STUDENT SPECIAL STUDY MATERIAL

Class 12

PHYSICS (Theory)

Session 2016-17

Kendriya Vidyalaya Sangathan

Regional Office

Guwahati

Our Source of Inspiration

CHIEF PATRON

Shri. Santosh Kumar Mall

IAS Commissioner

Kendriya Vidyalaya Sangathan

New Delhi

PATRONS

Shri. Chandra P. Neelap

Deputy Commissioner

Kendriya Vidyalaya Sangathan

Guwahati Region

Smt. Anjana Hazarika &Shri. D. Patle

Assisstant Commissioners

Kendriya Vidyalaya Sangathan

Guwahati Region

CONVENOR

Shri. Dhirendra Kumar Jha

Principal

Kendriya Vidyalaya, Air Force Station, Borjhar

Guwahati

PREPARED BY:

Kiran Kumari Soren PGT, Physics

Devendra Kumar PGT, Physics

Kendriya Vidyalaya, Air Force Station, Borjhar

Preface There is no substitute as such for hard work. However, planned study and a bit of

smart work can do the trick. With planning I mean prioritizing. When your days are

numbered you just can't go through everything. It is therefore advisable not to panic

and study steadily giving priority to the topics most likely to appear in the

examination.

When it comes to AISSCE, nothing is guaranteed. No one can predict anything

precisely. But, there exist concepts that can enable students to score more with

minimal of efforts.

This study module is aimed at ensuring at least pass mark in the board exams and is prepared

using the available study materials of KVS but in a concise manner. This also includes previous

years CBSE questions and marking scheme so that students will have idea on what type of

questions can come from a particular chapter and what points need to be in their answers to get

marks. Hope this module will boost your confidence both during the preparatory stage as well as

during the examinations.

Students can go through in this way:

(i) Go through this material on this basis of weightage of the units like optics,

electrostatics, electromagnetic induction, AC and EM waves, Dual nature and

radiation, Which carrying weightage of around 35 marks.

(ii) Focus on the topics which you like most and Make the target of completing

those chapters by going through this material.

(iii) Make practice of diagrams and graphical representation during preparation

of topics.

(iv) Some common topics to score easy marks are (a) principles of the devices

involved (ii)Graphical variations specially(a) temperature vs resistance

graph, (b)V-I graph, (c)photocurrent vs potential graph ,(d)frequency vs

stopping potential

(v) Practice well on circuit diagrams in chapter 14 and block diagrams of

chapter 15 communication systems.

NOTE: It is advisible to practice those methods of answering which you are

already practiced,do not change the method at this moment of time.

While attempting numerical questions pay attention to (i) Write the given

part of question. Do not forget to write the formula .

GENERAL QUESTIONS

(I) Write the principle of

(a) Meter bridge

(b) Potentiometer

(c) Cyclotron

(d) Moving coil galvanometer

(e) AC Generator

(f) Transformer

(g) Compound microscope

(h) Telescope

(i) Rectifier

(II) Show the graphical variations of Electric field vs distance for (i) point

charge (ii) dipole (iii) linear charge (iv) surface charge (v) hollow

charged sphere

(III) Show the graphical variations of Electric Potential vs distance for (i)

point charge (ii) dipole (iii) hollow charged sphere

(IV) Show the variation of frequency vs reactance for a capacitor and

inductor.

(V) Draw a graph showing variation between current amplitude and

frequency also mark resonance frequency on the graph.

(VI) Draw I-V characteristics for (i) photodiode(ii) solar cell(iii)zener diode

and output characteristics of a transistor.

(VII) Draw input and output wave form of half weave and full wave rectifier.

(VIII) Draw input and output waveform of CE Amplifier.

UNIT – 1 ELECTROSTATICS

Q1. Write statement of coulumb’s law.

Ans. Coulumb’s Law:- It states that the electro-static force of attraction or repulsion between two charged bodies id

directly proportional to the product of their charges and varies inversely as the square of the distance between the two

bodies.

F = kq1 q2/r2

Here, k = 1/4πε 0 = 9 X 109 Nm2C2 ( in free space)

Q2 Write definition of electric field intensity.also write its unit.

Ans. Electric field Intensity or Electric field: Electric field strength at a point in an electric field is the electrostatic force

per unit positive charge acting on a small positive test charge placed at that point.

E = F / q0 it is a Vector quantity. Its unit is N/C.

Electric field intensity due to a point charge(Q)-

E=kQ/r2 Where k = 9 x 109 Nm2 / C2

Q3 Define electric field lines. Mention their properties.

Ans. Electric field lines – An electric line of force is defined as the path, straight or curved, along which a unit positive

charge is urged to move when free to do so in an electric field. The direction of motion of unit positive charge gives the

direction of line of force.

Properties of field lines :-

(a) The lines of force are directed away from a positive charge abd are directed towards a negative charge.

(b) A line of force starts from a positive charge and ends on a negative charge. This signifies line of force starts from

higher potential and ends on lower potential. These are continuous curves.

(c) Field lines do not cross each other because E cannot have two directions at a given point.

Q4 write formula for electric dipole moment. Calculate electric field due to an electric dipole at its axial position.

Ans. Electric dipole moment P = (2l)q; direction of P from negative to positive charge of dipole.

Electric field intensity due to a point charge:- E = (1/4πε0) (q/r2)

Electric field intensity due to an electric dipole at an axial point

Net electric field due to at point P is the vector sum of EA and Eb

E = (1/4πε0) q/(r-a)2 – (1/4πε0) q/(r+a)2 = (1/4πε0) 2p/r2 Where p = 2aq(assuming a<<r)

Q5 Calculate electric field due to an electric dipole at its equatorial position.

Ans. Electric field intensity due to at a point on the equatorial line (perpendicular bisector):

E+ = kq/(x+d)2 and E+ = kq/(x+d)2

Net electric field E = E+Cosθ

= 2E+Cosθ

=2 kq/(x+d)2 . a/(x+d)1/2

= (1/4πε0) p/x3 (assuming d<<x)

Q6 An electric dipole is placed in a uniform electric field. Calculate torque experienced by it . Also discuss its

potential energy for stable and unstable equillibrum.

Ans. Torque on an electric dipole in an electric field:-

Dipole in a uniform field

Magnitude of net torque = dFSinθ + dFSinθ

=2dFSinθ

=2dqESinθ

=pESinθ,

As the two forces are equal and opposite, net force on the dipole=O

Electric potential energy of an electric dipole in an electric field:-

Potential energy of an electric dipole, in an electrostatic field, is defined as the work done in rotating the dipole from

zero energy position (900) to the desired position(θ) in the electric field.

U=-pECosθ

(a) If θ = 900,then U = 0

(b) If θ = 00, then U = -Pe, (stable equillibrum)

(c) If θ = 1800,then U = Pe, (unstable equillibrum)

Q7 Write statement of Gauss’s law. Calculate electric field due to a linearly distributed charge using this law.

Ans. Gauss Law: Net electric flux through a closed surface = 𝒒

∈𝟎Electric firld due to a linear charge distribution

Net flux through the Gaussian surface =𝑞

∈0

Q8. Calculater electric field due to a uniformly charged spherical shell using Gauss’s Law.

Ans. Electric field due to charged spherical shell

Net flux through the Gaussian surface = 𝑞

∈0

E.4πr2=𝑞

∈0

Hence E=(1/4πε0) (q/r2) for r > R, and r= R

For a point inside the shell E=0, as charge enclosed is zero.

Q9. Calculate electric field due to a uniformly charged Plane sheet using law.

Ans. Electric field due to a plane sheet of charge.

Net flux = EA+EA

By Gauss’ Law the net flux = qenc/ε0

Short answer questions

1. An electric dipole is kept inside a cube. What is the net electric flux through the cube?

Ans- As the net change is zero, net electric flux is zero.

2. A test charge experiences a force F near a large plane sheet of chage. What will be the force experienced by the

charge if the distance is double ?

Ans- As the electric field near the sheet is constant; the force will remain the same .

3. A point charge is kept inside a spherical surface. If the radius of the surface is doubled, what will be the effect on

the flux passing through the surface?

Ans- the electric flux through the surface depends on the charge inside it. Hence the flux will be the same.

4. A short electric dipole produces an electric field E at a distance r on the equatorial line. What will be the electric

field if the distance is doubled?

Ans- It will be E/8.

5. Why two field lines cannot cross each other?

Ans- Because E cannot have two directions at a point.

6. Find the work done to rotate the dipole from the most stabe orientation to the most unstable orientation in an

external electric field?

Ans- W= final potential energy-initial potential energy.

=Pe-(-Pe)=2Pe

7. What is the effect on electric field due to presence of a dielectric medium?

Ans- electric field becomes E/K, where K is the dielectric constant of the medium.

8. Find the expression for E at the centre pf am e;ectric dipole of dipole length 2a.

E=kq/a2 + kq/a2 =2kq/a2

ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential due to a point charge=work done in bringing a unit test charge from infinity to that point .

V=KQ/r It is a scalar . Its unit is J/C=Volt.

Electric potential due to an electric dipole at an axial point V= 1/4πε0(𝑞

𝑟1 -

𝑞

𝑟2)

Putting r1 =x-a and r2=r+a and simplifying above V=(1/4πε0)(q/r)

Electric potential due to an electric dipole at an equatorial point V = V++V-=0

ELECTRIC POTENTIAL ENERGY of a pair charges = work done in bringing the charges from infinity to their

respective locations.

U=kq1q2/r Here, k=1/4πε0 =9x109 Nm2C2(in free space)

CAPACITANCE

Capacitance=charge/Potential

Unit, 1 farad= 1 coulomb/1 volt

Capactance of a parallel plate capacitor

Inside the plates E=ϭ/2εo + ϭ/2εo=Q/Aεo

Potential difference V=Ed=Qd/AεO

Capacitance Q/V=Aεo/d

With dielectric filled inside the capacitor, the electric field decreases by k times hence, C=KAεo/d

Where k is the dielectric constant of the medium

Capacitance in parallel combination,1/CS=1/C1+1/C2

Capacitance in parallel combination, CP=C1+C2

Energy stored in a capacitor =work done to charge the capacitor=1/2CV2=1/2QV=1/2Q2/C

Short answer questions

1. What is an equipotential surface?

ANS-A surface of constant electric potential is called equipotential surface.

2. How much work is done to move a charge on an equipotential surface

ANS-Zero

3. What happens to electric potential due to a positive charge if the distance from the charge is increased?

ANS-As the value of v increases, becomes zero at infinity

4. What happens to electric potential due to a negative charge if the distance from the charge is increased?

ANS-The value of V increases, becomes zero at infinity.

5. A capacitor is charged and disconnected from the battery. It is now filled with a dielectric . What will be the

effect on the energy stored?

ANS-Capacitance becomes k times, charge is same. Hence energy will be reduced by k times.

CURRENT ELECTRICITY

Q.1. Write statement of Ohm’s law.

ANS- Ohm’s law: current through a conductor is directly propotional to the potential difference across the ends

of the conductor provided the physical conditions remains constant.

Mathematically V=IR , R is the resistance of the conductor.

Q.2. Write formula for resistivity or specific resistance of a conductor.

ANS-Resistance R=ρl/A where ρ is the resistivity of the material of the conductor .

Q.3. Draw V-I graph of ohmic and non ohmic conductors.

Q.4. A copper wire of resistance R and resistivity ρ is stretched to double its length. What will be the new

resistance and resistivity?

ANS. No change in resistivity, because it does not depend on length or radius. By stretching the wire its length

becomes double and area of cross section becomes half. Now according to formula. R = ρl/A , new resistance will

increase 4 times.

Q.5. What is drift velocity of electrons in a conductor.

ANS. Drift velocity is the average velocity of all electrons in the conductor under the influence of applied electric

field. Drift velocity vd = (e E /m)Ԏ ; where e is the charge of electron, E is electric field, m is mass of electron and

Ԏ is relaxation time.

Q.6. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature?

ANS. Decrease with increase in temperature

Q.7. How is drift velocity changed if (i) length of the conductor is doubled ,(ii) Radius/area of cross section is

doubled?

ANS. According to formula drift velocity Vd = (e E /m)Ԏ

(i) If length is doubled then E will becomes half (because E= potential difference/ length) , therefore Vd

will also becomes half.

(ii) No change on drift velocity

.

Q.8. What is emf of a cell? Write points of comparision between emf and terminal potential difference of a cell.

The electro motive force is the maximum potential difference between the electrodes of the cell when no current is

drawn from the cell.

Comparision of EMF and P.D

EMF POTENTIAL DIFFERENCE

1 EMF is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell i.e when the circuit is open.

P.D is the difference of potentials between any two points in a closed circuit.

2 It is independent of the resistance of the circuit.

It is proportional to the ressistance between the given points.

3 The term ‘emf’ is used only for the source of emf.

It is measured between any two points of the circuit.

4 It is greater than the potential difference between any two points in a circuit

However p.d is greater than emf when the cell is being charged.

ELECTOMEGNETIC INDUCTION & ALTERNATING CURRENT

1. A bar magnet is moved in a direction indicated by the arrow between two coils PQ and CD. Predict the direction of the

induced current in each coil. (AI 2012 1 mark)

Ans : The direction of the current is from Q to P and from C to D .

This is due to the fact that the coil PQ will have its south pole at Q end and on the other hand the coil CD will have its

south pole near C end.

2. The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the

cause of this damping? ( A I 2013 , 1 mark)

Ans: This is due to the eddy current produced in the copper plate.

3. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.

( A I 2013 , 1 mark)

Ans : As the switch is on , eddy current is produced in the metal disc.

4. How does the mutual inductance of a pair of coil changes when

(i) Distance between the coil is increased and

(ii) Number of turns of the coil is increased ( A I 2013 , 1 mark)

Ans: (i) Decreased, as Ф = M I (flux decreased as distance is increased)

(ii) Increased, as M α n1 n2

5. What are eddy current? Write their two applications. ( A I 2012 , 2 mark)

Ans: Eddy current are the current induced in a conductor when placed in a changing magnetic field. The two applications

(i) Electromagnetic braking and

(ii) Induction furnace

6. A rectangular coil LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of

the conductor. When the arm MN of length 20 cm is moved towards left with a velocity of 10 ms-1, calculate the emf

induced in the arm. Given the resistance of the arm to be 5 ohm ( assuming that the other arms are of negligible

resistance) find the value of the current in the arm. ( A I 2012)

Ans:

Given , B = 0.5 T ; L = 20 cm = 0.2 m ; v = 10 ms-1 ; R = 5 ohm ; I = ?

Using the expression,

Ε = - B l v = - 0.5 x 0.2 x 10 = - 1 V

I = E/R = 1/5 = 0.2 A

7. A wheel with 8 metallic spokes each 50 cm long is rotated with the speed of 120 rev/min in a plane normal to the

horizontal component of the earth’s magnetic field. The Earth’s magnetic field at the place is 0.4 G and the angle of dip is

60 0 . calculate the emf induced between the axle and the reem of the wheel. How will the value of emf be affected if the

number of spokes were increased? (2013)

Ans:

Given L = 50 cm = 0.5 m ; f = 120 rev/min = 120/60 rps = 2 rps ; B = 0.4 G = 0.4 x 10 -4 T

Dip = 60 0 ; B H = B cos 60 0 = 0.2 x 10 -4 T ; E = ?

Using the relation , E = ½ B w L2 , we have,

E = ½ 0.2 x 10 -4 x 2 x 3.14 x 2 x (0.5 )2

= 3.14 x 10 -5 V

8. (a) When a bar magnet is pushed towards ( or away) from the coil connected to a galvanometer, the pointer of the

galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and

direction of the deflection depends. State the law describing this phenomenon.

(b) Sketch the change in flux, emf and force when a conducting rod PQ of resistance R and length l moves freely to and fro

between A and C with speed v on a rectangular conductor placed in a uniform magnetic field as shown in the figure.

(AI 2016 , 5 marks)

Ans: (a) Phenomenon : Electromagnetic Induction

Factors : Strength of the magnetic field of the magnet

Speed of motion of the bar magnet

Direction depends on the north /south polarity of the magnet and the direction of motion

of the magnet .

(b)

9. (a) Define self inductance of a coil. Obtain the expression for the energy stored in a solenoid of self inductance ‘L’ when

the current through it grows from zero to ‘I’.

(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as

shown in the figure. The loop is pulled with a constant velocity of 20 cm per second till it goes out of the field.

(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long the current in the loop

persist?

(ii) Plot the graph showing the variation of the magnetic flux and induced emf as a function of time

Ans: (a) Self inductance is numerically equal to the magnetic flux linked with the coil when unit current flows through it.

We know that,

E = - L( dI/dt)

Or dW = E I dt = L I dI

Now ,

W = ∫dW = ∫ L I dI = ½ L I 2

This is the energy

(b) (i) Direction of the current is clockwise (MNOP) and the duration of the induced current is 1 s.

(ii) The graph is as shown

ELECTROMAGNETIC WAVE

1. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of

propagation of EMW? 1 (2012)

Ans- These waves are perpendicular to each other and perpendicular to the direction of propagation

2. Name the physical quantity which remains same for microwaves of wavelength 1mm and UV radiation of 1600𝐴° in

vacuum. 1 (2012)

Ans- velocity(3 × 108𝑚/𝑠) as both are electromagnetic wave

3. The speed of electromagnetic wave in a material medium is given by 𝑣 = 1

√𝜇𝜖.How does its frequency change?

Ans- does not change

4. Name the electromagnetic waves, which (i) maintain the earth’s warmth and (ii) are used in aircraft navigation.

Ans- (i) infrared (ii) Microwave

5. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic

radiations. Name the radiations and write the range of their frequency.

Ans- UV Radiation, frequency range (1015 − 1017) Hz

6. Why are infra-red radiations referred to as heat waves? Name the radiations in the electromagnetic spectrum

having (i) shorter wave length (ii) longer wave length

Ans- IR are produced by hot bodies and molecules.

Visible ,UV, X-ray ,𝛾 − 𝑟𝑎𝑦 (ii) Microwave, radio wave

7. To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019𝐻𝑧 belong? Ans- X-ray or gamma ray

8. A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac

source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.

Ans- 𝐼𝑑 =∈0𝑑∅𝐸

𝑑𝑡=∈0

𝑑

𝑑𝑡(

𝑞

∈0) =

𝑑𝑞

𝑑𝑡

9. Answer the following questions :

(a) Name the electromagnetic waves, which are produced during radioactive decay of a nucleus. Write their

frequency range.

(b) (b) Welders wear special goggles while working. Why? Explain.

(c) Why are infrared waves often called as heat waves? Give their one application

answer

(a) gamma rays , frequency range (1019 − 1023) Hz (b) to protect eyes from UV-radiations (c) IR are produced by hot bodies and molecules

And used in green houses to warm the plants

10. Answer the following

(a) Name the em waves which are used for the treatment of certain forms of cancer. write their frequency range.

(b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?

(c) Why is the amount of the momentum transferred by the em waves incident on the surface so small?

Answer

(a)X-ray of gamma rays , range :1018 -1022

(b)Ozone layers absorb the ultraviolet radiation from the sun and prevent it from reaching the earth’s surface.

(c) Momentum transferred , p=u/c Where u=energy transferred , and c= speed of light due to the large value

of speed of light (c) the amount of momentum transferred by the em waves incident on the surface is small.

11. State two properties of electromagnetic waves. How can we show that electromagnetic waves carry momentum?

Ans-Transverse nature does not deflected by electric and magnetic fields, same speed in vacuum for all waves, no

material medium required for propagation and they get diffracted, refracted and polarized.

Momentum transferred, p=u/c Where u=energy transferred, and c= speed of light

12. Name the parts of electromagnetic spectrum which is

(i) suitable for radar systems used in aircraft navigation.

(ii) (ii) used to treat muscular strain.

(iii) (iii) used as a diagnostic stool in medicine.

Write in brief, how these waves can be produced.

Ans- (i) microwave (ii) infrared (iii) X-ray

Microwaves are produced by klystroms, magnetrons and gunn diodes

Infrared are produced by the vibrating molecules and atoms in hot bodies.

X-rays are produced by bombardment of high energy electrons on a metal target of high atomic weight like tungsten.

Communication system

VERY SHORT ANSWER TYPE QUESTIONS [1 MARK]

1. What is the function of a ‘Repeater’ in a communication system?

2. What is the function of a transmitter in a communication system?

3. What is the function of a Receiver in a communication system?

4. Which part of the electromagnetic spectrum is used in satellite communication?

5. What is sky wave propagation?

6. What is ground wave propagation?

7. What is space wave propagation?

8. State the reason why microwaves are best suited for long distance transmission of signals.

9. Give the reason why transmission of T.V. signals via sky waves is not possible.

10. What is the purpose of modulating a signal in transmission?

11. What should be the length of dipole antenna for a carrier wave of frequency 6x108 HZ ?

12. A T.V. tower has a height of 71 m. what is the maximum distance upto which T.V. transmission can be

received? Given that the radius of the earth= 6.4x 106 m

13. Suggest a possible communication channel for the transmission of a message signal which has a bandwidth

of 5 MHz.

14. Name the type of communication in which the signal is a discrete and binary coded version of the message

of information.

15. What is the length of a dipole antenna to transmit signals of frequency 200 MHz?.

16. Name the type of communication systems according to the mode of transmission.

17. Name an appropriate communication channel needed to send a signal of band-width 100 KHz over a

distance of 8 km.

18. What is transponder?

19. How does the effective power radiated by an antenna vary with wavelength?

SHORT ANSWER TYPE QUESTIONS [2 MRKS]

1. Distinguish between ‘Analog and Digital signals’

2. Mention the function of any two of the following used inm communication system:

(i) Transducer

(ii) Repeater

(iii) Transmitter

(iv) Bandpass Filter

3. (i) Define modulation index.

(ii) Why is the amplitude of modulating signal kept less than the amplitude of

carrier wave?

4. What is sky wave communication? Why is this mode of propagation restricted

to the frequencies only up to few MHz?

5. What is ground wave communication? On what factors does the maximum

range of propagation in this mode depend?

6. What is space wave communication? Write the range of frequencies suitable

for space wave communication.

7. For an amplitude modulated wave, the maximum amplitude is found to be 10

V while the minimum amplitude is 2 V. Calculate the modulation index. Why

is modulation index generally kept less than one ?

8. Draw a block diagram showing the important components in a communication

system. What is the function of a transducer?

9. ATV lower has a height of 80 m at a given place. Calculate the coverage range,

assuming the radius of the Earth to be 6400 km.

10. The transmission tower at a particular station has a height of 125 m. Calculate the

population covered by the transmission if the average population density around

the tower is 1000 km -2.

11. Explain the function of a repeater in a communication system.

12. What is range of frequencies used for TV transmission? What is common between

these waves and light waves?

13. Write two factors justifying the need of modulating a signal.

A carrier wave of peak voltage 12 V is used to transmit a message signal. What

should be the peak voltage of the modulating signal in order to have a modulation

index of 75% ?

14. In standard AM broadcast, what mode of propagation is used for transmitting a

signal? Why is this mode of propagation limited to frequencies upto a few MHz.

15. By what percentage will the transmission range of a TV tower be affected when

the height of the tower is increased by 21% ?

16. Why are high frequency carrier waves used for transmission?

17. What is meant by term ‘modulation’? Draw a block diagram of a simple modulator

for obtaining an AM signal.

18. Write the function of (i) Transducer and (ii) Repeater in the context of

communication system.

19. Write two factors justifying the need of modulation for transmission of a signal.

20. (i) What is line of sight communication?

(ii) Why is it not possible to use sky wavers propagation for transmission of TV

signals?

Optics

OPTICAL INSTRUMENTS

1.

2.

3.

4..

5.

6.

7. Which of the following waves can be polarized (i) heat waves (ii) sound waves? Give

reason to support your answer. (2013)

Ans: heat waves can be polarized as they are transverse in nature (1M)

8. What will be the effect on interference fringes if red light is replaced by blue light? What

will be the effect on interference fringes if red light is replaced by blue light?

(2013)

Ans: β=Dλ/d, the wavelength of blue light is less than that of red light, hence if red light

is replaced by blue light, the fringes width decreases i.e. fringes come closer.

9. How does the angular separation between fringes in single-slit diffraction experiment

change when the distance of separation between the slit and screen is doubled?

(2012)(1M)

ANs: Angular separation is θ= β/D = λ/d

Since θ is independent of D, angular separation would remain same.

10. In a single-slit diffraction experiment, the width of the slit is made double the original

width. how does this affect the size and intensity of the central diffraction band?

Ans: in single slit diffraction experiment fringe width is β=2Dλ/d. if d is doubled, the width

of central maxima is halved. Thus size of central maxima is reduced to half. Intensity of

diffraction pattern varies square of slit width. so when the slit gets doubled, it makes the

intensity four times.

11. Define the term ‘wavefront’.

Ans: the wavefront is defined as the locus of all the particles of a medium, which are

vibrating in the same phase. (1M)(2014)

12. Draw the shape of the wavefront coming out of a convex lens when a plane wave is

incident on it. (2014)(1M)

Ans:

13. Draw the shape of the wavefront coming out of a concave mirror when a plane wave is

incident on it. (2014)(1M)

Two mark questions:

14. Laser light of wavelength 640 nm incident on a pair of slits produces an interference

pattern in which the bright fringes are separated by 7.2 mm. calculate the wavelength of

another source of light which produces interference fringes separated by 8.1 mm using

same arrangement. Also find the minimum value of the order ‘n’ of the bright fringe of

shorter wavelength which coincides with that of the longer wavelength. (2012)

Ans: distance between two bright fringes= fringe width

Β=𝜆𝐷

𝑑

For same values of D and d, we have

β1/β2=λ1/λ2 or 7.2

8.1=

640

𝜆2

or 0.8λ2=576 or λ2=720 nm

Calculation of minimum value of order: for n to be minimum (n+1)th maxima of shorter

wavelength should coincide with the nth maxima of longer wavelength

(n+1)x 640= n x 720

n= 8

minimum order of shorter wavelength=(n+1)=(8+1)=9

15. Yellow light (λ=6000Å) illuminates a single slit of width 1x10-4m. Calculate (i) the distance

between the two dark lines on either side of the central maximum, when the diffraction

pattern is viewed on a screen kept 1.5m away from the slit, (i) the angular spread of the

first diffraction minimum. (2012)

Ans: (i) Distance between two dark lines, on either side of central maxima= 2𝜆𝐷

𝑑

=(2x60000x10-10x1.5)/91x10-4)= 18mm

(ii) Angular spread of the first diffraction minimum (on either side)

=θ =𝜆

𝑎=(6x10-7)/(1x10-4) = 6x 10-3 radians (2M)

16. (a) Write two characteristic features distinguishing the diffraction pattern from the

interference fringes obtained in Young’s double slit experiment. (2013)

Ans: Diffraction: (i) width of principal maxima is twice the width of the other fringes.

(ii)Intensity goes on decreasing as the order of the diffraction bands increasing

Interference: (i) width of all the fringes is the same. (ii) All the fringes are of same

intensity.

17. A parallel beam of light of 500nm falls on a narrow slit and the resulting diffraction

pattern is observed on a screen 1m away. It is observed that the first minimum is at a

distance of 2.5mm from the screen. Calculate the width of the slit. (2013) (2M)

Ans: xnd/D=nλ D=1 n=1

d= 2 x 10-4 m

18. A parallel beam of light of 600nm falls on narrow slit and the resulting diffraction

pattern is observed on a screen 1.2 m away. It is observed that the first minimum is

at a distance of 3mm from the centre of the screen. Calculate the width of the

slit.(2013)(2M)

Ans: λ=600nm D=1.2m

θ1=x1 /D θ1=2.5 x 10-3 rad

asin θ1=nλ

a=0.24 mm

19. Find an expression for the intensity of transmitted light when a Polaroid sheet is rotate

between two crossed polaroids. In which position of the Polaroid sheet will be

transmitted intensity be maximum.

Ans: Let the rotating Polaroid sheet make an angle θ with the first Polaroid.

The angle with the other Polaroid will be (90-θ)

Applying Malu’s law between

P1 and P3

I’=I0cos2θ

Between P3 and P2

I’’=(I0cos2θ) cos2 (90- θ)

I’’=I0/4xsin2θ

Θ=π/4

Three mark questions:

20. (a) Why photoelectric effect cannot be explained on the basis of wave nature of light?

Give reason.

(b) Write the basic features of photon picture of electromagnetic radiation on which

Einstein’s photoelectric equation is based. (2013)

Ans: (a) (i) The maximum kinetic energy of the emitted electron should be directly

proportional to the intensity of incident radiations but it is not observed experimentally.

Also maximum kinetic energy of the emitted electrons should not depend upon the

incident frequency according to wave theory, but it is not so.

(ii) According to wave theory, threshold frequency should not exist. Light of all

frequencies should emit electrons provided intensity of light is sufficient for electrons to

eject.

(iv) According to wave theory, photoelectric effect should not be instantaneous. Energy

of wave cannot be transferred to a particular electron but will be distributed to all

the electrons present in the illuminate portion. Hence, there has to be a time lag

between incident of radiation and the emission of electrons.

(b)

Basic features of photon picture of electromagnetic radiation:

(i) Radiation behaves as if it is made of particles like photons. Each photons has energy E

=hµ and the momentum p = h/λ.

(ii) Intensity of radiation can be understood in terms of number of photons falling per

second on the surface. Photon energy depends only on frequency and is independent of

intensity.

(iii)Photoelectric effect can be understood as the result of the one to one collision

between an electron and a photon.

(iv)When a photon of frequency (µ) is incident on a metal surface, a part of its energy is

used in overcoming the work function and other part is used in imparting kinetic energy,

so KE=h(µ-µo)

WAVE OPTICS QUESTIONS

1 What is the geometrical shape of the wave front when a plane wave passes through a

convex lens? 2008

2 How would the angular separation of interference fringes in Young’s double slit experiment

change when the distance between the slits and screen is doubled? 2009

3 How does the fringe width, in Young’s double-slit experiment, change when the distance of

separation between the slits and screen is doubled? 2012

4 Define the term ‘linearly polarised light’. When does the intensity of transmitted light

become maximum, when a Polaroid sheet is rotated between two crossed Polaroid? 2009

8 State clearly how an unpolarised light gets linearly polarised when passed through a

Polaroid.

a) Unpolarised light of intensity Io is incident on Polaroid P1 which is kept near another

Polaroid P2 whose pass axis is parallel to that of P1. How will the intensities I1 and I2,

transmitted by the Polaroids P1 and P2 respectively change on rotating P1 without

disturbing P2?

b) Write the relation between the intensities I1 and I2. 2015

9 Use Huygens` principle to show how a plane wave front propagates from a denser to rarer

medium. Hence verify snell`s law of refraction. 2015

10 Answer the following : (a) When a tiny circular obstacle is placed in the path of light from a

distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain,

why? (b) How does the resolving power of a microscope depend on (i) the wave length of

the light used and (ii) the medium between the object and the objective lens?

11 (a) State Huygens` principle. Using this principle explain how a diffraction pattern is

obtained on a screen due to a narrow slit on which a narrow beam coming from a

monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central

fringe.

(c) If a monochromatic source of light is replaced by white light, what change would you

observe in the diffraction pattern?

14 (a) Using the phenomenon of polarisation, show how transverse nature of light can be

demonstrated.

(b)Two polaroids P1 and P2are placed with their pass axes perpendicular to each other.

Unpolarised light of intensity I0is incident on P1 . A third polaroid P3is kept in between

P1and P2 such that its pass axis makes an angle of 300 with that of P1. Determine the

intensity of light transmitted through P1,P2 and P3. 2014

15 (a) 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. Describe, with the help of a suitable diagram,

the basic phenomenon/process which occurs to explain this observation.

(b) Show how light reflected from a transparent medium gets polarized. Hence deduce

Brewster’s law . 2015

16 (a) Define a wave front.

(b) Using Huygens` principle, draw diagrams to show the nature of the wave fronts when an

incident plane wave front gets

(i) reflected from a concave mirror,

(ii) refracted from a convex lens. 2015

17 18 State the condition under which the phenomenon of diffraction of light takes place.

Derive the expression for the width of the central maximum due to diffraction of light at a

single slit. A slit if width ‘a’ is illuminated by a monochromatic light of wavelength 700 nm at

normal incidence. Calculate the value of ‘a’ for position of (i) First minimum at an angle of

diffraction of 30o. (ii) First maximum at an angle of diffraction of 30o. 19 (a) In a single slit

diffraction experiment, a slit of which ‘d’ is illuminated by red light of wavelength 650 nm.

For what value of ‘d’ will: (i) The first minimum fall at an angle diffraction of 30o, and (ii)

The first maximum fall at an angle of diffraction 30o? (b) Why does the intensity of the

secondary maximum become less as compared to the central maximum? 2009

20 In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by

monochromatic light of wavelength 450 nm. The screen is 0.1 m away from the slits. (a) Find

the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b)

How will the fringe pattern change if the screen is moved away from the slits? 2010

21 State the importance of coherent sources in the phenomenon of interference. In Young’s

double slit experiment to produce interference pattern, obtain the conditions for

constructive and destructive interference. Hence deduced the expression for the fringe

width. How does the fringe width get affected, if the entire experimental apparatus of

Young is immersed in water? 2011

22 1. How does an unpolarised light incident on a polaroid get polarised? Describe briefly, with

the help of the necessary diagram, the polarisation of light by refection from a transparent

medium. 2. Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third

Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by

Polaroid ‘B’ reduces to 1/8th of the intensity of unpolarised light incident on A?

23 (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are

obtained on the screen kept in front of a double slit. Hence obtain the expression for the

fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young’s double slit

experiment is 9:25. Find the ratio of the widths of the two slits. 2014

24 (a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow

slit illuminated by a mono-chromatic source of light. Hence obtain the conditions for the

angular width of secondary minima.

(b) Two wave lengths of sodium light of 590 nm and 596 nm are used in turn to study the

diffraction taking place at a single slit of aperture 2x 10-6 m. The distance between the slit

and the screen is 1.5 m. Calculate the separation between positions of first maxima of the

diffraction pattern obtained in the two cases. 2014

25 What is the effect on the interference fringes in Young’s double slit experiment when

(i) the width of the slit is increased ;

(ii) the monochromatic source of light is replaced by a source of white light?

26 (a) Using Huygens` construction of secondary wavelets explain how a diffraction pattern is

obtained on a screen due to a narrow slit on which a narrow beam coming from a

monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central

fringe. (c) Explain why the maxima becomes weaker and weaker with increasing n. 2015

QUESTIONS HAVE BEEN ASKED TWO TIMES

1 In what way is diffraction from each slit related to the interference pattern in a double slit

experiment? [2013, 2015]

2 In Young’s double slit experiment, derive the condition for (i) Constructive interference and

(ii) Destructive interference at a point on the screen. [2011, 2012]

QUESTIONS HAVE BEEN ASKED THREE TIMES OR MORE

1 State Huygens’ principle. With the help of a suitable diagram, prove Snell’s law of refraction

using Huygens’ principle. 3 [2006 , 2013, 2015]

2 In Young’s double slit experiment, deduce the conditions for (i) constructive, and (ii)

destructive interference at a point on the screen. Draw a graph showing variation of the

resultant intensity in the interference pattern against position ‘x’ on the screen. 3 [2006 ,

2011, 2012]

WAVE OPTICS EXPECTED QUESTIONS FOR AISSCE 17

1 State Huygens’ principle. With the help of a suitable diagram, prove Snell’s law of refraction

using Huygens’ principle. 3

2 State Huygens’ principle. With the help of a suitable diagram, prove the laws of reflection

using Huygens’ principle. 3

3 In Young’s double slit experiment, deduce the conditions for (ii) constructive, and (ii)

destructive interference at a point on the screen. Draw a graph showing variation of the

resultant intensity in the interference pattern against position ‘x’ on the screen.

4 State the importance of coherent sources in the phenomenon of interference. In Young’s

double slit experiment to produce interference pattern, obtain the conditions for

constructive and destructive interference. Hence deduced the expression for the fringe

width. How does the fringe width get affected, if (i) the entire experimental apparatus of

Young is immersed in water? (ii) The wavelength of light is increased? (iii) Separation

between the two slits decreased? (iv) Monochromatic light is replaced by white light? (v)

Distance of the screen is increased? 5

5 (a) Using Huygens` construction of secondary wavelets explain how a diffraction pattern is

obtained on a screen due to a narrow slit on which a narrow beam coming from a

monochromatic source of light is incident normally. (b) Show that the angular width of the

first diffraction fringe is half of that of the central fringe.

(c) Explain why the maxima at becomes weaker and weaker with increasing n.

6. How does an unpolarised light incident on a polaroid get polarised?

Describe briefly, with the help of the necessary diagram, the polarisation of light by

refection from a transparent medium.

7. Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third Polaroid ‘C’ be

placed between them so that the intensity of polarised light transmitted by Polaroid ‘B’

reduces to 1/8th of the intensity of unpolarised light incident on A?

8. (a) 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. Describe, with the help of a suitable diagram,

the basic phenomenon/process which occurs to explain this observation.

(b) Show how light reflected from a transparent medium gets polarized. Hence deduce Brewster’s

law

8 (a) Define a wave front.

(b) Using Huygens` principle, draw diagrams to show the nature of the wave fronts when an

incident plane wave front gets

(i) reflected from a concave mirror,

(ii) refracted from a convex lens.

2016

Distinguish between polarized and unpolarized light. Does the intensity of polarized light

emitted by a polaroid depend on its orientation ? Explain briefly. The vibrations in a beam

of polarized light make an angle of 60° with the axis of the polaroid sheet. What percentage

of light is transmitted through the sheet ?

(i) State the essential conditions for diffraction of light.

(ii) Explain diffraction of light due to a narrow single slit and the formation of pattern of

fringes on the screen.

(iii) Find the relation for width of central maximum in terms of wavelength ‘λ’, width of slit

‘a’, and separation between slit and screen ‘D’.

(iv) If the width of the slit is made double the original width, how does it affect the size and

intensity of the central band ?

OR

(i) Draw a labelled schematic ray diagram of astronomical telescope in normal adjustment.

(ii) Which two aberrations do objectives of refracting telescope suffer from ? How are these

overcome in reflecting telescope ?

(iii) How does the resolving power of a telescope change on increasing the aperture of the

objective lens ? Justify your answer.

2015

(a) The ratio of the widths of two slits in Young’s double slit experiment is 4 : 1. Evaluate the

ratio of intensities at maxima and minima in the interference pattern. (b) Does the

appearance of bright and dark fringes in the interference pattern violate, in any way,

conservation of energy ? Explain.

Write the factors by which the resolving power of a telescope can be increased. (b)

Estimate the angular separation between first order maximum and third order minimum of

the diffraction pattern due to a single slit of width 1 mm, when light of wavelength 600 nm

is incident normal on it.

Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses.

Justifying your answer. (b) Two polaroids P1 and P2 are placed in crossed positions. A third

polaroid P3 is kept between P1 and P2 such that pass axis of P3 is parallel to that of P1. How

would the intensity of light (I2) transmitted through P2 vary as P3 is rotated ? Draw a plot of

intensity ‘I2’ Vs the angle ‘θ’, between pass axes of P1 and P3.

Define a wavefront. How is it different from a ray ? (b) Depict the shape of a wavefront in

each of the following cases. (i) Light diverging from point source. (ii) Light emerging out of

a convex lens when a point source is placed at its focus. (iii) Using Huygen’s construction of

secondary wavelets, draw a diagram showing the passage of a plane wavefront from a

denser into a rarer medium.

OR

(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain

expression for total magnification when the image is formed at infinity. (b) How does the

resolving power of a compound microscope get affected, when (i) focal length of the

objective is decreased. (ii) the wavelength of light is increased ? Give reasons to justify

your answer.

2014

In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained

on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young’s double slit

experiment is 9 : 25. Find the ratio of the widths of the two slits.

OR

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow

slit illuminated by a monochromatic source of light. Hence obtain the conditions for the

angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the

– 6 m. The distance between the slit

and the screen is 1.5 m. Calculate the separation between the positions of first maxima of

the diffraction pattern obtained in the two cases.

2013

Which of the following waves can be polarized (i) Heat waves (ii) Sound waves?

(a) In what way is diffraction from each slit related to the interference pattern in a double

slit experiment?

(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the

diffraction taking place at a single slit of aperture 4 2 10− × m. The distance between the slit

and the screen is 1.5.m. Calculate the separation between the positions of the first maxima

of the diffraction pattern obtained in the two cases.

(a) State Huygen’s principle. Using this principle draw a diagram to show how a plane wave

front incident at the interface of the two media gets refracted when it propagates from a

rarer to a denser medium. Hence verify Snell’s law of refraction.

(b) When monochromatic light travels from a rarer to a denser medium, explain the

following, giving reasons:

(i) Is the frequency of reflected and reflected light same as the frequency of Incident light?

(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?

2012

(a) In Young’s double slit experiment, derive the condition for (i) constructive interference

and (ii) destructive interference at a point on the screen. (b) A beam of light consisting of

two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s

double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28

mm, calculate the least distance from the central bright maximum where the bright fringes

of the two wavelengths coincide.

OR

(a) How does an unpolarised light incident on a polaroid get polarised? Describe briefly, with

the help of a necessary diagram, the polarisation of light by reflection from a transparent

medium.

(b) Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be

placed between them so that the intensity of polarised light transmitted by polaroid B

reduces to 1/8th of the intensity of unpolarised light incident on A?

How does the fringe width, in Young’s double-slit experiment, change when the distance of

separation between the slits and screen is doubled?

Ans- (distance between slits and screen) is doubled, then fringe width will be doubled

ATOMS AND NUCLIE Q. No.

Question and Answers.

Years

1. Define ionization energy. How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?

2016

Ans: The ionisation energy is defined as the amount of energy required to remove the most loosely bound electron, the valence electron of an isolated gaseous atom to form a cation. Since, total energy is directly proportional to the mass of electron, so the ionisation energy becomes 200 times on replacing an electron by a particle of mass 200 times that of the electron but having the same charge.

2. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. In which region (infrared, visible, ultraviolet) of hydrogen spectrum does this wavelength lie? (Given, Rydberg constant, R = 107𝑚−1 )

2016

Ans: The wavelength associated with Balmer series is: 1

= 𝑅 (

1

22 − 1

𝑛2). For shortest

wavelength, n = . So the shortest wavelength of the spectral lines emitted is: 1

=

𝑅

4

Or, = 4

𝑅 =

4

107 = 4 × 10−7m. = 4000 𝐴0.

It falls in visible light region of em wave spectrum.

3. Write the basic nuclear process involved in the emission of 𝛽+in a symbolic form by a radioactive nucleus.

2016

Ans: The basic nuclear process involved in the emission of 𝛽+in a symbolic form by a radioactive nucleus

𝑋𝑍𝐴 → 𝑌𝑍−1

𝐴 + 𝛽+ + 𝜈 During the emission of 𝛽+one proton within the nucleus of X changes to neutron, so that the total number of nucleon remains same but the number of proton decreases by one and neutron increases by one.

4. In the reactions given below, find the value of x, y and z and a, b and c. (a) 𝐶6

11 → 𝐵𝑦𝑧 + 𝑥 + 𝜈

(b) 𝐶612 + 𝐶6

12 → 𝑁𝑒𝑎20 + 𝐻𝑒𝑏

𝑐

2016

Ans: (a) According to equation, 𝐶6

11 → 𝐵𝑦𝑧 + 𝑥 + 𝜈

Comparing with 𝛽+decay equation: 𝑋𝑍𝐴 → 𝑌𝑍−1

𝐴 + 𝛽+ + 𝜈. We find – x is 𝛽+, y is 5 and z is 11.

(b) According to equation, 𝐶6

12 + 𝐶612 → 𝑁𝑒𝑎

20 + 𝐻𝑒𝑏𝑐 .

In the above nuclear reaction a = 10, b = 2 and c = 4.

5. A nucleus with the mass number A = 240 and BE/A = 7.6 MeV breaks into fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the energy released.

2016

Ans: According to question: P → Q + Q BE/A of P = 7.6 MeV/A and mass number(A) of P = 240 So, BE of P = 7.6 × 240 MeV. = 1824 MeV. BE/A of Q = 8.5 MeV/A and mass number (A) of Q = 120 So, BE of Q = 8.5 × 120 MeV = 1020 MeV. Now, energy released = 2 (BE of Q) – BE of P. = 2 × 1020 – 1824 = 2040 – 1824 = 216 MeV.

6. Calculate the energy in the fusion reaction: 𝐻1

2 + 𝐻12 → 𝐻𝑒2

3 + 𝑛, where BE of 𝐻12 = 2.23𝑀𝑒𝑉 𝑎𝑛𝑑 𝑜𝑓 𝐻𝑒2

3 = 7.73 𝑀𝑒𝑉. 2016

Ans: According to question: 𝐻12 + 𝐻1

2 → 𝐻𝑒23 + 𝑛.

Energy of fusion = BE of 𝐻𝑒23 - 2 × 2 𝐵𝐸 𝑜𝑓 𝐻1

2 . = 7.73 – 2 × 2.23 = 3.27 MeV.

7. State Bohr’s quantisation condition for defining stationary orbits. How does de-Broglie’s hypothesis explain the stationary orbits?

2016

Ans: Bohr’s quantization principle states that electrons revolve in a stationary orbit of which energy and momentum are fixed. The momentum of electron in the fixed orbit

is given by 𝑛ℎ

2𝜋, where n is principal quantum number.

According to de-Broglie hypothesis, the electron is associated with wave character. Hence, a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-Broglie wavelengths, i.e.

2𝜋𝑟 = n.

Or, 2𝜋𝑟 = n × ℎ

𝑝,

Or, 2𝜋𝑟 = n× ℎ

𝑚𝑣.

Or, mv𝑟 = n× ℎ

2.

Or, L = n× ℎ

2.

This proves that angular momentum of the electron in a orbit is integral

multiple of ℎ

2.

8. Show that the radius of the orbit in hydrogen atom varies as 𝑛2, where ‘n’ is the principal quantum number of the atom.

2015

Ans: Electron revolves around the nucleus of an atom in circular orbit. The columbic force on the electron by the nucleus provides the required centripetal force. So we have- 𝑚𝑣2

𝑟 = k

𝑒2

𝑟2

Or, 1

𝑟 = k

𝑚𝑒2

𝑚2𝑟2𝑣2

Or, 1

𝑟 = k

𝑚𝑒2

𝐿2. ( L = mrv = n

ℎ

2𝜋 )

Or, r = 1

𝑘𝑚𝑒2 × 𝐿2 =

1

𝑘𝑚𝑒2 × (

𝑛ℎ

2𝜋)

2 =

1

𝑘𝑚𝑒2 ×

ℎ2

4𝜋2 × 𝑛2

Or, r = constant × 𝑛2 This shows that the radius of the orbit in hydrogen atom varies as 𝑛2, where ‘n’ is the principal quantum number of the atom.

9. In the study of Geiger-Marsden experiment on scattering of α – particles by a thin foil of gold, draw the trajectory of α – particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.

From the relation R = 𝑅0𝐴1

3⁄ , where 𝑅0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.

2015

Ans: In the study of Geiger-Marsden experiment on scattering of α – particles by a thin foil of gold, the trajectory of α – particle in the coulomb field of target nucleus is as follows.

The fact that only a small fraction of the number of incident particles rebound back

indicates that the number of 𝛼-particles undergoing head on collision is small. This, in

turn, implies that the mass of the atom is concentrated in a small volume. Rutherford

scattering therefore, is a powerful way to determine an upper limit to the size of the

nucleus.

Nuclear matter density = mass of nucleus/ volume of nucleus. = mass of a nucleon × No. of nucleons/ volume of nucleus

= m × A/ 4

3 𝜋 𝑅3 = m × A/

4

3 𝜋 𝑅𝑜

3 A = 3𝑚

4𝑅𝑜3 = constant.

This show that nuclear matter density is independent of A.

10. Write the three characteristic properties of nuclear force. 2015

Ans: The three characteristic properties of nuclear force are: i) It is the strongest force in nature. ii) It is a short range force. iii) It is a saturated force.

11. Draw a plot of potential energy of a pair of nucleons as a function their separation. Write two important conclusions that can be drawn from the graph.

2015

Ans:

Potential energy of a pair of nucleons as a function of their separation.

Two important conclusions that can be drawn from the graph are: i) The nuclear force is the strongest force in nature which can produce

potential energy of range of MeV. ii) The nuclear force is attractive for small separation but for large separation

the force is repulsive.

12. Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.

2015

Ans: Nuclear Fission Nuclear Fusion

It is the phenomenon of breaking of heavy nucleus to form two or more lighter nuclei.

It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus.

Radioactive waste are released. No radioactive waste are released.

Energy released is lesser than that released in nuclear fusion.

Energy released is greater than that released in nuclear fission.

Ex: 𝑛01 + 𝑈 92

235 → 𝑋𝑒54140 + 𝑆𝑟38

94 + 2 𝑛01

+ 200.4MeVn

Ex: 𝐻11 + 𝐻1

1 → 𝐻12 + 𝑒+ + 𝜈, +

0.42MeV

In both the processes, a certain mass (∆m) disappears, which appears in the form of

energy as per Einstein equation: E = (∆m)𝑐2.

13. Calculate the energy release in MeV in the deuterium-tritium fusion reaction: 𝐻1

2 + 𝐻13 → 𝐻2

4 + 𝑛 Using the data: m( 𝐻1

2 ) = 2.014102 u, m( 𝐻13 ) = 3.016049 u, m( 𝐻𝑒2

4 ) = 4.002603 u &𝑚𝑛 = 1.008665 u 1u = 931.5 MeV/𝑐2.

2015

Ans: According to question, the reaction is: 𝐻1

2 + 𝐻13 → 𝐻2

4 + 𝑛 Mass defect (∆m) = ((2.014102 + 3.016049) – (4.002603 + 1.008665))u = 0.018883u. Energy released = ∆m × 931.5 MeV = 0.018883 × 931.5 MeV = 17.589 MeV.

14. A 12.9 eV beam of electrons is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Lyman series and first member of Balmer series.

2014

Ans: The energy of gaseous hydrogen atom at room temperature are: 𝐸1 = −13.6 𝑒𝑉, 𝐸2 = −3.4 𝑒𝑉, 𝐸3 = −1.51 𝑒𝑉 and 𝐸4 = −0.85 𝑒𝑉. So the differences are: 𝐸2 − 𝐸1 = −3.4 + 13.6 = 10.2𝑒𝑉. 𝐸3 − 𝐸1 = −1.51 + 13.6 = 12.09 and 𝐸4 − 𝐸1 = −0.85 + 13.6 = 12.75 𝑒𝑉. As the energy given to the electron is more than 𝐸4 − 𝐸1, electron will jump to the 4th orbit.

For Lyman series, we have - 1

= 𝑅 (

1

12 − 1

𝑛2). For first member of the series n = 2.

Thus we get, 1

= 𝑅 (

1

12 − 1

22) = R × 3

4.

Or, = 4

3𝑅 =

4

3𝑋

1

1.097 × 107 = 1.215 × 10−7m.

For Balmer series, we have - 1

= 𝑅 (

1

22 − 1

𝑛2). For first member of the series n = 3.

Thus we get, 1

= 𝑅 (

1

2−

1

92) = R ×

5

36.

Or, = 36

5𝑅 =

36

5𝑋

1

1.097 × 107 = 6.56 × 10−7m.

15. Draw a plot of BE/A – mass number A for 2 ≤ 𝐴 ≤ 170. Use the graph to explain the release of energy in the process of nuclear fusion of two light nuclei and fission of a heavy nucleus into two lighter nuclei..

2014

Ans: A plot of BE/A – mass number A for 2 ≤ 𝐴 ≤ 170 is as follows:

Explanation of release of energy in the process of nuclear fusion of two light nuclei is as follows. Let us Consider two very light nuclei (A<10) joining to form a heavier nucleus. The

binding energy per nucleon of the fused heavier nuclei is more than the binding energy

per nucleon of the lighter nuclei. This means that the final system is more tightly

bound than the initial system. Again energy would be released in such a process of

fusion. A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to

that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120

nuclei, nucleons get more tightly bound. This implies energy would be released in the

process. It has very important implications for energy production through

Fission.

16. Write the relation for Binding Energy (BE) (in MeV) of a nucleus of mass number (A) in terms of the masses of its constituents namely neutrons and protons.

2014

Ans: The relation is:

BE = [𝑍𝑚𝑝 + (𝐴 − 𝑍)𝑚𝑛 − 𝑀𝑍𝐴 ] × 𝑐2.

Where, M is mass of nucleus, 𝑚𝑝 mass of proton and 𝑚𝑝 mass of neutron.

17. Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?

2014

Ans: The electrostatic force of attraction, Fe between the revolving electrons and the

nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus,

for a dynamically stable orbit in a hydrogen atom.

Thus the relation between the orbit radius and the electron velocity is:

The kinetic energy (K) and electrostatic potential energy (U) of the electron in

hydrogen atom are

Thus the total energy E of the electron in a hydrogen atom is:

The total energy of the electron is negative. This implies the fact that the electron is

bound to the nucleus. If E were positive, an electron will not follow a closed orbit

around the nucleus.

18. Using Bohr’s postulate of the atomic model, derive the expression for the total energy of the electron in the stationary orbit of hydrogen atom.

2014

The electrostatic force of attraction, Fe between the revolving electrons and the

nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus,

for a dynamically stable orbit in a hydrogen atom.

The relation between 𝑣𝑛and 𝑟𝑛 is:

Combining it with the eq:

,

Thus we get the following expressions for 𝑣𝑛and 𝑟𝑛.

and

The total energy of the electron in the stationary states of the hydrogen atom can be

obtained by substituting the value of orbital radius:

19. Using Bohr’s postulate of the atomic model, derive the expression for the radius of nth electron orbit. Hence, obtain the expression for Bohr’s radius.

2014

Ans: The electrostatic force of attraction, Fe between the revolving electrons and the

nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus,

for a dynamically stable orbit in a hydrogen atom.

The relation between 𝑣𝑛and 𝑟𝑛 is:

Combining it with the eq:

,

Thus we get the following expressions for 𝑣𝑛and 𝑟𝑛.

and

20. The value of ground state energy of hydrogen atom is -1.6 eV. (a) Find the energy required to move an electron from the ground state to the

first excited state of the atom. (b) Determine the kinetic energy and orbit radius in the first excited state of the

atom. ( Bohr’s radius is 0.53 𝐴0)

2014

Ans: a) Energy of first excited state is: 𝐸2 = −3.4 𝑒𝑉. So the energy required to move an electron from the ground state to the first excited state of the atom is: ∆𝐸 = 𝐸2 − 𝐸1 = −3.4 + 13.6 = 10.2𝑒𝑉.

b) Kinetic energy in the first excited state of the atom is 3.4eV and the orbit radius in the first excited state of the atom is 4 × 5.29 × 10−11𝑚.

21. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10−11m. calculate its radius in n=3 orbit.

2014

Ans: The radius of the n=3 electron orbit of a hydrogen atom is 32 ×5.3 × 10−11m. That is 𝑟3= 47.5 × 10−11m.

22. The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its kinetic and potential energy in this state.

2014

Ans: Kinetic energy = 3.4eV and potential energy = - 6.8 eV.

23. When is the 𝐻𝛼- line of the Balmer series in the emission spectrum of hydrogen atom obtained?

2013

Ans: The 𝐻𝛼- line of the Balmer series in the emission spectrum of hydrogen atom is obtained when the electron jumps from third orbit to second orbit of hydrogen atom.

24. In the ground state of hydrogen atom, its Bohr radius is 5.3 × 10−11m. The atom is excited such that the radius becomes 21.2 × 10−11m. Find (a) the value of the principle quantum number and (b) the total energy of the atom in this excited state.

2013

Ans: (a) The value of the principle quantum number 𝑛2 =21.2

5.3 = 4. So, n = 2

(b) the total energy of the atom in this excited state is E = −13.6

4 eV = - 3.4eV.

25. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence, draw the energy level diagram showing how the spectra corresponding to Balmer series occur due to transition between energy levels.

2013

Ans: The electrostatic force of attraction, Fe between the revolving electrons and the

nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus,

for a dynamically stable orbit in a hydrogen atom.

The relation between 𝑣𝑛and 𝑟𝑛 is:

Combining it with the eq:

,

Thus we get the following expressions for 𝑣𝑛and 𝑟𝑛.

and

The total energy of the electron in the stationary states of the hydrogen atom can be

obtained by substituting the value of orbital radius:

The energy level diagram of hydrogen atom Different spectral series obtained are

shown below:

26. The number of nuclei of a given radioactive sample at time t=0 and t=T are 𝑁0 and 𝑁0 𝑛⁄ , respectively. Obtain an expression for the half – life (𝑁1

2⁄ ) of the nucleus in

terms of n and T.

203

Ans: According to law of radioactivity, we have-

N= 𝑁0𝑒−𝑡 and at t = T, N = 𝑁0

𝑛 .

So, 𝑁0

𝑛 = 𝑁0𝑒−𝑡. Or, n = 𝑒−𝑡.

Or, = 𝑙𝑜𝑔(𝑛)

𝑇.

Again, 𝑇12⁄ =

0.693

=

0.693 𝑇

𝑙𝑜𝑔(𝑛).

27. Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the region where potential energy is (i) positive and (ii) negative.

2013

Ans: Graph showing the variation of potential energy between a pair of nucleons as a function of their separation

Here potential energy is positive in between 0 and the vertical dotted line and negative beyond that.

28. Write the basic nuclear process of nucleus undergoing 𝛽 − decay. Why is the detection of neutrinos found very difficult?

2013

Ans: During 𝛽 , either a proton or a neutron changes to a neutron or a proton respectively. The basic processes are as follows: In beta-minus decay, a neutron transforms into a proton within the nucleus according

to

whereas in beta-plus decay, a proton transforms into neutron (inside the nucleus) via

Neutrinos are neutral particles with very little mass, so they interact with matters very feebly. So detection of neutrinos found very difficult.

29. Why is the classic (Rutherford) model for an atom of electron orbiting around the nucleus not able to explain the atomic structure?

2012

Ans: According to classical electromagnetic theory, an accelerating charged particle emits

radiation in the form of electromagnetic waves. The energy of an accelerating electron

should therefore, continuously decrease. The electron would spiral inward and

eventually fall into the nucleus. Thus, such an atom cannot be stable. Further,

according to the classical electromagnetic theory, the frequency of the electromagnetic

waves emitted by the revolving electrons is equal to the frequency of revolution. As

the electrons spiral inwards, their angular velocities and hence their frequencies would

change continuously, and so will the frequency of the light emitted. Thus, they would

emit a continuous spectrum, in contradiction to the line spectrum actually observed.

Clearly Rutherford model tells only a part of the story implying that the classical ideas

are not sufficient to explain the atomic structure.

30. In hydrogen atom, an electron undergoes transition from second excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong. Find out the ratio of the wavelength of the emitted radiation in the two cases.

2012

Ans: Transition from second to first orbit results to Balmer series and again transition from first to ground state results to Lyman series.

The ratio of the wavelength of the two transition is: 1

2=

𝐸3− 𝐸2

𝐸2− 𝐸1 =

1.9

10.2 =

19

102

31. Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n-times the de-Broglie wavelength associated with it.

2012

Ans: From Bohr’s second postulate of quantization of orbital angular momentum we have-

L = nrv = n ℎ

2𝜋.

Or, 2𝜋𝑟 = n ℎ

𝑚𝑣 = n

ℎ

𝑝 = n .

This shows that the circumference of the electron in the nth orbital state in hydrogen atom is n-times the de-Broglie wavelength associated with it

32. The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

2012

Ans: No. of spectral lines emitted is given by the formula: N = 𝑛(𝑛−1)

2.

Here n = 3, so, N = 3(3−1)

2 =

3 ×2

2 = 3.

Three spectral lines which can be emitted when it finally moves to the ground state.

33. The ground state energy of hydrogen atom is -13.6 eV. If the makes a transition from an energy level -0.86 eV to -1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

2012

Ans: We know, ℎ𝑐

= ∆𝐸 = (−0.86 + 1.51) × 1.6 × 10−19) = 0.69 × 1.6 × 10−19. .

Or, = 6.63 × 10−34 × 3 × 108

0.69 ×1.6 × 10−19 = 18.02 × 10−7m.

34. In the Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an ∝ -particle of 8 MeV energy impinges on it before it comes to momentarily rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the

2012

∝ -particle is doubled?

Ans: Thus the distance of closest approach d is given by-

The kinetic energy of -particles is 8 MeV or 12.8 × 10

–12 J. Since 1/4𝜀0= 9.0 × 10

9

N m2/C

2. Therefore with e = 1.6 × 10

–19 C, we have,

d = 2 ×80 ×1.6 ×1.6 × 10−38×9 × 109

12.8 × 10−12 = 3.19 × 10−14 𝑚 = 31.9 fm.

When the kinetic energy of the ∝ -particle is doubled the distance of closest approach reduces to half of its previous value.

35. In a given sample, two radio isotopes A and B are initially present in the ratio 1:4. The half-lives of A and B are 100 years and 50 years, respectively. Find the time after which the amount of A and B becomes equal.

2012

Ans: We have the condition: 𝑁𝐴 = 𝑁𝐵 → 𝑁0𝐴𝑒−𝐴𝑡 = 𝑁0𝐵𝑒−𝐴𝑡 → 1𝑒−𝐴𝑡 = 4 𝑒−𝐴𝑡

→ 4 = 𝑒−(𝐴− 𝐵)𝑡 → ln(4) = −(

𝐴− 𝐵)𝑡. ………….(i)

Again, 𝐴 = ln 2

𝑇 =

𝑙𝑛 2

100 𝑎𝑛𝑑 𝐵 =

ln 2

𝑇 =

ln 2

50

→ 𝐵 = 2 𝐴 ………………………………..(ii)

From equations (i) and (ii), we get- ln(4) = 𝐴𝑡 → 𝑡 = ln 4

ln 2× 100 = 200 𝑦𝑒𝑎𝑟𝑠.

36. Why is the binding energy per nucleon found to be constant for nuclei in the range of mass number (A) lying between 30 and 170?

2012

Ans: The constancy of the binding energy in the range 30 < A < 170 is a consequence of the

fact that the nuclear force is short-ranged. Let us consider a particular nucleon inside a

sufficiently large nucleus. It will be under the influence of only some of its neighbours,

which come within the range of the nuclear force. If any other nucleon is at a distance

more than the range of the nuclear force from the particular nucleon it will have no

influence on the binding energy of the nucleon under consideration. If a nucleon can

have a maximum of p neighbours within the range of nuclear force, its binding energy

would be proportional to p. Let the binding energy of the nucleus be pk, where k is a

constant having the dimensions of energy. If we increase A by adding nucleons they

will not change the binding energy of a nucleon inside. Since most of the nucleons in a

large nucleus reside inside it and not on the surface, the change in binding energy per

nucleon would be small. The binding energy per nucleon is a constant and is

approximately equal to pk. The property that a given nucleon influences only nucleons

close to it is also referred to as saturation property of the nuclear force.

37. When a heavy nucleus with mass number A=240 breaks into two nuclei, A=120, energy is released in the process.

2012

Ans: Let us take a nucleus with A = 240 breaking into two fragments each of A = 120.

Then 𝐸𝑏𝑛 for A = 240 nucleus is about 7.6 MeV,

𝐸𝑏𝑛 for the two A = 120 fragment nuclei is about 8.5 MeV.

Therefore, Gain in binding energy for nucleon is about 0.9 MeV.

Hence the total gain in binding energy is 240×0.9 or 216 MeV