placed 4 m below the surface of water of refractive index 5/3. The minimum diameter of a disc which should be placed over the source, on the surface of water to cut-off all light coming out of water is
(a) infinite
(b) 6 cm
(c) 4 cm
(d) 3 cm
Answer:
(b) 6 cm
Hint:
r = \(\frac{h}{\sqrt{\mu^{2}-1}}\) = \(\frac{4}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}\) = 3 cm

Question 22.
In optical fibres, propagation of light is due to
(a) diffraction
(b) total internal reflection
(c) reflection
(d) refraction
Answer:
(b) total internal reflection
Hint:
In optical fiberes, light propagates due to tatal internal reflection.

Question 23.
Sparkling of diamond is due to
(a) reflection
(b) dispersion
(c) total internal reflection
(d) high refractive index of diamond
Answer:
(c) total internal reflection

Question 24.
For a given lens, the magnification was found to be twice as large as when the object was 0.15m distant from it as when the distance was 0.2 m. The focal length of the lens is
(a) 1.5 m
(b) 0.20 m
(c) 0.10 m
(d) 0.05 m
Asnwer:
(c) 0.10 m
Hint:
Here m1 = 2m2
\(\frac { f }{ f-0.15 }\) = 2 \(\frac { f }{ f-0.20 }\)
2f – 0.30 = f- 0.20 ; f = 0.10 m

Question 25.
Two lenses of focal lengths f1 and f2 are kept in contact coaxially. The resultant power of combination will be
(a) \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { { f }_{ 1 }-f }_{ 2 } } \)
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
(c) \({ f }_{ 1 }{ +f }_{ 2 }\)
(d) \(\frac { { f }_{ 1 } }{ { f }_{ 2 } } +\frac { { f }_{ 2 } }{ { f }_{ 1 } } \)
Answer:
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
Hint:
P = \(\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \).

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Question 26.
Two lenses of power 3D and -ID are kept in contact. What is focal length and nature of combined lens?
(a) 50 cm, convex
(b) 200 cm, convex
(c) 50 cm, concave
(d) 200 cm, concave
Answer:
(a) 50 cm, convex
Hint:
P = P1 + P2 = 3 – 1 = 2D
F= \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 2 }\) m = 50 cm

Question 27.
If two thin lenses are kept coaxially together, then their power is proportional (R1,R2) being the radii of curved surfaces) to
(a) R1 + R2
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
(c) \(\left[\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1} \mathrm{R}_{2}}\right]\)
(d) None of these
Answer:
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
Hint:
P = \(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{2}{\mathrm{R}_{1}}+\frac{2}{\mathrm{R}_{2}}\) = 2\(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
or p ∝ \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)

Question 28.
A ray incident at 15° on one refracting surface of a prism of angle 60°, suffers a deviation of 55°. What is the angle of emergance?
(a) 95°
(b) 45°
(c) 30°
(d) none of these
Answer:
(d) none of these
Hint:
A + δ = i + each
60° + 55° = 15° + each e = 115 -15 =100°

Question 29.
Dispersion of light is caused due to
(a) Wavelength
(b) intensity of light
(c) density of medium
(d) none of these
Answer:
(a) Wavelength
Hint:
Dispersion is due to the dependence of the speed of a wave on its wavelength in any medium.

Question 30.
White light is incident on one of the refracting surfaces of a prism of angle 50°. If the refractive indices for red and blue colours are 1.641 and 1.659 respectively, the angular separation between these two colours when they emerge out of the prism is
(a) 0.9°
(b) 0.09°
(c) 1.8°
(d) 1.2°
Answer:
(b) 0.09°
Hint:
Angular dispersion,
δB – δR = (µB – µR) A
= (1.659 – 1.641) x 5° = 0.09°

Question 31.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 32.
A setting sun appears to be at an altitude higher than it really is. This is because of
(a) absorption of light
(b) reflection of light
(c) refraction of light
(d) dispersion of light
Answer:
(c) refraction of light
Hint:
This is due to refraction of light by the earth’s atmosphere.

Question 33.
The reddish appearance of rising and setting sun is due to
(a) reflection of light
(b) diffraction of light
(c) scattering of light
(d) interference of light
Answer:
(c) scattering of light
Hint:
The reddish appearance of the rising and the setting sun is due to scattering of light.

Question 34.
In the formation of a rainbow, the light from the sun on water droplets undergoes
(a) dispersion only
(b) only total internal reflection
(c) dispersion and total internal reflection
(d) none of the above
Answer:
(c) dispersion and total internal reflection
Hint:
Rainow is formed due to dispersion of sunlight by raindrops which also deviate the colours by total internal reflection.

Question 35.
The angular magnification of a simple microscope can be increased by increasing
(a) focal length of lens
(b) size of object
(c) aperture of lens
(d) power of lens
Answer:
(b) size of object

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Question 36.
For compound microscope f0 = 1 cm, fe = 2.5 cm. An object is placed at distance 1.2 cm from objective lens. What should be length of microscope for normal adjustment?
(a) 8.5 cm
(b) 8.3 cm
(c) 6.5 cm
(d) 6.3 cm
Answer:
(a) 8.5 cm
Hint:
In the normal adjustment of a compound microscope, L = υ0 + υe = \(\frac{υ_{0} f_{e}}{υ_{0}+f_{e}}\) + fe = \(\frac { 1.2×1 }{ -1.2+1 }\) + 2.5 = 6 + 25 = 8.5 cm

Question 37.
Magnifying power of an astronomical telescope for normal vision with usual notation is
(a) -f0 / fe
(b) -f0 x fe
(c) -f0 / f0
(d) -f0 + fe
Answer:
(a) -f0 / fe
Hint:
In normal adjustment of the telescope, m = -f0 / fe

Question 38.
F1 and F2 are focal length of objective and eyepiece respectively of the telescope. The angular magnification for the given telescope is equal to
(a) \(\frac {F_{1}}{F_{2}}\)
(b) \(\frac { { F }_{ 2 } }{ { F }_{ 1 } } \)
(c) \(\frac { { F }_{ 1 }{ F }_{ 2 } }{ { F }_{ 1 }+{ F }_{ 2 } } \)
(d) \(\frac { { F }_{ 1 }+{ F }_{ 2 } }{ { F }_{ 1 }{ F }_{ 2 } } \)
Answer:
(a) \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)
Hint:
In normal adjustment of the telescope, \(\left| m \right| \) = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) = \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)

Question 39.
Focal length of objective and eyepiece of telescope are 200 cm and 4 cm respectively. What is length of telescope for normal adjustment?
(a) 196 cm
(b) 204 cm
(c) 250 cm
(d) 225 cm
Answer:
(b) 204 cm
Hint L = \(-{ f }_{ 0 }+{ f }_{ e }\) = 200 + 4 = 204 cm

Question 40.
For normal vision, what is minimum distance of object from eye?
(a) 30 cm
(b) 25 cm
(c) Infinite
(d) 40 cm
Answer:
(b) 25 cm
Hint:
For normal eye, the least distance of distinct vision is 25 cm.

SamacheerKalvi.Guru

Question 41.
The focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5° ; the angle subtended by the moon’s image will be
(a) 10°
(b) 250
(e) 100°
(d) 75°
Answer:
b) 250
Hint ;
m = \(\frac { β }{ α }\) β = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) ; α = \(\frac { 100 }{ 2 }\) x 0.5° = 25°

Question 42.
A person cannot clearly see distance more than 40 cm. He is advised to use lens of power,
(a) – 2.5 D
(b) 2.5 D
(c) – 6.25 D
(d) 1.5 D
Answer:
(a) – 2.5 D
Hint;
For the remedial lens, u = ∞,
v = – 40 cm = – 0.40 m
∴ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ -0.40 }\) – \(\frac { 1 }{ ∞ }\) = -2.5 ⇒ P = 2.5D

Question 43.
The light gathering power of a camera lens depends on
(a) its diameter only
(b) ratio of diameter and focal length
(c) product of focal length and diameter
(d) wavelength of the light used
Answer:
(a) its diameter only
Hint:
The light gathering power of a camera lens is proportional to its area or to the square of its diameter.

Question 44.
Amount of light entering into the camera depends upon
(a) focal length of objective lens
(b) product of focal length and diameter of the objective lens
(c) distance of object from camera
(d) aperture setting of the camera
Answer:
(d) aperture setting of the camera
Hint:
The amount of light entering into the camera depends upon the aperture setting of the camera.

Question 45.
Line spectrum can be obtained from
(a) sun
(b) candle
(c) mercury vapour lamp
(d) electric bulb
Answer:
(c) mercury vapour lamp

SamacheerKalvi.Guru

Question 46.
The Production of band spectra is caused by
(a) atomic nuclei
(b) hot metals
(c) molecules
(d) electrons
Answer:
(c) molecules

Question 47.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed, n = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 48.
A point source kept at a distance of 1000 m has a illumination I. To change the illumination to 161, the new distance should become
(a) 250 m
(b) 500 m
(c) 750 m
(d) 800 m
Answer:
(a) 250 m
Hint:
\(\frac{I_{2}}{I_{1}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}\)
\(\frac{16 \mathrm{I}_{1}}{\mathrm{I}_{1}}=\frac{(1000)^{2}}{r_{2}^{2}}\) ; r2 \(\frac { 1000 }{ 4 }\) = 250 cm

Question 49.
A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual will be
(a) 22.5 cm
(b) 7.5 cm
(c) 30 cm
(d) 45 cm
Answer:
(b) 7.5 cm
Hint:
For virtual images, m = \(\frac { -v }{ u }\) = + 2 or v = – 2u
As \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\)
∴ \(\frac { 1 }{ u }\) – \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) or \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) ⇒ u = -7.5 cm

Question 50.
when a ray of light enters a glass slab, then
(a) its frequency and velocity change
(b) only frequency changes
(c) its frequency and wavelength change
(d) its frequency does not change
Answer:
(d) its frequency does not change
Hint:
When a ray of light enters a glass slab, its velocity and wavelength change while frequency does not change.

SamacheerKalvi.Guru

Question 51.
A light wave of frequency u and wavelength A travels from air to glass. Then,
(a) y changes
(b) y does not change, A changes
(c) A does not change
(d) y and A change
Answer:
(b) y does not change, A changes
Hint:
Same reasoning as in the above question.

Question 52.
In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium.
(a) the frequency is different
(b) the coefficient of elasticity is different
(c) the speed is different
(d) the amplitude is smaller
Answer:
(c) the speed is different
Hint:
Speed of light in second medium is different than that in first medium.

Question 53.
A ray of light having wavelength 720 nm enters in a glass of refractive index 1.5. The wavelength of the ray within the glass will be
(a) 360 nm
(b) 480 nm
(e) 720 nm
(d) 1080 nm
Answer:
(b) 480 nm
Hint:
\({ \lambda }_{ g }\) = \(\frac { { \lambda }_{ 0 } }{ \mu } \) = \(\frac { 790nm }{ 1.5 }\) = 480nm

Question 54.
Brilliance of a diamond is due to
(a) shape
(b) cutting
(c) reflection
(d) total internal reflection
Answer:
(d) total internal reflection
Hint:
Brilliance of a diamond is due to total internal reflection of light.

Question 55.
An endoscope is employed by a physician to view the internal parts of a body organ. If is based on the principle of
(a) refraction
(b) reflection
(c) total internal reflection
(d) dispersion
Answer:
(c) total internal reflection
Hint:
An endoscope is made of optical fibres which work on the principle of total internal reflection.

SamacheerKalvi.Guru

Question 56.
‘Mirage’ is a phenomenon due to
(a) reflection of light
(b) refraction of light
(c) total internal reflection of light
(d) diffraction of light Hint Mirage occurs due to total internal reflection of light.
Answer:
(c) total internal reflection of light

Question 57.
Two lenses of power +12D and -2D are combined together. What is their equivalent focal length?
(a) 10 cm
(b) 12.5 cm
(c) 16.6 cm
(d) 8.33 cm
Answer:
(a) 10 cm
Hint:
P = P1 + P2 = + 12 – 2 = 10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 10 }\)m = 10 cm

Question 58.
If two lenses of power + 1.5 D and + 1.0 D are placed in contact, then the effective power of combination will be
(a) 2.5 D
(b) 1.5 D
(c) 0.5 D
(d) 3.25 D
Answer:
(a) 2.5 D
Hint:
P = P1 + P2 = + 1.5 + 1.0 = + 2.5D

Question 59.
The angle of a prism is 6° and its refractive index for green light is 1.5. If a green ray passes through it, the deviation will be
(a) 30°
(b) 15°
(c) 3°
(d) 0°
Answer:
(c) 3°
Hint:
5 = (p – 1) A = (1.5 – 1) x 6 = 3°

Question 60.
Sky appears to be blue in clear atmosphere due to light’s
(a) diffraction
(b) dispersion
(c) scattering
(d) polarisation
Answer:
(c) scattering
Hint:
Sky appears blue due to scattering of light by atmospheric modecules.

SamacheerKalvi.Guru

Question 61.
One can not see through fog, because
(a) fog absorbs the light
(b) light suffers total reflection at droplets
(c) refractive index of the fog is infinity
(d) light is scattered by the droplets
Answer:
(d) light is scattered by the droplets

Question 62.
Fraunhofer lines of the solar system is an example of
(a) emission lines spectrum
(b) emission band spectrum
(c) continuous emission spectrum
(d) line absorption spectrum
Answer:
(d) line absorption spectrum
Hint:
Fraunhofer lines is an example of line absorption spectrum.

Question 63.
A person using a lens as a sample microscope sees an
(a) inverted virtual image
(b) inverted real magnified image
(c) upright virtual image
(d) upright real magnified image
Answer:
(d) upright real magnified image
Hint:
A person sees an upright virtual image in a simple microscope.

Question 64.
Four lenses of focal length + 10 cm, + 50 cm, + 100 cm and + 200 cm are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be
(a) + 10 cm
(b) + 50 cm
(c) + 100 cm
(d) + 200 cm
Answer:
(a) + 10 cm
Hint:
To produce the largest magnification, the eyepiece should have minimum focal length.

Question 65.
The camera lens has an aperture of f and the exposure time is 1/60 s. What will be the new exposure time if the aperture become 1.4f?
(a) \(\frac { 1 }{ 42 }\) s
(b) \(\frac { 1 }{ 56 }\) s
(c) \(\frac { 1 }{ 72 }\) s
(d) \(\frac { 1 }{ 31 }\) s
Answer:
(d) \(\frac { 1 }{ 31 }\) s
Hint:
Time of exposure ∝ (f – number)2
\(\frac{t}{\left(\frac{1}{60}\right)}=\left(\frac{1.4}{1}\right)^{2}\) ⇒ t = \(\frac { 1.4×1.4 }{ 60 }\) ≈ \(\frac { 1 }{ 31 }\) s.

SamacheerKalvi.Guru

Question 66.
For a person near point of vision is 100 cm. Then the power of lens he must wear so as have normal vision, should be
(a) + ID
(b) – ID
(c) + 3D
(d) – 3D
Answer:
(c) + 3D
Hint:
f = \(\frac { yD }{ y-D }\) = \(\frac { 100×25 }{ 100-25 }\) = \(\frac { 100 }{ 3 }\) cm = \(\frac { 1 }{ 3 }\) cm ; P = \(\frac { 1 }{ f }\) = +3d

Question 67.
Ray optics is valid, when characteristic dimension ions are
(a) much smaller than the wavelength of light
(b) much larger than the wavelength of light
(c) of the same order as the wavelength of light
(d) of the order of one millimetre
Hint:
Ray optics is valid, when characteristic dimensions are much larger than the wavelength of
light.

Question 68.
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirrorwillbe
(a) 12 feet
(b) 3 feet
(e) 6 feet
(d) any length
Hint:
inimum height of mirror required for seeing full image Height of the man = 3 feet.

Question 69.
The refractive index of water is 1.33. What will be the speed of light in water?
(a) 3 x 108 ms-1
(b) 2.26 x 108 ms-1
(c) 4 x 108 ms-1
(d) 1.33 x 108 ms-1
Answer:
(b) 2.26 x 108 ms-1
Hint:
As, μ = \(\frac { c }{ v }\) ⇒ v = \(\frac { c }{ μ }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 1.33 } \) = 2.26 x 108 ms-1

Question 70.
A beam of monochromatic light is refracted from vacuum into a medium of refractive index
1.5. The wavelength of refracted light will be
(a) same
(b) dependent on intensity of refracted light
(c) larger
(d) smaller
Answer:
(d) smaller
Hint:
As light enters the medium, its wavelength decreases and becomes equal to \(\frac { λ }{ μ }\).

SamacheerKalvi.Guru

Question 71.
Optical fibers are based on
(a) total internal reflection
(b) less scattering
(c) refraGtion
(d) less absorption coefficient
Answer:
(a) total internal reflection

Question 72.
A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the lens. Then, its focal length will
(a) become zero
(b) becomes infinite
(e) remain unchanged
(d) become small, but non-zero
Answer:
(b) becomes infinite
Hint:
\(\frac { 1 }{ { f }_{ 1 } } \) = \(\left(\frac{\mu_{g}}{\mu_{e}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = (1 – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = 0 ; f1 = ∞

Question 73.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power of the combination (in diopter) is
(a) zero
(b) 25
(c) 50
(d) infinite
Answer:
(a) zero
Hint:
\(\frac { 1 }{ F }\) = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 1 }{ +25 }\) + \(\frac { 1 }{ -25 }\) = 0 ; P = \(\frac { 1 }{ F }\) = 0

Question 74.
The focal length of a converging lens is measured for violet, green and red colours. If is fV, fG and fR respectively. We will get
(a) fV = fG
(b) fG = fR
(c) fV < fR
(d) fV > fR
Answer:
(c) fV < fR
Hint:
\(\frac { 1 }{ f }\) = (μ – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) i.e., f ∝ \(\frac { 1 }{ μ – 1 }\)
As μV > μR, so, fV < fR

Question 75.
Rainbow is formed due to combination of
(a) refraction and scattering
(b) refraction and absorption
(c) dispersion and total internal reflection
(d dispersion and focusing
Answer:
(c) dispersion and total internal reflection
Hint:
Rainbow is formed due to dispersion and total internal reflection of sunlight by raindrops.

SamacheerKalvi.Guru

Question 76.
The blue colour of the sky is due to the phenomenon of
(a) scattering
(b) dispersion
(c) reflection
(d) refraction
Answer:
(a) scattering
Hint:
The blue colour of the sky is due to the scattering of sunlight by atmospheric molecules.

Question 77.
An astronomical telescope often fold angular magnification has a length of 44 cm. The focal length of the object is
(a) 4 cm
(b) 40 cm
(c) 44 cm
(d) 440 cm
Answer:
(b) 40 cm
Hint:
Here, f0 + fe = 44 cm
m =\(\frac { { f }_{ 0 } }{ { f }_{ e } } \) (or) f0 = 10 fe
∴ 10 fe + 10 fe = 44 cm (or) fe = 4 cm
Hence, f0 = 10 x 4 (or) f0 = 40 cm

Question 78.
Exposure time of a camera lens at the \(\frac { f }{ 2.8 }\) Setting is \(\frac { 1 }{ 200 }\) second. The correct time of exposure at \(\frac { f }{ 5.6 }\) is
(a) 0.20 second
(b) 0.40 second
(c) 0.02 second
(d) 0.04 second
Answer:
(c) 0.02 second
Hint:
Time of exposure, t ∝ (f- number)2
∴ \(\frac{t}{\left(\frac{1}{200}\right)}=\left(\frac{5.6}{2.8}\right)^{2}\) = 4 ⇒ t = 0.02s

Question 79.
Which of the following is not due to total internal reflection?
(a) Working of optical fibre
(b) Difference between apparent and real depth of a pond
(c) Mirage on hot summer day
(d) Brilliance of diamond
Answer:
(b) Difference between apparent and real depth of a pond
Hint:
Difference between apparent and real depth of a pond is due to refraction of light and the other three phenomena involve total internal reflection.

Question 80.
An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be
(a) 0.25 m
(b) 0.5 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:
Distance between object and image = 0.5 + 0.5 = 1.0 m.

SamacheerKalvi.Guru

Question 81.
An observer moves towards a stationary plane mirror at a speed of 4 ms-1 with what speed – will his image move towards him?
(a) 2 ms-1
(b) 4 ms-1
(c) 8 ms-1
(d) the image will stay at rest
Answer:
(c) 8 ms-1
Hint:
Speed of the image towards the observer = 2 x 4 = 8 ms-1

Question 82.
If two mirrors are kept at 60° to each other and a body is placed at the middle, then total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(c) five
Hint:
Number of images formed = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 83.
If an object is placed at 10 cm infront of a concave mirror of focal length 15 cm. The magnification of image is
(a) -1.5
(b) 1.5
(c) -3
(d) 3
Answer:
d) 3
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { -15 }{ -15(-10) }\) = +3

Question 84.
An object of length 2.5 cm is placed at the principal axis of a concave mirror at a distance 1.5f. The image height is
(a) + 5 m
(b) -5 cm
(c) – 10 cm
(d) + 1 cm
Answer:
(b) -5 cm
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { f }{ f-1.5f}\) = -2cm
Height image = m x height of object -2 x 2.5 = – 5 cm

Question 85.
Which of the following mirror is used by a dentist to examine a small cavity?
(a) Concave mirror
(b) Convex mirror
(c) Combination of (a) and (b)
(d) None of these
Answer:
(a) Concave mirror
Hint:
A concave mirror, because it forms erect and enlarged image when held closer to the cavity.

SamacheerKalvi.Guru

Question 86.
When a ray of light enters from one medium to another, then which of the following does not change?
(a) Frequency
(b) Wavelength
(c) Speed
(d) Amplitude
Answer:
(a) Frequency
Hint:
Only frequency remains unchanged.

Question 87.
When light travels from one medium to the other medium of which the refractive index is different, then which of the following will change?
(a) Frequency, wavelength and velocity
(6) Frequency and wavelength
(c) Frequency and velocity
(d) Wavelength and velocity
Answer:
(d) Wavelength and velocity
Hint:
Wavelength and velocity will change while frequency remains unchanged.

Question 88.
The time taken by the light to cross a glass of thickness 4 mm and refractive index (μ = 3), will be
(a) 4 x 10-11 sec
(b) 16 x 10-11 sec
(c) 8 x 10-11 sec
(d) 24 x 10-10 sec
Answer:
(a) 4 x 10-11 sec
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-50
\(\frac { d }{ c/a }\) = \(\frac { μd }{ c }\) = \(\frac { 3×4×10^{-3} }{ 3×10^{8} }\) =  4 × 10-11 s

Question 89.
The critical angle of a medium with respect to air is 45°. The refractive index of medium is
(a) 1.41
(b) 1.2
(c) 1.5
(d) 2
Answer:
(a) 1.41
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { 1 }{ sin 45° }\) = \(\frac { 1 }{ 1/√2 }\) ≈ 1.41

Question 90.
If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in the medium is
(a) 6 x 108 m/sec
(b) 2 x 108 m/sec
(c) 3 x 108 m/sec
(d) 1.5 x 108 m/sec
Answer:
(c) 3 x 108 m/sec
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { c }{ v }\)

SamacheerKalvi.Guru

Question 91.
When a ray of light enter from one medium to another, its velocity is doubled. The critical angle for the ray for two internal reflection will be
(a) 30°
(b) 60°
(c) 90°
(d) Information is incomplete
Answer:
(a) 30°
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \) =\(\frac { { 2v }_{ 2 } }{ { v }_{ 2 } } \) = 2 ; sin ic = \(\frac { 1 }{ 2 }\) ∴ic = 30°

Question 92.
A driver at a depth 12 m inside water (μ = 4/3) see the sky in a cone of semi-vertical angle is
(a) sin-1\(\left( \frac { 4 }{ 3 } \right) \)
(b) tan-1\(\left( \frac { 4 }{ 3 } \right) \)
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
(d) 90°
Answer:
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
Hint:
Required semi vertical angle = Critical angle ic = sin-1 \(\frac { 1 }{ μ }\) = sin-1\(\left( \frac { 3 }{ 4 } \right) \)

Question 93.
The principle behind optical fibres is
(a) total internal reflection
(b) total external reflection
(c) both (a) and (b)
(d) diffraction
Answer:
(a) total internal reflection
Hint:
Optical fibres work on the principle of total internal reflection.

Question 94.
Air bubble in water behaves as
(а) some times concave, sometimes convex lens
(b) concave lens
(c) convex lens
(d) always refracting surface
Answer:
(b) concave lens

Question 95.
A convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is
(a) -1.5 D
(b) – 6.5 D
(c) + 6.6 D
(d) + 6.5 D
Answer:
(a) -1.5 D
Hint:
P = P1 + P2 = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 100 }{ 40 }\) + \(\frac { 100 }{ -25 }\) = -1.5d.

SamacheerKalvi.Guru

Question 96.
Two thin lenses, one of focal length + 60 cm and the other of focal length – 20 cm are put in contact, the combined focal length is,
(a) 15 cm
(b) – 15 cm
(c) – 30 cm
(d) 30 cm
Answer:
(c) – 30 cm
Hint:
F = \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { f }_{ 1 }{ +f }_{ 2 } } \) = \(\frac { 60 ×(-20) }{ 60-20 }\) = 30 cm

Question 97.
How does refractive index (μ) of a material vary with respect to wavelength (λ). (A and B are constants).
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
(b) μ = A + Bλ2
(c) μ = A + \(\frac { B }{ λ }\)
(d) μ = A + Bλ
Answer:
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
Hint:
According to cauchy’s relation, μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)

Question 98.
a prism of a refracting angle 60° is made with a material of refractive index p. For a certain wavelength of light, the angle of minimum deviation is 30°. For this wavelength, the value of p of material is
(a) 1.820
(b) 1.414
(c) 1.503
(d) 1.231
Answer:
(b) 1.414
Hint:
μ = \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin 30^{\circ}}\) = \(\frac { sin 45° }{ sin 30° }\) = \(\frac { 1 }{ √2 }\) x \(\frac { 2 }{ 1 }\) √2 = 1.414

Question 99.
Refractive index of red and violet light are 1.52 and 1.54 respectively. If the angle of prism is 10°, the angular dispersion will be
(a) 0.02°
(b) 0.20°
(c) 3.06°
(d) 30.6°
Answer:
(b) 0.20°
Hint:
Angular dispersion = δV – δR = A(μV – μV) = 10° (1.54 – 1.52) = 0.20°

Question 100.
In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then magnifying power is
(a) \(\frac { 25 }{ f }\)
(b) 1 + \(\frac { 25 }{ f }\)
(c) \(\frac { f }{ 25 }\)
(d) \(\frac { f }{ 25 }\) + 1
Answer:
(b) 1 + \(\frac { 25 }{ f }\)
Hint:
When the final image is formed at the least distance of distinct vision in a simple microscope,
m = 1 + \(\frac { 25 }{ f }\)

SamacheerKalvi.Guru

Question 101.
Magnification at least distance of distinct vision of a simple microscope of focal length 5 cm is
(a) 2
(b) 5
(c) 4
(d) 6
Answer:
(d) 6
Hint:
m = 1 + \(\frac { 25 }{ f }\) = 1 + \(\frac { 25 }{ 5 }\) = 6

Question 102.
Magnification of a compound microscope is 30. Focal length of eyepiece is 5 cm and the image is formed at a distance of distinct vision of 25 cm. The magnification of the objective
lens is
(a) 6
(b) 5
(c) 7.5
(d) 10
Answer:
(b) 5
Hint:
me = \(\frac { v }{ u}\) = \(\frac { D }{ { u }_{ c } } \) = 1 + \(\frac { D }{ { f }_{ e } } \) = 1 + \(\frac { 25 }{ 5 }\) = 6
For the compound microscope, m = m0 x me ⇒ 30 = m0 x 6 (or) m0 = 5

Question 103.
The astronomical microscope consists of objective and eyepiece. The focal length of the objective is
(a) equal to that of the eyepiece
(b) shorter than that of the eyepiece
(c) greater than that of the eyepiece
(d) five times shorter than that of eyepiece
Answer:
(c) greater than that of the eyepiece
Hint:
For producing large magnification,f0 > fe

Question 104.
The number of lenses in terrestrial telescope is
(a) 2
(b) 4
(c) 3
(d) 6
Answer:
(c) 3
Hint:
A terrestrial telescope consists of three lenses: objective, erecting lens and eyepiece.

Question 105.
An achromatic combination of lenses is formed by joining
(a) 2 convex lens
(b) 1 convex, 1 concave lens
(c) 2 concave lenses
(d) 1 convex and 1 plane mirror
Answer:
(b) 1 convex, 1 concave lens
Hint:
An achromatic doublet should satisfy the condition \(\frac { { w }_{ 1 } }{ { f }_{ 1 } } \) + \(\frac { { w }_{ 2 } }{ { f }_{ 2 } } \) = 0.

SamacheerKalvi.Guru

Question 106.
The amount of light received by a camera depends upon
(a) diameter only
(b) ratio of focal length and diameter
(c) product of focal length and diameter
(d) only one of the focal length
Answer:
(b) ratio of focal length and diameter
Hint:
The amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture.

Question 107.
Myopia is corrected by using a
(a) cylindrical lens
(b) bifocal lens
(c) convex lens
(d) concave lens
Answer:
(d) concave lens

Question 108.
The critical angle for total internal reflection in diamond is 24.5°. The angle refractive index of diamond is
(a) 2.41
(b) 1.41
(c) 2.59
(d) 1.59
Answer:
(a) 2.41
Hint:
μ = \(\frac { 1 }{ { sin i }_{ c } } \) = \(\frac { 1 }{ sin 24.5° }\) = 2.41

Question 109.
When a glass lens with μ = 1.47 is immersed in a trough of liquid, it looks to be disappeared. The liquid in the trough could be
(a) water
(b) kerosene
(c) glycerine
(d) alcohol
Answer:
(c) glycerine
Hint:
Glass lens will disappear, if µl = µg.

Question 110.
In optical fibres, the refractive index of the core is
(a) greater than that of the cladding
(b) equal to that of the cladding
(c) smaller than that of the cladding
(d) independent of that of the cladding
Answer:
(a) greater than that of the cladding
Hint:
In optical fibres, refractive index of core material > refractive index of the cladding.

SamacheerKalvi.Guru

Question 111.
For a wavelength of light ‘λ’ and scattering object of size ‘a’, all wavelength are scattered nearly equally, if
(a) a = λ
(b) a >> λ
(c) a << λ
(d) a ≥ λ
Answer: (b) a >> λ
Hint:
For a >> λ, the scattering power is not selective.

Question 112.
Two coherent monochromatic light beams of intensities I and 4I are supperposed. The maximum and minimum possible intensities in the resulting beams are
(a) 5I and I
(b) 9I and I
(c) 5I and 3I
(d) 9I and 3I
Answer:
(b) 9I and I
Hint:
Imax = \((\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}+\sqrt{1})^{2}\) = 9I
Imax = \((\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}-\sqrt{1})^{2}\) = I

Question 113.
screen is doubled. The fringe width is
(a) unchanged
(b) halved
(c) doubled
(d) quadrupled
Answer:
(d) quadrupled
Hint:
β = \(\frac { λD }{ d }\) ; β = \(\frac { λ ×2D }{ D/2 }\) = 4β

Question 114.
In a young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen, when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by
(a) 12
(b) 18
(c) 24
(d) 30
Answer:
(b) 18
Hint:
n1λ1= n2λ2 ⇒12 x 600 = n2 x 400 or n2 = 18

Question 115.
Consider ffaunhoffer diffraction pattern obtained with a single slit illuminated at normal incident. At the angular position of the first diffraction minimum the phase difference between the wavelets from the opposite edges of the slits is
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) 2π
(d) π
Answer:
(c) 2π
Hint:
At the angular position of first minimum wavelets from opposite edges of the slits have a path difference of X and a phase difference of 2π radian.

SamacheerKalvi.Guru

Question 116.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm
Answer:
(c) 2.4 cm
Hint:
Distance between the first dark fringes on either side = width of central maximum
= \(\frac { 2Dλ }{ d }\) = \(\frac{2 \times 2 \times 600 \times 10^{-9}}{1.00 \times 10^{-3}}\) m = 2.4 x 10-3 m = 2.4mm

Question 117.
A young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.
(a) hyperbola
(b) circle
(c) straight line
(d) parabola
Answer:
(a) hyperbola
Hint:
In young’s double slit experiment, the fringes obtained are hyperbolic in shape. But in a small interference pattern, the fringes appear straight.

Question 118.
The initial shape of the wavefront of the beam is
(a) planar
(b) convex
(c) concave
(d) convex near the axis and concave near the periphery
Answer:
(a) planar
Hint:
As the beam is initially parallel, the shape of wavefront is planar.

Question 119.
The angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index μ) is
(a) sin-1 (μ)
(b) sin-1 \(\left( \frac { 1 }{ μ } \right) \)
(c) tan-1 \(\left( \frac { 1 }{ μ } \right) \)
(d) tan-1 (μ)
Answer:
(d) tan-1 (μ)
According to Brewster’s law,
μ = tan ip ∴ ip = tan-1 (μ)

Question 120.
According to Huygen’s principle, light is a form of
(a) particle
(b) rays
(c) wave
(d) none of the above
Answer:
(c) wave
Hint:
According to Huygen’s principle, light travels in the form of a longitudinal wave.

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Question 121.
Which one of the following phenomena is not explained by Huygen’s construction of wavefront?
(a) refraction
(b) reflection
(c) diffraction
(d) origin of spectra
Answer:
(d) origin of spectra
Hint:
Huygen’s construction of wavefront cannot explain origin of spectra which can be explained on the basis of quantum theory.

Samacheer Kalvi 12th Physics Optics Additional Problems

Question 1.
Light from a point source in air falls on a convex spherical glass surface (n = 1.5, radius of curvature = 20 cm). The distance of light source from the glass surface is 100 cm. At
What position is the image formed?
Solution:
n1 = 1; n2 = 1.5
u = 100cm ; R = + 20 cm
(R is + Ve for a convex refracting surfce)
As \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}\) = \(\frac{n_{2}-n_{1}}{R}\)
\(\frac{1.5}{v}+\frac{1}{100}\) = \(\frac{1.5-1}{20}=\frac{1}{40}\)
\(\frac { 3 }{ 2v }\) = \(\frac{1}{40}-\frac{1}{100}=\frac{5-2}{200}=\frac{3}{200}\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ 100 }\)
v = 100 cm
Thus the image is formed at a distance of 100 cm from the glass surface in the direction of incident light.

Question 2.
Find the value of critical angle for a material of refractive index √3 .
Solution:
Here, n = √3
sin ic = \(\frac { 1 }{ n }\) = \(\frac { 1 }{ √3 }\) = \(\frac { √3 }{ 3 }\)
∴ critical angle, ic = 35.3°

Question 3.
The radius of curvature of each face of biconcave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air.
Solution:
Here n = 1.5 ; R1 = – 30 cm ; R2 = 30 cm
Using len’s maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5-1)\(\left[\frac{1}{-30}-\frac{1}{30}\right]\) = 0.5 × \(\left(\frac{-2}{30}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 30 }\)
f = -30

Question 4.
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length is 12 cm. What is the refractive index of glass?
Solution:
f = +12 cm ; R1 = 10 cm ; R2 = – 15 cm ; n = ?
As, \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\frac { 1 }{ 12 }\) = (n – 1) \(\left(\frac{1}{10}+\frac{1}{15}\right)\) = (n – 1) x latex]\frac { 5 }{ 30 }[/latex]
(n – 1) = \(\frac { 6 }{ 12 }\) = 0.5
n = 0.5 +1
n = 1.5

Question 5.
A double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height is placed at a distance of 10 cm from the lens. Find the position, nature and size of the image.
Solution:
Here n = 1.5 ; R1= +20 cm ; R2 = – 20 cm
Using lens maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
(1.5-1) = \(\left[\frac{1}{20}-\frac{1}{-20}\right]\) 0.5 x \(\left(\frac{2}{20}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 20 }\)
f = 20 cm
Now , u = -10 cm andf = + 20 cm
From thin lens formula, \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ 20 }\) – \(\frac { 1 }{ 10 }\) = \(\frac { 1 }{ 20 }\)
∴ v = -20 cm
Magnification, m = \(\frac{h_{2}}{h_{1}}\)= \(\frac { v }{ u }\)
\(\frac{h_{2}}{5cm}\) = \(\frac { -20 }{ -10 }\)
h2
Hence a virtual and erect image of height 10 cm is formed at a distance of 20 cm from the lens on the same side as the the object.

SamacheerKalvi.Guru

Question 6.
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.
Solution.
Heref= 20 cm, m = + 4 for a virtual image.
To calculate u, we have
m = \(\frac { f }{ u+f }\)

Question 7.
The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power.
Solution:
Here, n = 105 ; R1 = + 40cm = 0.40 m
R2 = – 40cm = – 0.40 m
Power (p) = \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5 – 1)\(\left[\frac{1}{0.40}-\frac{1}{(-0.40)}\right]\) = 0.5 x \(\frac { 2 }{ 0.40 }\)
p = 2.5 D

Question 8.
A ray of light incident on an equilateral glass prism shows minimum deviation of 30°. Calculate the speed of light through the prism.
Solution:
Here, A = 60° ; D = 30°
Refractive index, n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) = \(\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}\)
n = \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\) = √2
n = 1.414
Velocity or light in glass, v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^{8}}{1.414}\)
v = 2.12 x 108 ms-1

Question 9.
Two sources of intensity I and 41 are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase
(i) Zero
(ii) \(\frac { π }{ 2 }\)
(iii) π
Solution:
The resultant intensity at a point where phase difference is Φ is
IR+I1+ I2 2\(+\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) cos Φ
As I1 + I and I2 = 4I, therefore,
IR = I + 4I + 2\(\sqrt{I.4I}\) cos Φ
IR = 5I + 4I cos Φ
(i) When Φ = 0 ; IR = 5I + 4I cos 0 = 9I
(ii) When Φ = \(\frac { π }{ 2 }\) ; IR = 5I + 4I cos \(\frac { π }{ 2 }\) = 5I
(iii) When Φ = π ; IR = 5I + 4I cos π = 51 – 41 = I
Φ = 0 ; IR = 9I
Φ = \(\frac { π }{ 2 }\) ; IR = 5I
Φ = π ; IR = I

Question 10.
Assume that light of wavelength 600 A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Solution:.
The limit of resolution of a telescope, dθ = \(\frac { 1.22λ }{ D }\)
D = 100 inch = 254 cm
[∴ 1 inch = 2.54 cm]
λ = 6000 Å = 6000 x 10-10 m/s
dθ = \(\frac{1.22 \times 6000 \times 10^{-10}}{254 \times 10^{-2}}\) = 2.9 x 10-7
dθ = 2.9 x 10-7 rad.

SamacheerKalvi.Guru

Question 11.
Two polarising sheet have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?
Solution:
Here, I = \(\frac{I_{o}}{2}\)
Using Malus law,,
I = Io cos2 θ
\(\frac{I_{o}}{2}\) = Io cos2 θ
cos θ ± \(\frac { 1 }{ √2 }\)
θ = ± 45°, ± 135°

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Students can Download Chemistry Chapter 8 Ionic Equilibrium Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Samacheer Kalvi 12th Chemistry Chapter 8 Ionic Equilibrium Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Multiple Choice Questions

I. Choose the correct answer.

Question 1.
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.24 x 10-4 mol L-1 solubility product of Ag2C2O4 is ………………
(a) 2.42 x 10-8 mol3 L-3
(b) 2.66 x 10-12 12 mol3 L-3
(c) 45 x 10-11 mol3 L-3
(d) 5.619 x 10-12 mol3 L-3
Answer:
(d) 5.619 x 10-12 mol3 L-3
Ag2C2O4 2Ag+ + C2 O42-
[Ag+] = 2.24 x 10-4 mol L-1
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-1
= 1.12 x 10-4 mol L-1
Ksp = [Ag+]2 [C2O42-]
=(2.24 x 10-4 mol L-1)2 (1.12 x 10-4 mol L-1)
=5.619 x 10-12 mol3 L-3

Question 2.
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
(i) 60 mL \(\frac { M }{ 10 }\) HCI + 40 mL \(\frac { M }{ 10 }\) NaOH
(ii) 55 mL \(\frac { M }{ 10 }\) HCl + 45 mL \(\frac { M }{ 10 }\) NaOH
(iii) 75 mL \(\frac { M }{ 5 }\) HCI +25 mL \(\frac { M }{ 5 }\) MNaOH
(iv) 100 mL \(\frac { M }{ 10 }\) HCI+ 100 mL \(\frac { M }{ 10 }\) NaOH

pH of which one of them wilt be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii) 75 mL \(\frac { M }{ 5 }\) HCI + 25 mL \(\frac { M }{ 5 }\) NaOH
No of moles of HCl = 0.2 x 75 x 10-3 = 15 x 10-3
No of moles of NaOH = 0.2 x 25 x 10-3 = 5 x 1o-3
No of moles of HCl after mixing = 15 x 10-3 – 5 x 10-3
∴ Concentration of HCl
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-2
for (iii) solution, pH of 0.1 M HCI = – 1og10 (0.1) = 1.

Question 3.
The solubility of BaSO4 in water is 2.42 x 10-3 gL-1 at 298K. The value of its solubility product
(Ksp) will be …………………..
(Given molar mass of BaSO4 = 233g mol-1)
(a) 1.08 x 10-14 mol2L2
(b) 1.08 x 10-12 mol2L2
(c) 1.08 x 10-10 mol2 L2
(d) 1.08 x 10-8 mol2L-2
Answer:
(c) 1.08 x 10-10 mol2 L2
BaSO4 \(\rightleftharpoons\) Ba2+ + SO42-
Ksp = (s) (s)
Ksp = (s)2
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-3

Question 4.
pH of a saturated solution of Ca(OH)2 is 9. The Solubility product (K) of Ca(OH)2 ………………..
(a) 0.5 x 10-15
(b) 0.25 x 10-10
(c) 0.125 x 10-15
(d) 0.5 x 10-10
Answer:
(a) 0.5 x 10-15
Ca(OH)2 \(\rightleftharpoons\) Ca2+ + 2OH
Given that pH = 9
pOH = 14 – 9 = 5
[p0K = – 1og10[OH]]
[OH] = 10-pOH
[OH] =10-5M
Ksp = [Ca2+] [OH]2
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-4
=0.5

Question 5.
Conjugate base for bronsted acids H2O and HF are ………………
(a) OH and H2FH+, respectively
(b) H3O+ and F, respectively
(c) OH and F, respectively
(d) H3O+ and H2F+, respectively
Answer:
(c) OH and F, respectively
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-5
∴ Conjugate bases are OH and F respectively

The Buffer Capacity is a measure of resistant a particular solution is resistant to change in pH when an acid or a base is added to it.

Question 6.
Which will make basic buffer?
(a) 50 mL of 0.1M NaOH + 25mL of 01M CH3COOH
(b) 100 mL of 0.1M CH3COOH + 100 mL of 0.1M NH4OH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH4OH
(d) 100 mL of 0.1M HCI + 100 mL of O.1 M NaOH
Answer:
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH4OH
Basic buffer is the solution which has weak base and its salt
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-6

Question 7.
Which of the following fluro – compounds is most likely to behave as a Lewis base?
(a) BF3
(b) PF3
(c) CF4
(d) SiF4
Answer:
(b) PF3
BF3 → electron deficient → Lewis acid
PF3 → electron rich → Lewis base
CF4 → neutral → neither lewis acid nor base
SiF4 → neutral → neither lewis acid nor base

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
Which of these is not likely to act as lewis base?
(a) BF3
(b) PF3
(c) CO
(d) F
Answer:
(a) BF3
BF3 → electron deficient → Lewis acid
PF3 → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F → unshared pair of electron → lewis base

Question 9.
What is the decreasing order of strength of bases?
OH, NH2, H – C = C and CH3 – CH2
(a) OH > NH2 > H – C = C > CH3 – CH2
(b) NH2 > OH > CH3 – CH2 > H – C = C
(c) CH3 – CH2, > NH2 > H – C = C > OH
(d) OH > H – C = C > CH3 – CH2 > NH2
Answer:
(c) CH3 – CH2, > NH2 > H – C = C > OH
Acid strength decreases in the order
HOH > CH = CH > NH3 > CH3CH3
Its conjucate bases arc in the reverse order
CH3 – CH2 > NH2 > H – C = C > OH

Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
Answer:
(b) basic, acidic, basic
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-7

Question 11.
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10M aqueous pyridine solution (Kb for C5H5N = 1.7 x 10-9) iS ……………..
(a) 0.006%
(b) 0.013%
(c) 0.77%
(d) 1.6%
Answer:
(b) 0.013%
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-8
Percentage of dissociation
= \(\sqrt { 1.7 }\) x 10-4 x 100 = 1.3 x 10-2 = 0.013%

Question 12.
Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H+ ion concentration in the mixture?
(a) 37 x 10-2
(b) 10-6
(c) 0.111
(d) none of these
Answer:
(a) 3.7 x 10-2
pH = – log10 [H+]
[H+] = 10-pH
Let the volume be x mL
V1M1 + V2M2 + V3M3 = VM
x mL of 10-1M + x mL of 10-2M + x mL of 10-3 M
= 3 x mL of [H+]
= 3 x mL of [H+]
[H+] =
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-9
= 0.037 = 3.7 x 10-2

Question 13.
The solubility of AgCl (s) with solubility product 1.6 x 10-10 in 0. 1 M NaCl solution would be ………….
(a) 1.26 x 10-5 M
(b) 1.6 x 10-9 M
(c) 1.6 x 10-11 M
(d) Zero
Answer:
(b) 1.6 x 10-9 M
AgCl (s) \(\rightleftharpoons\) Ag+(aq) + Cl(aq)
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-10
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl]
K = (s) (s+0.1)
0.1 >>s
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-11

Question 14.
If the solubility product of lead iodide is 3.2 x 10-8, its solubility will be …………..
(a) 2 x 10-3M
(b) 4 x 10-4 M
(c) l.6 x 10-5 M
(d) 1.8 x 10-5 M
Answer:
(a) 2 x 10-3M
PbI2 (s) → Pb2+ (aq) + 2I (aq)
Ksp = (s) (2s)2
3.2 x 10-8 = 4s3
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-12

Question 15.
Using Gibb’s free energy change, ∆G0 = 57.34 KJ mol-1, for the reaction, X2Y(g) \(\rightleftharpoons\) 2X+ + Y2-(aq), calculate the solubility product of X2Y in water at 300K (R = 8.3 J K-1 Mol-1) ……………….
(a) 10-10
(b) 10-12
(c) 10-14
(d) can not be calculated from the given data
Answer:
(a) 10-10
57.34 KJ mol-1 = – 2.303 x 8.3 JK-1 mol-1 x 300K log Ksp
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-13
log10Ksp = -10
∴ Ksp = 10-10
∆G0 = – 2.303 RT log Keq
X2Y(s) \(\rightleftharpoons\) 2X+(aq) +Y2-(aq)
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-14
Keq = [x+]2[Y2-] ( X2Y(s) = 1)
Keq = K
Question 16.
MY and NY3, are insoluble salts and have the same Ksp values of 6.2 x 10-13 at room temperature. Which statement would be true with regard to MY and NY3?
(a) The salts MY and NY3 are more soluble in O.5 M KY than in pure water
(b) The addition of the salt of KY to the suspension of MY and NY3 will have no effect on
(c) The molar solubities of MY and NY3 in water are identical
(d) The molar solubility of MY in water is less than that of NY3
Answer:
(d) The molar solubility of MY in water is less than that of NY3
Addition of salt KY (having a common ion Y) decreases the solubility of MY and NY3 due to common ion effect. Option (a) and (b) are wrong.
For salt MY, MY \(\rightleftharpoons\) M+ + Y
Ksp = (s) (s)
6.2 x 10-13 = s2
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-15
The molar solubility of MY in water is less than of NY3

Question 17.
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
Answer:
(d) 12.65
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 x x x 10-3 = 0.l x x 10-3
No. of moles of HCl = 0.01 x x x 10-3 = 0.01 x x 10-3
No. of moles of NaOH after mixing = 0.1x x 10-3 – 0.01x  x 10-3
= 0.09x x 10-3
Concentration of NaOH =
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-166
[OH] = 0.045
pOH = – log (4.5 x 10-2)
= 2 – log 4.5
= 2 – 0.65 = 1.35
pH = 14 – 1.35 = 12.65

Question 18.
The dissociation constant of a weak acid is 1 x 10-3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ………………..
(a) 4:3
(b) 3:4
(c) 10:1
(d) 1:10
Answer:
(d) 1:10
Ka = 1 x 10-3 ; pH = 4
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-16

Question 19.
The pH of 10-5 M KOH solution will be …………..
(a) 9
(b) 5
(c)19
(d) none of these
Answer:
(a) 9
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-17
[OH] = 10-5M.
pH = 14 – pOH .
pH = 14 – ( – log [OH])
= 14 + log [OH] = 14 + log 10-5
= 14 – 5 = 9

Question 20.
H2PO4 the conjugate base of …………….
(a) PO4
(b) P2O5
(c) H3PO4
(d) HPO42-
Answer:
(c) H3PO4
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-55
H2PO4 is the conjugate base of H3PO4

Question 21.
Which of the following can act as lowery – Bronsted acid well as base?
(a) HCl
(b) SO42-
(c) HPO42-
(d) Br
Answer:
(c) HPO42-
HPO42- can have the ability to accept a proton to form H2PO4.
It can also have the ability to donate a proton to form PO4-3.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 22.
The pH of an aqueous solution is Zero. The solution is ……………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(b) strongly acidic
pH = – log10[H+]
[H+] =10-pH
= 100 = 1
[H+] = 1 M
The, solution is strongly acidic

Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by ………………
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-18
Answer:
According to Henderson equation
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-19
According to Henderson equation
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-20

Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-21
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-22

Question 25.
Dissociation constant of NH4OH is 1.8 x 10-5 the hydrolysis constant of NH4Cl would be …………….
(a) 1.8 x 10-19
(b) 5.55 x 10-10
(c) 5.55 x 10-5
(d) 1.80 x 10-5
Answer:
(b) 5.55 x 1010
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-23

II. Answer the following questions.

Question 1.
What are lewis acids and bases? Give two example for each.
Answer:
1. Lewis acids:

  • Lewis acid is a species that accepts an electron pair.
  • Lewis acid is a positive ion (or) an electron deficient molecule.
  • Example, Fe2+, CO2, BF3, SiF4 etc…

2. Lewis bases:

  • Lewis base is a species that donates an electron pair.
  • Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons.
  • Example, NH3, F, CH2 = CH2, CaO etc….

Question 2.
Discuss the Lowry – Bronsted concept of acids and bases.
Answer:
According to Lowry – Bronsted concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance. When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCl behaves as an acid and H2O is base. The proton transfer from the acid to base can be represented as
HCl + H2O \(\rightleftharpoons\) H3O+ + Cl

When ammonia is dissolved in water, it accepts a proton from water. In. this case, ammonia (NH3) acts as a base and H2O is acid. The reaction is represented as
H2O + NH3 \(\rightleftharpoons\) NH4+ + OH
Let us consider the reverse reaction in the following equilibrium
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-24
H3O+ donates a proton to Cl to form HCI i.e., the products also behave as acid and base. In general, Lowry – Bronsted (acid – base) reaction is represented as
Acid1 + Base2 \(\rightleftharpoons\) Acid2 + Base1
The species that remains after the donation of a proton is a base (Base1)and is called the conjugate base of the Bronsted acid (Acid1). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs. Conjugate acid – base pair
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-25
HCl and Cl, H2O and H3O are two conjugate acid – base pairs. i.e., Cl is the conjugate base of the acid HCl (or) HCl is conjugate acid of Cl Similarly H3O is the conjugate acid of H2O. Limitations of Lowry – Bronsted theory. Substances like BF3 , AICl3 etc., that do not donate protons are known to behave as acids.

Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.

  1. HS (aq) + HF \(\rightleftharpoons\) F (aq) + H2S (aq)
  2. HPO42- + SO32- \(\rightleftharpoons\) PO43- + HSO3
  3. NH4+ + CO32- \(\rightleftharpoons\) NH3 + HCO3

Answer:
1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-26
• HF and F , HS and H2S are two conjugate acid – base pairs.
• F is the conjugate base of the acid HF (or) HF is the conjugate acid of F
• H2S is the conjugate acid of HS (or) HS is the conjugate base of H2S.

2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-27
• HPO42- and PO43-, SO32- and HSO3 are two conjugate acid – base pairs.
.PO43- is the conjugate base of the acid HPO42- (or) HPO42- is the conjugate acid of PO4.
•HSO3 is the conjugate acid of SO32- (or) SO32- is the conjugate base of HSO3.

3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-28
• NH+ and NH3, CO32- and HCO3 are two conjugate acid-base pairs.
• HCO3 is the conjugate of acid CO32- (or) CO32- is the conjugate bases of HCO3.
• NH3 is the conjugate base of NH4+ (or) NH4+ is the conjugate acid of NH3.

Question 4.
Account for the acidic nature of HCIO4. In terms of Bronsted – Lowry theory, identify its conjugate base.
Answer:
HClO4 \(\rightleftharpoons\) H+ + ClO4
1. According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

2. Let us consider the stabilities of the conjugate bases ClO4, ClO3, ClO2 and ClO formed from these acid HClO4, HClO3, HClO2, HOCl respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently, the corresponding acid will be strongest because weak conjugate base has strong acid and a strong conjugate base has weak acid.

3. The charge stabilization mercases in the order, ClO < ClO2 < ClO3 < ClO4.

This means ClO4 will have maximum stability and therefore will have a minimum attraction for W. Thus CIO4 will be weakest base and its conjugate acid HCIO4 is the strongest acid.

4. CIO4 is the conjugate base of the acid HClO4.

Question 5.
When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H2O)6]2+(aq) + 4NH3 (aq) \(\rightleftharpoons\) [Cu(NH3)4]2+ (aq), among H2O and NH3 which is stronger Lewis base.
Answer:
Copper (II) sulphate solution, for example contains the blue hexaaqua copper (II) complex ion. In the first stage of the reaction, the ammonia acts as a Bronsted – Lowry base. With a small amount of ammonia solution, hydrogen ions are pulled off two water molecules in the hexaaqua ion. This produces a neutral complex, one carrying no charge.

If you remove two positively charged hydrogen ions from a 2+ ion, then obviously there isn’t going to be any charge left on the ion. Because of the lack of charge, the neutral complex isn’t soluble in water and so you get a pale blue precipitate. [Cu(H2O)6]2+ + 2NH3 [Cu(H2O)4OH] + 2NH4+

This precipitate is often written as Cu(OH)2 and called copper (II) hydroxide. The reaction is reversible because ammonia is only a weak base. That precipitate dissolves if you add an excess of ammonia solution, giving a deep blue solution. The ammonia replaces four of the water molecules around the copper to give tetramminc diaqua copper (II) ions. The ammonia
uses its lone pair to form a coordinate covalent bond with the copper. It is acting as an electron pair donor – a Lewis base.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-29

Question 6.
The concentration of hydroxide ion in a water sample is found to be 2.5 x 10-6 M. Identify the nature of the solution.
Answer:
The concentration of OH ion in a water sample is found to be 2.5 x 10-6 M
pOH = – log10 [OH ]
pOH = – 1og10 [2.5 x 10-6]
= – log10 [2.5] – log10 [10-6]
= – 0.3979 – ( – 6)
= – 0.3979 + 6
pOH = 5.6
Since pOH is less than 7, the solution is basic

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H3O+] = 4 x 105 M. Is the solution neutral (or) acidic (or) basic.
Answer:
[H3O+] = 4 x M
pH = – log10 [H3O+]
pH = – 1og10[4 x 105]
pH = – log10 [4] – log10 [10-5]
pH = – 0.6020 – ( – 5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
Calculate the pH of 0.04 M HNO3 solution.
Answer:
Concentration of HNO3 = 0.04M
[H3O+] = 0.04 mol dm-3
pH = – 1og[H3O+]
= – log (0.04)
= – log(4 x 10-2)
= 2 – log4 = 2 – 0.6021
= 1.3979 = 1.40

Question 9.
Define solubility product.
Answer:
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-30

Question 10.
Define ionic product of water. Give its value at room temperature.
Answer:
1. The product of the concentration of H+ and OH ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature (25°C) is,
Kw = [H+] [OH+] (or)
Kw= [H3O+] [OH+]
Kw =(1 x 10-7) (1 x 10-7)
Kw= 1 x 10-14 mol2 dm-6

Question 11.
Explain common ion effect with an example.
Answer:
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in an aqueous solution and hence the following equilibrium exists.
CH3COOH (aq) \(\rightleftharpoons\) H+(aq)+ CH3COO (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na+ and CH3COO ion.
CH3COONa (aq) → Na+ (aq) + CH3COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s priñciple that when stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH3COO ions combine with H ions to produce much more unionized CH3COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 12.
Derive an expression for Ostwald’s dilution law.
Answer:
Ostwald’s dilution law:
It relates the dissociation constant of the weak acid (Ka) with its degree of dissociation (α) and the concentration (c). Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,
CH3COOH \(\rightleftharpoons\) CH3COO + H+
The dissociation constant of acetic acid is,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-31
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-32
Substituting the equilibrium concentration in equation
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-33
We know that weak acid dissociates only to a very small extent compared to one, a is so small.
equation (1) becomes,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-34
Similarly, for a weak base,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-35
The concentration of H can be calculated using the Ka value as below,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-36
Substituting a value in equation (2),
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-37
For weak base
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-38

Question 13.
Define pH.
Answer:
pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = – log10 [H3O] (or) pH = – log10 [H+]

Question 14.
Calculate the pH of 1.5 x 10-3 M solution of Ba(OH)2
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-39
[OH] = 3 x 103M.
[pH + pOH = 14]
pH = 14 – pOH
pH = 14 – ( – log [OH])
= 14 + log [OH]
= 14 + log (3 x 10-3)
= 14 + log 3 + log 10-3
= 11 + 0.4771
pH = 11.48

Question 15.
50 ml of 0.05 M HNO3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution.
Number of moles of HNO3 = 0.05 x 50 x = 2.5 x 10-3
Number of moles of KOH = 0.025 x 50 x 10-3 = 1.25 x 10-3
Number of moles of HNO3 after mixing = 2.5 x 10-3 – 1.5 x 10-3 = 1.25 x 10-3
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-40
After mixing, total volume = 100 ml = 100 x 10-3 L
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-41
pH = – log [H+]
pH = – log (1.25 x 10-2) = 2 – 0.0969
= 1.9031

Question 16.
The Ka value for HCN is 10-9. What is the pH of 0.4 M HCN solution?
Answer:
Ka =10-9
c = O.4M
pH = – log [H+]
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-42
∴ pH = – log(2 x 10-5)
= – log 2 – log (10-5)
= – 0.3010 + 5
pH = 4.699

Question17.
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that.
Ka = Kb = 1.8 x 10-5
Solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-43
= 0.7453 x 10-2
pH = \(\frac { 1 }{ 2 }\) pKw + \(\frac { 1 }{ 2 }\) pKa – \(\frac { 1 }{ 2 }\) pKb
Given that Ka = Kb = 1.8 x 10-5
if Ka = Kb, then. pKa = pKb
pH = \(\frac { 1 }{ 2 }\) pKw = \(\frac { 1 }{ 2 }\) (14) = 7

Question 18.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Answer:
Let us consider the reactions between a strong acid, HCl, and a weak base, NH4OH, to produce a salt, NH4Cl, and water.
HCl (aq) + NH4OH (aq) \(\rightleftharpoons\) NH4Cl (aq) + H2O (I)
NH4CI(aq) → NH4+ + Cl (aq)

NH4+ is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH from water to produce unionised NH4OH shown below.
NH4+ (aq) + H2O (1) \(\rightleftharpoons\) NH4OH (aq) + H+(aq)

There is no such tendency shown by Ct and therefore [H+] > [OH] the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the Ka and Kb as
Kh.Kb = Kw
Let us calculate the Kb value in terms of degree of hydrolysis (h) and the concentration of salt
Kh = h2 C and
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-44
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-45

Question 19.
The solubility product of Ag2CrO4 is 1 x 10-12. What is the solubility of Ag2CrO4 in 0.01 M AgNO3 solution?
Answer:
Solubility product of Ag2CrO4,
Ksp = 1 x 10-2
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-46
Ksp = [Ag+]2 [ CrO42-]
[Ag+] = 2s +0.01
0.01 >> 2s
[Ag+]  = 0.01M
[CrO4-2] = s
Ksp = (0.01)2. (s)
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-47

Question 20.
Write the expression for the solubility product of Ca3(PO4)2
Answer:
Ca3(PO4)2 (s) \(\rightleftharpoons\) 3Ca2+ (3s) + 2PO43- (2s)
The solubility of Ca3(PO4)2 is,
Ksp = [Ca2+]3 . [PO43-]2
Ksp = (3s)3 . (2s)2
Ksp= (27 s3) . (4s2)
Ksp = 108s5.

Question 21.
A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca2+] = 3.3 x 10-4 M. What is the Kspof CaF2?
Answer:
CaF2 (s) \(\rightleftharpoons\) Ca2+(aq) + 2F(aq)
[F] = 2 [Ca2+] = 2 x 33 x 10-4 M
= 6.6 x 10-4 M
= [Ca2+] [F]2
= (3.3 x 10-4) (6.6 x 10-4)2
= 1.44 x 10-10

Question 22.
Ksp of AgCl is 1.8 x 10-10. Calculate molar solubility in 1 M AgNO3
Answer:
AgCl(s) → Ag+(aq) + Cl(aq)
x = solubility of AgCl in 1M AgNO3
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-49
[Cl] = x
Ksp = [Ag+] [Cl]
1.8 x 10-10 = (1) (x)
x = 1.8 x 10-10M

Question 23.
A particular saturated solution of silver chromate Ag2CrO4 has [Ag+] = 5 x 10-5 and [CrO4]2- = 4.4 x 10 M. What is the value of for Ag2CrO4?
Answer:
Ag2CrO4 (s) \(\rightleftharpoons\) 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2 [CrO42-]
= (5 x 10-5)2 (4.4 x 10-4)
= 1.1 x 10-12

Question 24.
Write the expression for the solubility product of Hg2CI2.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-50

Question 25.
Ksp of Ag2CrO4 is 1.1 x 10-12 What is solubility of Ag2CrO4 in 0.1M K2CrO4.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-51
x is the solubility of Ag2CrO4 in 0.1 M K2CrO4
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-52

Question 26.
Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO3)2 and 0.100 L of 0.2 M NaCl are mixed? (PbCI2) = 1.2 x 10-5.
Answer:
When two or more solutions are mixed, the resulting concentrations are different from the original.
Total volume 0.250L
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-53
Precipitation of PbCI2 (s) occurs if [Pb2+][Cl]2 > Ksp
[Pb2+][Cl]2 = (0.06)(0.08)2
= 3.84 x 10-4
Since ionic product [Pb2+][Cl]2 > Ksp PbCl2 is precipitated.

Question 27.
of Al(OH)3 is 1 x 10-15 M. At what pH does 1.0 x 10-13 M AI3+ precipitate on the addition of buffer of NH4CI and NH4OH solution?
Answer:
Al(OH)3 ⇌ Al3+ (aq) + 3OH (aq)
Ksp = [Al3+] [OH]3
Al(OH)3 precipitates when
[Al3+] [OH]3 > Ksp
(1 x 10-3)[OH]3 > Ksp
[OH]3 > 1 x 10-12
[OH] > 1 x 10-4M
[OH] = l x 10-4 M
pOH = – 1og10[OH] = – log (1 x 10-4) = 4
pH = 14 – 4 = 10
Thus, Al (OH)3 precipitates at a pH of 10

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Evaluate Yourself

Question 1.
Classify the following as acid (or) base using Arrhenius concept

  1. HNO3
  2. Ba(OH)2
  3. HlPO4
  4. CH3COOH

Answer:
1. HNO3:
Nitric acid, dissociates to give hydrogen ions in water.
HNO3 is acid.

2. Ba(OH)2:
Barium hydroxide dissociates to give hydroxyl ions in water.
Ba(OH)2 is base.

3. H3PO4:
Orthophosphoric acid dissociates to give hydrogen ions in water.
H3PO4 is acid.

4. CH3COOH:
Acetic acid dissociates to give hydrogen ions in water.
CH3COOH is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid-base pairs.

  1. NH4
  2. H2SO4
  3. CH3COOH.

Answer:
1. NH4 + Conjugate acid-base pair
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-56
NH4+ and NH3, H2O and H3O+ are two conjugate acid – base pairs.

H2SO4 and CH3COO, H2O and H3O+ are two conjugate acid-base pairs.

2. H2SO4:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-57
H2SO4 and HSO4, H2O and H3O+ are two conjugate acid-base pairs.

3. CH3COOH:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-58
CH3COOH and CH3COO, H2O and H3O+ are two conjugate acid-base pairs.

Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-59

Answer:
(i). CaO + CO2 → CaCO3

  • CaO – Lewis base – All metals oxides are Lewis bases
  • CO2 – Lewis acid – CO2 contains a polar double bond.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-60

  • CH3 – O – CH3 – Lewis base – Electron rich species
  • AlCl3 – Lewis acid – AICI3 is electron deficient molecule.

Question 4.
H3BO3 accepts hydroxide ion from water as shown below
Answer:
H3BO3 (aq) + H2O(l) = B(OH)4 + H+
Predict the nature of H3BO3 using Lewis concept. Boric acid is also called as hydrogen borate or orthoboric acid. It is a weak mono basic Lewis acid of boron and it is written as B(OH)3. It accepts hydroxyl (OH) ion from water. It does not dissociate to give hydronium (H3O+) ion rather forms metaborate ion and this ions in turn give H3O ion.
B(OH)3 + H2O [B(OH)4] + H3O+ Hence it is considered as weak acid.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 5.
At a particular temperature, the Kw of a neutral solution was equal to 4 x 10-14. Calculate the concentration of [H3O+] and [OH].
Answer:
Given solution is neutral
[H3O+] = [OH]
Let [H3O+] = x ; then [OH] = x
Kw = [H3O+] [OH]
4 x 10-14 = x . x
x2 = 4 x 10-14
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-61

Question 6.

  1. Calculate pH of 10-8 M H2SO4
  2. Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4
  3. Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M NaOH

Answer:
1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-62
In this case the concentration of H2SO4 is very low and hence [H3O] from water cannot be neglected
[H3O+] = 2 x 10-8 (from H2SO4) + 10-7 (from water)
= 10-8(2+ 10)
= 12 x 10-8 = 1.2 x 10-7
pH = – log10[H3O+]
= – log10( 1.2 x 10-7)
= 7 – log101.2
= 7 – 0.0791 = 6.9209

2.
pH of the solution = 5.4
[H3O+] = antilog of (- pH)
= antilog of (- 5.4)
= antilog of (-6 + 0.6) = \(\overline{6} .6\)
= 3.981 x 10-6
i.e., 3.98 x 10-6 mol dm-3

3.
No. of moles of HCl = 0.2 x 50 x 10-3 = 10 x 10-3
No. of moles of NaOH =0.1 x 50 x 10-3 = 5 x 10-3
No. of moles of HCl after mixing = 10 x 10-3 – 5 x 10-3
= 5 x 10-3
after mixing total volume = 100 mL
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-63
[H3O+] = 5 x 10-2 M
pH = – log ( 5 x 10-2)
= 2 – log 5
= 2 – 0.6990
= 1.30

Question 7.
Kb for NH4OH is 1.8 x 10-5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-64

Question 8.
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.

2. Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4 M CH3COONa . What is the change in the pH after adding 0.01 mol of HCI to 500m1 of the above buffer solution.

Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105).
Answer:
1. Dissociation of buffer components
NH4OH (aq) \(\rightleftharpoons\) NH4+ (aq) + OH (aq)
NH4CI → NH4+ + Cl
Addition of OH
The added H+ ions are neutralized by NH4OH and there is no appreciable decrease in pH.
NH4OH(aq) + H+ \(\rightleftharpoons\) NH4+(aq) + H2O (1)
Addition of
NH4 (aq) + OH (aq) → NH4OH (aq)
The added OH ions react with NH4 to produce unionized NH4OH . Since NH4OH is a weak base, there is no appreciable increase in pH.

2. pH of buffer
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-65
Addition of 0.01 mol HCI to 500 ml of buffer
Added [H+]
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-66
∴ pH = – log (1.8 x 10-5) = 4.74
Addition of 0.01 mol HCl to 500ml of buffer
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-67

Question 9.
1. How can you prepare a buffer solution of pH9. You are provided with 0.1 M NH4OH solution and ammonium chloride crystals. (Given: pKb for NH4OH is 4.7 at 25°C)

2. What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pKa for formic acid is 3.75.)
Answer:
1.
SamacSamacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-68heer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-15
We know that pH + pOH = 14
9 + pOH = 14
= pOH = 14 – 9 = 5
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-69
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-70
[NH4Cl] = 0.1 M x 1.995
= 0. 1995 M
=0.2 M
Amount of NH4CI required to prepare 1 litre 0.2 M solution = Strength of NH4CI x molar
mass of NH4CI
= 0.2 x 535
= 10.70 g
10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH4OH solution and the prepared NH4CI solution will give a buffer solution with required pH value (pH = 9).

2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-71
[Sodium formate] = number of moles of HCOONa
= 0.6 x V x 10-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10-3
= 80 x 10-3
4 = 3.75 + log \(\frac { 0.6V }{ 80 }\)
0.25 = log \(\frac { 0.6V }{ 80 }\)
antilog of 0.25 = \(\frac { 0.6V }{ 80 }\)
0.6V = 1.778 x 80
= 1.78 x 80
= 142.4
V = \(\frac { 142.4 mL }{ 0.6 }\) = 237.33 mL

Question 10.
Calculate the

  1. hydrolysis constant
  2. degree of hydrolysis and
  3. pH of 0.05M sodium carbonate solution pKa for HCO3 is 10.26.

Answer:
1. Hydrolysis constant
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-72
Given
Kw = 1 x 10-14
c = 0.05 M
PKa = 10.26
Ka = – log Ka
Ka = antilog of (- pKa)
Ka = antilog of (- 10.26)
Ka = 5.49 x 10-11
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-73
h = 6.034 x 10-2

2. Degree of hydrolysis
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-74

3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-75

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Textbook Example problems solved

Question 1.
Identify the Lewis acid and the Lewis base in the following reactions.
Cr3+ + 6H2O → [Cr(H2O)6]3+
In the hydration of ion, each of six water molecules donates a pair of electron to Cr3+ to form the hydrated cation, Hexa aqua chromium (III) ion, thus, the lewis acid is Cr and the Lewis base H2O.

Question 2.
Calculate the concentration of OH in a fruit juice which contains 2 x 10 M, H3O+ ion. Identify the nature of the solution.
Answer:
Given that H3O+ = 2 x 10-3 M
Kw = [H3O+] [OH]
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-76
2 x 10-3 >> 0.5 x 10-11
i.e., [H3O+] >> [OH], hence the juice is acidic in nature

Question 3.
Calculate the pH of 0.001M HCI solution
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-77
H3O from the auto ionisation of H2O (10-7 M) is negligible when compared to the H3O from 10-3 M HCI.
Hence [H3O+] = 0.001 mol dm-3
pH = – log10 [H3O+]
= – log10 (0.001)
= – log10 (10-3) = 3

Question 4.
Calculate pH of 10-7 M HCI
Answer:
If we do not consider [H3O+] from the ionisation of H2O, then [H3O+] = [HCl] = 107M i.e., pH =7, which is a pH of a neutral solution. We know that HCI solution is acidic whatever may be the concentration of HCI i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10-7M). Hence, the H3O+ (10-7M) formed due to the auto ionisation of water cannot be neglected. So, in this case we should consider [H3O+] from ionisation of H2O
Answer:
[H3O+] = 10-7 (from HCl) + 10-7(from water)
= 10-7 (1+1)
= 2 x 10
pH = – log10 [H3O]
= – log10(2 x 107) = – [log2 + log 10-7]
= – log 2 – ( – 7). 1og10
= 7 – log 2
= 7 – 0.3010 = 6.6990
= 6.70

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 5.
A solution of 0.10 M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25°C. Find the dissociation constant of the acid.
Answer:
Given that α = 1.20% = \(\frac { 1.20 }{ 100 }\) = 1.2 x 10-2
Ka = α2c = (1.2 x 10-2)2 (0.1)
= 1.44 x 10-4 x 10-1
= 1.44 x 10-5

Question 6.
Calculate the pH of 0.1M CH3COOH solution. The dissociation constant of acetic acid is 1.8 x 10-5.
Answer:
pH = – log[H+]
For weak acids,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-78

Question 7.
Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 x 10-5
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-79
Given that Ka = 1.8 x 10-5
pKa = – log (1.8 x 10-5)
= 5 – log 1.8
= 5 – 0.26 = 4.74
pH = 4.74 + log \(\frac { 0.20 }{ 0.18 }\)
= 4.74 + log \(\frac { 10 }{ 9 }\)
= 4.74 + log 10 – log 9
= 4.74 + 1 – 0.95
= 5.74 – 0.95
= 4.79

Question 8.
What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate and making the volume equal to 500 ml. (Given: K for acetic acid is 8 x 10)
Answer:
According to Henderson – Hessalbalch equation,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-80
Given that Ka = 1.8 x 10-5
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-81
pH = 4.74 + log 1
pH = 4.74 + 0 = 4.74

Question 9.
Calculate

  • the hydrolysis constant,
  • degree of hydrolysis and
  • pH of O.1 M CH3COONa solution (pKa for CH3COOH is 4.74).

Answer:
CH3COONa is a salt of weak acid (CH3COOH) and a strong base (NaOH). Hence, the solutions is alkaline due to hydrolysis.
CH3COO (aq) + H2O (aq) \(\rightleftharpoons\) CH3COOH (aq) + OH (aq)
Give that pKa = 4.74
pKa = – log Ka
i.e., Ka = antilog of ( – PKa)
= antilog of ( – 4.74)
= antilog of( – 5 + 0.26)
10-5 x 1.8
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-82
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-83

Question 10.
Establish a relationship between the solubility product and molar solubility for the following

  1. BaSO4
  2. Ag2(CrO4)

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-84

Samacheer Kalvi 12th Chemistry Ionic Equilibrium Multiple Additional Questions

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 1 Marks Questions and Answers

I. Choose the best answer.

Question 1.
Which one of the following buffer is present in blood?
(a) CH3COOH + CH3COONa
(b) NH4OH + NH4Cl
(c) H2CO3 + NaHCO3
(d) HCI + NaCl
Answer:
(c) H2CO3 + NaHCO3

Question 2.
Which of the following is mostly used in the fertilizer industry?
(a) Lactic acid
(b) Sulphuric acid
(c) Tannic acid
(d) Carbonic acid
Answer:
(b) Sulphuric acid

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 3.
Which of the following is present in an antacid tablet?
(a) NaOH
(b) Mg(OH)2
(c) Al(OH)3
(d) either (b) or (c)
Answer:
(d) either (b) or (c)

Question 4.
The acid present in milk is …………..
(a) Lactic acid
(b) Tannic acid
(c) Tartaric acid
(a) Acetic acid
Answer:
(a) Lactic acid

Question 5.
Consider the following statements.
(i) Acid tastes sour
(ii) Acid turns red litmus to blue
(iii) Acid reacts with metals and liberates hydrogen gas

Which of the above statement is I are correct?
(a) (i) only
(b) (i) & (ii)
(c) (i) & (iii)
(d) (ii) only
Answer:
(c) (i) & (iii)

Question 6.
Consider the following statements.
(i) Acid tastes sour.
(ii) Acid turns blue litmus to red
(iii) Acid has a tendency to accept a proton from other substances.

Which of the above statement is I are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) only
(a) (ii) only
Answer:
(c) (iii) only

Question 7.
Which of the following can act as an acid as well as a base by Lowry – Bronsted theory?
(a) H2O
(b) NH3
(c) NH4OH
(d) Ca(OH)2
Answer:
(a) H2O

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
In the reaction HCI + H2O \(\rightleftharpoons\) H3O + Cl which one of the acid-base pair?
(a) HCl + H3O+
(b) HCI + CI
(c) H3O + Cl
(d) H2O + Cl
Answer:
(b) HCI + CI

Question 9.
Which of the following is considered Lewis acid?
(a) NH3
(b) BF3
(c) HF
(d) HCl
Answer:
(b) BF3

Question 10.
Which of the following is considered as Lewis base?
(a) BF3
(b) AICI3
(c) HCI
(d) NH3
Answer:
(d) NH3

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 11.
Consider the following statements.
(i) A Lewis acid is a species that accepts an electron pair.
(ii) A Lewis acid is a species that donates an electron pair.
(iii) The ligand act as Lewis base.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (I) & (ii)
(d) (ii) & (iii)
Answer:
(b) (ii) only

Question 12.
In [Cr(H2O)6]3+ which one of the following acts as Lewis acid?
(a) Cr
(b) Cr3+
(c) (HO)6
(d) Cr3-
Answer:
(b) Cr3+

Question 13.
In [Cr(H2O)6]3+ which one of the following acts as Lewis base?
(a) H2O
(b) H3O+
(c) Cr3+
(d) Cr
Answer:
(a) H2O

Question 14.
Among the following which is the strongest acid?
(a) Formic acid
(b) Acetic acid
(c) Hydrochloric acid
(d) Lactic acid
Answer:
(c) Hydrochloric acid

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 15.
Which of the following is the weak acid?
(a) HCl
(b) H2SO4
(c) HNO3
(d) CH3COOH
Answer:
(d) CH3COOH

Question 16.
Identify the weakest acid?
(a) H3O+
(b) H2 SO4
(c) OH
(d) CH3COOH
Answer:
(c) OH

Question 17.
Which one of the following is the very weak base?
(a) NO2
(b) NO3
(c) NH2
(d) O2-
Answer:
(b) NO3

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 18.
Which one of the following is the strong base?
(a) ClO4
(b) HSO4
(c) O2-
(d) F
Answer:
(c) O2-

Question 19.
Which of the following is the weak base?
(a) H
(b) OH
(c) HSO4
(d) F
Answer:
(d) F

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 20.
The value of the ionic product of water at 25°C is ………………..
(a) 1 x 10-7
(b) 1 x 107
(c) 1 x 10-14
(d) 1 x 1014
Answer:
(c) 1 x 10-14

Question 21.
Consider the following statements.
(i) The dissociation of water is an exothermic reaction.
(ii) With the increase in temperature, the ionic product of water value decreases.
(iii) With the increase in temperature, the ionic product of water value increases.
Which of the above statement is/are correct?
(a) (i) and (ii)
(b) (ii) only
(c) (iii) only
(d) (ii) & (iii)

Question 22.
Which of the following is a neutral solution?
(a) Aqueous NaCl solution
(b) Aqueous NaOH solution
(c) Aqueous HCl solution
(d) Aqueous NH3
Answer:
(a) Aqueous NaCl solution

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 23.
The pH of a neutral solution is ………..
(a) less than 7
(b) more than 7
(c) equal to 7
(d) 14
Answer:
(a) less than 7

Question 24.
If the pH of a solution is less than 7, it is called ………………… solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(b) Acidic

Question 25.
If the pH of a solution is more than 7, it is called ……………. solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(a) Basic

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 26.
The pH value of water is …………..
(a) 14
(b) 7
(c) 3
(d) 1
Answer:
(b) 7

Question 27.
The pH of drain cleaner is ………..
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(c) 14

Question 28.
The pH of the battery acid is ………….
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(d) 0

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 29.
The pH of 0.001 M HCI solution is ………
(a) 3
(b) 2
(c) 1
(d) 11
Answer:
(a) 3
Solution:
[H3O+] = 0.001 mol dm-3
pH = – log10 [H3O+] = – log10 [0.001]
= – 1og10 [10-3] = 3
pH = 3

Question 30.
The pH of 0.01 M HCl solution is …………..
(a) 3
(b) 2
(c) 1
(d) 10
Answer:
(b) 2
Solution:
[H3O+] = 0.01 M
pH = – 1og10 [H3O+] = – log10 [0.01]
= – 1og10[10-3] = 3
pH = 3

Question 31.
What is the pH of 0.1 M HCI solution?
(a) 1
(b) 2
(c) 13
(d) 3
Answer:
(a) 1
Solution:
[H3O+] = 0.1 M
pH = – log10[H3O+] = – log10[0.1]
= – log10[10-1]
pH = l

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 32.
Consider the following statements.
(i) Degree of dissociation (a) is the fraction of the total number of moles of a substance that dissociates at equilibrium.
(ii) When the dilution increases by 100 times, the dissociation increases by 100 times.
(iii) When the dilution increases by 100 times, the dissociation increases by 10 times.
Which of the above statement is I are correct?
(a) (ii) only
(b) (i) & (iii)
(c) (iii) only
(d) (I) only
Answer:
(a) (ii) only

Question 33.
Which of the following is not a buffer solution?
(a) CH3COOH + CH3COONa
(b) NH4OH + NH4Cl
(c) H2CO3 + NaHCO3
(d) NaOH + NaCI
Answer:
(d) NaOH + NaCI

Question 34.
The mathematical expression of buffer capacity is …………
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-85
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-158

Question 35.
Which one of the following is not correct?
(a) pH+pOH = 14
(b) pH = 7 +\(\frac { 1 }{ 2 }\) pKa – \(\frac { 1 }{ 2 }\) pKb
(c) pH x pOH = 1 x 1014
(d) pH = – log10 [H3O+]
Answer:
(c) pH x pOH = 1 x 1014

Question 36.
The chemical present in the kidney as stones is …………..
(a) CaCl2
(b) Ca(CO3)2
(c) Calcium nitrate
(d) Calcium oxalate
Answer:
(d) Calcium oxalate

Question 37.
If the pH of an aqueous solution is 7, the solution is …………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(c) neutral

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 38.
Cl is the conjugate base of …………..
(a) HClO4
(b) HCI
(c) ClO4
(d) HClO3
Answer:
(b) HCI

Question 39.
The conjugate base of H2O and H2SO4 are …………..
(a) OH and HSO4
(b) H4O and SO42-
(c) OH and SO42-
(d) H3O and HSO4
Answer:
(a) OH and HSO4

Question 40.
The Ksp of AgI is 1.5 x 10-16. On mixing equal volume of the following solutions, precipitation will occur only with ………….
(a) 10-7MAg+ and 10-19M I
(b) 10-8MAg+ and 10-8M I
(c) 10-16 M Ag+ and 10-16 M I
(d) 10-9 M Ag+ and 10-9 M I
Answer:
(b) 10-8MAg+ and 10-8M I
Ksp of AgI = 1.5 x 10-16
10-8 M Ag+ and 10-8 M I
Ionic product = 10-16 = Ksp

Question 41.
The strongest Bronsted base in the following anion is ……………
(a) ClO
(b) ClO2-
(c) ClO3-
(d) ClO4-
Answer:
(a) ClO
Solution:
HClO is the weakest acid and its conjugate base ClO is the strongest base.

Question 42.
Calculate the hydrolysis constant of the salt containing NO2.
Given the Ka for HNO2 = 4.5 x 10-10
(a) 2.22 x 10-5
(b) 2.02 x 105
(c) 4.33 x 104
(d) 3.03 x 10-5
Answer:
(a) 2.22 x 10-5
Solution:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-86

Question 43.
Electrophiles are usually ……….
(a) Lewis acid
(b) Lewis base
(c) Bronsted acid
(d) Bronted base
Answer:
(a) Lewis acid
Solution:
Lewis acid are electrophile because they accept electron pair.

Question 44.
Which one is a Lewis acid?
(a) CIF3
(b) H2O
(c) NH3
(d) OH
Answer:
(a) CIF3
Solution:
CIF3 has a vacant d-orbital in a central atom.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 45.
An aqueous solution of ammonium acetate is …………….
(a) faintly acidic
(b) faintly basic
(c) fairly acidic
(d) Almost neutral
Answer:
(d) Almost neutral
Solution:
It is a salt of a weak acid and weak base.

Question 46.
The dissociation constant of a weak acid is 1.0 x 10-10. The equilibrium constant for the reaction with a strong base is …………
(a) 1.0 x 10-5
(b) 1.0 x 10-9
(c) 1.0 x 109
(d) 1.0 x 1014
Answer:
(c) 1.0 x 109
Solution:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-87

Question 47.
Arrange the acids
(i) H2SO3
(ii) H3PO3 and
(iii) HClO3in the decreasing order of acidity.
(a) (i) > (iii) > (ii)
(b) (i) > (ii) > (iii)
(c) (ii) > (iii) > (I)
(d) (iii) > (i)> (ii)
Answer:
(d) (iii) > (i)> (ii)
Solution:
Acidity is directly proportional to oxidation number. As the oxidation number of S, P and Cl in H2SO3, H3PO3 and HCIO3 is +4, +3, +5 respectively. So decreasing order of acidity will
be (iii) > (I) > (ii)

Question 48.
The pH of 0.1 M solution of a weak monoprotic acid 1% ionised is ………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Solution:
Conc = 0.1 M
α = 1
= 0.1 x \(\frac { 1 }{ 100 }\) = 10-3
[H+] = 10-3
∴ pH = 3

Question 49.
Ksp for Cr(OH)3 is 2.7 x 10-3. What is the solubility in moles/litre?
(a) 1 x 10-8
(b) 8 x 10-8
(c) 1.1 x 10-8
(d) 0.18 x 10-8
Answer:
(b) 8 x 10-8
Solution:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-88

Question 50.
pKa of acetic acid is 4.74. The concentration of CH3COONa is 0.01 M. The pH of CH3COONa
is …………..
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a) 3.37
Solultion:
[H+] = c.α = \(\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
pH = – log \(\left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}\right)^{1 / 2}\)
= \(\frac { 1 }{ 2 }\) [-1ogKa – log c]
= \(\frac { 1 }{ 2 }\) [4.74 – log 10-2]
= \(\frac { 1 }{ 2 }\) [4.74 + 2] = 3.37
pH = 3.37

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 51.
One litre of water contains 10 mol hydrogen ions. The degree of ionisation in water will be …………….
(a) 8 x 10-7
(b) 0.8 x 10-9
(c) 3.6 x 10-7
(d) 3.6 x 10-9
Answer:
(a) 8 x 10-7
Solution:
1 litre of water contains 1000/18 mole.
So, degree of ionisation = \(\frac{10^{-7} \times 18}{1000}\) = 1.8 x 10-7

Question 52.
If the solubility product of lead iodide (PbI2) is 3.2 x 10-8. Then its solubility in moles/litre will be ………….
(a) 2 x 10-3.
(b) 4 x 10-4
(c) 1.6 x 10-5
(d) 1.8 x 10-5
Solution:
Ksp = 4s3
4s3 = 3.2 x 10-8
s = 2 x 10-3M

Question 53.
The pH of a soft drink is 3.82. It’s hydrogen ion concentration will be ……………
(a) 1.96 x 10-2mol / L
(b) 1.96 x 1o-3 mol / L
(c) 1.5 x 10-4 mol / L
(d) 1.96 x 10-1 mol / L
Answer:
(c) 1.5 x 10-4 mol / L
Solution:
pH = 3.82 = – log10[H+]
∴ [H+] = 1.5 x 10-4 mol / litre

Question 54.
The pH of a solution at 25°C containing 0.10 M sodium acetate and 0.03 M acetic acid is …………..
(pKa for CH3COOH = 4.57)
(a) 4.09
(b) 5.09
(c) 6.10
(d) 7.09
Answer:
(b) 5.09
Solution:
pH = pKa + log \(\frac { [salt] }{ [acid] }\)
= 4.57 + log \(\frac { 0.10 }{ 0.03 }\) = 5.09

Question 55.
A weak acid is 0.1% ionised in 0.1 M solution. Its pH is …………..
(a) 2
(b) 3
(c) 4
(d) 1
Answer:
(c) 4
Solution:
For a monobasic acid [H+] = c.α
= \(\frac { 1 }{ 2 }\) x 0.001
= 10-4
pH = – log10[10-4] = 4

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 56.
Which one of the following is not a buffer solution?
(a) 0.8 M H2S + 0.8 M KHS.
(b) 2 M C6H5NH2 + 2 M C6H5N
(c) 3 M H2CO3 + 3 M KHCO3
(d) 0.05 M KCIO4 + 0.05 M HCIO
Answer:
(d) 0.05 M KCIO4 + 0.05 M HCIO
Hint. HClO4 is a strong acid while buffer is a mixture of weak acid and its salt.

Question 57.
The pH of pure water or neutral solution at 50°C is …………… (pKw = 13.2613 at 50°C)
(a) 7.0
(b) 7.13
(c) 6.0
(d) 6.63
Answer:
(d) 6.63
Solution:
[H+] [OH] = 10-13.26
[H+] = [OH]
[H+] = \(\frac { { 10 }^{ \frac { -13.26 }{ 2 } } }{ 2 }\)
∴ pH = 6.63

Question 58.
Increasing order of acidic character would be ……………..
(a) CH3COOH < H2SO4 < HCO3
(b) CH3COOH < H2CO3 < H2SO
(c) H2CO3 < CH3COOH < H2SO4
(d) H2SO4 < H2CO3 < CH3COOH
Answer:
(d) H2SO4 < H2CO3 < CH3COOH

Question 59.
What is the pH of 1 M CH3COOH solution?. Ka of acetic acid is 1.8 x 10-5. K = 10-14 mol2 litre2.
(a) 9.4
(b) 4.8
(c) 3.6
(d) 2.4
Answer:
(a) 9.4
Solution:
CH3COO + H2O \(\rightleftharpoons\) CH3COOH + OH
[OH] = c x h
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-93
= 2.35 x 10-5
pOH = 4.62
pH + pOH = 14
pH = 14 – 4.62 = 9.38

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 60.
4Na + O2 → 2Na2O
Na2O + H2O → 2NaOH
In the given reaction, the oxide of sodium is …………..
(a) Acidic
(b) Basic
(c) Amphoteric
(d) Neutral
Answer:
(b) Basic
Solution.
Na2O form NaOH so that it is basic oxide.

Question 61.
The pH of 0.001 M NaOH will be ………….
(a) 3
(b) 2
(c) 11
(d) 12
Answer:
(c) 11
Solution:
0.001 M NaOH means [OH] 0.001 .
10-3 pOH = 3
pH + pOH = 14
pH = 14 – 3 = 11

Question 62.
The addition of pure solid sodium carbonate to pure water causes …………….
(a) an increase in hydronium ion concentration
(b) an increase in alkalinity
(c) No change in acidity
(d) A decrease in hydroxide ion
Answer:
(b) an increase in alkalinity
Hint.
Adding Na2CO3 to water makes the solution basic and hence pH increases from 7.

Question 63.
When solid potassium cyanide is added in water then ……………
(a) pH will increase
(b) pH will decrease
(c) pH will remain the same
(d) electrical conductivity will not change
Answer:
(a) pH will increase
Hint:
KCN + H2O \(\rightleftharpoons\) KOH + HCN.
KOH is a strong base and HCN is a weak acid. So pH will increase.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 64.
pH of a solution is 5. Its hydroxyl ion concentration is …………..
(a) 5
(b) 10
(c) 10-5
(d) 10-9
Answer:
(d) 10-9
Solution:
pH = 5 means [H+] = 10-5
pOH = 14 – pH = 14 – 5 = 9
[OH] = 10-pOH = 10-9

Question 65.
Which one of the following is a buffer?
(a) CH3COOH + CH3COONa
(b) CH3COOH + CH3COONH4
(c) NaOH + NaCI
(d) CH3COOH + NH4CI
Answer:
(a) CH3COOH + CH3COONa

Question 66.
Which will have maximum pH?
(a) Distilled water
(b) 1 M NH3
(c) 1 M NaOH
(d) Water saturated by chlorine
Answer:
(c) 1 M NaOH
Hint:
NaOH has maximum [OH] and minimum of [H+ ] and so maximum pH value.

Question 67.
pH of a solution is 9.5. The solution is …………..
(a) Neutral
(b) Acidic
(c) Basic
(d) Amphoteric
Answer:
(c) Basic
Solution:
If pH = 7 solution is neutral
pH < 7 solution is acidic
pH > 7 solution is basic

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 68.
A solution has pH = 5, it is diluted 100 times, then it will become ……………….
(a) Neutral
(b) Basic
(c) unaffected
(d) more acidic
Answer:
(a) Neutral
Solution:
pH = 5 means [H+] = 10-5
After dilution [H+] = 10-5 / 100 = 10-7 M
[H+] from H2O cannot be neglected.
Total [H+] = 10-7 + 10-7 = 2 x 10-7
pH = 7 – 0.3010 = 6.6990 = 7
pH = 7 (Neutral)

Question 69.
By adding a strong acid to the buffer solution, the pH of the buffer solution ……………..
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant

Question 70.
pH of human blood is 7.4. Then H+ concentration will be ……………..
(a) 4 x 10-8
(b) 2 X 10-8
(c) 4 x 10-4
(d) 2 x 10-4
Answer:
(a) 4 x 10-8
pH = – log [H+]
7.4 = – log [H+]
7.4 = log 1 – log [H+]
log [H+] = log 1 – 7.4
log [H+] = 8.6
[H+] Antilog of 8.6
= 4 x 10-8

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 71.
The highest pH 14 is given by ……………….
(a) 0.1 M H2SO4
(b) 0.1 M NaOH
(c) 1 N NaOH
(d) 1 N HCl
Answer:
(a) 0.1 M H2SO4
Solution:
[OH] = 1
pOH = 0
pH + pOH = 14
pH = 14 – 0 = 14

Question 72.
Which of the following is not a Bronsted acid?
(a) CH3NH4
(b) CH3COO
(c) H2O
(d) HSO4
Answer:
(b) CH3COO
Hint:
Those substances which give a proton is called Bronsted acid, while CH3COO doesn’t have a proton. So it is not a Bronsted acid.

Question 73.
Pure water is kept in a vessel and it remains exposed to atmospheric CO2 which is absorbed, then its pH will be
(a) greater than 7
(b) less than 7
(c) equal to 7
(d) depends on ionic production of water
Hint:
CO2 is acidic oxide which on dissolution in water develops acidic nature.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 74.
As the temperature increases, the pH of KOH solution ……………..
(a) will decrease
(b) will increase
(c) remains constant
(d) depends upon concentration of KOH solution
Answer:
(a) will decrease

Question 75.
The pH of millimolar HCl is ………….
(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(b) 3
Solution:
pH = – log [H+]
[H+] = 10-3
pH = log 1 – log [H+]
= log 1 – log 10-3 = 3

Question 76.
The unit of ionic product of water K is ……………
(a) mol-1 L-1
(b) mol-2 L-2
(c) mol-2 L-1
(d) mol2 L-2
Answer:
(d) mol2 L-2

Question 77.
Review the equilibrium and choose the correct statement.
HClO4 + H2O \(\rightleftharpoons\) H3O+ + ClO4+
(a) HClO4 is the conjugate acid of H2O
(b) H3O is the conjugate base of H2O
(c) H2O is the conjugate acid of H3O
(d) ClO4 is the conjugate base of HCIO4
Answer:
(d) ClO4 is the conjugate base of HCIO4

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 78.
Which of the following is the strongest conjugate base?
(a) CI
(b) CH3COO
(c) SO42-
(d) NO2
Hint:
CH3COO is a conjugate base of a weak acid.
CH3COOH \(\rightleftharpoons\) CH3COO + H+

Question 79.
Which one of the following substance has the highest proton affinity?
(a) H2O
(b) H2S
(c) NH3
(d) PH3
Answer:
(c) NH3

Question 80.
Which of the following is the strongest Lewis acid?
(a) BI3
(b) BBr3
(c) BCI3
(d) BF3
Hint:
Larger the size of the halogen atom less is the back donation of electrons into empty 2p orbital of B.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 81.
Which of the following is the weakest acid?
(a) HF
(b) HCI
(c) HBr
(d) HI
Answer:
(a) HF
Hint:
HF does not give proton easily.

Question 82.
Among the following, the weakest Lewis base is …………
(a) H
(b) OH
(c) CI
(d) HClO3
Hint:
CI is a conjugate base of strong acid HCI.

Question 83.
Which of the following is not a Lewis acid?
(a) BF3
(b) AlCI3
(c) HCl
(d) LiAIH4
Hint:
It is a nucleophile and capable of donating electron pair and so it can act as Lewis base.

Question 84.
Which one of the following is called amphoteric solvent?
(a) Ammonium hydroxide
(b) Chloroform
(c) Benzene
(d) Water
Answer:
(d) Water
Hint:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-163

Question 85.
Which of the following is non – electrolyte?
(a) NaCl
(b) CaCl2
(c) C12H22O11
(d) CH3COOH
Hint:
C12H22O11 is a sugar and non-electrolyte.

Question 86.
At infinite dilution, the percentage ionisation for both strong and weak electrolyte is ………….
(a) 1%
(b) 20%
(c) 50%
(d) 100%
Answer:
(d) 100%
Hint:
According to Ostwald’s dilution law, degree of ionisation is directly proportional to the dilution.

Question 87.
An acid HA ionises as HA \(\rightleftharpoons\) H+ + A The pH of 1.0 M solution is 5. its dissociation constant would be …………..
(a) 1 x 10-5
(b) 1 x 10-10
(c) 5
(d) 5 x 108
Answer:
(b) 1 x 10-10

Question 88.
Three reactions involving H2PO4 are given below.
(i) H3PO4 + H2O → H3O+ + H2PO4
(ii) H2PO4 + H2O → HPO42 + H3O+
(iii) H2PO4 + OH → H3PO4 + O2

In which of the above does H2PO4 act as an acid.
(a) (i) only
(b) (ii) only
(c) (i) & (iii)
(d) (iii) only
Answer:
(b) (ii) only

Question 89.
Which of the following is not a Lewis acid?
(a) CO
(b) SiCl4
(c) SO3
(d) Zn2+
Answer:
(c) SO3
CO does not contain vacant d-orbital.

Question 90.
A chemist dissolves an excess of BaSO4 in pure water at 25°C if its Ksp = 1 x 10-10 What is the concentration of Barium in the water?
(a) 10-14 M
(b) 10-5 M
(c) 10-15 M
(d) 10-6 M
Answer:
(d) 10-6 M

Question 91.
On addition of ammonium chloride to a solution of ammonium hydroxide ……………
(a) dissociation of NH4OH increases
(b) concentration of OH increases
(c) concentration of OH decreases
(d) concentration of NH4 and OH increases
Hint:
Due to common ion effect.

Question 92.
The solubility product of a salt having a general formula MX2 in water is 4 x 10-2. The concentration of M2+ ions in the aqueous solution of the salt is ………………….
(a) 2.0 x 10-6 M
(b) 1.0 X 10-4 M
(c) 1.6 x 10-4 M
(d) 4.0 x 10-2 M
Answer:
(b) 1.0 X 10-4 M
Solution:
MX2 \(\rightleftharpoons\) M2++ 2X
Ksp = (2s)2 (s) = 4s3
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-94
= 1.0 x 10-4M

Question 93.
The solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is ……………
(a) 4 x3
(b) 108 x5
(c) 27 x4
(d) 9 x
Answer:
(a) 4 x3
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-95

Question 94.
What is the correct representation of the solubility product constant of Ag2CrO4?
(a) [Ag+]2 [CrO4-2]
(b) [Ag+] [CrO4-2]
(c) [2Ag+] [CrO4-2]
(d) [2Ag+]2 [CrO4-2]
Answer:
(a) [Ag+]2 [CrO4-2]
Ag2CrO4 \(\rightleftharpoons\) [2 Ag+] [CrO4-2]
Hence, Ksp = [Ag+]2 + [CrO4-2]

Question 95.
What is the pH value of \(\frac { N }{ 100 }\) KOH solution?
(a) 10
(b) 3
(c) 2
(d) 11
Answer:
(d) 11
Solution.
10-3 N KOH will give [OH] = 10-3 M
pOH = 3
pH + pOH = 14
pH = 14 – 3 = 11

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 96.
Which pair will show common ion effect?
(a) BaCI2 + Ba(NO3)2
(b) NaCI + HCI
(c) NH4OH + NH4CI
(d) AgCN + KCN
Answer:
(c) NH4OH + NH4CI

Question 97.
The sotubility of AgCI will be minimum in ………….
(a) 0.00 1 M AgNO3
(b) pure water
(c) 0.01 M CaCI2
(d) 0.01 M NaCl
Answer:
(c) 0.01 M CaCI2
Solution.
0.01 M CaCI2 gives maximum CI ions to keep Ksp of AgCl constant, decrease in [Ag+] will be maximum.

Question 98.
Ionic product of water increases if ………….
(a) pressure is reduced
(b) H+ is added
(c) OH is added
(d) temperature increases
Answer:
(d) temperature increases
Solution:
Kw increases with increase ¡n temperature.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 99.
pH of water is 7. When a substance Y is added in water, the pH becomes 13. The substance Y is a salt of …………..
(a) strong acid and strong base
(b) weak acid and weak base
(c) strong acid and weak base
(d) weak acid and strong base
Answer:
(d) weak acid and strong base

Question 100.
Sodium chloride is purified by passing HCl gas in a impure solution of sodium chloride. It is based on ………………
(a) Buffer action
(b) Common ion effect
(c) Association of salt
(d) Hydrolysis of salt
Answer:
(b) Common ion effect

II. Fill in the blanks.

  1. …………. theory does not explain the behaviour of acids and base in non aqueous solvents.
  2. According to Lowry Bronsted theory, an acid is defined as a substance that has a tendency to …………. a proton and base is a substance that has a tendency to …………. a proton.
  3. HCI and Cl are called …………. pairs.
  4. A …………. acid is a positive ion or an electron deficient molecule.
  5. …………. is an anion or neutral molecule that donates one lone pair of electrons.
  6. The ligands act as …………. and the central metal atoms that accepts a pair of electrons behave as a ………….
  7. Carbonium ion act as …………. and carbanion act as ………….
  8. Acids with …………. greater than ten are considered as strong acids and less than one are called weak acids.
  9. OH and H2 are considered as ………….
  10. ClO4, Cl, HSO4, NO3 are considered as ………….
  11. …………. can act as an acid as well as base.
  12. At 25°C, the value of Kw is equal to ………….
  13. With the increase in temperature, Kw value is ………….
  14. The dissociation of water is an …………. reaction.
  15. Aqueous solution of HCl is …………. whereas aqueous solution of NH3 is ………….
  16. For neutral solutions, the concentration of [H3O+] as well as [OH] is equal to at 25°C.
  17. The pH of battery acid is equal to ………….
  18. The pH of drain cleaner is equal to ………….
  19. …………. is the fraction of the total number of moles of a substance that dissociates at equilibrium.
  20. When the dilution increases by 100 times, the dissociation increases by ………….
  21. When dilution…………., the degree of dissociation of weak electrolyte also increases.
  22. The buffer present in the blood is ………….
  23. …………. introduced a quantity called buffer index ? as a quantitative measure of the
  24. When an acid reacts with a base, a salt and water are formed and the reaction is called ………….
  25. …………. is the conjugate base of the weak acid CH3COOH.
  26. Kidney stones are developed over a period of time due to the precipitation of ………….
  27. The pH of sea water is …………. than 7.
  28.  O2- and H are ………….
  29. All metal ions (or) atoms are ………….
  30. All anions are ………….

Answers:

  1. Arrhenius
  2. donate, accept
  3. Conjugate acid-base
  4. Lewis
  5. Lewis base
  6. Lewis base, Lewis acid
  7. Lewis acid, Lewis base
  8. Ka value
  9. very weak acids
  10. very weak base
  11. Water
  12. 1 x 10-14
  13. increases
  14. endothermic
  15. acidic, basic
  16. 1 x 10-17
  17. zero
  18. 14
  19. degree of dissociation ?
  20. 10 times
  21. increases
  22. H2CO3 and NaHCO3
  23. Vanslyke, buffer capacity
  24. Neutralization
  25. CH3COO
  26. Calcium oxalate
  27. greater
  28. strong base
  29. Lewis acids
  30. Lewis bases

III. Match the following column – I & II using the correct code given below that.

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-96
Answer:
(a) 4 3 1 2

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-97
Answer:
(b) 3 1 4 2

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-98
Answer:
(c) 2 3 4 1

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-99
Answer:
(d) 4 3 2 1

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-100
Answer:
(a) 3 1 4 2

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-101
Answer:
(a) 2 3 4 1

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-102
Answer:
(b) 2 4 1 3

Question 8.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-103
Answer:
(c) 3 1 4 2

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-104
Answer:
(d) 3 4 1 2

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-105
Answer:
(a) 2 1 4 3

Question 11.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-106
Answer:
(a) 3 4 2 1

IV. Assertion and reasons.

Question 1.
Assertion(A): In the process of dissolution of HCl in water, HCl act as acid and H2O act as base.
Reason (R): When HCl is dissolved in water, it donates a proton to water.
(a) Both A and R are correct and R explains A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A

Question 2.
Assertion(A): When ammonia dissolved in water, H20 acts as an acid.
Reason (R): When ammonia is dissolved in water, it accepts a proton from water. According to Lowry – Bronsted theory, proton donor is acid and so water act as an acid.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct. .
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 3.
Assertion (A): In the reaction HCl + H2O \(\rightleftharpoons\) H3O+ + Cl, HCl and Cl are conjugate acid – base pair.
Reason (R): By Lowry – Bronsted theory, chemical species that differ only by a proton are called conjugate acid – base pair.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 4.
Assertion(A): BF3 is a Lewis acid.
Reason (R): Boron has a vacant 2p orbital to accept the lone pair of electrons donated by any substance to form a new coordinate covalent bond.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong .
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 5.
Assertion(A): in coordination compounds, the ligands acts as Lewis acid aid the central metal atom or ion act as Lewis base.
Reason (R): Ligands are capable of accepting of a pair of electrons donated by the central metal atom or ion.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is not the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are wrong

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 6.
Assertion(A): SiF4 can act as Lewis acid.
Reason (R): In SiF4, the central atom can expand its octet due to the availability of empty d – orbitais and can accept a pair of electrons.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 7.
Assertion(A): NH3, H2O, ROH all are examples of Lewis bases.
Reason (R): Molecules with one or more lone pairs of electrons act as Lewis bases.
(a) Both A and R are wrong
(b) A is correct and R is the correct explanation of A.
(c) A is wrong but R is correct.
(d) A is correct but R is wrong
Answer:
(b) A is correct and R is the correct explanation of A.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
Assertlon(A): HCl is an strong acid while HCOOH is a weak acid.
Reason (R): HCI is completely ionised in water whereas HCOOH is paritally ionised in water.
(a) Both A and R are wrong
(b) A is wrong but R is correct
(c) A is correct but R is wrong
(d) Both A and R are correct and R is the correct explanation of A
Answer:
(d) Both A and R are correct and R is the correct explanation of A

Question 9.
Assertion(A): With the increase in temperature, the ionic product of water also increases.
Reason (R): The dissociation of water is an endothermic reaction.
(a) Both A and R are correct and R is the correct explanation of A
(b) A is correct but R is wrong
(c) A is wrong but R is correct
(d) Both A and R are wrong
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 10.
Assertion(A): When dilution increases, the degree of dissociation of weak electrolyte also increases.
Reason (R): The degree of dissociation a is inversely proportional to concentration c. When the dilution increases by 100 times, the dissociation increases by 10 times.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 11.
Assertion(A): The addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid.
Reason (R): This is due to common ion effect. i.e., CH3COOH and CH3COONa both contains CH3COO ion as common.
(a) Both A and R are correct and R is the correct explanation of A
(b) BothA and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 12.
Assertion(A): The solution of NH4CI has pH value less than 7.
Reason (R): The salt of weak base (NH4OH) and strong acid (HCl) is acidic in nature, when dissolved in water. So pH value is less than 7.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 13.
Assertion(A): pH = 7 signifies pure water.
Reason (R): pH = 7 means it is a neutral solution where [H3O+] [OH]
(a) A is correct but R is wrong
(b) A is wrong but R is correct
(c) Both A and R are wrong
(d) A is correct and R does not explain A
Answer:
(b) A is wrong but R is correct

Question 14.
Assertion(A): A mixture of CH3COOH and CH3COONH4 is an acidic buffer.
Reason (R): An acidic buffer contains a weak acid and the salt of weak acid with strong base.
(a) A is correct but R is wrong.
(b) A is wrong but R is correct.
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are wrong
Answer:
(b) A is wrong but R is correct.

Question 15.
Assertion(A): Buffer mixture is the one whose pH remains constant even by addition of strong acid or strong base.
Reason (R): To resist changes in its pH on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid or base.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

V. Find the odd one out and give the reasons.

Question 1.
(a) HNO3
(b) Ba(OH)2
(c) H3PO4
(d) CH3COOH
Answer:
(b) Ba(OH)2
Reason: Ba(OH)2 is the base whereas the others are acids.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 2.
(a) NH3
(b) H2O
(c) RNH2
(d) BF3
Answer:
(d) BF3
Reason: BF3 is a Lewis acid whereas others are Lewis base.

Question 3.
(a) SiF4
(b) SF4
(c) FeCl3
(d) NH3
Answer:
(d) NH3
Reason: NH3 is a Lewis base whereas others are Lewis acid.

Question 4.
(a) HCl
(b) H2SO4
(c) CH3COOH
(d) HNO3
Answer:
(c) CH3COOH
Reason: CH3COOH is a weak acid whereas others are strong acids.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 5.
(a) HCOOH
(b) CH3COOH
(c) Lactic acid
(d) HCI
Answer:
(d)HCI
Reason: HCl is a strong acid whereas others are weak acids.

Question 6.
(a) HClO4
(b) HCI
(c) HSO4
(d) H2SO4
Answer:
(c) HSO4.
Reason: HSO4 is a very weak base whereas others are strong acid.

Question 7.
(a) NH2
(b) O2
(c) H
(d) OH
Answer:
(d) OH
Reason: OH is a very weak acid whereas others are strong bases.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
(a) HNO2
(b) HF
(c) H2SO4
(d) CH3COOH
Answer:
(c) H2SO4
Reason: H2SO4 is a strong acid whereas others are weak acids.

Question 9.
(a) F
(b) CH3COO
(c) O2
(d) NO2
Answer:
(c) O2-
Reason: O2- is a strong base whereas others are weak bases.

Question 10.
(a) Vinegar
(b) Black coffee
(c) Sea water
(d) Orange juice
Answer:
(c) Sea water
Reason: Sea water is basic and has pH > 7 whereas others are acidic and have pH < 7.

Question 11.
(a) Baking soda
(b) Tomato
(c) Soapy water
(d) Drain cleaner
Answer:
(b) Tomato
Reason: Tomato has pH less than 7 and it is acidic whereas others have pH greater than 7 and they are basic.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 12.
(a) CH3COOH + CH3COONa
(b) NH4OH + NH4CI
(c) H2CO3 + NaHCO3
(d) NaOH + NaCl
Answer:
(d) NaOH + NaCl
Reason: NaOH + NaCl is not a buffer mixture whereas others are buffer mixtures.

VI. Find out the incorrect pair.

Question 1.
(a) HNO3, H2SO4
(b) Al(OH)3 , Mg (OH)2
(c) CH3COOH, HCOOH
(d) H2O, OH
Answer:
(d) H2O, OH

Question 2.
(a) HCl , Cl
(b) H2O, H3O+
(c) HNO3 , HNO2
(d) H2SO4, HSO4
Answer:
(c) HNO3 , HNO2

Question 3.
(a) NH3, H2O
(b) ROH, ROR
(c) CN, SCN
(d) BF3, H2O
Answer:
(d) BF3, H2O

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 4.
(a) BF3 , BF2
(b) Fe2, Fe3
(c) CaO, Mg(OH)2
(d) SiF4, SF4
Answer:
(c) CaO, Mg(OH)2

Question 5.
(a) Orange juice, Tomato juice
(b) Soapy water, Sea water
(c) Water, H3O
(d) Bleach , Ammonia solution
Answer:
(c) Water, H3O

VII. Find out the correct pair.

Question 1.
(a) HNO3, Ba(OH)2
(b) CH3COOH , HCI
(c) H3O+, Cl
(d) HCl + H2SO4
Answer:
(d) HCl + H2SO4.

Question 2.
(a) BF3, NH4+
(b) CH2 , CH3+
(c) Fe2+, Fe3+
(d) (CH3)3C+ , CH2 = CH2
Answer:
(c) Fe2+, Fe3+

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 3.
(a) H3O+, HCI
(b) HSO4, NO2
(c) HNO2, H2
(d) HCl, Cl
Answer:
(d) HCI, CI

Question 4.
(a) Orange, Black coffee
(b) Baking soda,, Water
(c) Ammonia, Stomach acid
(d) Bleach , Tomato
Answer:
(a) Orange , Black coffee

Question 5.
(a) NH4OH + NaOH
(b) NaOH + NaCl
(c) CH3COOH + CH3COONa
(d) CH3COOH + CH3COONH4
Answer:
(c) CH3COOH + CH3COONa

VIII. Answer the following.

Question 1.
What are the general characteristics of acid and base?
Answer:

  1. Acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas.
  2. Base tastes bitter, turns the red litmus to blue and soapy to touch.

Question 2.
Explain the Arrhenius concept of acid and base with example.
Answer:
1. According to Arrhenius concept, an acid is a substance that dissociates to give hydrogen ions in water. For example,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-107

2. Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-108

Question 3.
What are the limitations of Arrhenius concept?
Answer:
1. Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.

2. This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 4.
What is meant by strong acid and weak acid? Explain with example.
Answer:
1. A strong acid is the one that is almost completely dissociated in water.
HCl + H2O → H3O+ + Cl

2. A weak acid is the one that is partially dissociated in water.
CH3COOH +H2O \(\rightleftharpoons\) H3O + CH3COO

Question 5.
Give two examples for

  1. Strong acid
  2. Strong base

Answer:

  1. Strong acid: HCIO4, H2SO4
  2. Strong base: NH2, O2-

Question 6.
Give two examples for

  1. Very weak acid
  2. Very weak base

Answer:

  1. Very weak acid: OH,H2
  2. Very weak base: Cl, ClO4

Question 7.
Give two examples for

  1. Weak acid
  2. Weak base

Answer:

  1. Weak acid: HF, CH3COOH
  2. Weak base: F, CH3COO

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 8.
What is meant by auto ionisation of water?
Answer:
Pure water has a little tendency to dissociate. i:e., one water molecule donates a proton to another water molecule. This is known as auto ionisation of water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-109

Question 9.
Define – ionic product of water.
Answer:
H2O + H2O \(\rightleftharpoons\) H3O + OH
The dissociation constant for the above ionisation is given as,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-110
The concentration of pure water is one.
i.e., [H2O]2 = 1
K = [H3O+] [OH]
Kw = ionic product of water.
Kw = 1 x 10-14 at 25°C

Question 10.
Kw = 1 x 10-14 at 25°C. Justify this statement.
Answer:
1. Experimentally found that the concentration of H3O in pure water is 1 x 10-7 at 25°C.

2. Since the dissociation of water produces equal number of H3O4 and OH, the concentration of OH is also equal to 1 x 10 at 25°C. The ionic product of water at 25°C is
Kw = [H3O+] [OH]
= [1 x 10-7] [1 x 10-7]
Kw = [1 x 10-14]

Question 11.
With increase in temperature, Kw also increases. Why?
Answer:

  1. All equilibrium constant Kw. is also a constant at a particular temperature.
  2. The dissociation of water is an endothermic reaction.
  3. With the increase in temperature, the concentration of H3O+ and OH also increases and hence the ionic product also increases.

Question 12.
Aqueous HCl is an acidic solution whereas aqueous NH3 is a basic solution. Justify this statement.
Answer:
HCI + H2O \(\rightleftharpoons\) H3O+ + Cl in this case, in addition to auto ionisation of water, HCI molecule also produces H3O ion by donating a proton to water and hence [H3OJ> [OH]. It means that the aqueous HCI solution is acidic. Similarly in basic solution such as aqueous NH3, [OH] > [H3O+] and it is basic.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 13.
What is the statement of Ostwald’s dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolyte also increases.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-111
This statement is known as Ostwald’s dilution law.

Question 14.
Define – Salt hydrolysis.
Answer:
Salts completely dissociate in aqueous solution to give their constituent ions. The ions so produced are hydrated in water. ¡n certain cases, the cation, anion ?r both react with water and the reaction is called salt hydrolysis. e.g.,
NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(1)

Question 15.
What is meant by conjugate acid-base pair? Find the conjugate acid / base for the following species
Answer:
HNO2, CH, HCIO4, OH, CO32-, S2-
An acid-base pair which differs by a proton only (HA \(\rightleftharpoons\) A + H+) is known as conjugate acid-base pair.
Conjugate acid: HCN, H2O, HCO3, HS.
Conjugate base: NO2, ClO4, O2.

Question 16.
Which of the following are Lewis Acids?
H2O, BF3, H+ and NH4+
Answer:
BF3, H+ ions are Lewis acids.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 17.
What will be the conjugate bases for the Bronsted acids? HF, H2SO4 and H2CO3?
Answer:
Conjugate bases: F, HSO4, HCO3.

Question 18.
Write the conjugate acids for the following Bronsted bases:
NH2, NH3 and HCOO
Answer:
NH3, NH4 and HCOOH

Question 19.
The species H2O, HCO3, HSO4 and NH3 can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-112

Question 20.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid I Lewis base?
(a) OH ions
(b) F
(c) H+
(d) BCI3
Answer:
(a) OH ions can donate an electron pair and act as Lewis base.
(b) F ions can donate an electron pair and act as Lewis base.
(a) H+ ions can accept an electron pair and act as Lewis base.
(b) BCl3 can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

Question 21.
Predict the acidic, basic or neutral nature of the solutions of the following salts:
NaCI, KBr, NaCN, NH4NO3, NaNO2, KF.
Answer:
NaCN, NaNO2, KF solutions are basic, as they are salts of strong base, weak acid.
NaCl, KBr solutions are neutral, as they are salts, of strong acid, strong base.
NH4NO3 solution is acidic, as it is a salt of strong acid, weak base.

Question 22.
Ionic product of water at 310 K is 2.7 x 10-14 What Is the pH of neutral water at this temperature?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-113

Question 23.
The aqueous solution of sugar does not conduct electricity whereas when sodium chloride is added to water, it conducts electricity. Justify this statement.
Answer:
1. Sugar is a non electrolyte and when it dissolves in water, there will be no ionisation takes place. If there is no free ions, it does not conduct electricity.

2. When sodium chloride is added to water, it is completely ionised to give Na ions and Cl ions. Due to the presence of ions, they will be possibility of electhcal conductance. Because ions are carriers of electric current.

Question 24.
A reaction betwen ammonia and boron trifluoride is given below
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-114
Identify the acid and base in the reaction. Which theory explain it?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-115
1. In the above reaction BF3 is an acid and NH3 is the base.

2. Lewis concept explain it as follows
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-116

3. A Lewis acid is an electron deficient molecule and capable of accepting a pair of electrons and a Lewis base is electron rich molecule and capable of donating a pair of electrons.

Question 25.
The salt of strong acid and strong base does not undergo hydrolysis. Explain.
Answer:
1. In this case, neither the cations nor the anions undergo hydrolysis. Therefore the solution remains neutral.

2. For example, in the aqueous solution of NaCl, its ions Na+ and Cl ions have no tendency to react with H+ or OH ions of water.

This is because the possible products of such reaction are NaOH and HCI which are completely dissociated. As a result, there is no change in the concentration of W and OH ions and hence the solution continues to remain neutral.

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 3 Mark Questions ans Answers

Question 1.
Explain Lowry – Bronsted theory of acid and base.
Answer:
1. According to Lowry-Bronsted theory, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance.

2. An acid is a proton donor and a base is a proton acceptor.

3. When HCI is dissolved in H2O, HCI donates a proton to H2O. Thus HCI behaves as an acid and H2O is a base.
HCI + H2O \(\rightleftharpoons\) H3O+ + Cl

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 2.
Explain the reaction of water with ammonia by proton theory.
Answer:
1. When ammonia dissolved in water, it accepts a proton from water. In this case, ammonia (NH3) acts as a base and H2O is acid.

2. The reaction is represented as
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-117

3. The species that remains after the donation of a proton is a base (Base1) and is called the conjugate base of Bronsted acid (Acid1). In other words, chemical species that differ only by a proton are called conjugate acid base pairs Conjugate acid – base pair
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-118

Question 3.
Explain about the strength of acids on the basis of Ka value.
Answer:
1. Ka is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid.

2. Acids such as HCI, HNO3 are almost completely? onised and hence they have high Ka value i.e., Ka for HCI at 25°C is 2 x 106.

3. Acids such as formic acid and acetic acid are partially ionised in solution and have low Ka value. i.e., Ka for acetic acid 1.8 x 10-5 at 25°C

4. Acids with Ka value greater than ten are considered as strong acids and less than one considered as weak acids.

Question 4.
Write 3 formulas of strong acids, strong bases and weak acids.
Answer:

  1. HClO4, HCI, H2SO4 – are strong acids
  2. NH2, O2-, H – are strong bases
  3. HNO2, HF, CH3COOH are weak acids

Question 5.
pH of a neutral solution is equal to 7. Prove it.
Answer:
1. in neutral solutions, the concentration of [H3O+] as well as [OH] are equal to 1 x 10-7M at 25°C.

2. The pH of a neutral solution can be calculated by substituting this [H3O+] çoncentration in the expression
pH = – log10 [H3O+]
= – log10 [1 x 10-7]
= – ( – 7)log \(\frac { 1 }{ 2 }\) = + 7 (l) = 7

3. pH = 7 for a neutral solution

Question 6.
Derive the relation between pH and pOH
Answer:
pH = – log10 [H3O] ………………(1)
pOH = – log10 [OH] ……………(2)
Adding equations (1) and (2),
pH + pOH = (- log10 [H3O+) + ( – log10[OH])
= – [(log10[H3O]) + (log10 [OH)]
pH + pOH = – log10[H3O+] [OH]
[H3O] [OH] = Kw
pH + pOH = – log Kw
pH + pOH = pKw
[pKw = – 1og10Kw]
At 25°C, the ionic product of water Kw = 1 x 10-14.
pKw = – 1og1010-14 = 14 log1010 = 14
pKw = 14
pH + pOH = 14 at 25°C.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 7.
When the dilution increases by 100 times, the dissociation increases by 10 times. Justify this statement.
Answer:
(i). Let us consideran acid with Ka value 4 x 104. We are calculating the degree of dissociation of that acid at two different concentration 1 x 10-2 M and 1 x 10-4 M using Ostwalds dilution law
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-119

(iv) i.e., when the dilution increases by 100 times (concentration decreases from 1 x 10-2M to 1 x 10-4M), the dissociation increases by 10 times.

(v) When dilution increases, the degree of dissociation of weak electrolyte also increase. (Ostwalsd’s dilution law).

Question 8.
What Is buffer solution? Give an example for an acidic buffer and a basic buffer.
Answer:

  1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.
  2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.
  3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH4O and NH4Cl.

Question 9.
Define buffer capacity and buffer index.
Answer:

  1. The buffering ability of a solution can be measured in terms of buffer capacity.
  2. Buffer index ?, as a quantitative measure of the buffer capacity.
  3. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.
  4. β = \(\frac { dB }{ d(pH) }\). dB = number of gram equivalents of acid / base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.

Question 10.
How is solubility product is used to decide the precipitation of ions?
Answer:
1. When the product of molar concentration of the constituent ions i.e., ionic product exceeds the solubility product then the compound gets precipitated.

2. When the
ionic Product > Ksp precipitation will occur and the solution is super saturated. ionic Product < Ksp no precipitation and the solution is unsaturated. ionic Product = Ksp equilibrium exist and the solution ¡s saturated.

3. By this way, the solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contain the constituent ions are mixed.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 11.
Derive the value of solubility product from molar solubility.
Answer:
1. Solubility can be calculated from molar solubility.i.e., the maximum number of moles of the solute that can be dissolved in one litre of the
solution.

2. For a solute XmYn
Xm Yn(s) \(\rightleftharpoons\) mXn+(aq) + n Ym-(aq)

3. From the above stoichiometrically balanced equation, it is clear that I mole of Xm Yn(s) dissociated to furnish ‘m’ moles of x and ‘n’ moles of Y. If’s’ is the molar solubility of Xm Ynthen
Answer:
[Xn+] = ms and [Ym-] = ns
Ksp = [Xn+]m [Ym-]n
Ksp = (ms)m (ns)n
Ksp = (m)m (n)n (s)m+n

Question 12.
The concentration of hydrogen ions ¡n a sample of soft drink is 3.8 x 10-3m. What is the. pH value? Whether the soft drink is acidic (or) basic?
Answer:
pH = – log10 [H3O+]
= – log10 [3.8 x 10]
= – log 3.8 + 3
= 3 – 0.5798 = 2.4202
pH = 2.42
When pH < 7, the soft drink is acidic.

Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

Question 13.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
pH = – log10 [H3O+]
= – log10 = – pH = – 3.76
= \(\overline{4}\).24
[H3O+] = antilog \(\overline{4}\).24
= l.738 x 10-4
[H3O+] = 1.74.x 10-4M

Question 14.
The ionisation constant of HF, HCOOH, HCN at 298 K are 6.8 x 10-4, 1.8 x 10-4 and 4.8 x 10-9 respectively. Calculate the ionisation constant of the corresponding conjugate base.
Answer:
1. HF, conjugate base is F
Kb = Kw/Ka = \(\frac{1 \times 10^{-4}}{6.8 \times 10^{-4}}\) = l.47 x 10-11 = l.5 x 10-11

2. for HCOO
Kb = \(\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10-11

3. for CN
Kb = \(\frac{1 \times 10^{-14}}{4.8 \times 10^{-4}}\) = 2.8 x 10 -6

Question 15.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
HCNO \(\rightleftharpoons\) H+ + CNO
pH = 2.34 means – log [H+] = 2.34 or log [H+] = – 2.34 = 3.86
or
[H+] = Antilog 3.86 = 4.57 x 10-3 M
[CNO] = [H+] = 4.57 x 10-3 M
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-123

Question 16.
The Ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Sodium mtrite is a salt of weak acid, strong base. Hence,
Kh = 2.22 x 10-11 Kw/Kb = 10-14/(4.5x 10-4)
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-124
[OH] = ch = 0.04 x 2.36 x 10-5 = 944 x 10-7
pOH = – log (9.44 x 10-7) = 7 – 0.9750 = 6.03
pH = 14 – pOH = 14 – 6.03 = 7.97

Question 17.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K. For calcium sulphate, Ksp = 9.1 x 10-6.
Answer:
CaSO4(s) Ca2(aq) + SO2-4(aq)
If ‘s’ is the solubility of CaSO4 in moles L, then Ksp = [Ca2+] x [SO42-] = s2
or
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-125
= 3.02 x 10-3 x 136gL-1 = 0.411 gL-1
(Molar mass of CaSO4 = 136 g mol-1)
Thus, for dissolving 0.441 g, water required = I L
For dissolving 1g, water required = \(\frac { 1 }{ 0.411 }\)L = 2.43L

Question 18.

  1. Point out the differences between ionic product and solubility product.
  2. The solubllity of AgCI in water at 298 K is 1.06 x 10-5 mole per litre. Calculate is solubility product at this temperature.

Answer:
1. Ionic product

  1. It is applicable to all types of solutions.
  2. Its value changes with the change in con centration of the ions.

Solubility product

  1. It is applicable to the saturated solutions.
  2. It has a definite value for an electrolyte at a constant temperature.

2. The solubility equilibrium in the saturated solution is
AgCl (s) \(\rightleftharpoons\) Ag+(aq) + Cl (aq)
The solubility of AgCl is 1.06 x 10-5 mole per litre.
[Ag+(aq)] = 1.06 x 10-5 mol L-1
[Cl (aq)] = 1.06 x 10-5 mol L-1
Ksp = [Ag+(aq)] [Cl (aq)]
= (1.06 x 10-5 mol L-1) x (1.06 x 10-5 mol L-1)
= 1.12 x 10-2 moI2 L-2

Question 19.
The value of K of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 x 10-15 and 6 x 10-17 respectively. Which salt is more soluble? Explain.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-126
Ni(OH)2 is more soluble than AgCN.

Question 20.
If 0.561 g KOH is dossolved in water to give 200 mL of solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-159

Samacheer Kalvi 12th Chemistry Ionic Equilibrium 5 Marks Questions and Answers

Question 1.
Differentiate Lewis acids and Lewis bases.
Answer:
Lewis Acids

  1. Lewis acids are substances that can accept one or more lone pair of electrons.
  2. All metal ions (or) atoms can act as Lewis acids. Examples: Fe2+, Fe3+, Cu2+, Cr3+
  3. Molecules that contain a polar double bond can act as Lewis acids. Examples: SO2, CO2, SO3
  4. Molecules in which the central atom can expand its act due to the availability of empty d-orbitais can act as Lewis acid. Example: SiF4, SF4, FeCI3
  5. Carbonium ion (CH3)3C+ can act as Lewis acid
  6. Electron deficient molecules such as BF3, AlCl3, BeF2 act as Lewis acid (electron pair acceptors)

Lewis Bases

  1. Lewis bases are substances that can donate one or more lone pair of electrons.
  2. All anions can act as Lewis bases. Examples: F, Cl, CN, SO42-
  3. Molecules that contain carbon-carbon multiple bond. Example: CH2 = CH2, CH = CH
  4. All metal oxides can act as Lewis bases. Examples : CaO, MgO, Na2O
  5. CH2 carbanion cari act as Lewis acid
  6. Electron rich molecules such as NH3, H2O, ROH, R – O – R, R – NH2 act as Lewis base (Electron pair donors)

Question 2.
Explain about the ionisation of weak acid and how K2 is derived?
Answer:
1. Weak acids are partially dissociated ¡n water and there is an equilibrium between the undissociated acid and its dissociated ions.

2. Consider the ionisation of weak monobasic acid HA in water
HA + H2O \(\rightleftharpoons\) H3O+ + A …………….(1)

3. Applying law of chemical equilibrium, the equilibrium constant Kc is given by the expression
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-127

4. In dilute solutions, water is present in large excess, hence its concentration may be taken as constant say K. Further H3O+ indicates hydrated hydrogen ions, for simplicity, it may be replaced by H+. So the equation (2) becomes
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-128

5. The product of two constants K and K gives another constant. Let is be K2
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-129
The constant Ka is called dissociation constant of weak acid.

Question 3.
Explain Buffer action with suitable example.
Answer:
Buffer action:
1. Let us consider buffer solution containing CH3COOH and CH3COO Na. The dissociation the buffer components occur as below.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-130

2. If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO to form undissociated weak acid. i.e., the increase in the concentration of H+ does not reduce the pH significantly.
CH3COO(aq) + H+(aq) → CH3COOH(aq)

3. If a base is added, it will be neutralised by H3O and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not altered.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-131

4. On the addition of an acid (or) base to a buffer solution, there will be no change in its pH value. Because the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid (or) base at the same time, these components should not consume each other.

Question 4.
Prove the buffer action of acetic acid and sodium acetate by the addition of 0.01 mol of solid sodium hydroxide.
Answer:
1. Consider one litre of buffer solution containing 0.8 m CH3COOH and O.8m CH3COONa. Assume that the volume change due to the addition of 0.01 mol of solid NaOH is negligible. Ka for CH3COOH is l.8 x 10-5.

2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-132

3. The dissociation constant for CH3COOH is given by
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-133
The above expression shows that the concentration of H+ is directly proportional to
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-134
degree of dissociation of CH3COOH = α

4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-135

5. Given that Ka for CH3COOH is 1.8 x 10-5
[H+] = 1.8 x 10-5
pH = – log [H+]
= – log [ 1.8 x 10-5]
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

6. After adding 0.01 moI NaOH to I litre of buffer. Given that volume change due to the addition of NaOH is negligible. [OH] = 0.01 M. The consumption of OH are expressed by the following equation.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-136
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-137

7. The addition of a strong base (0.01 M NaOH) increased the pH only slightly i.e., from 4.74 to 4.75. So the buffer action is verified.

Question 5.
DerIve Henderson – Hasselbalch equation
Answer:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution. i.e.,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-138

2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-139

3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.

4. Taking logarithm on both sides of the equation
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-140

5. reverse the sign on both sides
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-141
Eqaution (6) & (7) are called Henderson – Hasselbalch equations.

Question 6.
Explain about the hydrolysis of salt of strong acid and a strong base with a suitable example.
Answer:
1. Let us consider the neutralisation reaction between NaOH and HNO3 to give NaNO3 and water.
NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(1)

2. The salt NaNO3 completely dissociates in water to produce Na+ and NO3 ions
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-142

3. Water dissociates to a small extent as
H2O(1) H+(aq) + OH(aq)
Since [H+] = [OH], water is neutral.

4. NO3 ion is the conjugate base of strong acid HNO3 and hence it has no tendency to react withH+,

5. Similarly Na is the conjugate acid of the strong base NaOH and it has no tendency to react with OH

6. It means that there is no hydrolysis. In such cases [H+] (OH), pH is maintained and there fore the solution is neutral.

Question 7.
Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of Kh for that reaction.
Answer:
1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water.
NaOH(aq) + CH3COOH(aq) \(\rightleftharpoons\) CH3COONa(aq) + H2O(1)

2. In aqueous solution, CH3COONa is completely dissociated as follows.
CH3COONa(aq) CH3COO(aq) + Na+(aq)

3. CH3COO is a conjugate base of the weak acid CH3COOH and it has a tendency to react with H+ from water to produce unionised acid. But there is no such tendency for Na+ to react with OH

4. CH3COO(aq) + H2O(1) CH3COOH(aq) + OH3 and therefore [OH] > [H+], in such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-143
Equation (1) x (2)
Kh.Ka = [H+] [OH]
[H+] [OH] = Kw
Kh.Ka = Kw
Kh value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law Kh = h2C and [OH] =Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-144

Question 8.
DerIve the value of pH of salt solution in terms of Ka and concentration of electrolyte.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-145

Question 9.
Explain about the hydrolysis of salt of strong acid and weak base. Derive Kh and pH for that solution.
Answer:
1. Consider a reaction between strong acid HCl and a weak base NH4OH to produce a salt NH4CI and water
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-160

2. NH4 is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH- from water to produce unionised NH4 as below,
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-161

3. There is no such tendency shown by Cl and therefore [H+] > [OH] the solution is acidic and the pH is less than 7.

4. In the salt hydrolysis of strong base and weak acid, we have to derive a relationship between Kh and Kb as
Kh . Kb = Kw

5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-146

Question 10.
Discuss about the hydrolysis of salt of weak acid and weak base and derive pH value for the solution.
Answer:
1. Consider the hydrolysis of ammonium acetate
CH3COONH4(aq) → CH3COO(aq) + NH+4(aq)

2. In this case both the cation (NH4+) and (CH3COO) anion have the tendency to react with water.
CH3COO + H2O \(\rightleftharpoons\) CH3COOH + OH
NH4+ + H2O \(\rightleftharpoons\) NH4OH + H3

3. The nature of the solution depends on the strength of acid (or) base i.e., if Ka > Kb, then the solution is acidic and pH < 7, if Ka < Kb then the solution is basic and pH > 7. If Ka = Kb, then the solution is neutral.

4. The relation between the dissociation constant Ka, Kb and hydrolysis constant is given by the following expression.
Ka . Kb. Kh = Kw

5. pH of the solution
pH = 7 + \(\frac { 1 }{ 2 }\)pKa – \(\frac { 1 }{ 2 }\)pKb

Question 11.
It has been found that the pH of a 0.01 M solution of an organic acid Is 4.15. Calculate the concentration of the anion, the Ionization constant of the acid and its pKa.
Answer:
HA \(\rightleftharpoons\) H+
pH = log [H+] or log [H+] = – 4.15 = 5.85
[H+] = 7.08 x 10-5 M = 7.08 x 10-5 M
[A] = [H+] = 7.08 x 10-5M
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-162
pKa = – logKa = – log (5.0 x 10-7) = 7 – 0.699 = 6.301

Question 12.
Assuming complete dissociation, calculate the pH of the following solutions.
(i) 0.003 M HCl
(ii) 0.005 M NaOH
(iii) 0.002 M HBr
(iv) 0.002 M KOH
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-148

Question 13.
What ¡s the pH of 0.001 M aniline solution? The ionisation constant of aniline is 4.27 x 10-10. Calculate degree of ionization of aniline in the solution. Also calculate the ionisation constant of the conjugate acid of anile.
Answer:
1. C6H5NH2 + H2O \(\rightleftharpoons\) C6H5NH3 + OH
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-149

2. Also
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-150

3. pKa + pKb = 14 (for a pair of conjugate acid and base)
pKb = – log (4.27 x 10-10) = 10 – 0.62 = 9.38
pKa = 14 – 9.38 = 4.62
i.e., – log Ka 4.62 or log Ka = – 4.62 =
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-151

Question 14.
Calculate the degree of ionization of 0.05 M acetic acid If its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains

  1. 0.01 M
  2. 0.1 M HCI

Answer:
PKa = i.e., – log Ka = 4.74
or log Ka = 4 . 74 = 5.26
Ka = 1.82 x 10-5
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-152
In presence of HCI, due to high concentration of H+ ion, dissociation equilibrium will shift backward, Le., dissociation of acetic acid will decrease.
1. In presence of 0.01 M HCI, if x is the amount dissociated, then
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-153

2. In the presence of 0.1 M HCl, if y is the amount of acetic acid dissociated, then at equilibrium
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-154
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-155

Question 15.
The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-156

Common Errors

  1. Acid and Base – Definition
  2. pH value
  3. Buffer mixture
  4. Conjugate Acid-base pair

Rectifications

  1. Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium-157
  2. pH neutral
  3. pH less than 7 – Acid
  4. pH more than 7 – Base
  5. Always either a weak acid and its salt (or) weak base and it’s salt.
  6. They differ by H+. For e.g., CH3COOH. Its conjugate base is CH3COO. H2O – Acid and its conjugate base is OH

Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics

Students can Download Physics Chapter 8 Atomic and Nuclear Physics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Multiple Choice Questions

Question 1.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
(a) 14.4\(\frac { Z }{ V }\) Å
(b) 14.4\(\frac { V }{ Z }\) Å
(c) 1.44\(\frac { Z }{ V }\) Å
(d) 14.4\(\frac { V }{ Z }\) Å
Answer:
(c) 1.44\(\frac { Z }{ V }\) Å

Question 2.
In a hydrogen atom, the electron revolving in the fourth orbit, has angular momentum equal to
(a) h
(b) \(\frac { h }{ π }\)
(c) \(\frac { 4h }{ π }\)
(d) \(\frac { 2h }{ π }\)
Answer:
(d) \(\frac { 2h }{ π }\)
Hint:
Angular momentum of an electron is an integral multiple of \(\frac { h }{ 2π }\)
According to Bohr atom model,
Angular momentum of an electron mvr = \(\frac { nh }{ 2π }\)
n = 4th orbit = \(\frac { 4h }{ 2π }\)
mvr = \(\frac { 2h }{ π }\)

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Question 3.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Hint:
The ionisation energy of a hydrogen atom is, IE = \(\frac {{ 13.6z }^{2}}{{ n }^{2}}\)
∴ z2 = \(\frac{I E \times n^{2}}{13.6}\) = \(\frac{122.4 \times(1)^{2}}{13.6}\) = 9

Question 4.
The ratio between the first three orbits of hydrogen atom is
(a) 1 : 2 : 3
(b) 2 : 4 : 6
(c) 1 : 4 : 9
(d) 1 : 3 : 5
Answer:
(c) 1 : 4 : 9
Hint:
En = \(\frac {{ -13.6×z }^{2}}{{ n }^{2}}\) eV / atom
n = 1; E1 = 13.6 eV / atom
n = 2; E2 = 3.4 eV / atom
n = 3; E3 = 151 eV / atom
The ratio of theree orbits
E1 : E2 : E3 = 13.6 : 3.4 : 1.51
= 1 : 4 : 9

Question 5.
The charge of cathode rays is
(a) positive
(b) negative
(c) neutral
(d) not defined
Answer:
(b) negative

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Question 6.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint:
In the condition of no deflection \(\frac { e }{ m }\) = \(\frac {{ E }^{2}}{{ 2vB }^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt { 208 } \) = 14.4 times

Question 7.
The ratio of the wavelengths for the transition from n =2 to n = 1 in Li++, He+ and H is
(a) 1 : 2 : 3
(b) 1 : 4 : 9
(c) 3 : 2 : 1
(d) 4 : 9 : 36
Answer:
(d) 4 : 9 : 36
Hint:
According to Rydberg formula, the wavelength
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-1

Question 8.
The electric potential between a proton and an electron is given by V = V0 In \(\left( \frac { r }{ { r }_{ 0 } } \right) \), where r0 is a constant. Assume that Bohr atom model is applicable to potential, then variation of radius of nth orbit rn with the principal quantum number n is
(a) rn ∝\(\frac { 1 }{ n }\)
(b) rn ∝ n
(c) rn ∝ \(\frac { 1 }{{ n }^{2}}\)
(d) rn ∝ n2
Answer:
(b) rn ∝ n
Hint:
Electric potential between proton and electron in nth orbit is given as,
V = V0 In \(\left( \frac { { r }_{ n } }{ { r }_{ 0 } } \right) \)
Thus the coulomb force |Fc| = e \(\left( \frac { { dv } }{ dr } \right) \) = e \(\left( \frac { { V }_{ 0 } }{ { r }_{ n } } \right) \)
This coulomb force is balance by the centripetal force
\(\frac {{ mv }^{2}}{{r}_{n}}\) = e \(\left( \frac { { V }_{ 0 } }{ { r }_{ n } } \right) \left( \frac { { dv } }{ dr } \right) \) ⇒ V = \(\sqrt { \frac { e{ V }_{ 0 } }{ m } } \)
Now from
mvrn = \(\frac { nh }{2π}\)
rn ∝ n

Question 9.
If the nuclear radius of 27 Al is 3.6 fermi, the approximate unclear radius of64 Cu is
(a) 2.4
(b) 1.2
(c) 4.8
(d) 3.6
Answer:
(c) 4.8
Hint:
\(\frac {{ R }_{Al}}{{ R }_{Cu}}\) = \(\frac{(27)^{1 / 3}}{(64)^{1 / 3}}\) = \(\frac { 3 }{ 4}\)
Rcu = \(\frac { 4 }{ 3}\) RAl = \(\frac { 4 }{ 3}\) x 3.6 fermi
Rcu = 4.8 fermi

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Question 10.
The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as
(a) A2/3
(b) A4/3
(c) A1/3
(d) A5/3
Answer:
(a) A2/3
Hint:
Volume of nucleus is proportional to mass number 4
\(\frac { 4 }{ 3 }\) πR3 ∝ A = R0 A1/3
So, πR2 = RR0 A2/3 ⇒ 4πR2 ∝ A2/3
Surface area is proportional to (mass number)2/3

Question 11.
The mass of a \(_{ 3 }^{ 7 }{ Li }\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of \(_{ 3 }^{ 7 }{ Li }\) nucleus is nearly
(a) 46 MeV
(b) 5.6 MeV
(c) 3.9 MeV
(d) 23 MeV
Answer:
(b) 5.6 MeV
Hint:
If w = 1 u, C = 3 x 108 ms-1 then, E = 931 MeV
1 u = 931 Mev
Binding energy = 0. 042 x 931
= 39. 10 MeV
∴ B.E 39.10
Binding energy per nucleon = \(\frac { B.E }{ A }\) = \(\frac { 39.10 }{ 7 }\) = 5.58 = 5.6 MeV

Question 12.
denotes the mass of the proton and Mn denotes mass of a neutron. A given nucleus of binding energy B, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by(where c is the speed of light)
(a) M (N, Z) = NMn + ZMp – Bc2
(b) M (N, Z) = NMn + ZMp + Bc2
(c) M (N, Z) = NMn + ZMp – B / c2
(d) M (N, Z) = NMn + ZMp + B / c2
Answer:
(c) M (N, Z) = NMn + ZMp – B / c2
Hint:
Binding energy, B = [ZMp + NMn – M (N, Z)] C2
M(N,Z) = ZMp + NMn – \(\frac { B }{{ C }^{ 2 }}\)

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Question 13.
A radioactive nucleus (initial mass number A and atomic number Z) emits 2α and 2 positrons. The ratio of number of neutrons to that of proton in the final nucleus will be
(a) \(\frac { A-Z-4 }{ Z-2 }\)
(b) \(\frac { A-Z-2 }{ Z 6 }\)
(c) \(\frac { A-Z-4 }{ Z-6 }\)
(d) \(\frac { A-Z-12 }{ Z-4 }\)
Answer:
(b) \(\frac { A-Z-2 }{ Z 6 }\)

Question 14.
The half-life period of a radioactive element A is same as the mean life time of another radioactive element B. Initially both have the same number of atoms. Then
(a) A and B have the same decay rate initially
(b) A and B decay at the same rate always
(c) B will decay at faster rate than A
(d) A will decay at faster rate than B.
Answer:
(c) B will decay at faster rate than A
Hint:
(t1/2)A = (tmean )B
\(\frac { 0.6931 }{{ λ }_{A}}\) = \(\frac { 1 }{{ λ }_{B}}\)
λA = 0.6931 λB
λA < λB

Question 15
A system consists of N0 nucleus at t = 0. The number of nuclei remaining after half of a half-life (that is, at time t =\(\frac { 1 }{ 2 }\) T\(\frac { 1 }{ 2 }\))
(a) \(\frac {{ N }_{0}}{ 2 }\)
(b) \(\frac {{ N }_{0}}{ √2 }\)
(c) \(\frac {{ N }_{0}}{ 4 }\)
(d) \(\frac {{ N }_{0}}{ 8 }\)
Answer:
(b) \(\frac {{ N }_{0}}{ √2 }\)
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-2

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Short Answer Questions

Question 1.
What are cathode rays?
Answer:
A cathode ray is a stream of electrons that are seen in vaccum tubes. It is called a “cathode ray” because the electrons are being emitted from the negative charged element in the vaccum tube called the cathode.

Question 2.
Write the properties of cathode rays.
Answer:

  • Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 107 m s-1
  • It can be deflected by application of electric and magnetic fields.
  • When the cathode rays are allowed” to fall on matter, they produce heat.
  • They affect the photographic plates and also produce fluorescence when they fall on certain crystals and minerals.
  • When the cathode rays fall on a material of high atomic weight, x-rays are produced.
  • Cathode rays ionize the gas through which they pass.
  • The speed of cathode rays is up to( \(\frac { 1 }{ 10 }\))th

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Question 3.
Give the results of Rutherford alpha scattering experiment.
Answer:

  • Most of the alpha particles are undeflected through the gold foil and went straight.
  • Some of the alpha particles are deflected through a small angle.
  • A few alpha particles (one in thousand) are deflected through the angle more than 90°.
  •  Very few alpha particles returned back (back scattered) -that is, deflected back by 180°.

Question 4.
Write down the postulates of Bohr atom model.
Answer:
Postulates of Bohr atom model:

  1. The electron in an atom moves around nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction. This Coulomb force gives necessary centripetal force for the electron to undergo circular motion.
  2. Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy. Only those discrete orbits allowed are stable orbits.

Question 5.
What is meant by excitation energy?
Answer:
The energy required to excite an electron from lower energy state to any higher energy state is known as excitation energy.

Question 6.
Define the ionization energy and ionization potential.
Answer:
The ionization energy and ionization potential are:

  1. Ionization energy: The minimum energy required to remove an electron from an atom in the ground state is known as binding energy or ionization energy.
  2. Ionization potential: Ionization potential is defined as ionization energy per unit charge.

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Question 7.
Write down the draw backs of Bohr atom model.
Answer:
Limitations of Bohr atom model:
The following are the drawbacks of Bohr atom model:

  1. Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for complex atoms.
  2. When the spectral lines are closely examined, individual lines of hydrogen spectrum is accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
  3. Bohr atom model fails to explain the intensity variations in the spectral lines.
  4. The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 8.
What is distance of closest approach?
Answer:
The minimum distance between the centre of the nucleus and the alpha particle just before it gets reflected back through 180° is defined as the distance of closest approach r0 (also known as contact distance).

Question 9.
Define impact parameter.
Answer:
The impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance.

Question 10.
Write a general notation of nucleus of element X. What each term denotes?
Answer:
The nucleus of any element, we use the following general notation \(_{ Z }^{ A }X\)
where X is the chemical symbol of the element, A is the mass number and Z is the atomic number.

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Question 11.
What is isotope? Give an example.
Answer:
Isotopes are atoms of the same element having same atomic number Z, but different mass number A. For example, hydrogen has three isotopes and they are represented as \(_{ 1 }^{ 1 }H\) (hydrogen), \(_{ 1 }^{ 2 }H\) (deuterium),and \(_{ 1 }^{ 3 }H\) (tritium).

Question 12.
What is isotone? Give an example.
Answer:
Isotones are the atoms of different elements having same number of neutrons. \(_{ 5 }^{ 12 }B\) and \(_{ 6 }^{ 13 }B\) are examples of isotones which 7 neutrons.

Question 13.
What is isobar? Give an example.
Answer:
1. Isobar: Isobars are the atoms of different elements having the same mass number A, but different atomic number Z.
2. For example \(_{ 16 }^{ 40 }S\), \(_{ 17 }^{ 40 }Cl\), \(_{ 18 }^{ 40 }Ar\),\(_{ 19 }^{ 40 }K \) and \(_{ 20 }^{ 40 }Ca\) are isobars having same mass number 40 and different atomic number.

Question 14.
Define atomic mass unit u.
Answer:
One atomic mass unit (u) is defined as the 1/12th of the mass of the isotope of carbon \(_{ 6 }^{ 12 }C\).

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Question 15.
Show that nuclear density is almost constant for nuclei with Z > 10.
Answer:
Nuclear density,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-3
The expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z > 10) have the same density and it is an important characteristics of the nuclei.

Question 16.
What is mass defect?
Answer:
The mass of any nucleus is always less than the sum of the mass of its individual constituents. The difference in mass Am is called mass defect.
∆m = (Zmp + Nmn) – M.

Question 17.
What is binding energy of a nucleus? Give its expression.
Answer:
when Z protons and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding energy is released. This is called the binding energy of the nucleus (BE) and is equal to (∆m) c2.
BE = (Zmp + Nmn – M ) c2

Question 18.
Calculate the energy equivalent of 1 atomic mass unit.
Answer:
We take, m = 1 amu = 1.66 x 10-27 kg
c = 3 x 108ms-1
Then, E = mc2 = 1.66 x 10-27 x (3 x 108)2 J
\(\frac{1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} e \mathrm{V}\)
E ≈ 981 MeV
∴ 1 amu = 931 MeV.

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Question 19.
Give the physical meaning of binding energy per nucleon.
Answer:
The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.

Question 20.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 21.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Answer:
1. Alpha decay:
The alpha decay process symbolically in the following way
\(_{ Z }^{ A }X\) → \(_{ Z-2 }^{ A-4 }Y\) + \(_{ 2 }^{ 4 }He\)

2. Beta decay:
β decay is represented by \(_{ Z }^{ A }X\) → \(_{ Z-1 }^{ A }Y\) +e+ + v

3. Gamma decay:
The gamma decay is given by \(_{ Z }^{ A }{{ X }^{ * }}\) → \(_{ Z }^{ A }X\) + gamma (γ ) rays

Question 22.
In alpha decay, why the unstable nucleus emits \(_{ 2 }^{ 4 }He\) He nucleus? Why it does not emit four separate nucleons?
Answer:
After all \(_{ 2 }^{ 4 }He\) He consists of two protons and two neutrons. For example, if \(_{ 92 }^{ 238 }U\) nucleus decays into \(_{ 90 }^{ 234 }U\) Th by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent (\(_{ 92 }^{ 238 }U\)) nucleus. This kind of process cannot occur in nature because it would violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

Question 23.
What is mean life of nucleus? Give the expression.
Answer:
The mean life time of the nucleus is the ratio of sum or integration of life times of all nuclei to the total number nuclei present initially.
The expression for mean life time, τ = \(\frac { 1 }{ λ }\).

Question 24.
What is half-life of nucleus? Give the expression.
Answer:
The half-life T1/2 is defined as the time required for the number of atoms initially present to reduce to one half of the initial amount.
T1/2 = \(\frac { In 2 }{ λ }\) = \(\frac { 0.6931 }{ λ }\).

Question 25.
What is meant by activity or decay rate? Give its unit.
Answer:
The activity (R) or decay rate is defined as the number of nuclei decayed per second and it is denoted as R = \(\left| \frac { dN }{ dt } \right| \)
The SI unit of activity R is Becquerel.

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Question 26.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 1010 decays/s.

Question 27.
What are the constituent particles of neutron and proton?
Answer:
Protons and neutrons are Baryon which are made up of three Quarks. According to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quarks.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Long Answer Questions

Question 1.
Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-4
A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.

Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays:
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O.
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This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then
eE = eBv
⇒ v = \(\frac { E }{ B }\) ….. (1)

(ii) Determination of specific charge:
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
eV = \(\frac { 1 }{ 2 }\) mv2 ⇒ \(\frac { e }{ m }\) = \(\frac {{ v }^{ 2 }}{ 2V }\)
Substituting the value of velocity from equation (1), we get
\(\frac { e }{ m }\) = \(\frac { 1 }{ 2V }\) = \(\frac {{ E }^{ 2 }}{{ B }^{ 2 }}\) …… (2)
Substituting the values of E, B and V, the specific charge can be determined as
\(\frac { e }{ m }\) = 1.7 x 1011 C kg-1

(iii) Deflection of charge only due to uniform electric field:
When the magnetic field is turned off, the deflection is only due to electric field. The deflection in vertical direction is due to the electric force.
Fe = eE ….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion, acceleration of the electron is
ae = \(\frac { 1 }{ m }\) Fe …. (4)
Substituting equation (4) in equation (3),
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-6
ae = \(\frac { 1 }{ m }\) eE = \(\frac { e }{ m }\) E
Lety be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be u = 0 before entering the parallel electric plates. Let t be the time taken by the cathode rays to travel in electric field. Let t be the length of one of the plates, then the time taken is
t = \(\frac { 1 }{ v }\) ….. (5)
Hence, the deflection y’ of cathode rays is (note : u = 0 and ae = \(\frac { e }{ m }\) E)
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Therefore, the deflection y on the screen is
y ∝ y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get
y = C\(\frac { 1 }{ 2 }\) \(\frac { e }{ m }\) \(\frac{l^{2} B^{2}}{E}\) …… (7)
Rerranging equation (7) as
\(\frac { e }{ m }\) = \(\frac{2 y E}{C l^{2} B^{2}}\) ……. (8)
Substituting the values on RHS, the value of specific charge is calculated as
\(\frac { e }{ m }\) = 1.7 x 1011 Ckg-1

(iv) Deflection of charge only due to uniform magnetic field:
Suppose that the electric field is switched off and only the magnetic field is switched on. Now the deflection occurs only due to magnetic field. The force experienced by the electron in uniform magnetic field applied perpendicular to its path is
Fm = evB (in magnitude)
Since this force provides the centripetal force, the electron beam undergoes a semicircular path. Therefore, we can equate Fm to centripetal force
\(\frac {{ mv }^{2}}{ R }\)
Fm = evB = m \(\frac {{ v }^{2}}{ R }\)
where v is the velocity of electron beam at the point where it enters the magnetic field and R is the radius of the circular path traversed by the electron beam.
eB = m \(\frac { v }{ R }\) ⇒ \(\frac { e }{ m }\) = \(\frac { v }{ BR }\) …… (9)
Further, substituting equation (1) in equation (9), we get
\(\frac { e }{ m }\) = \(\frac{E}{B^{2} R}\) ……. (10)
By knowing the values of electric field, magnetic field and the radius of circular path, the value of specific charge\(\left( \frac { e }{ m } \right) \) can be calculated.

SamacheerKalvi.Guru

Question 2.
Discuss the Millikan’s oil drop experiment to determine the charge of an electron.
Answer:
Millikan’s oil drop experiment is another important experiment in modem physics which is used to determine one of the fundamental constants of nature known as charge of an electron. By adjusting electric field suitably, the motion of oil drop inside the chamber can be controlled – that is, it can be made to move up or down or even kept balanced in the field of view for sufficiently long time.
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1. The apparatus consists of two horizontal circular metal plates A and B each with diameter around 20 cm and are separated by a small distance 1.5 cm. These two parallel plates are enclosed in a chamber with glass walls. Further, plates A and B are given a high potential difference around 10 kV such that electric field acts vertically downward.

2. A small hole is made at the centre of the upper plate A and atomizer is kept exactly above the hole to spray the liquid. When a fine droplet of highly viscous liquid (like glycerine) is sprayed using atomizer, it falls freely downward through the hole of the top plate only under the influence of gravity.

3. Few oil drops in the chamber can acquire electric charge (negative charge) because of friction with air or passage of x-rays in between the parallel plates. Further the chamber is illuminated by light which is passed horizontally and oil drops can be seen clearly using microscope placed perpendicular to the light beam. These drops can move either upwards or downward.

4. Let m be the mass of the oil drop and q be its charge. Then the forces acting on the droplet are
(a) gravitational force Fg = mg
(b) electric force Fe = qE
(c) buoyant force Fb

(a) Determination of radius of the droplet: When the electric field is switched off, the oil drop accelerates downwards. Due to the presence of air drag forces, the oil drops easily attain its terminal velocity and moves with constant velocity. This velocity can be carefully measured by nothing down the time taken by the oil drop to fall through a predetermined distance. The free body diagram of the oil drop), we note that viscous force and buoyant force balance the gravitational force.
Let the gravitational force acting on the oil drop (downward) be Fg = mg.
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Let us assume that oil drop to be spherical in shape. Let ρ be the density of the oil drop, and r be the radius of the oil drop, then the mass of the oil drop can be expressed in terms of its density as
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The gravitational force can be written in terms of density as
Fg = mg ⇒ Fg = ρ \(\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)g
Let σ be the density of the air, the upthrust force experienced by the oil drop due to displaced air is
Fb = σ \(\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)g
Once the oil drop attains a terminal velocity υ, the net downward force acting on the oil drop is equal to the viscous force acting opposite to the direction of motion of the oil drop. From Stokes law, the viscous force on the oil drop is
Fr = 6πr vη
From the free body diagram as shown in Figure (a), the force balancing equation is Fg = Fb + Fv
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Thus, equation (1) gives the radius of the oil drop.

(b) Determination of electric charge: When the electric field is switched on, charged oil drops experience an upward electric force (qE). Among many drops, one particular drop can be chosen in the field of view of microscope and strength of the electric field is adjusted to make that particular drop to be stationary. Under these circumstances, there will be no viscous force acting on the oil drop. Then, from the free body diagram, the net force acting on the oil droplet is
Fe = Fb + Fg
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Substituting equation (1) in equation (2), we get
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Millikan repeated this experiment several times and computed the charges on oil drops. He found that the charge of any oil drop can be written as integral multiple of a basic value, -1.6 x 10-19C, which is nothing but the charge of an electron.

Question 3.
Derive the energy expression for hydrogen atom using Bohr atom model.
Answer:
The energy of an electron in the nth orbit
Since the electrostatic force is a conservative force, the potential energy for the nth orbit is
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The kinetic energy for the nth orbit is
KEn = \(\frac { 1 }{ 2 }\) \({ mv }_{ n }^{ 2 }\) \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac{Z^{2}}{n^{2}}\)This implies that Un = -2 KEn. Total energy in the nth orbit is
En = kEn + Un = KEn – 2KEn = – KEn
En = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac{Z^{2}}{n^{2}}\)
For hydrogen (Z = 1),
En = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac { 1 }{{ n }^{ 2 }}\) joule ….. (1)
where n stands for principal quantum number. The negative sign in equation (1) indicates that the electron is bound to the nucleus.

Substituting the values of mass and charge of an electron (m and e), permittivity’ of free space ε0and Planck’s constant h and expressing in terms of eV, we get
En = -13.6 \(\frac { 1 }{{ n }^{ 2 }}\) eV
For the first orbit (ground state), the total energy of electron is E1 = – 13.6 eV. For the second orbit (first excited state), the total energy of electron is E2 = -3.4 eV. For the third orbit (second excited state), the total energy of electron is E3 = -1.51 eV and so on.

Notice that the energy of the first excited state is greater than the ground state, second excited state is greater than the first excited state and so on. Thus, the orbit which is closest to the nucleus (r1) has lowest energy (minimum energy compared with other orbits). So, it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (-13.6 eV ) is used as a unit of energy called Rydberg (1 Rydberg = -13.6 eV). The negative value of this energy is because of the way the zero of the potential energy is defined. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.

SamacheerKalvi.Guru

Question 4.
Discuss the spectral series of hydrogen atom.
Answer:
The spectral lines of hydrogen are grouped in separate series. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit. These series are named as Lyman series, Balmer series, Paschen series, Brackett series, Pfund series, etc. The wavelengths of these spectral lines perfectly agree with the equation derived from Bohr atom model.
\(\frac { 1 }{ λ }\) R \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \) … (1)
where \(\bar { v } \) is known as wave number which is inverse of wavelength, R is known as Rydberg constant whose value is 1.09737 x 107 m-1 and m and n are positive integers such that m > n. The various spectral series are discussed below:

(a) Lyman series:
Put n = 1 and m = 2, 3, 4 …..in equation (1). The wave number or wavelength of spectral lines of Lyman series which lies in ultra-violet region is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(b) Balmer series:
Put n = 2 and m = 3, 4, 5 …. in equation (1). The wave number or wavelength of spectral lines of Balmer series which lies in visible region is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{2}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(c) Paschen series:
Put n = 3 and m = 4, 5, 6…. in equation (1). The wave number or wavelength of spectral lines of Paschen series which lies in infra-red region (near IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{3}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(d) Brackett series:
Put n = 4 and m = 5, 6, 7 ….in equation (1). The wave number or wavelength of spectral lines of Brackett series which lies in infra-red region (middle IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{4}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(e) Pfund series:
Put n = 5 and m = 6, 7, 8 …. in equation (1). The wave number or wavelength of spectral lines of Pfund series which lies in infra-red region (far IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{5}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

Question 5.
Explain the variation of average binding energy with the mass number by graph and discuss its features.
Answer:
We can find the average binding energy per nucleon \(\overline { BE } \). It is given by
\(\overline { BE } \) = \(\frac{\left[Z m_{H}+N m_{n}-M_{\mathrm{A}}\right] c^{2}}{\mathrm{A}}\)
The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. \(\overline { BE } \) is plotted against A of all known nuclei.
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Important inferences from of the average binding energy curve:

(i) The value of \(\overline { BE } \) rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.

(ii) The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.

(iii) For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.
If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.

(iv) If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

Question 6.
Explain in detail the nuclear force.
Answer:
Nucleus contains protons and neutrons. From electrostatics, we leamt that like charges repel each other. In the nucleus, the protons are separated by a distance of about a few Fermi (1 0-15 m), they must exert on each other a very strong repulsive force. For example, the electrostatic repulsive force between two protons separated by a distance 10-15 m
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The acceleration experienced by a proton due to the force of 230 N is
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This is nearly 1028 times greater than the acceleration due to gravity. So if the protons in the nucleus experience only the electrostatic force, then the nucleus would fly apart in an instant. From this observation, it was concluded that there must be a strong attractive force between protons to overcome the repulsive Coulomb’s force. This attractive force which holds the nucleus together is called strong nuclear force. A few properties of strong nuclear force are

(i) The strong nuclear force is of very short range, acting only up to a distance of a few Fermi. But inside the nucleus, the repulsive Coulomb force or attractive gravitational forces between two protons are much weaker than the strong nuclear force between two protons. Similarly, the gravitational force between two neutrons is also much weaker than strong nuclear force between the neutrons. So nuclear force is the strongest force in nature.

(ii) The strong nuclear force is attractive and acts with an equal strength between proton-proton, proton-neutron, and neutron – neutron.

(iii) Strong nuclear force does not act on the electrons. So it does not alter the chemical properties of the atom.

SamacheerKalvi.Guru

Question 7.
Discuss the alpha decay process with example.
Answer:
When unstable nuclei decay by emitting an α-particle (\(_{ 2 }^{ 4 }{ He }\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. We write the alpha decay process symbolically in the following way
\(_{ Z }^{ A }{ X }\) → \(_{ Z-2 }^{ A-4 }{ Y}\) +\(_{ 2 }^{ 4 }{ He }\)
Here X is called the parent nucleus and Y is called the daughter nucleus.

Example:
Decay of Uranium \(_{ 92 }^{ 238 }{ U }\) to thorium \(_{ 92 }^{ 234 }{ Th }\)with the emission of \(_{ 2 }^{ 4 }{ He }\) nucleus (α-particle)
\(_{ 92 }^{ 238 }{ U }\) → \(_{ 92 }^{ 234 }{ Th }\) + \(_{ 2 }^{ 4 }{ He }\)
As already mentioned, the total mass of the daughter nucleus and \(_{ 2 }^{ 4 }{ He }\) nucleus is always less than that of the parent nucleus. The difference in mass Q = (∆mx – my – mα) is released as energy called disintegration energy Q and is given by Q = (∆mx – my – mα) c2

Note that for spontaneous decay (natural radioactivity) Q > 0. In alpha decay process, the disintegration energy is certainly positive (Q > 0). In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the 2 He nucleus. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.

Question 8.
Discuss the beta decay process with examples.
Answer:
In beta decay, a radioactive nucleus emits either electron or positron. If electron (e) is emitted, it is called β decay and if positron (e+) is emitted, it is called p+ decay. The positron is an anti-particle of an electron whose mass is same as that of electron and charge is opposite to that of electron – that is, +e. Both positron and electron are referred to as beta particles.

1. β decay:
In β decay, the atomic number of the nucleus increases by one but mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\) → \(_{ Z+12 }^{ A }{ Y}\) + e + \(\bar { v } \) …(1)
It implies that the element X becomes Y by giving out an electron and antineutrino (\(\bar { v } \)). In otherwords, in each β decay, one neutron in the nucleus of X is converted into a proton by emitting an electron (e) and antineutrino. It is given by
n → p + e + \(\bar { v } \)
Where p -proton, \(\bar { v } \) -antineutrino. Example: Carbon (\(_{ 6 }^{ 14 }{ C }\)) is converted into nitrogen (\(_{ 7 }^{ 14 }{ N }\)) through β- decay.
\(_{ 6 }^{ 14 }{ C }\) → \(_{ 7 }^{ 14 }{ N }\) + e + \(\bar { v } \)

2. β+ decay:
In p+ decay, the atomic number is decreased by one and the mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\) → \(_{ Z-12 }^{ A }{ Y}\) + e+ + v
It implies that the element X becomes Y by giving out an positron and neutrino (v). In otherwords, for each β+ decay, a proton in the nucleus of X is converted into a neutron by emitting a positron (e+) and a neutrino. It is given by
p → n + e+ + v

However a single proton (not inside any nucleus) cannot have β+ decay due to energy conservation, because neutron mass is larger than proton mass. But a single neutron (not inside any nucleus) can have β decay.
Example: Sodium (\(_{ 11 }^{ 23 }{ Na }\)) is converted into neon (\(_{ 10 }^{ 22 }{ Ne }\)) decay.
\(_{ 11 }^{ 23 }{ Na }\) → \(_{ 10 }^{ 22 }{ Ne }\) + e+ + v

Question 9.
Discuss the gamma decay process with example.
Answer:
In a and p decay, the daughter nucleus is in the excited state most of the time. The typical life time of excited state is approximately 10-11 s. So this excited state nucleus immediately returns to the ground state or lower energy state by emitting highly energetic photons called 7 rays. In fact, when the atom is in the excited state, it returns to the ground state by emitting photons of energy in the order of few eV. But when the excited state nucleus returns to its ground state, it emits a highly energetic photon (γ rays) of energy in the order of MeV. The gamma decay is given by
\(_{ Z }^{ A }{ { X }^{ * } }\) → \(_{ Z }^{ A }{ X}\) + gamma (γ) rays
Here the asterisk (*) means excited state nucleus. In gamma decay, there is no change in the mass number or atomic number of the nucleus.
Boron (\(_{ 5 }^{ 12 }{ B }\)) has two beta decay modes:

(i) it undergoes beta decay directly into ground state carbon by emitting an electron of maximum of energy 13.4 MeV.

(ii) it undergoes beta decay to an excited state of carbon (\(_{ 6 }^{ 12 }{{ C}^{ * }}\)) by emitting an electron of maximum energy 9.0 MeV followed by gamma decay to ground state by emitting a photon of energy 4.4 MeV.
It is represented by
\(_{ 5 }^{ 12 }{ B }\) → \(_{ 6 }^{ 12 }{ C }\) + e+ + \(\bar { v } \)
\(_{ 6 }^{ 12 }{{ C }^{ * }}\) → \(_{ 6 }^{ 12 }{ C }\) + γ

Question 10.
Obtain the law of radioactivity.
Answer:
Law of radioactive decay:
At any instant t, the number of decays per unit time, called rate of decay \(\left( \frac { dN }{ dt } \right) \) is proportional to the number of nuclei (N) at the same instant.
\(\frac { dN }{ dt } \) ∝ N
By introducing a proportionality constant, the relation can be written as
\(\frac { dN }{ dt } \) = -λN …… (1)
Here proportionality constant λ is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. By rewriting the equation (1), we get
dN = -λNdt …… (2)
Here dN represents the number of nuclei decaying in the time interval dt. Let us assume that at time t =0 s, the number of nuclei present in the radioactive sample is N0. By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t. From equation (2), we get
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Taking exponentials on both sides, we get
N = N0 e-λt ….. (4)
[Note: eInx = ey ⇒ x = ey]
Equation (4) is called the law of radioactive decay. Here N denotes the number of undecayed nuclei present at any time t and N0 denotes the number of nuclei at initial time t = 0. Note that the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite. Equation (4) is plotted.
We can also define another useful quantity called activity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as R = \(\left| \frac { dN }{ dt } \right| \).
Note: that activity R is a positive quantity. From equation (4), we get
R = \(\left| \frac { dN }{ dt } \right| \) = λ N0 e-λt ….. (5)
R = R0 e-λt ….. (6)
Where R = λ N0
The equation (6) is also equivalent to radioactive law of decay. Here R0 is the activity of the sample at t = 0 and R is the activity of the sample at any time t. From equation (6), activity also shows exponential decay behavior. The activity R also can be expressed in terms of number of undecayed atoms present at any time t. From equation (6), since N = N0 e-λtwe write
R = λ N …… (7)
Equation (4) implies that the activity at any time t is equal to the product of decay constant and number of undecayed nuclei at the same time t. Since N decreases over time, R also decreases.

SamacheerKalvi.Guru

Question 11.
Discuss the properties of neutrino and its role in beta decay.
Answer:
Neutrino:
Initially, it was thought that during beta decay, a neutron in the parent nucleus is converted to the daughter nuclei by emitting only electron as given by
\(_{ Z }^{ A }{ X }\) → \(_{ Z+1 }^{ A }{ X}\) Y+e

1. But the kinetic energy of electron coming out of the nucleus did not match with the experimental results. In alpha decay, the alpha particle takes only certain allowed discrete energies whereas in beta decay, it was found that the beta particle (i.e, electron) have a continuous range of energies.

2. But the conservation of energy and momentum gives specific single values for electron energy and the recoiling nucleus Y. It seems that the conservation of energy, momentum are violated and could not be explained why energy of beta particle have continuous range of values. So beta decay remained as a puzzle for several years.

3. After a detailed theoretical and experimental study, in 1931 W. Pauli proposed a third particle which must be present in beta decay to carry away missing energy and momentum. Fermi later named this particle the neutrino (little neutral one) since it has no charge, have very little mass.

4. For many years, the neutrino (symbol v , Greek nu) was hypothetical and could not be verified experimentally. Finally, the neutrino was detected experimentally in 1956 by Fredrick Reines and Clyde Cowan. Later Reines received Nobel prize in physics in the year 1995 for his discovery.
The neutrino has the following properties

  • It has zero charge
  • It has an antiparticle called anti-neutrino.
  • Recent experiments showed that the neutrino has very tiny mass.
  • It interacts very weakly with the matter. Therefore, it is very difficult to detect. In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

Question 12.
Explain the idea of carbon dating.
Answer:
Carbon dating:
1. The interesting application of beta decay is radioactive dating or carbon dating. Using this technique, the age of an ancient object can be calculated. All living organisms absorb carbon dioxide (CO2) from air to synthesize organic molecules. In this absorbed CO2, the major part is \(_{ 6 }^{ 12 }{ C }\) and very small fraction (1.3 x 10-12) is radioactive \(_{ 6 }^{ 14 }{ C }\) whose half-life is 5730 years.
Carbon-14 in the atmosphere is always decaying but at the same time, cosmic rays from outer space are continuously bombarding the atoms in the atmosphere which produces \(_{ 6 }^{ 14 }{ C }\). So the continuous production and decay of \(_{ 6 }^{ 14 }{ C }\) in the atmosphere keep the ratio of
\(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) always constant.

2. Since our human body, tree or any living organism continuously absorb CO2 from the atmosphere, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in the living organism is also nearly constant. But when the organism dies, it stops absorbing C2.

3. Since \(_{ 6 }^{ 14 }{ C }\) starts to decay, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in a dead organism or specimen decreases over the years. Suppose the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 14 }{ C }\) in the ancient tree pieces excavated is known, then the age of the tree pieces can be calculated.

Question 13.
Discuss the process of nuclear fission and its properties.
Answer:
1. When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-19

2. The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.

3. The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.

4. Uranium undergoes fission reaction in 90 different Neutrons ways. The most common fission reactions of \(_{ 92 }^{ 235 }{ U }\) nuclei are shown here.
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5. Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state \(_{ 92 }^{ 236 }{{ U}^{ * }}\). But this excited state does not last longer than 10-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 14.
Discuss the process of nuclear fusion and how energy is generated in stars.
Answer:
Nuclear Fusion:
1. When two or more light nuclei (A < 20) combine to form a heavier nucleus, then it is called nuclear fusion.

2. In the nuclear fusion, the mass of the resultant nucleus is less than the sum of the masses of original light nuclei. The mass difference appears as energy. The nuclear fusion never occurs at room temperature unlike nuclear fission. It is because when two light nuclei come closer to combine, it is strongly repelled by the coulomb repulsive force.

3. To overcome this repulsion, the two light nuclei must have enough kinetic energy to move closer to each other such that the nuclear force becomes effective. This can be achieved if the temperature is very much greater than the value 107 K. When the surrounding temperature reaches around 107 K, lighter nuclei start fusing to form heavier nuclei and this resulting reaction is called thermonuclear fusion reaction.

Energy generation in stars:
1. The natural place where nuclear fusion occurs is the core of the stars, since its temperature is of the order of 107 K. In fact, the energy generation in every star is only through thermonuclear fusion. Most of the stars including our Sun fuse hydrogen into helium and some stars even fuse helium into heavier elements.

2. The early stage of a star is in the form of cloud and dust. Due to their own gravitational pull, these clouds fall inward. As a result, its gravitational potential energy is converted to kinetic energy and finally into heat.

3. When the temperature is high enough to initiate the thermonuclear fusion, they start to release enormous energy which tends to stabilize the star and prevents it from further collapse.

4. The sun’s interior temperature is around 1.5 x 107 K. The sun is converting 6 x 1011 kg hydrogen into helium every second and it has enough hydrogen such that these fusion lasts for another 5 billion years.

5. When the hydrogen is burnt out, the sun will enter into new phase called red giant where helium will fuse to become carbon. During this stage, sun will expand greatly in size and all its planets will be engulfed in it.

6. According to Hans Bethe, the sun is powered by proton-proton cycle of fusion reaction. This cycle consists of three steps and the first two steps are as follows:
\(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 1 }{ H }\) → \(_{ 1 }^{ 2 }{ H }\) + e+ + v …… (1)
\(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 2 }{ H }\) → \(_{ 2 }^{ 3 }{ H }\) + γ …… (2)
A number of reactions are possible in the third step. But the dominant one is
\(_{ 2 }^{ 3 }{ H }\) + \(_{ 12}^{ 3 }{ H }\) → \(_{ 2 }^{ 4}{ H }\) + \(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 1 }{ H }\)…… (3)
The overall energy production in the above reactions is about 27 MeV. The radiation energy we received from the sun is due to these fusion reactions.

Question 15.
Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:
1. Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.

2. The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:
1. The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only 0.7% of \(_{ 92 }^{ 235 }{ U }\) and 99.3% are only If \(_{ 92 }^{ 238 }{ U }\). So the \(_{ 92 }^{ 238 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\).

2. In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:
1. The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced

2. Most of the reactors use water, heavy water (D2O) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-21

Control rods:
1. The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.

2. Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.

3. If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:
1. For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.
Cooling system:

2. The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.

3. This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

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Question 16.
Explain in detail the four fundamental forces.
Answer:
Fundamental forces of nature:

  1. It is known that there exists gravitational force between two masses and it is universal in nature. Our planets are bound to the sun through gravitational force of the sun.
  2. The force between two charges there exists electromagnetic force and it plays major role in most of our day-today events.
  3. The force between two nucleons, there exists a strong nuclear force and this force is responsible for stability of the nucleus.
  4. In addition to these three forces, there exists another fundamental force of nature called the weak force. This weak force is even shorter in range than nuclear force. This force plays an important role in beta decay and energy production of stars.
  5. During the fusion of hydrogen into helium in sun, neutrinos and enormous radiations are produced through weak force.
  6. Gravitational, electromagnetic, strong and weak forces are called fundamental forces of nature.

Question 17.
Briefly explain the elementary particles of nature.
Answer:
Elementary particles:
1. An atom has a nucleus surrounded by electrons and nuclei is made up of protons and neutrons. Till 1960s, it was thought that protons, neutrons and electrons are fundamental building blocks of matter.

2. In 1964, physicist Murray Gellman and George Zweig theoretically proposed that protons and neutrons are not fundamental particles; in fact they are made up of quarks. These quarks are now considered elementary particles of nature. Electrons are fundamental or elementary particles because they are not made up of anything.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-22
In the year 1968, the quarks were discovered experimentally by Stanford Linear Accelerator Center (SLAC), USA. There are six quarks namely, up, down, charm, strange, top and bottom and their antiparticles. All these quarks have fractional charges.
For example, charge of up quark is +\(\frac { 2 }{ 3 }\)e and that of down quark is –\(\frac { 1 }{ 3 }\)e.

3. According to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quarks.

4. The study of elementary particles is called particle physics.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Exercises

Question 1.
Consider two hydrogen atoms HA and HB in ground state. Assume that hydrogen atom HA is at rest and hydrogen atom HB is moving with a speed and make head-on collide on the stationary hydrogen atom HA. After the strike, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom HB, such that any one of the hydrogen atoms reaches one of the excitation state.
Solution:
Collision between hydrogen HA and hydrogen HB atom will be inelastic if a part of kinetic energy is used to excite atom.
If u1 and u2 are speed of HA and HB atom after collision, then
mu = mu1 + mu2 …… (1)
\(\frac { 1 }{ 2 }\) mu2 = \(\frac { 1 }{ 2 }\) \({ mu }_{ 1 }^{ 2 }\) + \(\frac { 1 }{ 2 }\) \({ mu }_{ 2 }^{ 2 }\) + ∆ E …… (2)
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-23
The minimum K.E of the moving hydrogen atom HB is 20.4 eV.

Question 2.
In the Bohr atom model, the frequency of transitions is given by the following expression υ = Rc \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\), Where n < m,
Consider the following transitions:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-24
Show that the frequency of these transitions obey sum rule (which is known as Ritz combination principle)
Solution:
In the Bohr atom model, the frequency of transition
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-25
IIIrd transition, m = 3 and n = 1
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-26
According to Ritz combination principle, the frequency transition of single step is the sum of frequency transition in two steps
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-27

Question 3.
(a) A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.
(b) Show that the total number of lines in emission spectrum is \(\frac { n(n-1) }{ 2 }\) and compute the total number of possible lines in emission spectrum.
Solution:
(a) Wavelength, λ = 97.5 nm = 97.5 x 10-9 m
Principle quantum number n = ?
According to Bohr atom model,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-28

(b) A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level
So total number of lines in emission spectrum is \(\frac { n(n-1) }{ 2 }\)
= \(\frac { (4(4-1) }{ 2 }\) = \(\frac { 4×3) }{ 2 }\) = 6
So the total number of possible lines in emission spectrum is 6.

Question 4.
Calculate the radius of the earth if the density of the earth is equal to the density of the nucleus. [mass of earth 5.97 x 1024 kg].
Solution:
The density of the nucleus of an atom
ρN = 2.3 x 1017 kg m-3
ρN = ρE = 2.3 x 1017 kg m-3
Mass of the earth ME = 5.97 x 1024 kg
Density of the earth,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-29
r3 = 0.62155 x 107 m3
r3 = 183.85 m
r ≈ 180 m.

Question 5.
Calculate the mass defect and the binding energy per nucleon of the \(_{ 47 }^{ 108 }{ Ag }\) nucleus, [atomic mass of Ag = 107.905949]
Solution:
Mass of proton, mp = 1.007825 amu
Mass of neutron, mn = 1.008665 amu
Mass defect, ∆m = Zmp + Z mN – MN
= 47 x 1.007825 + 61 x 1.008665 – 107.905949
= 108.89634- 107.905949
∆m = 0.990391 u
Binding energy per nucleon of the \(_{ 47 }^{ 108 }{ Ag }\) nucleus
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-30

Question 6.
Half lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. Initially, the samples have equal number of nuclei. Calculate the ratio of decayed numbers of A and B nuclei after 80 minutes.
Solution:
80 minutes = 4 half lives of A = 2 half live of B
Let the initial number of nuclei in each sample be N.
NN after 80 minutes = \(\frac { N }{{ 2 }^{ 4 }}\)
Number of A nuclides decayed = \(\frac { 15 }{16}\)N
NB after 80 minutes = \(\frac { N }{{ 2 }^{ 4 }}\)
Number of B nuclides decayed = \(\frac { 3 }{4}\)N
Required ratio = \(\frac { 15 }{16}\) x \(\frac { 4 }{3}\) = \(\frac { 5 }{4}\)
NN : NB = 5 : 4.

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Question 7.
On your birthday, you measure the activity of the sample 210Bi which has a half-life of 5.01 days. The initial activity that you measure is lμCi . (a) What is the approximate activity of the sample on your next birthday? Calculate (b) the decay constant (c) the mean life (d) initial number of atoms.
Solution:
(a) A year of 365 days is equivalent to 365 d/5.01 d ≈ 73 half-lives. Thus, the activity will be reduced after one year to approximately (1/2)73 (1.000 μCi) ~ 10-22 μCi.

(b) Initial measure R0 = 1.000 μCi
= 10-6 x 3.7 x 1010
= 3.7 x 104 Bq
After 1 year, the measure R = 10-22 μCi.
= 10-22 x 10-6 x 3.7 x 1010
= 3.7 x 10-18 Bq
decay constant,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-31

(c) Mean life
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-32

(d) Initial number of atoms
R0 = λN ; N = \(\frac {{ R }_{ 0 }}{ λ }\)
= \(\frac{3.7 \times 10^{4}}{1.6 \times 10^{-6}}\) ; N = 2.31 x 1010

Half Life Formula. Half-life is the time required for the amount of something to fall to half its initial value.

Question 8.
Calculate the time required for 60% of a sample of radon undergo decay. Given T1/2 of radon = 3.8 days.
Solution:
Here consider Rn – 222 with a half life of 3.823 days.
From decay equation,
Current amount = Initial amount x (2)-n
N = N0 (2)-n
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-33

Question 9.
Assuming that energy released by the fission of a single \(_{ 92 }^{ 235 }{ U }\) nucleus is 200MeV, calculate the number of fissions per second required to produce 1 watt power.
Solution:
The fission of a single \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy
Energy released in the fission is given by the formula,
E = \(\frac { Pt }{ n }\) ⇒ \(\frac { n }{ t }\) = \(\frac { P }{ E }\)
E = 200 MeV = 200 x 106 x 1.6 x 10-19
E = 3.2 x 10-11 J
\(\frac { n }{ t }\) = \(\frac { P }{ E }\) = \(\frac{1}{3.2 \times 10^{-11}}\) = 0.3125 x 1011 = 3.125 x 1010
\(\frac { n }{ t }\) = 3.125 x 1010

Question 10.
Show that the mass of radium (\(_{ 88 }^{ 226 }{ Ra }\)) with an activity of 1 curie is almost a gram. Given T1/2 = 1600 years.
Solution:
The activity of the sample at any time t
R = λN
Here, λ = \(\frac{0.6931}{\mathrm{T}_{1 / 2}}\)
R = 1 Ci = 3.7 x 1010 dis s-1
T1/2 = 1600 year = 1600 x 3.16 x 107 dis
∴ The amount of radium,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-34
= 26990.62 x 1017
N = 2.7 x 1021 atoms
As 226 g of radium contains 6.023 x 1023 atoms so the amount of required strength.
= \(\frac{226 \times 2.7 \times 10^{21}}{6.023 \times 10^{23}}\)
= 101.311 x 10-2
= 1.013 g ≈ 1 g.

SamacheerKalvi.Guru

Question 11.
Characol pieces of tree is found from an archeological site. The carbon-14 content of this characol is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree?
Solution:
R0 = 100%
R = 17.5%
λ = \(\frac{0.6931}{\mathrm{T}_{1 / 2}}\)
T1/2 = 5730 years
According to radioactive law
R = R0 e-λt
e-λt = \(\frac {{ R }_{ 0 }}{ R }\)
Talking log on both sides
t = \(\frac {1}{ λ }\) in \(\left( \frac { { R }_{ 0 } }{ R } \right) \)
Half life of carbon, T1/2 = 5730 years
t = \(\frac{\mathrm{T}_{1 / 2}}{0.6931}\) In \(\left(\frac{1}{0.175}\right)\)
= \(\frac { 5730 years }{ 0.6931 }\) x 1.74297
= 14409.49 years
t = 1.44 x 104 years.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Additional Questions

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Multiple Choice Questions

Question 1.
The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is
(a) 2 x 1016
(b) 5 x 1018
(c) 1 x 1017
(d) 4 x 105
Answer:
(a) 2 x 1016
Hint:
n = \(\frac { It }{ e }\) = \(\frac{3.2 \times 10^{-3} \times 1}{1.6 \times 10^{-19}}\) = 2 x 1016.

Question 2.
The allowed energy for the particle for a particular value of n is proportional to
(a) a-2
(b) a-3/2
(c) a-1
(d) a2
Answer:
(a) a-2
Hint:
For the standing wave, a = n \(\frac { λ }{ 2 }\) or λ= \(\frac { 2a }{ n }\)
P = \(\frac {h}{ λ }\) = \(\frac { nh }{ 2a }\) ; E = \(\frac {{ p }^{2}}{ 2m}\) = \(\frac{n^{2} h^{2}}{2 a^{2} m}\) ; E ∝ a-2.

Question 3.
A diatomic molecular has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-35
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-36
Hint:
Angular momentum, L = \(\frac { nh }{ 2π }\)
Rotation K.E = \(\frac {{ L }^{2}}{ 2I}\) = \(\frac{n^{2} h^{2}}{8 \pi^{2} I}\).

SamacheerKalvi.Guru

Question 4.
The speed of the particle, that can take discrete values is proportional to
(a) n-3/2
(b) n-1
(c) n1/2
(d) n
Answer:
(d) n
Hint:
P = mv = \(\frac { nh }{ 2a }\) ; V ∝ n.

Question 5.
If 13.6 eV energy is required to 10 is the hydrogen atom, then energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV
Answer:
(c) 3.4 eV
Hint:
En = \(\frac { 13.6 }{ n }^{2}\)eV
∴ ∆E = E – E2 = 0 + \(\frac { 13.6 }{ n }^{2}\) = 3.4 eV.

Question 6.
Which of the following transitions in hydrogen atoms emits photon of highest frequency?
(a) n = 1 to n = 2
(b) n = 2 to n = 6
(c) n = 6 to n = 2
(d) n = 2 to n = 1
Answer:
(d) n = 2 to n = 1
Hint:
The energy difference E2 – E1 is maximum as calculated in the above problem.

Question 7.
The wavelengths involved in the spectrum of deuterium \(_{ 1 }^{ 2 }{ H }\) are slightly different from that of hydrogen spectrum because
(a) sizes of the two nuclei are different
(b) masses of the two nuclei are different
(c) attraction between the electron and the nucleus is different in the two cases
(d) nuclear forces are different in the two cases
Answer:
(b) masses of the two nuclei are different
Hint:
It is because the masses of the two nuclei are different.

Question 8.
Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is
(a) 12.1 eV
(b) 36.3 eV
(b) 36.3 eV
(c) 108.8 eV
Answer:
(c) 108.8 eV
Hint:
En = – 13.6 \(\frac { { Z }^{ 2 } }{ { n }^{ 2 } } \)
∆E = E3 – E2 = 13.6 (3)2 \(\left[ \frac { 1 }{ { 1 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right] \)
= \(\frac { 13.6×9×8 }{ 9 } \) = 108.8 eV.

Question 9.
Minimum energy required to take out the only one electron from ground state of He+ is
(a) 13.6 eV
(b) 54.4 eV
(c) 27.2 eV
(d) 6.8 eV
Answer:
(b) 54.4 eV
Hint:
Ionisation energy, E = 13.6 Z2 eV
Fe He+, Z = 2
∴ E= 13.6 x (2)2 = 13.6 x 4 = 54.4 eV.

Question 10.
Energy of characteristic X-ray is a consequence of
(a) energy of projectile electron
(b) thermal energy of target
(c) transition in target atoms
(d) none of the above
Answer:
(c) transition in target atoms.

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Question 11.
How much energy is needed to excite an electron in H-atom from ground state to first excited state?
(a) – 13.6 eV
(b) – 10.2 eV
(c) + 10.2 eV
(d) + 13.6 eV
Answer:
(c) + 10.2 eV
Hint:
E1 = – 13.6 eV,
E2 = – 13.6/222 = – 3.4 eV
Required excitation energy
= E2 – E2 = – 3.4 + 13.6 = + 10.2 eV.

Question 12.
For an electron in the second orbit of hydrogen, what is the moment of momentum as per the Bohr’s model?
(a) 2πh
(b) πh
(c) h / π
(d) 2h / π
Answer:
(c) h / π
Hint:
In second orbit of hydrogen, L = 2 \(\left( \frac { h }{ 2\pi } \right) \) = \(\frac { h }{ π }\).

Question 13.
The total energy of an electron in the first excited state of hydrogen atom is about -3.4 eV. Its kinetic energy in this state is
(a) 3.4 eV
(b) 6.8 eV
(c) – 3.4 eV
(d) – 6.8 eV
Answer:
(a) 3.4 eV
Hint:
K.E = – Total energy = +3.4 eV.

Question 14.
The energy of the ground electronic state of hydrogen atom is 13.6 eV. The energy of the first excited state will be
(a) – 27.2 eV
(b) – 52.4 eV
(c) – 3.4 eV
(d) – 6.8 eV
Answer:
(c) – 3.4 eV
Hint:
For the first excited state, n = 2
∴ E2 = \(\frac {{ E }_{ 1 }}{{ E }_{ 2 }}\) = \(\frac {-13.6 eV}{4}\) = -3.4 eV.

Question 15.
The total energy of electron in the ground state of hydrogen atom is – 13.6 eV. The kinetic energy of an electron in the first excited state is
(a) 6.8 eV
(b) 13.6 eV
(c) 1.7 eV
(d) 3.4 eV
Answer:
(d) 3.4 eV
Hint:
Total energy in the first excited state,
E2 = \(\frac {{ E }_{ 1 }}{{ E }_{ 2 }}\) = \(\frac {{ E }_{ 1 }}{{ 2 }^{ 2 }}\) = \(\frac {-13.6 }{4}\) = -3.4 eV
K.E = -E2 = 3.4 eV.

Question 16.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint:
Bohr theory could not explain the five structure of hydrogen spectrum.

Question 17.
In Bohr’s model of an atom, which of the following is an integral multiple of \(\frac { h }{ 2\pi } \) ?
(a) Kinetic energy
(b) Radius of an atom
(c) Potential energy
(d) Angular momentum
Answer:
(d) Angular momentum
Hint:
L = mvr = \(\frac { nh }{ 2\pi } \).

SamacheerKalvi.Guru

Question 18.
According to Bohr’s theory, relation between n and radius of orbit is:
(a) r ∝ \(\frac { 1 }{ n } \)
(b) r ∝ n
(c) r ∝ n2
(d) r ∝ \(\frac { 1 }{{ n }^{2}} \)
Answer:
(c) r ∝ n2
Hint:
r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K Z e^{2}}\) i.e., r ∝ n2.

Question 19.
In Bohr’s model of hydrogen atom, the radius of the first electron orbit is 0.53 Å. What will be the radius of the third orbit?
(a) 4.77 Å
(b) 47.7 Å
(c) 9 Å
(d) 0.09 Å
Answer:
(a) 4.77 Å
Hint:
r3 = (3)2 r1 = 9 x 0.53 = 4.77 Å.

Question 20.
In Bohr model of hydrogen atom, which of the following is quantised?
(a) linear velocity of electron
(b) angular velocity of electron
(c) linear momentum of electron
(d) angular momentum of electron
Answer:
(d) angular momentum of electron.

Question 21.
In Bohr’s model, the atomic radius of the first orbit is r0. Then, the radius of the third orbit is
(a) r0/9
(b) r0
(c) 9r0
(d) 3r0
Answer:
(c) 9r0
Hint:
rn = r1 n2, where r1 = r0
∴ v3 = r0 (3)2 9r0

Question 22.
What is ratio of Bohr magneton to the nuclear magneton?
(a) \(\frac {{ m }_{ p }}{{ m }_{ e }}\)
(b) \(\frac{m_{p}^{2}}{m_{e}^{2}}\)
(c) 1
(d) \(\frac {{ m }_{ e }}{{ m }_{ p }}\)
Answer:
(a) \(\frac {{ m }_{ p }}{{ m }_{ e }}\)
Hint:
Bohr magneton, μB = \(\frac {eh}{{ 2m }_{ e }}\)
Nuclear magneton, μN = \(\frac {eh}{{ 2m }_{ p }}\)
∴ \(\frac {{ μ }_{ B }}{{ μ }_{ N }}\) = \(\frac {{ m }_{ p }}{{ m }_{ e }}\).

SamacheerKalvi.Guru

Question 23.
In terms of Bohr radius a0, the radius of the second Bohr orbit of a hydrogen atom is given by
(a) 4a0
(b) 8a0
(c) √2a0
(d) 2a0
Answer:
(a) 4a0
Hint:
rn = r1 n2
r2 = a0 (2)2 =4a0

Question 24.
If an a-particle collides head on with a nucleus, what is impact parameter?
(a) zero
(b) infinite
(c) 10-10 m
(d) 1010 m
Answer:
(a) zero

Question 25.
One femtometre is equivalent to
(a) 1015 m
(b) 10-15 m
(c) 10-12 m
(d) 1011 m
Answer:
(b) 10-15 m

Question 26.
Wavelength of Kα line of X-ray spectra varies with atomic number as
(a) λ ∝ Z
(b) λ ∝ √Z
(c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\)
(d) λ ∝ \(\frac { 1 }{ √Z }\)
Answer:
(c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\)
Hint:
ccording to moseley’s law, √V = a(Z – b) or V = \(\frac { c }{ λ }\) = a2 (Z – b)2
∴ (c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\).

Question 27.
The shortest wavelength of X-rays, emitted from a X-ray tube, depend upon
(a) current in the tube
(b) voltage applied to the tube
(c) nature of glass material in the tube
(d) atomic number of the target material
Answer:
(b) voltage applied to the tube
Hint:
λmin = \(\frac { 12375 }{V (volt)}\) Å ; λmin ∝ \(\frac { 1 }{ V }\).

Question 28.
During X-ray formation, if voltage is increased
(a) minimum wavelength decreases
(b) minimum wavelength increases
(c) intensity decreases
(d) intensity increases
Answer:
(a) minimum wavelength decreases
Hint:
As λmin ∝ \(\frac { 1 }{ V }\) if voltage is increased, the minimum wavelength of X-rays emitted decreases.

Question 29.
What would be the radius of second orbit of He+ ions?
(a) 1.058 Å
(b) 3.023 Å
(c) 2.068 Å
(d) 4.458 Å
Answer:
1.058 Å
Hint:
rn = \(\frac {{ n }^{2}}{ Z }\) r1
For He+ ion, n = 2, Z = 2
∴ r2 = \(\frac {4}{ 2 }\) x 0.59 Å = 1.058 Å.

SamacheerKalvi.Guru

Question 30.
The minimum wavelength of the X-rays produced by electrons accelerated through a potential difference of V volts is directly proportional to
(a) \(\frac { 1 }{ √V }\)
(b) \(\frac { 1 }{ V }\)
(c) √V
(d) V2
Answer:
(b) \(\frac { 1 }{ V }\)
Hint:
\(\frac { hc }{ λ }\) =eV or λ = \(\frac { hc }{ eV }\), i.e., λ ∝ \(\frac { 1 }{ V }\).

Question 31.
Which source is associated with a line emission spectrum?
(a) Electric fire
(b) Neon street sign
(c) Red traffic light
(d) Sun
Answer:
(b) Neon street sign

Question 32.
Which one of the relation is correct between time period and number of orbits while an electron is revolving in a orbit?
(a) T ∝ \(\frac { 1 }{{ n }^{2}}\)
(b) T ∝ n2
(c) T ∝ n3
(d) T ∝ \(\frac { 1 }{{ n }^{2}}\)
Answer:
(c) T ∝ n3
Hint:
In Bohr’s atomic model, T ∝ n3.

Question 33.
The size of atom is proportional to
(a) A
(b) A1/3
(c) A2/3
(d) A-1/3
Answer:
(b) A1/3

Question 34.
If an electron jumps from 1st orbit to 3rd orbit, then it will
(a) not lose energy
(b) not given energy
(c) release energy
(d) absorb energy
Answer:
(d) absorb energy
Hint:
Only by absorbing energy, an electron jumps from first orbit to third orbit.

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Question 35.
According to uncertainty principle for an electron, time measurement will become uncertain if following is measured with high certainty
(a) energy
(b) momentum
(c) location
(d) velocity
Answer:
(a) energy
Hint:
According to uncertainty principle, ∆E.∆t ≥ \(\frac { h }{ 2π }\).

Question 36.
According to Rutherford’s atomic model, the electrons inside an atom are
(a) stationary
(b) centralized
(c) non-stationary
(d) none of these
Answer:
(c) non-stationary
Hint:
According to Rutherford model, the electron inside an atom cannot be stationary.

Question 37.
Wavelength of a light emitted from second orbit to first orbit in a hydrogen atom is
(a) 1.215 x 10-7 m
(b) 1.215 x 10-5 m
(c) 1.215 x 10-4 m
(d) 1.215 x 10-3 m
Answer:
(a) 1.215 x 10-7 m
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-37

Question 38.
In terms of Rydberg constant R, the wave number of the first Balmer line is
(a) R
(b) 3R
(c) \(\frac { 5R }{ 36 }\)
(d) \(\frac { 8R }{ 9 }\)
Answer:
(c) \(\frac { 5R }{ 36 }\)
Hint:
For the first Balmer line, \(\bar { v } \) =\(\frac { 1 }{ λ }\) = R\(\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)\) =\(\frac { 5R }{ 36 }\).

Question 39.
The K X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atom is about
(a) 0.51 MeV
(b)1.2MeV
(c) 59 keV
(d) 136
Answer:
(c) 59 keV
Hint:
E = \(\frac { hc }{ λ }\) = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.021 \times 10^{-9}}\) eV = 589.3 x 102 eV ≈ 59 KeV.

Question 40.
The radius of an electron orbit in a hydrogen atom is of the order of
(a) 10-8 m
(b) 10-9 m
(c) 10-11 m
(d) 10-13 m
Answer:
(c) 10-11 m

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Question 41.
Which of the following atoms has the lowest ionisation potential?
(a) \(_{ 7 }^{ 14 }{ N }\)
(b) \(_{ 55 }^{ 133 }{ Cs }\)
(c) \(_{ 18 }^{ 40 }{ Ar }\)
(d) \(_{ 8 }^{ 16 }{ O }\)
Answer:
(b) \(_{ 55 }^{ 133 }{ Cs }\)
Hint:
In \(_{ 55 }^{ 133 }{ Cs }\), the outermost electron is farthest from the nucleus and so minimum energy is required to remove this electron from the atom. Hence \(_{ 55 }^{ 133 }{ Cs }\) has lowest concision potential.

Question 42.
The transition from the state n = 4 to n = 3 in a hydrogen like atom result in ultraviolet radiation. Infrared radiation will be obtained in the transition from
(a) 2 → 1
(b) 3 → 2
(c) 4 → 2
(d) 5 → 4
Answer:
(d) 5 → 4
Hint:
The energy gap between 4th and 3rd states is more than the gap between 5th and 4th states.

Question 43.
The number of waves, contained in unit length of the medium, is called
(a) elastic wave
(b) wave number
(c) wave pulse
(d) electromagnetic wave
Answer:
(b) wave number
Hint:
The number of waves contained in a unit length of the medium is called a wave number.

Question 44.
When hydrogen atom is in its first excited level, its radius is
(a) sarhe
(b) half
(c) twice
(d) four times
Answer:
(d) four times
Hint:
r2 = r1 (2)2 = 4r1

Question 45.
The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the electron in this state?
(a) 0 eV
(b) -27.2 eV
(c) 1 eV
(d) 2 eV
Answer:
(b) -27.2 eV
Hint:
PE = 2 x Total energy = 2 x (-13.6) = – 27.2 eV.

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Question 46.
For ionising an excited hydrogen atom, the energy required (in eV) will be
(a) a little less than 13.6
(b) 13.6
(c) more than 13.6 eV
(d) 3.4 or less
Answer:
(d) 3.4 or less
Hint:
The energy of the electron is – 3.4 eV in first excited state and the its magnitude is less for higher excited state.

Question 47.
What is the energy of He+ electron in first order?
(a) 40.8 eV
(b) -27.2 eV
(c) -54.4 eV
(d)-13.6eV
Answer:
(c) -54.4 eV
Hint:
For hydrogen like atoms or ions, En = \(\frac{-13.6 Z^{2}}{n^{2}}\) eV
For He+, Z = 2 and n = 1
E1 = \(\frac{-13.6 \times 2^{2}}{12}\) 54.4 eV.

Question 48.
If voltage across on X-ray tube is doubled, then energy of X-ray emitted by
(a) be doubled
(b) be quadrupled
(c) become half
(d) remain the same
Answer:
(d) remain the same
Hint:
The energy of the X-rays depends on the nature of the target material. Thus the energy of the X-rays remain the same.

Question 49.
When hydrogen atom is in its first excited level, its radius is of the Bohr radius.
(a) twice
(b) 4 times
(c) same
(d) half
Answer:
(b) 4 times
Hint:
For first excited level, n = 2
r2 = (2)2 r0 = 4r0

Question 50.
The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of a singly ionsed helium atom would be
(a) 13.6 eV
(b) 27.2 eV
(c) 6.8 eV
(d) 54.4 eV
Answer:
(d) 54.4 eV
Hint:
\({ E }_{ 2 }^{ 1 }\) = (2)2 E1 = 4 x 13.6 = 54.4 eV.

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Question 51.
When an electron makes transition from n = 4 to n = 2, then emitted line spectrum will be
(a) first line of lyman series
(b) second line of Balmer series
(c) first line of paschen series
(d) second line of paschen series
Answer:
(b) second line of Balmer series
Hint:
The transition from n = 4 to n = 2 emits second line of Balmer series.

Question 52.
Maximum frequency of emission is obtained for the transition
(a) n = 2 to n = 1
(b) n = 6 to n = 2
(c) n = 1 to n = 2
(d) n = 2 to n = 6
Answer:
(a) n = 2 to n = 1
Hint:
The energy difference E2 – E1 is maximum, so photon of maximum frequency is emitted in transition n = 2 to n = 1.

Question 53.
Hydrogen atoms are excited from ground state to the state of principle quantum number 4. Then the number of spectral lines observed will be
(a) 3
(b) 6
(c) 5
(d) 2
Answer:
(b) 6
Hint:
Here n = 4
∴ The number of spectral lines emitted \(\frac { n(n-1) }{ 2 }\) = \(\frac { 4×3 }{ 2 }\) = 6

Question 54.
The radius of hydrogen atom, in the ground state is of the order of
(a) 10-18 cm
(b) 10-7 cm
(c) 10-6 cm
(d) 10-4 cm
Answer:
(a) 10-18 cm
Hint:
Radius of first orbit of H-atom = 0.53 Å ≈ 10-8 cm.

Question 56.
According to Bohr’s theory of the hydrogen atom, the speed vn of the electron in a stationary orbit is related to the principal quantum number n as (c is a constant)
(a) vn = c/n2
(b) vn = c/n
(c) vn = c x n
(d) vn = c x n2
Answer:
(b) vn = c/n
Hint:
Speed of electron in nth orbit, υn= c/n.

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Question 57.
Out of the following which one is not possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model?
(a) 13.6 eV
(b) 0.65 eV
(c) 1.9 eV
(d) 11.1 eV
Answer:
(d) 11.1 eV
Hint:
For no two energy levels of hydrogen atom, E2 – E1 = 11.1 eV.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Short Answer Questions

Question 1.
Write down the drawbacks of Rutherford model.
Answer:
1. Drawbacks of Rutherford model:
Rutherford atom model helps in the calculation of the diameter of the nucleus and also the

2. Size of the atom but has the following limitations:
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom. According to classical electrodynamics, any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy.

Hence, it can no longer sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen. Hence, Rutherford model could not account for the stability of atoms.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

Question 2.
Define excitation potential.
Answer:
Excitation potential is defined as excitation energy per unit charge.

Question 3.
What is meant by atomic number?
Answer:
The number of protons in the nucleus is called the atomic number and it is denoted by Z.

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Question 4.
What is meant by neutron number?
Answer:
The number of neutrons in the nucleus is called neutron number (N).

Question 5.
What is meant by mass number?
Answer:
The total number of neutrons and protons in the nucleus is called the mass number and it is denoted by A. Hence, A = Z + N.

Question 6.
Write down the properties of neutrino.
Answer:
The neutrino has the following properties:

  1. It has zero charge
  2. It has an antiparticle called anti-neutrino.
  3. Recent experiments showed that the neutrino has very tiny mass.
  4. It interacts very weakly with the matter. Therefore, it is very difficult to detect. In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Numerical Problems

Question 1.
What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus.
Solution:
q1 = ze
q2 = e
At the distance r0 of closest approach,
K.E of a Proton = P.E. of proton and the gold nucleus
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-38

Question 2.
Calculate the impact parameter of a 5 MeV particle scattered by 90° when it approaches.
Solution:
KE = 5 MeV = 5 x 106 x 1.6 x 10-19 J
θ = 90°
For gold, Z = 79
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-39

Question 3.
What is the angular momentum of an electron in the third orbit of an atom?
Solution:
Here n = 3; h = 6.6 x 10-34 Js
Angular momentum,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-40

Question 4.
Write down the expression for the radii of orbits of hydrogen atom. Calculate the radius of the smallest orbit.
Solution:
The radius of the nth orbit of a hydrogen atom is given by
r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K e^{2}}\)
Radius of innermost orbit, called Bohr’s radius, is obtained by putting n = 1. It is denoted by r0
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-41
r0 = 0.53 x 10-10 m = 0.53 A°.

Question 5.
Calculate the frequency of the photon, which can excite the electron to – 3.4 eV from -13.6 eV.
Solution:
Energy of photon, hυ = E2 – E1
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-42
υ = 2.47 x 1015 Hz

Question 6.
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -0.85 eV to -1.51 eV, Calculate the wavelenth of the spectral line emitted. To which series of hydrogen spectrum does this wavelenth belong?
Solution:
Here ∆E = E2 – E1 = -0.85-(-1.51).
= 0.66 eV
∆E = 0.66 x 1.6 x 10-19 J
λ = \(\frac { hc }{ ∆E }\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.66 \times 1.6 \times 10^{-19}}\)
= 18.84 x 10-7
λ = 18840 Å
This wavelength belongs to the Pachen series of the hydrogen spectrum.

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Question 7.
Express 16 rag mass into equivalant energy in eV.
Solution:
Here m = 16 mg = 16 x 10-16 kg, C = 3 x 108 ms-1
Equivalent energy, E = mc2
= 16 x 10-16 x (3 x 108)2 J
= \(\frac{16 \times 10^{-6} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} \mathrm{eV}\)
E = 9 x 1030 eV.

Question 8.
The nuclear mass of \(_{ 26 }^{ 56 }{ Fe }\) is 55.85 amu. Calculate its nuclear density.
Solution:
Here MFe = 55.85 amu = 55.85 x 1.66 x 10-27 kg
= 9.27 x 10-26 kg
Nuclear Mass = R0 A1/3 = 1.1 x 10-15 x (56)1/3 m
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-43
ρnu = 2.9 x 1017 kg m-3.

Question 9.
Calculate the density of hydrogen nuclear in SI units. Given R0 = 1.1 fermi and mp = 1.007825 amu.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-44
ρ = 2.98 x 1017 kg m-3.

Question 10.
Express one atomic mass unit in energy units, first in Joules and then in MeV. Using this, express the mass defect of \(_{ 8 }^{ 16 }{ O }\) in MeV.
Solution:
We have, m = 1 amu = 1.66 x 10-27 kg, c = 3 x 108 ms-1
E = mc2 = 1.66 x 10-27 x (3 x 108)2
= 14.94 x10-11 J
= \(\frac{1.494 \times 10^{-10}}{1.6 \times 10^{-13}} \mathrm{MeV}\) [ 1 MeV = 1.6 x 10-13]
= 931.5 MeV
The \(_{ 8 }^{ 16 }{ O }\) nucleus contains 8 protons and 8 neutrons
Mass of 8 protons = 8 x 1.00727 = 8. 05816 amu
Mass of 8 neutrons = 8 x 1.00866 = 8. 06928 amu
Total Mass = 16.12744 amu
Mass of \(_{ 8 }^{ 16 }{ O }\) nucleus = 15.99053 amu
Mass defect = 0.13691 amu
∆Eb = 0.13691 x 931.5 Mev
∆Eb = 127.5 Mev

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Question 11.
The decay constant, for a given redioactive sample is 0.3465 / day. What percentage of this sample will get decayed in a period of 4 years?
Solution:
Here λ, = 0.3465/day; t = 4 years
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-45
Hence sample left undecayed after a period of 4 years,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-46

Question 12.
If 200 MeV energy is released in the fission of a single nucleus of \(_{ 92 }^{ 235 }{ U }\), how many fissions must occur to produce a power of 1 kW?
Solution:
Let the number of fissions per second be n.
Then, Energy released per second = n x 200 MeV
= n x 200 x 1.6 x 10-13 J
Energy required per second = Power x Time
= 1kW x 1 s = 1000 J
Energy released = Energy required
n x 200 x 1.6 x 10-13 = 1000
n = 3.125 x 10-13

Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy

Students can Download Chemistry Chapter 1 Metallurgy Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy

Samacheer Kalvi 12th Chemistry Metallurgy TextBook Evalution

I. Choose the correct answer.

Question 1.
Bauxite has the composition ………………
(a) Al2O3
(b) Al2O3.nH2O
(c) Fe2O3.2H2O
(d) None of these
Answer:
(b) Al2O3.nH2O

Question 2.
Roasting of sulphide ore gives the gas (A). (A) is a colourless gas. Aqueous solution of (A) is acidic. The gas (A) is ………………
(a) CO2
(b) SO3
(c) SO2
(d) H,S
Answer:
(c) SO2

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Question 3.
Which one of the following reaction represents calcination?
(a) 2Zn + O2 → 2ZnO
(b) 2ZnS + 3O2 → 2ZnO + 2SO2
(c) MgCO3 → MgO + CO2
(d) Both (a) and (c)
Answer:
(c) MgCO3 → MgO + CO2

Question 4.
The metal oxide which cannot be reduced to metal by carbon is ………………
(a) PbO
(b) Al2O3
(C) ZnO
(d) FeO
Answer:
(b) Al2O3

Question 5.
Which of the metal is extracted by Hall-Herold process?
(a) Al
(b) Ni
(c) Cu
(d) Zn
Answer:
(a) Al

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Question 6.
Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true?
(a) ∆Gf° of sulphide is greater than those for CS2 and H2S.
(b) ∆Gr° is negative for roasting of sulphide ore to oxide.
(c) Roasting of the sulphide to its oxide is thermodynamically feasible.
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
Answer:
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.

Question 7.
Match items in column -1 with the items of column – II and assign the correct code:
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-1
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-2
Answer:
(c) A – (iv), B – (ii), C – (iii), D – (i)

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Question 8.
Wolframite ore is separated from tinstone by the process of ………………
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer:
(d) Electromagnetic separation

Question 9.
Which one of the following is not feasible?
(a) Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
(b) Cu(s) + Zn2++(aq) → Zn(s) + Cu2+(aq)
(c) Cu(s) + 2Ag+(aq) → Ag(s) + Cu2+(aq)
(d) Fe(s) + Cu2+(aq) → Cu(s) + Fe2+(aq)
Answer:
(b) Cu(s) + Zn2++(aq) → Zn(s) + Cu2+(aq)

Question 10.
Electrochemical process is used to extract ………………
(a) Iron
(b) Lead
(c) Sodium
(d) Silver
Answer:
(c) Sodium

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Question 11.
Flux is a substance which is used to convert ………………
(a) Mineral into silicate
(b) Infusible impurities to soluble impurities
(c) Soluble impurities to infusible impurities
(d) All of these
Answer:
(b) Infusible impurities to soluble impurities

Question 12.
Which one of the following ores is best concentrated by froth – floatation method?
(a) Magnetite
(b) Hematite
(c) Galena
(d) Cassiterite
Answer:
(c) Galena

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Question 13.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to ………………
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Question 14.
Zinc is obtained from ZnO by ………………
(a) Carbon reduction
(b) Reduction using silver
(c) Electrochemical process
(d) Acid leaching
Answer:
(a) Carbon reduction

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Question 15.
Cupellation is a process used for the refining of ………………
(a) Silver
(b) Lead
(c) Copper
(d) Iron
Answer:
(a) Silver

Question 16.
Extraction of gold and silver involves leaching with cyanide ion. Silver is later recovered by ………………
(a) Distillation
(b) Zone refining
(c) Displacement with zinc
(d) liquation
Answer:
(c) Displacement with zinc

Question 17.
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
(a) Fe
(b) Cu
(c) Mg
(d) Zn
Answer:
(c) Mg

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Question 18.
The following set of reactions are used in refining Zirconium
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-3
This method is known as
(a) Liquation
(b) Van Arkel process
(c) Zone refining
(d) Monds process
Answer:
(b) Van Arkel process

Question 19.
Which of the following is used for concentrating ore in metallurgy?
(a) Leaching
(b) Roasting
(c) Froth floatation
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

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Question 20.
The incorrect statement among the following is ………………
(a) Nickel is refined by Monds process
(b) Titanium is refined by Van Arkels process
(c) ZinC blende is concentrated by froth floatation
(d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution
Answer:
(d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution

Question 21.
In the electrolytic refining of copper, which one of the following is used as anode?
(a) Pure copper
(b) Impure copper
(c) Carbon rod
(d) Platinum electrode
Answer:
(b) Impure copper

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Question 22.
Which of the following plot gives Ellingham diagram?
(a) ∆S Vs T
(b) ∆G° Vs T
(c) ∆G° Vs
(d) ∆G° Vs T
Answer:
(b) ∆G° Vs T

Question 23.
In the Ellingham diagram, for the formation of carbon monoxide
(a) \(\left( \frac { \triangle { S }^{ 0 } }{ \triangle T } \right) \)is negative
(b) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is positive
(c) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative
(d) initially \(\left( \frac { \triangle T }{ \triangle { G }^{ 0 } } \right) \)is positive, after 700°C, \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative
Answer:
(c) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative

Question 24.
Which of the following reduction is not thermodynamically feasible?
(a) Cr2O3 → Al2O3 + 2Cr
(b) Al2O3 → Cr2O3 + 2Al
(c) 3TiO2 + 4Al → 2Al2O3 + 2Al
(d) none of these
Answer:
(b) Al2O3 → Cr2O3 + 2Al

Question 25.
Which of the following is not truc with respect to Ellingham diagram?
(a) Free energy changes follow a straight line. Deviation occurs when there is a phase change.
(b) The graph for the formation of CO2 is a straight line almost parallel to free energy axis.
(c) Negative slope of CO shows that it becomes more stable with increase in temperature.
(d) Positive slope of metal oxides shows that their stabilities decrease with increase in temperature.
Answer:
(b) The graph for the fonnation of CO2 is a straight line almost parallel to free energy axis.

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II. Answer the following questions:

Question 1.
What is the difference between minerals and ores?
Answer:
Naturally occurring substances obtained by mining which contain the metals in a free state or in the form of compounds like oxides, sulphides etc. are called minerals.

Minerals that contain a high percentage of metal from which it can be extracted conveniently and economically are called ores.

All ores are minerals but all minerals are not ores.

Question 2.
What are the various steps involved in the extraction of pure metals from their ores?
Answer:
The extraction of pure metals from the concentrated ores is carried out in two steps:

  1. Conversion of the ore into oxides of the metal of interest.
  2. Reduction of the metal oxides to elemental metals.

Question 3.
What is the role of Limestone in the extraction of Iron from its oxide Fe2O3?
Answer:
In the extraction of iron, a basic flux limestone is used. Limestone decomposes to form CaO which reacts with silica gangue present in the iron ore is acidic in nature to form calcium silicate (slag).
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-4

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Question 4.
Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
Answer:
Sulphide ores can be concentrated by the froth floatation method.
(eg) Galena (PbS), Zinc blende (ZnS).

Question 5.
Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why?
Answer:
Coke (C) is a better reducing agent for the reduction of ZnO. Because, when we use coke, the reduction can be easily carried out at 673 K. Thus Carbon (Coke) reduces zinc oxide more easily than carbon monoxide (CO). From the Ellingham diagrams, it is quite clear that the reduction of zinc oxide is more favourable using coke ∆G for the formation of carbon monoxide from carbon is more negative).

Question 6.
Describe a method for refining nickel.
Answer:
The impure nickel is heated in a stream of carbon monoxide at around 350K. The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left behind.
Ni (s) + 4 CO (g) → Ni(CO)4(g)

On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal.
Ni(CO)4 (g) → Ni (s) + 4 CO (g)

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Question 7.
Explain zone refining process with an example using the Ellingham diagram given below.
Answer:
Zone refining:

  • The principle is fractional crystallization.
  • When an impure metal is melted and allowed to solidify, the impurities will prefer to remain in the molten region, ie; impurities are more soluble in the melt than in the solid-state metal.
  • In this process, the impure metal is taken in the form of a rod. One end of the rod is heated using a mobile induction heater, melting the metal on that portion of the rod.
  • When the heater is slowly moved to the other end pure metal crystallizes while impurities will move on to the adjacent molten zone formed due to the movement of the heater.
  • As the heater moves further away, the molten zone containing impurities also moves along with it.
  • This process is repeated several times by moving the heater in the same direction again and again to achieve the desired purity level.
  • This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
  • Germanium, Silicon, and gallium which are used as semiconductors are refined by this process.

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Question 8.
1. Predict the conditions under which
(a) Aluminium might be expected to reduce magnesia.
(b) Magnesium could reduce alumina.
2. Carbon monoxide is a more effective reducing agent than carbon below 983 K but, above this temperature, the reverse is true -Explain.
3. it is possible to reduce Fe2O3 by coke at a temperature around 1200 K.
Answer:
1. The conditions under which:
(a) Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the Ellingham diagram. In the Ellingham diagram, for the formation of magnesia (magnesium oxide) occupy lower position than aluminum oxide. Therefore aluminium cannot be used to reduce the oxides of magnesium (magnesia). Above 1623K, A1 can reduce MgO to Mg, so that ArG° becomes negative and the process becomes thermodynamically feasible.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-5

(b)

  • \(\left( \frac { 4 }{ 3 } \right) \)Al + O2 → \(\left( \frac { 2 }{ 3 } \right) \)Al2O3
  • 2Mg + O2 → 2MgO

At the point of intersection of the Al2O3 and MgO curves in Ellingham diagram. ∆G°
becomes zero for the reaction:

\(\left( \frac { 2 }{ 3 } \right) \)Al2O3 → 2MgO + \(\left( \frac { 4 }{ 3 } \right) \)Al
Below that point magnesium can reduce alumina.

2. From the Ellingham diagram, we find that at 983 K, the curves intersect.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-5

The value of ∆G° for change of C to CO2 is less than the value of ∆G° for change of CO to CO2. Therefore, coke (C) is a better reducing agent than CO at 983K or above temperature. However, below this temperature (e.g. at 673K), CO is a more effective reducing agent than C.

3. Yes, it is possible to reduce Fe2O3 by coke at a temperature around 1200 K. In the Ellingham diagram, carbon line cuts across the lines of many metal oxides and hence it can reduce all those metal oxides at sufficiently high temperature. Ellingham diagram for the formation of Fe2O3 and CO intersects around 1000 K.

Below this temperature, the carbon line lies above the iron line which indicates that Fe2O3 is more stable than CO and hence at this temperature range the reduction is not thermodynamically feasible. However above 1000 K carbon line lies below the iron line and hence we can use coke as a reducing agent around 1200 K. Around 1200 K, coke is better reducing agent because above 1000 K, Gibb’s free energy for the formation of Fe2O3 is more than the formation of CO2 from C.

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Question 9.
Give the uses of zinc.
Answer:
Applications of Zinc (Zn):

  1. Metallic zinc is used in galvanising metals such as iron and steel structures to protect them from rusting and corrosion.
  2. Zinc is also used to produce die-castings in the automobile, electrical and hardware industries.
  3. Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles arid electrical equipment. Zinc sulphide is used in making luminous paints, fluorescent lights and x-ray screens.
  4. Brass an alloy of zinc is used in water valves and communication equipment as it is highly resistant to corrosion.

Question 10.
Explain the electrometallurgy of aluminium.
Answer:
Electrochemical extraction of aluminium Hall-Herold process:
In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as an anode. A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cryolite and is taken in the electrolysis chamber. About 10%, calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K. The chemical reactions involved in this process are as follows:

Ionisation of alumina:
Al2O3 → 2Al3 + 3OO2-

Reaction at cathode:
2Al3+ (melt) + 3e → Al(l)

Reaction at anode:
2O2- (melt) → O2 + 3e

Since carbon acts as anode the following reaction also takes place on it.

  • C (s) + O2- (melt) → CO + 2e
  • C (s) + 2O2- (melt) → CO2 + 4e

Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction can be written as follows:
4Al3+ (melt) + 6O2- (melt) + 3C(s) → 4A(l) + 3CO2(g)

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Question 11.
Explain the following terms with suitable examples.

  1. Gangue
  2. Slag

Answer:
1. Gangue:
The impurities associated with the minerals are known as Gangue or Matrix.

2. Slag:
A compound formed when gangue is combined with flux is called slag.
Flux + Gangue → Slag

For example, the oxide of iron can be reduced by carbon monoxide as follows:
Fe2O3 + 3CO → 2Fe + 3CO2

In this extraction a basic flux, limestone is used.
Since the silica gangue present in the ore is acidic in nature, the limestone combines with it to form Calcium silicate (Slag).
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-6

Question 12.
Give the basic requirement for vapour phase refining.
Answer:
The two requirements for vapour phase refining are:

  1. The metal should form a volatile compound with a suitable reagent.
  2. The volatile compound is decomposed to give the pure metal.

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Question 13.
Describe the role of the following in the process mentioned.

  1. Silica in the extraction of copper.
  2. Cryolite is the extraction of aluminium.
  3. Iodine in the refining of Zirconium.
  4. Sodium cyanide in froth floatation.

Answer:
1. The role of silica in the extraction of copper is to remove the iron oxide obtained during the process of roasting as slag. If the sulphide ore of copper contains iron, the silica (SiO2) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO3 (Slag).

2. Cryolite reduces the melting point of Al2O3 and increases its electrical conductivity. Aluminium is produced by the electrolytic reduction of fused alumina in the electrolytic cell. Alumina is not an electrolyte. So it is made as an electrolyte by dissolving it in the fused cryolite. The function of cryolite is to lower the fusion temperature.

3. Zirconium crude metal is heated with iodine in an evacuated vapour to separate from impurities and this decomposes at 1800 K to give a pure zirconium metal and iodine. Initially, iodine is heated with zirconium to form a volatile compound.

4. Sulphide ores are concentrated by the froth floatation process. Depressants are used to prevent a certain types of particles from forming the froth. NaCN act as a depressant to separate ZnS from PbS.

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Question 14.
Explain the principle of electrolytic refining with an example.
Answer:
The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing an aqueous solution of the salts of the metal of interest. The rods of impure metal is used as anode and thin strips of pure metal are used as a cathode.

The metal of interest dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud. Let us understand this process by considering electrolytic refining of silver as an example.

Cathode:
Pure silver

Anode:
Impure silver rods

Electrolyte:
An acidified aqueous solution of silver nitrate. When a current is passed through the electrodes the following reactions will take place Reaction at the anode.
2Ag (s) → Ag+(aq) + 1 e

The reaction at the cathode:
Ag+ (aq) + 1 e → Ag (s)

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

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The factor pairs of 96 are, how you find them and list them out for you to prove the calculation works.

Question 15.
The selection of reducing agent depends on the thermodynamic factor. Explain with an example.
Answer:
From the Ellingham diagram, it is clear that metals for which the standard free energy of formation (∆fG0) of their oxides is more negative can reduce the metal oxides for which the standard free energy of formation (∆fG0) of oxides is less negative.

The thermodynamic factor has a major role in selecting the reducing agent for a particular reaction. Only that reagent will be preferred which will lead to a decrease in the free energy (AG°) at a certain specific temperature.
E.g – Carbon reduce ZnO to Zn but not CO.

  • ZnO + C → Zn + CO …………..(1)
  • ZnO + CO → Zn + CO2 ………………(2)

In the first case, there is increase in the magnitude of ∆S° while in the second case, it almost remains the same. In other words, ∆G° will have more negative value in the first case, when C is the reducing agent then in the second case when CO acts as the reducing agent. Therefore, C is a better reducing agent.

Question 16.
Give the limitations of Ellingham diagram.
Answer:
Limitations of Ellingham diagram:
1. Ellingham diagram is constructed based only on thermodynamic considerations. It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about the rate of the reaction. Moreover, it does not give any idea about the possibility of other reactions that might be taking place.

2. The interpretation of ∆G is based on the assumption that the reactants are in equilibrium with the product which is not always true.

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Question 17.
Write a short note on electrochemical principles of metallurgy.
Answer:
Electrochemical principles also find applications in metallurgical process. The reduction of oxides of active metals such as sodium, potassium etc., by carbon is thermodynamically not feasible. Such metals are extracted from their ores by using electrochemical methods. In this technique, the metal salts are taken in a fused form or in solution form. The metal ion present can be reduced by treating it with some suitable reducing agent or by electrolysis. Gibbs free energy change for the electrolysis process is given by the following expression
∆G° = -nFE°

Where n is number of electrons involved in the reduction process, F is the Faraday and E° is the electrode potential of the redox couple. If E° is positive then the ∆G is negative and the reduction is spontaneous and hence a redox reaction is planned in such a way that the e.m.f of the net redox reaction is positive. When a more reactive metal is added to the solution containing the relatively less reactive metal ions, the more reactive metal will go into the solution. For example,

  • Cu (s) + 2Ag+(s) → Cu2+ (aq) + 2Ag (s)
  • Cu2+ (aq) + Zn (s) → Cu (s) + Zn+(aq)

Evaluate Yourself

Question 1.
Write the equation for the extraction of silver by leaching with sodium cyanide and show that the leaching process is a redox reaction.
Answer:
The crushed ore of argentite (Ag2S) is leached with sodium cyanide solution. This reaction forms sodium Argento cyanide
Na[Ag(CN)2]

Step 1:
Ag2S + 4NaCN \(\rightleftharpoons\) 2Na[Ag(CN)2] + Na2S
The solution of sodium Argento cyanide combines with zinc dust and forms sodium tetra cyano zincate and precipitated silver.

Step 2:
Zn + 2Na[Ag(CN)2] → Na2[Ag(CN)4] + 2 Ag↓
In the step 2, redox reaction take place.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-6

Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-7

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Question 2.
Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction.
Answer:
Magnesite is a carbonate of magnesium. Magnesite when heated at 800°C to 1000°C at the CO2 content in it is driven off. The residue so obtained is known as calcined magnesite.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-8

Question 3.
Using Ellingham diagram indicate the lowest temperature at which ZnO can be reduced to zinc metal by carbon. Write the overall reduction reaction at this temperature.
Answer:
Ellingham diagram shows variation in standard Gibbs free energy change with temperature for the formation of the oxide. The Ellingham diagram shows straight line upward slope with the formation of oxide, but in case of ZnO there is a sudden change. Ellingham diagram helps in selecting the suitable reducing agent.

By seeing the Ellingham diagram, the free energy formation (∆fG°) of CO from C becomes lower temperatures above 1120 K while that of CO2 from C becomes lower above 1323 K than ∆fG° of ZnO. As ∆fG° of C02 from CO is always higher than that of ZnO. So C can reduce ZnO to Zn but not CO. Thus carbon is better reducing agent than CO for ZnO.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-9

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Question 4.
Metallic sodium is extracted by the electrolysis of brine (aq. NaCl). After electrolysis, the electrolytic solution becomes basic in nature. Write the possible electrode reactions.
Answer:
Brine is a solution of sodium chloride (molten state):
The process of electrolysis involves using an electric current to bring about a chemical change and make new chemicals. In the electrolysis of brine, sodium ions migrate to the cathode, where electrons enter the melt and are reduced to sodium metal.
Na+ + e → Na (at cathode)

Chloride ions migrate the other way toward the anode. They give up their electrons to the anode and are oxidised to chlorine gas.
Cl → \(\frac { 1 }{ 2 } \)Cl2 + e (at anode)

Overall reaction:
2NaCl → 2Na(s) + Cl2 (g)

For aqueous solution of NaCl:

  • H2O + 2e → H2\(\uparrow \)+ 20H (at cathode)
  • Cl → \(\frac { 1 }{ 2 } \)Cl2 + e (at anode)

Overall reaction:
NaCl (aq) + H2O(1) → Na+(aq) + OH(aq) + H2(g) + \(\frac { 1 }{ 2 } \)Cl2(g)
After electrolysis the electrolytic solution becomes basic in nature. [Due to formation of hydroxide (OH) ion].

Samacheer Kalvi 12th Chemistry Metallurgy Additional Questions

Samacheer Kalvi 12th Chemistry Metallurgy 1 Mark Questions and Answers

I. Choose the best answer:

Question 1.
The chemical formula of bauxite is ……………
(a) Fe2O3
(b) Fe2O3. FeO
(c) Al2O3 . H2O
(d) Al2O3.2H2O
Answer:
(d) Al2O3.2H2O

Question 2.
An example of an oxide ore is ……………
(a) malachite
(b) bauxite
(c) galena
(d) zinc blende
Answer:
(b) bauxite

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Question 3.
Froth floatation process involves the ……………
(a) treatment of the ore with water and pine oil
(b) washing of the ore with a steam of water
(c) owing off the ore over a conveyor belt rolling over magnetic roller
(d) melting of ore
Answer:
(a) treatment of the ore with water and pine oil

Question 4.
In the froth floatation process for the purification of ores, the particles float because ……………
(a) they are light
(b) their surface is not easily wetted by water
(c) they bear electrostatic charge
(d) they are insoluble
Answer:
(b) their surface is not easily wetted by water

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Question 5.
In a metallurgical process, an acid flux is used for removing ……………
(a) Slag
(b) basic flux
(c) acidic gangue
(d) basic gangue
Answer:
(d) basic gangue

Question 6.
The process of the removal of impurities from a crude metal is called ……………
(a) concentration
(b) calcination
(c) refining
(d) roasting
Answer:
(c) refining

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Question 7.
Which of the following metal is obtained by self reduction method?
(a) Fe
(b) Cu
(c) Ag
(d) Mg
Answer:
(b) Cu

Question 8.
Which one of the following ore is best concentrated by froth floatation method?
(a) Magnetite
(b) Malachite
(c) Galena
(d) Haematite
Answer:
(c) Galena

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Question 9.
Heating mixture of Cu2O and Cu2S will give ……………
(a) Cu + SO2
(b) Cu + SO3
(c) CuO + CuS
(d) Cu2SO3
Answer:
(a) Cu + SO2

Question 10.
Which of the following pairs of metals is purified by VanArkel method?
(a) Ag and Au
(b) Ni and Fe
(c) Ga and In
(d) Zr and Ti
Answer:
(d) Zr and Ti

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Question 11.
Aluminium is extracted from alumina (Al2O3) by electrolysis of a molten mixture of ……………
(a) (Al2O3) + KF + Na3AlF6
(b) (Al2O3) + HF + NaAlF4
(c) (Al2O3) + CaF2 + NaAlF4
(d) (Al2O3) + Na3AlF6 + CaF2
Answer:
(d) (Al2O3) + Na3AlF6 + CaF2

Question 12.
The ore which contains both copper and iron ……………
(a) Cuprite
(b) Haematite
(c) Copper pyrite
(d) Malachite
Answer:
(c) Copper pyrite

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Question 13.
Match the extraction processes listed in Column-I with metals listed in Column-II.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-10

(a) A – 2, B – 1, C – 4, D – 3
(b) A – 3, B – 4, C – 1, D – 2
(c) A – 3, B – 1, C – 2, D – 4
(d) A – 4, B – 2, C – l, D – 3
Answer:
(c) A – 3, B – 1, C – 2, D – 4

Question 14.
Chief ore of aluminium is ……………
(a) bauxite
(b) clay
(c) haematite
(d) magnetite
Answer:
(a) bauxite

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Question 15.
Which one of the following metal having least chemical reactive?
(a) Na
(b) Mg
(c) Al
(d) Au
Answer:
(d) Au

Question 16.
Pick out the more reactive metal.
(a) Cu
(b) Ag
(c) Au
(d) Na
Answer:
(d) Na

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Question 17.
Consider the following statements.
(i) All ores are minerals but all minerals are not ores.
(ii) Bauxite is an ore of aluminium while clay is not.
(iii) Extraction of aluminium form clay is profitable one.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(c) (iii) only

Question 18.
Match the List-I and List-II correctly using the code given below.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-11
Answer:
(c)  3, 4, 2, 1

Question 19.
The impurities associated with ores is ……………
(a) flux
(b) slag
(c) gangue
(d) metal
Answer:
(c) gangue

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Question 20.
Oxide ores are concentrated by ……………
(a) handpicking
(b) hydraulic washing
(c) froth floatation process
(d) Magnetic separation process
Answer:
(b) hydraulic washing

Question 21.
Haematite and tin stone are purified by ……………
(a) gravity separation process
(b) magnetic separation process
(c) froth floatation process
(d) handpicking
Answer:
(a) gravity separation process

Question 22.
Froth floatation process is applicable for ……………
(a) oxide ores
(b) carbonate ores
(c) chloride ores
(d) sulphide ores
Answer:
(d) sulphide ores

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Question 23.
Copper pyrite and zinc blende are purified by ……………
(a) gravity separation process
(b) magnetic separation process
(c) froth floatation process
(c) handpicking
Answer:
(c) froth floatation process

Question 24.
Frothing agent used in froth floatation process is ……………
(a) pine oil
(b) olive oil
(c) mustard oil
(d) neem oil
Answer:
(a) pine oil

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Question 25.
Depressing agents used to separate ZnS from PbS is ……………
(a) NaCN
(b) NaCl
(c) NaNO3
(d) NaNO2
Answer:
(a) NaCN

Question 26.
Leaching is also called as ……………
(a) hand picking
(b) Electrolysis
(c) Chemical process
(d) magnetic separation process
Answer:
(c) Chemical process

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Question 27.
In the leaching process, the metal present in the ore is converted into ……………
(a) soluble salt
(b) soluble complex
(c) insoluble complex
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 28.
Gold ore is concentrated by ……………
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) handpicking
Answer:
(a) cyanide leaching

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Question 29.
Bauxite is purified by ……………
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) handpicking
Answer:
(b) alkali leaching

Question 30.
Which type of leaching process convert insoluble sulphide ores into soluble sulphates?
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) handpicking
Answer:
(c) acid leaching

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Question 31.
A chemist involves mining process and he got tow ores mixed together. If one is tinstone and another one is chromite, which type of process will be used to separate that two ores?
(a) Leaching process
(b) Froth floatation process
(c) Zone refining process
(d) Magnetic separation process
Answer:
(d) Magnetic separation process

Question 32.
The process in which the concentrated ore is strongly heated in the absence of air is called as ……………
(a) Roasting
(b) Calcination
(c) Smelting
(d) Leaching
Answer:
(b) Calcination

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Question 33.
A chemical substance that forms an easily fusible slag with gangue is called as ……………
(a) flux
(b) pure metal
(c) ore
(d) impure metal
Answer:
(a) flux

Question 34.
Blistered copper is ……………
(a) 98% pure copper
(b) 96% pure copper
(c) 97% pure copper
(d) 88% pure copper
Answer:
(a) 98% pure copper

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Question 35.
Ignition mixture used in aluminothermic process is ……………
(a) Mg + BaO2
(b) MgO + BaO
(c) Al2O3 + Mg
(d) Al2O3 + BaO2
Answer:
(a) Mg + BaO2

Question 36.
For spontaneous reaction, the change in free energy should be ……………
(a) positive
(b) negative
(c) zero
(d) neutral
Answer:
(b) negative

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Question 37.
The change in Gibbs free energy for a reaction is expressed by ……………
(a) ∆G = ∆H + T∆S
(b) ∆G = ∆H – TS
(c) G = H – TS
(d) ∆G = ∆H – T∆S
Answer:
(d) ∆G = ∆H – T∆S

Question 38.
Relationship between ∆G° and Kp is ……………
(a) ∆G° = – RT In Kp
(b) ∆G° = – R In Kp
(c) ∆G° = – T In Kp
(d) ∆G° = RT In Kp
Answer:
(a) ∆G° = – RT In Kp

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Question 39.
Consider the following statements.
(i) Ellingham drawn on a plot by considering the temperature in the x-axis and the standard free energy change for the formation of metal oxide in y-axis.
(ii) The resultant plot is straight line.
(ii) In the Ellingham diagram, ∆H as slope and ∆S as y-intercept.

Which of the above statement(s) is / are not correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) only
(d) (iii) only
Answer:
(d) (iii) only

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Question 40.
Which of the following oxides is unstable j at moderate temperature?
(a) Al2O3
(b) Cr2O3
(c) MgO
(d) HgO
Answer:
(d) HgO

Question 41.
The oxides will decompose on heating even in the absence of a reducing agent is ……………
(a) Ag2O
(b) HgO
(c) MgO
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

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Question 42.
Consider the following statements.
(i) Ellingham diagram gives information about the thermodynamic feasibility of a reaction.
(ii) It explains the rate of the reaction.
(iii) Below 1000 K temperature, FeO is more stable than CO.

Which of the above statements (s) is / are not correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(d) (ii) only

Question 43.
Gibbs free energy change for the electrolysis process is expressed by ……………
(a) ∆G° = – nFE°
(b) ∆G° = nF
(c) ∆G° = – nE°
(d) ∆G° = nFE°
Answer:
(a) ∆G° = – nFE°

Question 44.
The technique used to refining zinc and mercury is ……………
(a) Liquation
(b) Distillation
(c) Zone refining
(d) Van-Arkel method
Answer:
(b) Distillation

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Question 45.
Which of the following is not purified by zone refining process?
(a) Ge
(b) Si
(c) Ga
(d) Al
Answer:
(d) Al

Question 46.
Nickel is purified by ……………
(a) Mond process
(b) Van-Arkel method
(c) Zone refining
(d) Electrolytic refining
Answer:
(a) Mond process

Question 47.
Titanium is purified by ……………
(a) Mond process
(b) Van-Arkel method
(c) Zone refining
(d) Electrolytic refining
Answer:
(b) Van-Arkel method

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Question 48.
The metal used for galvanisation of iron is ……………
(a) Al
(b) Zn
(c) Cu
(d) Au
Answer:
(b) Zn

Question 49.
Which metal alloy is used in design of aeroplane parts?
(a) Al
(b) Zn
(c) Cu
(d) Au
Answer:
(a) Al

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Question 50.
Which metal is used for making coins and ornaments along with gold and other metals?
(a) Zn
(b) Al
(c) Cu
(d) Fe
Answer:
(c) Cu

II. Fill in the blanks:

  1. Minerals that contains a high percentage of metal from which it can be extracted conveniently and economically are called ……………
  2. …………… helps us to select a suitable reducing agent and appropriate temperature range for reduction.
  3. If E° is positive, then the ∆G is ……………
  4. Chemical formula of cuprite is ……………
  5. Gravity separation is also called as ……………
  6. …………… ore is purified by gravity separation process.
  7. In froth floatation process …………… acts as a collector.
  8. Sodium cyanide is added to depresses the floatation property of ZnS by forming a layer of ……………
  9. The process of gold reduced to its elemental state is called ……………
  10. Leaching process is a …………… reaction.
  11. Magnesite is calcined to give ……………
  12. …………… a chemical substance that forms an easily fusible slag with gangue.
  13. Flux + Gangue → ……………
  14. In the extraction of iron …………… is removed as slag.
  15. In the extraction of copper …………… is removed as slag.
  16. Cr2O3 can be reduced by …………… an process.
  17. …………… is used as a reducing agent for the reduction of chromic oxide.
  18. …………… is purified by zone refining.
  19. In the mond process, impure nickel is heated with …………… compound.
  20. …………… filament is used to decompose titanium tetraoxide.
  21. …………… is the most abundant metal.
  22. …………… is used in packing materials for food items.
  23. …………… is used in the manufacture of paints, rubber and cosmetics.
  24. …………… is the first metal used by the humans.
  25. …………… is one of the expensive and precious metals.
  26. …………… is used for increasing the efficiency of solar cells and also used as catalysts.
  27. …………… ore is concentrated by froth floatation process.
  28. Zinc blende is concentrated by …………… process.
  29. Gold ore is leached by adding of ……………

Answer:

  1. Ores
  2. Ellingham diagram
  3. negative
  4. Cu2O
  5. Hydraulic washing
  6. oxide
  7. Sodium ethyl xanthate
  8. Na2[Zn(CN)4] – Sodium negative
  9. Cementation
  10. Redox
  11. Magnesia
  12. Flux
  13. slag
  14. Calcium silicate (CaSiO3)
  15. Ferrous silicate (FeSiO3)
  16. Aluminothermic process
  17. Aluminium
  18. semi-conductior
  19. carbon monoxide
  20. Tungsten
  21. Aluminium
  22. Aluminium foils
  23. Aluminium
  24. Zinc oxide
  25. Copper
  26. Gold
  27. Gold nanoparticle
  28. Sulphide
  29. Froth floatation
  30. NaCN

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III. Match the following:

Question 1.
(i) Haematite – (a) Fe3O4
(ii) Siderite – (b) Fe3O3
(iii) Iron pyrite – (c) Fe3O3
(iv) Magnetite – (d) FeS2
Answer:
(i) – (b)
(ii) – (c)
(iii) – (d)
(iv) – (a)

Question 2.
(i) Copper glance – (a) CuCO3. Cu(OH)2
(ii) Malachite – (b) Cu2S
(iii) Copper pyrite – (c) 2CuCO3. Cu(OH)2
(iv) Azurite – (d) CuFeS2
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

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Question 3.
(i) Zinc blende – (a) Al203.2H2O
(ii) Bauxite – (b) ZnC03
(iii) Zincite – (c) ZnS
(iv) Calamine – (d) ZnO
Answer:
(i) – (c)
(ii) – (a)
(iii) – (d)
(iv) – (b)

Question 4.
(i) Tin stone – (a) AgCl
(ii) Argentite – (b) Ag3SbS3
(iii) Ruby silver – (c) Ag2S
(iv) Horn silver – (d) SnO2
Answer:
(i) – (d)
(ii) – (c)
(iii) – (b)
(iv) – (a)

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Question 5.
(i) Oxide ore – (a) Zinc blende
(ii) Carbonate ore – (b) Horn silver
(Hi) Sulphide ore – (c) Haematite
(iv) Chloride ore – (d) Calamine
Answer:
(i) – (c)
(ii) – (d)
(iii) – (a)
(iv) – (b)

Question 6.
(i) Tin stone – (a) Magnetic separation process
(ii) Copper pyrite – (b) Leaching process
(Hi) Bauxite – (c) Froth floatation process
(iv) Chromite – (d) Hydraulic washing process
Answer:
(i) – (d)
(ii) – (c)
(iii) – (b)
(iv) – (a)

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Question 7.
(i) Aluminium – (a) cosmetics
(ii) Zinc oxide – (b) gas pipelines
(iii) Iron – (c) making coins
(iv) Copper – (d) bicycle chains
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

Question 8.
(i) Gold nanopaticle – (a)Cast iron
(ii) Cast iron – (b) Cooking vessels
(iii) Metallic zinc – (c) Solar cells
(iv) Aluminium – (d) Pump stoves
Answer:
(i) – (c)
(ii) – (d)
(iii) – (a)
(iv) – (b)

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IV. Assertion and reasons:

Note:
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) A and R are correct, R explains A
(b) A and R are correct, R does not explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct

Question 1.
Assertion(A) – Clay is an ore of aluminium while bauxite is not.
Reason (R) – Aluminium can be economically extracted from bauxite not from clay.
Answer:
(d) A is wrong but R is correct

Question 2.
Assertion(A) – Haematite and tin stone are concentrated by hydraulic washing process.
Reason (R) – Oxide ores are heavy and have high specific gravity.
Answer:
(a) A and R are correct, R explains A .

Question 3.
Assertion(A) – Sulphide ores are concentrated by froth floatation process.
Reason (R) – Sulphide ores are preferentially wetted by oil can be separated from gangue.
Answer:
(a) A and R are correct, R explains A

Question 4.
Assertion(A) – A suitable reducing agent is selected based on the thermodynamic consideration.
Reason (R) – The reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is positive.
Answer:
(c) A is correct but R is wrong

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Question 5.
Assertion(A) – Aluminium can be used as a reducing agent for the reduction of chromic oxide.
Reason (R) – In the Ellingham diagram, formation of chromium oxide lies above that of the aluminium, therefore Al2O3 is more stable than Cr2O3.
Answer:
(a) A and R are correct, R explains A

Question 6.
Assertion(A) – Zone refining is based on the principles of fractional crystallisation,
Reason (R) – This process is carried out in an inert gas temperature.
Answer:
(b) A and R are correct, R does not explain A

Question 7.
Assertion(A) – Aluminium is used in the j design of chemical reactors.
Reason (R) – Aluminium has high resistance to corrosion.
Answer:
(a) A and R are correct, R explains A

V. Find the odd one out and give the reasons:

Question 1.
(a) Nickel
(b) Silicon
(c) Germanium
(d) Galium
Answer:
(a) Nickel
Reason:
Si, Ge and Ga are purified by I zone refining while Ni is not.

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Question 2.
(a) Copper
(b) Zinc
(c) Silver
(d) Galium
Answer:
(d) Galium
Reason:
Cu, Zn and Ag are purified by electrolytic refining while Ga is not.

Question 3.
(a) ∆G = -ve
(b) ∆H = -ve
(c) ∆S = +ve
(d) ∆S = -ve
Answer:
(d) ∆S = -ve

Reason:
∆G=-ve, ∆H=-ve, ∆S=+ve are the conditions for spontaneous reaction, ∆S = -ve is condition for non-spontaneous process.

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Question 4.
(a) Haematite
(b) Siderite
(c) Limonite
(d) Azunte
Answer:
(a) Azurite

Reason:
Azurite is ore of copper but others arc ore of iron.

Question 5.
(a) Copper glance
(b) Zinc blende
(c) Argentite
(d) Magnetite
Answer:
(d) Magnetite

Reason:
Magnetite is oxide ore but others are sulphide ores.

Question 6.
(a) Silver
(b) Gold
(c) Sodium
(d) Platinum
Answer:
(c) Sodium

Reason:
Sodium is more reactive element than Ag, Au and Pt.

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VI. Find out the correct pair:

Question 1.
(a) Metallurgy – Extraction of metals
(b) Clay – Ore of Al
(c) Na – Native element
(d) Ore – Gangue
Answer:
(a) Metallurgy Extraction of metals

Question 2.
(a) Bauxite – Iron
(b) Siderite – Aluminium
(c) Malachite – Copper
(d) Argentite – Gold
Answer:
(c) Malachite – Copper

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Question 3.
(a) Limonite – Sulphide ore
(b) Cuprite – Oxide ore
(c) Calamine – Suiphide ore
(d) Horn silver – Oxide ore
Answer:
(b) Cuprite – Oxide ore

Question 4.
(a) Oxide ore – Froth floatation process
(b) Suiphide ore – Gravity separation process
(c) Gold ore – Leaching method
(d) Oxide ore – Magnetic separation process
Answer:
(c) Gold ore – Leaching method

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Question 5.
(a) Aluminium – Galvanising metals
(b) Zinc – Cooking vessels
(c) Iron – Cutlery
(d) Copper – Dental fillings
Answer:
(c) Iron -Cutlery

VII. Find out the incorrect pair:

Question 1.
(a) Copper – Least reactive
(b) Clay – Mineral of Al
(c) Bauxite – Mineral of Al
(d) Gangue – Impurity
Answer:
(c) Bauxite – Mineral of Al

Question 2.
(a) Aluminium – Corundum
(b) Limonite – Iron
(c) Galena – Lead
(d) Tin – Siderite
Answer:
(d) Tin – Siderite

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Question 3.
(a) Magnetite – FeO4
(b) Malachite – CuCO3 .Cu(OH)2
(c) Horn silver – Ag2S
(d) Stefinite – Ag2SbS4
Answer:
(c) Horn silver – Ag2S

Question 4.
(a) Tin stone – Oxide ore
(b) Copper pyrite – Oxide ore
(c) Zincite – Oxide ore
(d) Bauxite – Oxide ore
Answer:
(b) Copper pyrite Oxide ore

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Question 5.
(a) Haematitc – Gravity separation process
(b) Copper pyrite – froth floatation process
(c) Bauxite – Leaching process
(a) pyrolusite – Magnetic separation process
Answer:
(b) Copper pyrite – froth floatation process

Question 6.
(a) Gold ore – Cyanide leaching
(b) Nickel ore – Ammonia leaching
(c) Aluminium ore – Alkali leaching
(d) Silver ore – Acid leaching
Answer:
(a) Silver ore – Acid leaching

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Question 7.
(a) Aluminium – Design of aeroplane
(b) Zinc oxide – Cosmetics
(c) Zinc suiphide – X-ray screens
(d) Iron – Artifici al limb joints
Answer:
(d) Iron – Artificial limb joints

Samacheer Kalvi 12th Chemistry Metallurgy 2 Mark Questions and Answers

VIII. Answer the following:

Question 1.
Define Metallurgy.
Answer:
The various steps involved in the extraction of metals from their ores as well as refining of crude metals are collectively known as metallurgy.

Question 2.
What are minerals?
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds like oxides, sulphides etc… is called as mineral. Bauxite and clay are minerals of Aluminium.

Question 3.
What are ores?
Answer:
A mineral from which the metal can be extracted easily and economically are called ores. Bauxite is an ore of Aluminium.

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Question 4.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

  1. Concentration of the ore
  2. Extraction of crude metal
  3. Refining of crude metal

Question 5.
What is Gangue?
Answer:
Generally, the ores are associated with nonmetallic impurities, rocky materials and siliceous matter which arc collectively known as gangue.

Question 6.
Give the depressing agents used in the froth floatation process and why we use depressing agents in that process?
Answer:
Generally sodium cyanide and sodium carbonate are used as an depressing agents for froth floatation process. These reagents are used to selectively prevent other metal sulphides from coming to the froth. For example, when impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the flotation property of ZnS by forming a layer of zinc complex Na2[Zn(CN)4] on the surface of zinc sulphide.

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Question 7.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 8.
Explain Cyanide leaching.
Answer:
In the concentration of gold ore, the crashed ore of gold is leached with aerated dilute solution of sodium cyanide. Gold is converted into a soluble cyanide complex. The gangue, aluminosilicate remains insoluble.
4Au (s) + 8CN (aq) + O2 (g) + 2H2O (1) → 4[Au(CN)2](aq) + 4OH (aq)

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Question 9.
What is Cementation?
Answer:
Gold can be recovered by reacting the deoxygenated leached solution with zinc. In this process the gold is reduced to its elemental state (zero oxidation sate) and the process is called
Zn (s)+ 2[Au(CN)2] (aq) → [Zn(CN)4]2-(aq)+ 2Au (s)

Question 10.
What is Ammonia leaching.
Answer:
When a crashed ore containing nickel, copper and cobalt is treated with aqueous ammonia . under suitable pressure, ammonia selectively leaches these metals by forming their soluble complexes viz. [Ni(NH3)6]2+, [Cu(NH3)4]2+, and [Co(NH3)5H2O]3+ respectively from the ore leaving behind the gangue, iron(III) oxides/hydroxides and aluminosilicate.

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Question 11.
What is Acid leaching?
Answer:
Leaching of sulphide ores such as ZnS, PbS etc,, can be done by treating them with hot aqueous sulphuric acid.
2ZnS(s) + 2H2SO4(aq) + O2 (g) → 2ZnSO4 (aq) + 2S (s) + H2O
In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur.

Question 12.
What is roasting?
Answer:
The process of heating an ore (generally, sulphide are) is strongly below its melting point in the presence of an excess of air is called roasting.

Question 13.
Define Calcination.
Answer:
It is the process in which the concentrated ore is strongly heated in the absence of air. This method can also be carried out with a limited supply of air.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-12

Question 14.
How will you manage sulphur dioxide produced during roasting process?
Answer:
The sulphur dioxide produced during roasting process is harmful to the environment. In modem metallurgical factories, this by product is trapped and converted into sulphuric acid to avoid air pollution.

Question 15.
What is smelting?
Answer:
Smelting is a process of reducing the roasted metallic oxide to metal in molten condition. In this process, impurities are removed by the addition of flux as slag.

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Question 16.
Explain Auto reduction.
Answer:
Simple roasting of some of the ores give the crude metal. In such cases, the use of reducing agents is not necessary. For example, mercury is obtained by roasting of its ore cinnabar (HgS)
HgS (s) + O2(g) → Hg (l) + SO2 \(\uparrow \)

Question 17.
What is Ellingham diagram?
Answer:
The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.

Question 18.
CO is more stable at higher temperature. Why?
Answer:
In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

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Question 19.
Ag2(g)O and HgO are unstable at moderate temperature and they will decompose on heating even in the absence of a reducing agent. Why?
Answer:
Ellingham diagram for the formation of Ag2(g)O and HgO is at upper part of the diagram and their decomposition temperatures are 600 and 700 K respectively. It indicates that these oxides are unstable at moderate temperatures and will decompose on heating even in the absence of a reducing agent.

Question 20.
What is refining process?
Answer:
Generally the metal extracted from its ore contains some impurities such as unreacted oxide ore, other metals, nonmetal etc. Removal of such impurities associated with the isolated crude metal is called refining process.

Question 21.
List out the common refining methods.
Answer:

  1. Distillation
  2. Liquation
  3. Electrolytic refining
  4. Zone refining
  5. Vapour phase method

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Question 22.
Explain Distillation process with suitable example.
Answer:
This method is employed for low boiling volatile metals like zinc (boiling point 1180 K) and mercury (630 K). In this method, the impure metal is heated to evaporate and the vapours are condensed to get pure metal.

Question 23.
Write the applications of copper.
Answer:

  1. Copper is the first metal used by the humans and extended use of its alloy bronze resulted in a new era, Bronze age.
  2. Copper is used for making coins and ornaments along with gold and other metals.
  3. Copper and its alloys are used for making wires, water pipes and other electrical parts.

Question 24.
Why aluminium cannot be extracted by reducing alumina with carbon?
Answer:
Alumina (Al2O3) cannot be reduced by using carbon because aluminum has more affinity for oxygen than carbon. Therefore, aluminium cannot be extracted by reducing alumina with carbon.

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Question 25.
Write the names of two ores of copper.
Answer:
Main ores of copper are:

  1. Copper pyrites – CuFeS2
  2. Copper glance – Cu2S
  3. Malachite – CuCO3. Cu(OH)2

Question 26.
Explain the role of carbon monoxide in the purification of nickel?
Answer:
During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl. Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

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Question 27.
ZnO can be reduced to the metal by heating with carbon but not Cr2O3. Justify your answer.
Answer:
Carbon has more affinity for oxygen than zinc, whereas chromium has higher affinity for oxygen than zinc. Hence ZnO can be reduced to the metal by heating with carbon but not Cr2O3.

Question 28.
Name the method used for the refining of

  1. Nickel
  2. Zirconium

Answer:

  1. Mond’s process
  2. Van Arkel’s method

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Question 29.
Give one example for

  1. Acidic flux
  2. Basic flux

Answer:

  1. Acidic flux: SiO2
  2. Basic flux: CaCO3, MgCO3

Question 30.
Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?
Answer:
The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (∆S) of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus, the value of ∆G° becomes more on negative side and the reduction becomes easier.

Samacheer Kalvi 12th Chemistry Metallurgy 3 Mark Questions and Answers

IX. Answer the following questions:

Question 1.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 2.
Bauxite is an ore of aluminium while china clay is not. Why?
Answer:
Bauxite and china clay, both are minerals of aluminium. However, aluminium can be commercially extracted from bauxite while extraction from china clay is not a profitable one. Hence the mineral, bauxite is an ore of aluminium while china clay is not.

Question 3.
Explain Gravity separation process or Hydraulic washing process?
[OR]
How will you concentrate oxide ores?
[OR]
Explain the suitable method to concentrate hematite and tinstone ores.
Answer:
In this method, the ore having high specific gravity is separated from the gangue that has low specific gravity by simply washing with running water. Ore is crushed to a finely powdered form and treated with rapidly flowing current of water. During this process the lighter gangue particles are washed away by the running water. This method is generally applied to concentrate the native ore such as gold and oxide ores such as hematite (Fe2O3), tin stone (SnO2) etc.

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Question 4.
Write a notes on alkali leaching process?
Answer:
In this method, the ore is treated with aqueous alkali to form a soluble complex. For example, bauxite, an important ore of aluminium is heated with a solution of sodium hydroxide or sodium carbonate in the temperature range 470 – 520 K at 35 atm to form soluble sodium meta- aluminate leaving behind the impurities, iron oxide and titanium oxide.
Al2O3 (S) + 2NaOH (aq) + 3H2O (l) → 2Na[Al(OH)4] (aq)

The hot solution is decanted, cooled, and diluted. This solution is neutralised by passing CO2 gas, to the form hydrated Al2O2 precipitate.
2Na[Al(OH)4](aq) + CO2(g) → Al2O3. xH2O(s) + 2NaHCO3 (aq)
The precipitate is filtered off and heated around 1670 K to get pure alumina Al2O3.

Question 5.
Complete the following reaction.

  1. ZnS + O2 \(\underrightarrow { \Delta } \) ?
  2. P4+ O2 \(\underrightarrow { \Delta } \)?
  3. CaCO3 \(\underrightarrow { \Delta } \) ?

Answer:

  1. 2ZnS + 3O2 \(\underrightarrow { \Delta } \) 2ZnO + 2SO2 \(\uparrow \)
  2. P4+ 5O2 \(\underrightarrow { \Delta } \) P4O10 \(\uparrow \)
  3. CaCO3 \(\underrightarrow { \Delta } \) CaO + C02 \(\uparrow \)

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Question 6.
Explain Aluminothermic process.
Answer:
Metallic oxides such as Cr2O3 can be reduced by an aluminothermic process. In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To initiate the reduction process, an ignition mixture (usually magneisium and barium peroxide) is used.
BaO2 + Mg → BaO + MgO

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is – 852 kJ mol ) which facilitates the reduction of Cr2O3 by aluminium power.
Cr2O3 + 2Al \(\underrightarrow { \Delta } \) 2Cr + Al2O3

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Question 7.
Why aluminium can be used as a reducing agent for the reduction of chronic oxide?
Answer:
Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the diagram. For example, in the Ellingham diagram, for the fonnation of chromium oxide lies above that of the aluminium, meaning that Al2O3, is more stable than Cr2O3. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide. However, it cannot be used to reduce the oxides of magnesium and calcium which occupy lower position than aluminium oxide.

Question 8.
Write notes on liquation.
Answer:
Liquation:
This method, is employed to remove the impurities with high melting points from metals having relatively low melting points such a stin(Sb;mp=904K), lead(Pb;mp=600K), mercury(Hg; mp = 234 K), and bismuth (Bi; mp = 545K). In this process, the crude metal is heated to form fusible liquid and allowed to flow on a sloping surface.

The impure metal is placed on sloping hearth of a reverberatory furnace and it is heated just above the melting point of the metal in the absence of air, the molten pure metal flows down and the impurities are left behind. The molten metal is collected and solidified.

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Question 9.
Write the application of Iron (Fe).
Answer:
1. Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.

2. Cast iron is used to make pipes, valves and pumps stoves etc.

3. Magnets can be made of iron and its alloys and compounds.

4. An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and curshing machines

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Question 10.
Mention the uses of Gold (Au).
Answer:
1. Gold, one of the expensive and precious metals. It is used for coinage, and has been used as standard for monetary systems in some countries.

2. It is used extensively in jewellery in its alloy form with copper. It is also used in electroplating to cover other metals with a thin layer of gold which are used in watches, artificial limb joints, cheap jewellery, dental fillings and electrical connectors.

3. Gold nanoparticles are also used for increasing the efficiency of solar cells and also used an catalysts.

Question 11.
The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equation.
Answer:

  • 4Au (s) + 8CN(aq) + 2H2O2(aq) + O2(g) → 4[Au(CN)2] (aq) + 4OH(aq)
  • 2[Au(CN)2](aq) + Zn(s) → 2Au(s)+ [Zn(CN)4]-2(aq)

In the first reaction Au changes into Au+, i.e. its oxidation takes place. In the second reaction:
Au+ → Au°
(i.e.) reduction takes place.

Question 12.
Account for the following facts:

  1. The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
  2. The reduction of Cr2O3 with Al is thermodynamically feasible, yet it does not occur at room temperature.
  3. Pine oil is used in froth floatation method.

Answer:

  1. In liquid state entropy is higher than the solid form, this makes ∆fG more negative.
  2. By increasing the temperature, fraction of activated molecules increases, which helps in crossing over the energy barrier.
  3. Pine oil enhances the non-wetting property of the ore particles and also acts as the froth collector.

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Question 13.
Write the chemical reactions for purification of Zirconium by Van Arkel’s process.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-13

Samacheer Kalvi 12th Chemistry Metallurgy 5 Mark Questions and Answers

X. Answer the following questions:

Question 1.
Explain froth floatation process.
(Or)
How will you concentrate sulphide ores?
(Or)
Explain-the concentration of copper pyrites and galena ores.
Answer:
Froth flotation:
This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc blende (ZnS) etc. In this method, the metallic ore particles which are preferentially wetted by oil can be separated from gangue. In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. A small quantity of sodium ethyl xanthate which acts as a collector is also added. A froth is generated by blowing air through this mixture.

The collector molecules attach to the ore particle and make them water repellent. As a result, ore particles, wetted by the oil, rise to the surface along with the froth. The froth is skimmed off and dried to recover the concentrated ore. The gangue particles that are preferentially wetted by water settle at the bottom.

When a sulphide ore of a metal of interest contains other metal sulphides as impurities, depressing agents such as sodium cyanide, sodium carbonate etc are used to selectively prevent other metal sulphides from coming to the froth. For example, when impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the flotation property of ZnS by forming a layer of zinc complex Na2[Zn(CN)4] on the surface of zinc sulphide.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-14

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Question 2.
Discuss the magnetic separation process.
(Or)
How will you separate magnetic ores from non-magnetic ores?
Answer:
Magnetic separation:
This method is applicable to ferromagnetic ores and it is based on the difference in the magnetic properties of the ore and the impurities. For example tin stone can be separated from the wolframite impurities which is magnetic. Similarly, ores such as chromite, pyrolusite having magnetic property can be removed from the non magnetic siliceous impurities.

The crushed ore is poured on to an electromagnetic separator consisting of a belt moving over two rollers of which one is magnetic. The magnetic part of the ore is attracted towards the magnet and falls as a heap close to the magnetic region while the nonmagnetic part falls away from it as shown in the figure.
Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-15

Question 3.
Write a note on thermodynamic principle of metallurgy.
Answer:
Thermodynamic principle of metallurgy:
The extraction of metals from their oxides can be carried out by using different reducing agents. For example, consider the reduction of a metal
oxide MxOy.
\(\frac { 2 }{ y } \)MxOy(s) → \(\frac { 2x }{ y } \)M(s) + O2(g) ……………..(1)

The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either CO or CO2.

  • C + O2 → CO2(g) ……….(2)
  • 2C + O2 → 2CO(g) …………(3)

Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Metallurgy img-16
If carbon monoxide is used as a reducing agent, it is oxidised to CO2 as follows,
2CO + O2 → 2CO2(g)  ……………(4)

A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) & (2), (1) & (3) or (1) & (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.

Ellingham diagram:
The change in Gibbs free energy (∆G) for a reaction is given by the expression.
∆G = ∆H – T∆S ……….(1)
where, ∆H is the enthalpy change , T the temperature in kelvin and ∆S the entropy change. For an equilibrium process, ∆G° can be calculated using the equilibrium constant by the following expression ∆G° = – RT lnKp

Harold Ellingham used the above relationship to calculate the ∆G° values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the x-axis and the standard free energy change for the formation of metal oxide in y-axis. The resultant plot is a straight line with ∆S as slope and ∆H as y-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.

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Question 4.
Explain the observations from the Ellingham diagram.
Answer:
1. For most of the metal oxide formation, the slope is positive. It can be explained as follows. Oxygen gas is consumed during the formation of metal oxides which results in the decrease in randomness. Hence, ∆S becomes negative and it makes the term, T∆S positive in the straight line equation.

2. The graph for the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas. It indicates that CO is more stable at higher temperature.

3. As the temperature increases, generally ∆G value for the formation of the metal oxide become less negative and becomes zero at a particular temperature. Below this temperature, ∆G is negative and the oxide is stable and above this temperature ∆G is positive. This general trend suggests that metal oxides become less stable at higher temperature and their decomposition becomes easier.

4. There is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO. This is due to the phase transition (melting or evaporation).

Question 5.
Discuss.the applications of the Ellingham diagram:
Answer:
Ellingham diagram helps us to select a suitable reducing agent and appropriate temperature range for reduction. The reduction of a metal oxide to its metal can be considered as a competition between the element used for reduction and the metal to combine with oxygen.

If the metal oxide is more stable, then oxygen remains with the metal and if the oxide of element used for reduction is more stable, then the oxygen from the metal oxide combines with elements used for the reduction. From the Ellingham diagram, we can infer the relative stability of different metal oxides at a given temperature.

1. Ellingham diagram for the formation of Ag2O and HgO is at upper part of the diagram and their decomposition temperatures are 600 and 700 K respectively. It indicates that these oxides are unstable at moderate temperatures and will decompose on heating even in the absence of a reducing agent.

2. Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the diagram. For example, in the Ellingham diagram, for the formation of chromium oxide lies above that of the aluminium, meaning that Al2O3 is more stable than Cr2O3. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide. However, it cannot be used to reduce the oxides of magnesium and calcium which occupy lower position than aluminium oxide.

3. The carbon line cuts across the lines of many metal oxides and hence it can reduce all those metal oxides at sufficiently high temperature. Let us analyse the thermodynamically favourable conditions for the reduction of iron oxide by carbon. Ellingham diagram for the formation of FeO and CO intersects around 1000 K.

Below this temperature the carbon line lies above the iron line which indicates that FeO is more stable than CO and hence at this temperature range, the reduction is not thermodynamically feasible. However, above 1000 K carbon line lies below the iron line and hence, we can use coke as reducing agent above this temperature. The following free energy calculation also confirm that the reduction is thermodynamically favoured.

From the Ellingham Diagram at 1500 K:

  • 2Fe(s) + O2(g) → 2FeO(g)
  • 2C(s) + O2(g) → 2CO(g)
  • ∆G1 = – 350 kJ mol-1 …………(5)
  • ∆G2 = – 480 kJ mol-1 ………….(6)

Reverse the reaction (1)

  • 2FeO(s) → 2Fe(s) + O2(g)
  • – ∆G1= +350 kJ mol-1 ………. (7)

Now couple the reactions (2) and (3)

  • 2FeO(s) + 2C 2Fe (l,s) + 2CO(g)
  • ∆G3 = -130 kJ mol-1 ……………. (8)

The standard free energy change for the reduction of one mole of FeO is:
∆G3/2 = -65 kJ mol-1.

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Question 6.
Explain the method to purify Titanium metal.
[OR]
Explain Van-Arkel method for refining Titanium.
[OR]
How will you purify metals by using iodine?
Answer:
This method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Titanium and zirconium can be purified using this method. For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide.(TiI4). The impurities are left behind, as they do not react with iodine.
Ti(s) + 2I2(s) → TiI4 (vapour)

The volatile titanium tetraiodide vapour is passed over a tungsten filament at a temperature around 1800 K. The titanium tetraiodide is decomposed and pure titanium is deposited on the filament. The iodine is reused.
Til4 (vapour) → Ti(s)+ 2I2(s)

Question 7.
Mention the applications of Aluminium.
Answer:
Aluminium is the most abundant metal and is a good conductor of electricity and heat. It also resists corrosion. The following are some of its applications.

  1. Many heat exchangers/sinks and our day to day cooking vessels are made of aluminium.
  2. It is used as wraps (aluminium foils) and is used in packing materials for food items,
  3. Aluminium is not very strong, However, its alloys with copper, manganese, magnesium and silicon are light weight and strong and they are used in design of aeroplanes and other forms of transport.
  4. As Aluminium shows high resistance to corrosion, it is used in the design of chemical reactors, medical equipments,refrigeration units and gas pipelines.
  5. Aluminium is a good electrical conductor and cheap, hence used in electrical overhead electric cables with steel core for strength.

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Question 8.
Complete the following reactions,

  1. Cr2O3 + A1 \(\underrightarrow { \Delta } \) ?
  2. B2O2 + Na \(\underrightarrow { \Delta } \) ?
  3. ThO2 + Ca \(\underrightarrow { \Delta } \) ?
  4. Mn3O4 + C \(\underrightarrow { \Delta } \) ?
  5. Ag2O + H2 \(\underrightarrow { \Delta } \) ?

Answer:

  1. Cr2O3 + 2Al \(\underrightarrow { \Delta } \) 2Cr +Al2O3
  2. B2O3 + 6Na \(\underrightarrow { \Delta } \) 2B +3Na2O
  3. ThO2 + 2Ca \(\underrightarrow { 1250K }\) Th + 2CaO
  4. Mn3O4 + 4C \(\underrightarrow { \Delta } \) 3Mn + 4CO
  5. Ag2O + H2 \(\underrightarrow { \Delta } \) 2Ag + H2O

Common Errors and its Rectifications

Common Errors:

  1. In balancing inorganic reactions, they may struggle.
  2. Ores formula they may forget.
  3. In metallurgy, the steps are very important.

Rectifications:

  1. Students are advised to balance first the negative charged atoms like oxygen, chlorine in the equation.
  2. Simple way is to remember one oxide ore (or) one sulphide ore from which the metal is extracted.
  3. Students should practice to write steps headings first along with explanation.

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