NEET-UG / JEE (Main) Challenger Physics Vol. - 2

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© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Printed at: India Printing Works, Mumbai

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For all Agricultural, Medical, Pharmacy and Engineering Entrance Examinations held across India.

Challenger

PHYSICS Vol. II

NEET – UG & JEE (Main)

Salient Features • Exhaustive coverage of MCQs under each sub-topic. • ‘1829’ MCQs including questions from various competitive exams. • Includes solved MCQs from NEET (UG), MHT-CET, JEE (Main) and

various entrance examinations from year 2015 to 2018. • Concise theory for every topic. • Hints provided wherever deemed necessary. • Test papers for thorough revision and practice. • Important inclusions: Mind Over Matter and Problems to Ponder.

TEID: 12821_JUP

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PREFACE Target’s ‘Challenger Physics: Vol-II’ is a compact guidebook, extremely handy for preparation of various competitive exams like NEET, JEE (Main). Features of each chapter: Theoretical Concepts presented in the form of pointers, tables and diagrams that form a vital part of any

competitive examination. Multiple Choice Questions segregated into two sections. Concept Building Problems – Contains questions of

various difficulty range and pattern. Practice Problems – Contains ample questions for thorough revision. Formulae section for each chapter according to its relevance for quick revision. Shortcuts section to help students save time while dealing with lengthy questions. Mind Over Matter: Tips/clues to help students understand and focus on the key-concept that is involved in

solving certain brain-racking questions. Problems to Ponder: Multiple questions, passages, MCQs / non- MCQs of different pattern created with the

primary objective of helping students to understand the application of various concepts of Physics. Two Model Test Papers are included to assess the level of preparation of the student on a competitive level. MCQs have been created and complied with the following objective in mind – to help students solve complex problems which require strenuous effort and understanding of multiple-concepts. The MCQs are a mix of questions based on high order thinking, theory, numerical, graphical, multiple concepts. The level of difficulty of the questions is at par with that of various competitive examinations like CBSE, AIIMS, CPMT, JEE, AIEEE, TS EAMCET (Med. and Engg.), BCECE, Assam CEE, AP EAMCET (Med. and Engg.) and the likes. Also to keep students updated, questions from most recent examinations such as AIPMT/NEET (UG), MHT-CET, K CET, GUJ CET, WB JEEM, JEE (Main) of years 2015, 2016, 2017 and 2018 are covered exclusively. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops. From, Publisher Edition: Second

Disclaimer This reference book is based on the NEET-UG syllabus prescribed by Central Board of Secondary Education (CBSE). We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the National Council of Educational Research and Training (NCERT). Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

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No. Topic Name Page No.

1 Electrostatics 1

2 Capacitors 57

3 Current Electricity 90

4 Magnetic Effect of Electric Current 138

5 Magnetism 188

6 Electromagnetic Induction and Alternating Current 222

7 Electromagnetic Waves 269

8 Ray Optics 289

9 Wave Optics and Interference of light 350

10 Diffraction and Polarisation of light 385

11 Dual Nature of matter and radiation 409

12 Atoms and Nuclei 436

13 Electronic Devices 474

14 Communication system 522

Model Test Paper - I 541

Model Test Paper – II 545

Note: ** marked section is not for JEE (Main)

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Why Challenger Series? Gradually, every year the nature of competitive entrance exams is inching towards conceptual

understanding of topics. Moreover, it is time to bid adieu to the stereotypical approach of solving a problem using a single conventional method. To be able to successfully crack the NEET and JEE (Main) examination, it is imperative to develop skills such as data interpretation, appropriate time management, knowing various methods to solve a problem, etc. With Challenger Series, we are sure, you’d develop all the aforementioned skills and take a more holistic approach towards problem solving. The way you’d tackle advanced level MCQs with the help of hints, tips, shortcuts and necessary practice would be a game changer in your preparation for the competitive entrance examinations.

What is the intention behind the launch of Challenger Series? The sole objective behind the introduction of Challenger Series is to severely test the student’s

preparedness to take competitive entrance examinations. With an eclectic range of critical and advanced level MCQs, we intend to test a student’s MCQ solving skills within a stipulated time period.

What do I gain out of Challenger Series? After using Challenger Series, students would be able to: a. assimilate the given data and apply relevant concepts with utmost ease. b. tackle MCQs of different pattern such as match the columns, diagram based questions, multiple

concepts and assertion-reason efficiently. c. garner the much needed confidence to appear for various competitive exams. Can the Questions presented in Problems to Ponder section be a part of the NEET/JEE (Main)

Examination? No, the questions would not appear as it is in the NEET/JEE (Main) Examination. However, there are fair chances that these questions could be covered in parts or with a novel question construction.

Why is then Problems to Ponder a part of this book? The whole idea behind introducing Problems to Ponder was to cover an entire concept in one question.

With this approach, students would get more variety and less repetition in the book.

Best of luck to all the aspirants!

Frequently Asked Questions

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Chapter 11: Dual Nature of Matter and Radiation

Dual nature of radiation: i. In various phenomena, radiations either

possess wave nature or quantum (particle) nature, this is known as dual nature of radiation.

ii. The phenomena such as interference, diffraction and polarisation can be explained on the basis of wave nature of radiation (light).

iii. The phenomena such as photoelectric effect, Compton effect etc. can be explained on the basis of quantum nature of radiation (light).

Electron emission: i. The phenomenon of emission of electrons

from the surface of a metal is called electron emission.

ii. Metals have free electrons which help them conduct. However these electrons need external energy to break the bond completely.

iii. This minimum energy required to free the electron from metal surface can be provided by following processes.

a. Thermionic emission: When metal undergoes heating at high temperatures, the electrons escape from the metal surface. This process is known as thermionic emission.

b. Field emission: Emission of electrons by applying strong electric field (of the order of 108 V/m) is termed as field emission.

c. Photoelectric emission: Electrons are emitted from the metal surface by illuminating surface by light of suitable frequency. This process is called as photoelectric emission.

Photoelectric effect: i. The emission of electrons from a metal

surface when illuminated by light or any other radiation of suitable wavelength (or frequency) is called photoelectric effect.

ii. The emitted electrons are called as photoelectrons and the current so produced is called photoelectric current or photocurrent.

iii. It is based on the principle of conservation of energy.

Experimental study of photoelectric effect: i. S is a monochromatic source of light of

known frequency and intensity I. ii. As the radiation from S is incident on the

emitter electrode E, the photoelectric current starts flowing through the circuit due to emission of photoelectrons.

iii. The emitted photoelectrons (e) are collected by the electrode C. Both the electrodes are enclosed in an evacuated glass envelope tube G with a quartz window (W) to permit the incidence of radiations.

11.0 Introduction 11.1 Photoelectric effect 11.2 Einstein’s photoelectric equation: Energy quantum of radiation 11.3 Wave nature of matter 11.4 Davisson and Germer experiment

Experimental arrangement

S

CA

V

E

W G

Rh

RK

B

+ +

+

e

Dual Nature of Matter and Radiation 11

Introduction 11.0 Photoelectric effect11.1

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Challenger Physics Vol - II (Med. and Engg.)

410

iv. A potential difference is applied between emitter E and collector C by the potential divider arrangement consisting of the battery B and rheostat Rh.

v. The reversing key RK enables to reverse the polarities of the two electrodes. The microammeter (A) measures the

photoelectric current Ip and the voltmeter (V) measures the potential of the collector C with respect to the emitter E.

vi. The photoelectric current can be increased or decreased by varying potential of the collector plate C.

Effect of frequency, intensity and potential difference on photoelectric current

Effect of frequency Effect of intensity Effect of potential difference

i. A suitable constant positive potential is applied to collector plate. The intensity of incident radiation is kept constant and the frequency of incident light is gradually increased from its minimum value.

ii. It is observed that no photoelectric current is recorded till a certain frequency 0 is reached.

iii. Threshold frequency (0): The minimum frequency of

incident radiation required to eject the electrons from the photosensitive material is called threshold frequency.

iv. Threshold wavelength (0): The maximum wavelength

required to release photo electrons from the surface of photosensitive material is called threshold wavelength.

v. Relation between threshold frequency and threshold wavelength is given by,

00

c

where, c = speed of light.

i. Keeping the frequency of incident radiation and accelerating potential fixed, the intensity of light is varied and the resulting photoelectric current is measured.

ii. The photoelectric current is directly proportional to the number of photoelectrons emitted per second.

iii. This implies that the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation.

i. The threshold frequency and intensity of incident radiation, both are kept constant at suitable value.

ii. It is found that photoelectric current increases with increase in positive (accelerating) potential.

iii. At some stage for certain positive potential of plate C, all the emitted electrons are collected by the plate C and photoelectric current becomes maximum. This maximum value of photoelectric current is called saturation current.

iv. After saturation state, the negative potential applied to the plate C with respect to plate E is reversed (using reversing key RK) and the applied negative potential of plate C is gradually increased till the photoelectric current reduces to zero. So, it is a negative potential at which photoelectric current becomes zero and is called as stopping potential.

v. Photoelectric current becomes zero when the stopping potential is sufficient to repel the most energetic photoelectrons with the maximum kinetic energy.

K.Emax = 12

m 2maxv = eV0

where, vmax = maximum velocity of

photoelectron, m = mass of electron, e = magnitude of charge on an

electron, V0 = stopping potential.

Phot

oele

ctric

cur

rent

0 Frequency () Intensity of light

Phot

oele

ctric

cur

rent

0

Collector plate potentialV0 0

Y

X X

Phot

oele

ctric

cur

rent

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Effect of frequency and intensity of stopping potential:

i. For a given frequency of incident radiation, the stopping potential and the maximum kinetic energy of photoelectrons are independent of the intensity of incident radiation [refer figure (a)].

ii. For different frequencies of incident

radiations, the different values of stopping potential are obtained but the saturation current remains invariant [refer figure (b)].

iii. The stopping potential becomes more

negative when the frequency of incident radiation is increased above threshold frequency.

i.e., V0

Note: K.E.max eV0 and V0

Thus, K.E.max Effect of photometal on stopping potential: i. For different photometals, the stopping

potential (V0) versus frequency () curves are parallel straight lines intercepting frequency axis at different points.

ii. The stopping potential and the threshold

frequency are different for different photometals.

iii. The slope (V0/) for all the metals is same and hence it is a universal constant.

Characteristics of photoelectric emission: i. The emission of photoelectrons starts as

soon as (within 108 s) light falls on metal surface.

ii. The emission of photoelectrons takes place only when the frequency of the incident radiation is above a certain critical value i.e., threshold frequency which is the characteristic frequency of that metal.

iii. The number of photoelectrons emitted from a metal surface (or photocurrent) depends only on the intensity of incident light and is independent of its frequency.

iv. The maximum kinetic energy with which photoelectrons are emitted from a metal surface depends only upon the frequency of the incident light and is independent of its intensity.

Photoelectric cell: An optical device which converts light energy

into electrical energy is called photoelectric cell or photocell.

Photocell is based on the principle of photoelectric effect.

Symbolically, it is represented as Construction: i. It consists of an evacuated glass tube or

bulb in which two electrodes, a cathode C and an anode A are fixed.

ii. The cathode is semicylindrical in shape and its concave surface is coated with a photosensitive material like sodium or caesium.

C A

Collector plate potential Figure (a)

I1

I2

I3

I1 < I2 < I3

Phot

oele

ctric

cu

rrent

V0 0

Y

XX

Phot

oele

ctric

cu

rrent

X X

Y

01V

3

1<2 <3

Collector plate potentialFigure (b)

0 02V03

V

Saturation current

2 1

Metal 1

Metal 2

0 2 0

V0

01V

02V

01

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412

iii. The anode is in the form of platinum rod or wire which is fixed along the axis of the cathode.

iv. The cathode is connected to the negative

terminal and the anode to the positive terminal of a strong d.c. source as shown in the figure.

Working: i. When a suitable radiation falls on the

concave surface of the cathode, it emits photoelectrons. These photoelectrons are attracted by the anode which is at a positive potential with respect to the cathode.

ii. The electrons pass into the external circuit causing an electric current which can be measured from the microammeter (A) connected in series with the anode.

Applications of photocell: i. In exposure meter. ii. In burglar alarm. iii. In lux meter. iv. In smoke detector. Failure of wave theory of light to explain

photoelectric effect: i. Wave theory fails to explain that in

photoelectric effect, K.E.max eV0 and is independent of intensity of incident radiations.

ii. Wave theory cannot explain the experimentally confirmed existence of threshold frequency (0) in photoelectric effect.

iii. Wave theory fails to explain that the photoelectric emission is (nearly) an instantaneous process (which is an experimentally observed fact).

Particle nature of light: The photon i. As the wave theory of light fails to explain

photoelectric effect satisfactorily, the particle nature of light is considered.

ii. It is assumed that the energy of light radiations is not continuous but it is discrete i.e., distributed in the form of quantas (packets of energy) known as photons.

iii. The energy possessed by all the photons in the radiation of a particular frequency (or wavelength) is same irrespective of intensity of the radiation.

K.E.max iv. Thus, existence of threshold frequency is

supported by particle nature of light. v. As the photon of radiation with suitable

frequency falls on metal surface, it is absorbed by the electron immediately and photoelectron is ejected out of the metal instantaneously (within 10–8 s).

Characteristics of photon: i. A photon has energy E (= h = hc/) and

momentum p (= h/c = h/). where, h = Planck’s constant. Energy of photon is expressed in electron

volt (eV). ii. The photons are electrically neutral. iii. The photons travel with the speed of light. iv. The velocity of photon is different in

different media. v. Frequency of photon remains invariant in

different media. Einstein’s photoelectric equation: i. Einstein assumed that the photoelectric

effect results from the collision between photon of a incident radiation and electron of a material.

ii. During this collision, total energy and total momentum remains conserved but the numbers of photons are not conserved.

iii. The photon loses its energy (E = h) in two parts, firstly in liberation of electron from metal surface and secondly in the form of kinetic energy of liberated electrons.

iv. The minimum energy required to free (liberate) an electron from the metal surface is called photoelectric work function () of that metal surface.

v. The law of conservation of energy gives, Photon energy = Maximum K.E. of

photoelectron + work function of metal surface.

A C

E

light

A +

+

Photoelectric cell Rh

Einstein’s photoelectric equation: Energy quantum of radiation

11.2

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h = K.E.max +

h = 12

2maxmv + h0

12

2maxmv = h( 0) ….(i)

or eV0 = h( 0) .…(ii)

But = c

; 0 = 0

c

eV0 = hc 0

1 1

….(iii)

Matter waves: i. The waves associated with every moving

particle are called as matter waves (or de-Broglie waves).

ii. According to de-Broglie, a moving particle always has a wave associated with it and the particle is controlled by the wave (in a manner similar to that a photon is controlled by the waves).

iii. Like radiations, matter also has a dual characteristic (i.e., wave-like and particle-like).

De-Broglie wave relation: i. The de-Broglie wavelength () associated

with the particle is given by,

= hp

=

hmv

For photon, = hmc

where, p = momentum of particle, m = mass of particle, v = velocity of particle. ii. The matter waves are associated with

material particles only if they are in motion.

i.e., 1v

, if v = 0, =

iii. Smaller the mass of the particle, higher is the wavelength associated with it.

i.e., 1m

iv. Wavelength associated with a material particle is independent of the charge of the particle.

De-Broglie wavelength of an electron: i. Energy (E) acquired by an electron of mass

m and charge e, when accelerated through a potential difference V, is given by

E = eV ….(i)

ii. If v is the velocity of electron, then

E = 12

mv2

or v = 2Em

….(ii)

iii. The de-Broglie wavelength (e) of electron is given by,

e = hmv

= h2Emm

or

e = h2mE

….(iii)

iv. Substituting E = eV from (i), we get

e = h2meV

….(iv)

v. Using, m = 9.1 1031 kg, e = 1.6 1019 C and h = 6.62 1034 Js, we get

e = 12.27V

1010 m

e = 12.27V

Å ….(v)

vi. Equations (iii), (iv) and (v) are different forms of de-Broglie wavelength of an electron.

Ratio of de-Broglie wavelengths of photon and electron:

i. The de-Broglie wavelength of a photon (ph) of energy Eph and moving with speed of light (c) is given by,

ph = ph

hcE

ii. The de-Broglie wavelength of an electron (e) of energy Ee and mass m is given by,

e = e

h2mE

iii. The ratio of de-Broglie wavelengths is,

ph

e

=

ph

cE e2mE =

2e

2ph

2mc EE

Characteristics of matter waves: i. Matter wave represents the probability of

finding a particle in space. ii. Matter waves are not electro-magnetic in

nature. iii. De-Broglie or matter wave is independent

of the charge on the material particle. iv. Practical observation of matter waves is

possible only when the de-Broglie wavelength is of the order of the size of the particles.

v. The phase velocity of the matter waves can be greater than the speed of the light.

Wave nature of matter11.3

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414

vi. As matter waves can propagate in vacuum, they are not mechanical waves.

vii. The number of de-Broglie waves associated with nth orbital electron is n.

viii. Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with orbital electron.

Davisson and Germer Experiment: i. The experimental arrangement used to

scatter the slow moving electrons from a crystal is shown in figure (a) below.

ii. The intensity of scattered beam is

measured for different values of scattering angle (the angle between incident and scattered electron beam) by varying accelerating voltage from 44 V to 68 V.

iii. As shown in the graph, the intensity peak

goes on increasing as the accelerating voltage is increased.

iv. At accelerating voltage 54 V, there is a bump or sharp peak in the intensity of the scattered electrons for scattering angle = 50.

v. The appearance of sharp peak in a particular direction is due to constructive interference of electrons scattered from different regularly spaced layers of atoms of the crystal i.e., the diffraction of electrons also takes place.

vi. From figure (a),

= 12

(180 )

For = 50,

= 12

(180 50) = 65

vii. According to Bragg’s law, for first order diffraction maxima (n = 1),

2d sin = d can be found by X-Ray reflection by

nickel-crystal. Here, d = 0.91 Å = 2 0.91 sin 65 = 1.65 Å viii. For V = 54 V, de Broglie wavelength will

be,

= 12.27V

= 12.2754

= 1.67 Å

ix. The theoretical and experimental values of de Broglie wavelengths of an electron are approximately similar. Hence, Davisson and Germer experiment proved the existence of matter waves.

1. Maximum K.E. of electrons: 2

max1 mv2

= eV0

2. Einstein’s photoelectric equation: max 0 0K.E. eV h( )

2 0max max 0

0 0

1 1 1K.E. mv eV hc hc2

3. Maximum velocity of photoelectrons:

0max

2h( )vm

= 0

0

2hc( )m

4. Work function of a metal: = h K.E.max 5. Energy of photon: E = h 6. Effective (kinetic) mass of photon: M = 2

Ec

= 2

hc = h

c

7. Momentum of photon: p = mc

p = Ec

= hc = h

8. Number of emitted photons per second: n = P

E = P

h = P

hc

where, P = power (of source used for illumination),

E = Energy of each photon.

Davisson and Germer Experiment11.4

Formulae

I

50 Figure (b)

0

+

Diffracted electron beam

Electron beam

To galvanometer

Movable collector

Electron gun

S1 S2 Nickel Target

L.T

H.T

Figure (a)

Electron detector

Evacuatedchamber

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9. de-Broglie wavelength of matter waves:

= hp

= hmv

and 1v

, 1m

, 1p

10. de-Broglie wavelength of an electron:

e =h

2mE= h

2meV=

12150 Å

V

= 12.27V

Å 11. Ratio of wavelength of photon to wavelength

of electron:

ph

e

= e

ph

c 2mEE

= 2

e2ph

2mc EE

12. Bragg’s law: 2d sin = n where, n = order of diffraction For n = 1, 2d sin = 1. The work function of metal decreases as the

atomic number of element increases. 2. When temperature increases, work function

decreases. 3. For alkali metals, the threshold frequency lies in

the visible region. For zinc, it lies in the ultraviolet region.

4. Intensity of incident radiation is inversely

proportional to the square of distance between source of light and photosensitive plate

I 2

1d

But photoelectric current, i I

i 2

1d

5. Photoelectric current can be increased by filling

an inert gas into the glass tube. The photoelectrons emitted by cathode ionize the gas by collision and hence the current is increased.

6. Infrared radiations cannot eject photoelectrons while X-rays always eject photoelectrons.

7. The photoelectrons emitted from the metallic

surface have different kinetic energies even when the incident photons have same energy.

8. The de-Broglie wavelength associated with the orbital electron in the nth orbit of hydrogen atom is given by,

0 = 12.27V

= 2

12.2713.6n

= n 12.2713.6

= 3.3(n) Å

9. The de-Broglie wavelength associated with gas molecules varies inversely as the square root of

temperature 1T

.

10. Expression for the wavelength associated with

charged particles accelerated through a potential difference V.

i. Electron: e = 12.27V

Å

ii. Proton : p = 0.286V

Å

iii. Deuteron : d = 0.202V

Å

iv. -particle : = 0.101V

Å

11. The de-Broglie wavelength of the neutron

having kinetic energy K electron volt is given by

0 = 0.286K

Å 12. The amplitude of the matter waves gives the

probability of existence of the particle at that point.

13. Matter waves travel as a wave packet with

decreasing amplitude on either side of the present position.

14. Matter waves propagate as wave packet with

group velocity vg = d/dk, where = 2 and k = 2/. Note that for electromagnetic waves, the wave velocity or the phase velocity, c = /k.

15. Out of photon and electron having same de-

Broglie wavelength, the total energy of electron is greater than photon and kinetic energy of photon is greater than that of electron.

16. The value of i. for h = 6.63 1034 Js,

hce

= 1243.1 109 V-m

ii. for h = 6.6 1034 Js ,

hce

= 1237.5 109 V-m

17. The slope of the graph between stopping

potential and frequency gives the ratio of Planck’s constant to electronic charge (h/e).

18. The slope of the graph between kinetic energy

and frequency gives the Planck’s constant (h).

Shortcuts

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416

1. In photoelectric experiment, if the intensity of

the incident light is doubled and the frequency is quadrupled, then the saturation current

(A) remains same. (B) decreases.

(C) increases. (D) cannot be predicted.

2. When yellow light is incident on a surface, no electrons are emitted while green light can emit. If red light is incident on the surface, then

(A) No electrons are emitted (B) Photons are emitted (C) Electrons of higher energy are emitted (D) Electrons of lower energy are emitted 3. Assertion: If the intensity of incident light is

doubled, then the number of photoelectrons emitted per unit time would not be doubled.

Reason: The value of photoelectric efficiency is equal to 1.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 4. The variation of photo-current with collector

potential for different frequencies of incident radiation 1, 2, and 3 is as shown in the graph, then [K CET 2016]

(A) 1 = 2 = 3 (B) 1 > 2 > 3

(C) 1 < 2 < 3 (D) 3 = 1 2

2

5. Following graphs show the variation of stopping potential corresponding to the frequency of incident radiation (F) for a given metal. The correct variation is shown in graph (0 = Threshold frequency)

[MHT CET 2018] (A) (1) (B) (2) (C) (3) (D) (4) 6. The anode voltage of a photocell is kept fixed.

The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows: (A) (B)

(C) (D) 7. The maximum kinetic energy of

photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 3 eV. The stopping potential in volts is

(A) 2 (B) 3 (C) 6 (D) 9 8. A photo cell is receiving light from a source

placed at a distance of 1 m. If the same source

is to be placed at a distance of 12

m, then the

number of ejected electrons (A) will reduce to zero. (B) will remain same. (C) will decrease. (D) will increase.

Concept Building Problems

I

O

O

I

Collector Potential

3 2 1

V03 V02 V01

Saturation Current

Retarding Potential

Photo Current

The key to crack this question lies incomprehending that, photoelectric efficiencyis the ratio of number of photoelectronsemitted per second to the number of photonsfalling per second.

Mind over Matter

Photoelectric effect 11.1

O

I

O

I

0 0 F

stopp

ing

pote

ntia

l

(1) 0 F

stopp

ing

pote

ntia

l

(2)

0 0 F

stopp

ing

pote

ntia

l

(4) F

stopp

ing

pote

ntia

l

(3) 0

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9. If E and P are the energy and the momentum of a photon respectively, then on reducing the wavelength of photon [BCECE 2015]

(A) both P and E will decrease (B) both P and E will increase (C) P will increase but E will decrease (D) P will decrease but E will increase 10. _______ is the wavelength of photon of

energy 35 keV. [GUJ CET 2018] (A) 35 10–12 m (B) 35 Å (C) 3.5 nm (D) 3.5 Å 11. If the energy of photons corresponding to

wavelength of 6000 Å is 3.2 10–19 J. The photon energy for wavelength of 4000 Å will be _______. [GUJ CET 2018]

(A) 4.44 10–19 J (B) 2.22 10–19 J (C) 1.11 10–19 J (D) 4.80 10–19 J 12. The threshold wavelength for photoelectric

emission from a material is 5000 Å. 1.25 1020 photoelectrons will be emitted per second when this metal is illuminated with monochromatic radiation from

(A) 50 watt infrared lamp (B) 50 watt ultraviolet lamp (C) 1 watt infrared lamp (D) 1 watt ultraviolet lamp 13. The work function for aluminium surface is

4.2 eV and that for sodium surface is 2.0 eV. The two metals were illuminated with appropriate radiations so as to cause photo emission. Then

(A) both aluminium and sodium will have the same threshold frequency.

(B) the threshold frequency of aluminium will be more than that of sodium.

(C) the threshold frequency of aluminium will be less than that of sodium.

(D) the threshold wavelength of aluminium will be more than that of sodium.

14. Match the following Columns: [ = Work function, I = Intensity, = Frequency, V0 = Stopping potential]

I II i. If 0 increases, and I

are kept constant then a. V0 remains

same ii. If I increases, and 0

are kept constant then b. V0 increases

iii. If increases, I and 0 are kept constant then

c. V0 decreases

(A) (i) c; (ii) b; (iii) a (B) (i) b; (ii) a; (iii) c (C) (i) c; (ii) a; (iii) b (D) (i) b; (ii) c; (iii) a 15. For a metal M, value of work function is 0

and the threshold wavelength is 0 = 620 nm. For which of the incident radiations given below, there will not be any photoemission possible?

(A) Infrared radiation (B) Red coloured visible radiations (C) Ultraviolet radiations (D) Microwave radiations. 16. Light of wavelength ‘λ’ which is less than

threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased then stopping potential will

[MHT CET 2016] (A) increase (B) decrease (C) be zero (D) become exactly half 17. For photoelectric emission from certain metal,

the cut-off frequency is . If radiation of frequency 4 impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

(A) hν(3m)

(B) hνm

(C) 2hνm

(D) 6hνm

18. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelength of light cannot be used for photoelectric effect.

(Consider, Planck constant = 4 × 1015 eVs, velocity of light = 3 × 108 m/s)

[WB JEEM 2015] (A) 510 nm (B) 650 nm (C) 400 nm (D) 570 nm 19. The photoelectric work function of a surface is

2.2 eV. The maximum kinetic energy of photoelectrons emitted when light of wavelength 6200 Å is incident on the surface

[BCECE 2015] (A) 0.4 eV (B) 1.2 eV (C) 1.6 eV (D) Photoelectrons are not emitted


11.2

SAMPLE C

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418

20. Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is [NEET P-II 2016]

(A) 3 V (B) +3 V (C) +4 V (D) 1 V 21. On a photosensitive material, when frequency

of incident radiation is increased by 30%, kinetic energy of emitted photoelectrons increases from 0.4 eV to 0.9 eV. The work function of the surface is [MHT CET 2017]

(A) 1 eV (B) 1.267 eV (C) 1.4 eV (D) 1.8 eV 22. The work function of a metal is 1 eV. Light of

wavelength 4000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be

(A) 0.86 m/s (B) 0.86 103 m/s (C) 0.86 104 m/s (D) 0.86 106 m/s 23. The photoelectric threshold wavelength of

silver is 3250 1010 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 1010 m is: (Given h = 4.14 1015 eVs and c = 3 108 ms1) [NEET (UG) 2017]

(A) 6 106 ms1 (B) 0.6 106 ms1 (C) 61 103 ms1 (D) 0.3 106 ms1 24. The radiation of wavelength 332 nm is

incident on a metal of work function 1.7 eV. The value of the stopping potential will be

(A) 0.70 V (B) 1.14 V (C) 1.68 V (D) 2.04 V 25. Light of two different frequencies whose

photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

[K CET 2015] (A) 1 : 2 (B) 1 : 5 (C) 1 : 1 (D) 1 : 4 26. When the light of frequency 20 (where 0 is

threshold frequency), is incident on a metal plate, then the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 50, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is [NEET (UG) 2018]

(A) 1 : 2 (B) 1 : 4 (C) 4 : 1 (D) 2 : 1

27. Radiation of wavelength , is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3

4 ,

the speed of the fastest emitted electron will be [JEE (Main) 2016]

(A) < v244

3

(B) = v124

3

(C) = v123

4

(D) >v244

3

28. The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency of the incident radiation is doubled. The work function of the metal is [Assam CEE 2015]

(A) 0 (B) h / 3 (C) h / 2 (D) 2h / 3 29. When light of frequency 1 is incident on a

metal with work function (where h1 > ), the photocurrent falls to zero at a stopping potential of V1. If the frequency of light is increased to 2, the stopping potential changes to V2. Therefore, the charge of an electron is given by [WB JEE 2017]

(A) 2 1

1 2 2 1V V

(B) 2 1

1 1 2 2V V

(C) 2 1

1 2 2 1V V

(D) 2 1

2 2 1 1V V

30. A photoelectric surface is illuminated

successively by monochromatic light of

wavelength λ and 2

. If the maximum kinetic

energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is:

(h = Planck’s constant, c = speed of light) [AIPMT Re-Test 2015]

(A) hc3

(B) hc2

(C) hc

(D) 2hc

31. When a metallic surface is illuminated with radiation of wavelength , the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2, the stopping potential is V

4. The threshold

wavelength for metallic surface is [NEET P-I 2016]

(A) 52 (B) 3 (C) 4 (D) 5

SAMPLE C

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32. A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength for this surface for photoelectric effect is [AIPMT 2015]

(A) 6 (B) 4

(C) 4 (D)

6

33. The de-Broglie hypothesis of matter waves

supports (A) Bohr’s concept of stationary orbits. (B) Kepler’s concept of elliptical orbits. (C) Einstein’s concept of energy–mass

relation. (D) None of the above. 34. The energy that should be added to an electron

to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is [TS EAMCET (Engg.) 2017]

(A) four-times the initial energy (B) equal to the initial energy (C) two-times the initial energy (D) three-times the initial energy 35. Which of the following figures represent the

variation of particle momentum and the associated de-Broglie wavelength?

[AIPMT 2015] (A) (B) (C) (D) 36. An electron of mass m with an initial velocity

v

= v0 i (v0 > 0) enters an electric field

E

= E0 i (E0 = constant > 0) at t = 0. If 0 is its de-Broglie wavelength at time t is

[NCERT Exemplar; NEET (UG) 2018]

(A) 0

0

0

eE t1m v

(B) 0 0

0

eE t1mv

(C) 0 (D) 0t

37. An electron is moving with an initial velocity

v

= v0 i

and is in a magnetic field B

= B0 j

. Then, its de-Broglie wavelength

[NCERT Exemplar] (A) remains constant. (B) increases with time. (C) decreases with time. (D) increases and decreases periodically. 38. An electron beam is accelerated by a potential

difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If min is the smallest possible wavelength of X-ray in the spectrum, the variation of log min with log V is correctly represented in: [JEE (Main) 2017]

(A) (B) (C) (D) 39. Electrons of mass m with de-Broglie

wavelength fall on the target in an X-ray tube. The cutoff wavelength (0) of the emitted X-ray is [NEET P-II 2016]

(A) 0 = (B) 0 = 22mc

h

(C) 0 = 2hmc

(D) 0 = 2 2 3

2

2m ch

p

p


p

p

log min

log V

log min

log V

log min

log V

log min

log V

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420

Mind over Matter

The key to crack this question lies incomprehending that, kinetic energy of thermalneutrons is KBT.

40. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

[NCERT Exemplar, K CET 2017] (A) H (B) H1/2 (C) H0 (D) H1/2

41. A photo sensitive metallic surface emits

electrons when X-rays of wavelength ‘’ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, m is mass of the electron. h-Planck’s constant, c-velocity of light) [AP EAMCET (Engg.) 2016]

(A) 2mch

(B) h2mc

(C) mch

(D) hmc

42. The energy of an electron having de-Broglie

wavelength ‘’ is (h = Planck’s constant, m = mass of electron) [MHT CET 2018]

(A) h2m

(B) 2

2h

2m

(C) 2

2 2h

2m (D)

2

2h

2m 43. Find the de-Broglie wavelength of an electron

with kinetic energy of 120 eV. [K CET 2015] (A) 112 pm (B) 95 pm (C) 124 pm (D) 102 pm 44. If an electron has an energy such that its

de-Broglie wavelength is 5500Å, then the energy value of that electron is

(h = 6.6 1034 Js, mc = 9.1 1031 kg) [TS EAMCET (Engg.) 2015] (A) 8 1020 J (B) 8 1010 J (C) 8 J (D) 8 1025 J 45. If the momentum of an electron is increased

by Pm then the de Broglie wavelength associated with it changes by 0.5 %. Then the initial momentum of the electron is

[AP EAMCET (Med.) 2016] (A) 100Pm (B) mP

100

(C) 200Pm (D) mP200

46. According to de-Broglie hypothesis, the wavelength associated with moving electron of mass ‘m’ is ‘e’. Using mass energy relation and Planck’s quantum theory, the wavelength associated with photon is ‘P’. If the energy (E) of electron and photon is same then relation between ‘e’ and ‘P’ is

[MHT CET 2017]

(A) p e (B) p 2

e

(C) p e

(D) p e

1

47. An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is

[NEET P-I 2016]

(A) 12c 2mE (B)

121 2m

c E

(C) 121 E

c 2m

(D) 12E

2m

(c being velocity of light) 48. If an alpha particle and a deuteron move with

velocity v and 2v respectively, the ratio of their de-Broglie wavelength will be _______.

[GUJ CET 2015] (A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 2 : 1 49. An electron of mass ‘m’ has de-Broglie

wavelength ‘λ’ when accelerated through potential difference ‘V’. When proton of mass ‘M’, is accelerated through potential difference ‘9V’, the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)

[MHT CET 2016]

(A) λ M3 m

(B) λ M.3 m

(C) λ m3 M

(D) λ m.3 M

50. The de-Broglie wavelength of thermal neutrons at 27 C will be [BCECE 2015]

(A) 1.77 Å (B) 1.77 mm (C) 1.77 cm (D) 1.77 m 51. The energy that should be added to an electron

to reduce its de Broglie wavelength from 4.5 1010 m to 1.5 1010 m will be

(A) Four times the initial energy (B) Thrice the initial energy (C) Equal to the initial energy (D) Eight times the initial energy

SAMPLE C

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52. The de Broglie wavelength of an electron

accelerated to a potential of 400 V is approximately [K CET 2016]

(A) 0.03 nm (B) 0.04 nm (C) 0.12 nm (D) 0.06 nm 53. In Davisson-Germer experiment the decrease

of the wavelength of the electron wave was done by [TS EAMCET (Med.) 2015]

(A) keeping more distance between the anode and filament.

(B) keeping the same potential difference between anode and filament.

(C) decreasing the potential difference between anode and filament.

(D) increasing the potential difference between anode and filament.

54. Consider figure given below. Suppose the

voltage applied to A is increased. The diffracted beam will have the maximum at a value of that [NCERT Exemplar]

(A) will be larger than the earlier value. (B) will be the same as the earlier value. (C) will be less than the earlier value. (D) will depend on the target. 55. The distance between two parallel planes of a

crystal is 3 Å. What should be the wavelength of X-ray used for getting the first order Bragg reflection at a glancing angle of 14.7?

[Assam CEE 2015] (A) 1 Å (B) 1.3 Å (C) 2.5 Å (D) 1.5 Å

56. The number of photons falling per second on a

completely darkened plate to produce a force of 6.62 10–5 N is ‘n’. If the wavelength of the light falling is 5 10–7 m, then

n = _______ 1022. (h = 6.62 10–34 J-s) [K CET 2018] (A) 1 (B) 5 (C) 0.2 (D) 3.3 57. The de-Broglie wavelength of a neutron in

thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is:

[NEET (UG) 2017, Similar in GUJ CET 2015]

(A) hmkT

(B) h3mkT

(C) 2h3mkT

(D) 2hmkT

58. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is: [AIPMT Re-Test 2015]

(A) ≤ 2.8 × 10–12 m (B) < 2.8 × 10–10 m (C) < 2.8 × 10–9 m (D) ≥ 2.8 × 10–9 m 59. The number of de-Broglie wavelengths

contained in the second Bohr orbit of hydrogen atom is [WB JEE 2016]

(A) 1 (B) 2 (C) 3 (D) 4 60. A particle A of mass m and initial velocity v

collides with a particle B of mass m2

which is

at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths A to B after the collision is: [JEE (Main) 2017]

(A) A

B

23

(B) A

B

12

(C) A

B

13

(D) A

B

2

61. An electron microscope uses electrons

accelerated by 100 kV. If other factors (such as numerical aperture, etc) are unchanged, then the ratio of limit of resolution of electron microscope to that of optical microscope which uses redlight of 646 nm is approximately

(A) 6 10–6 (B) 4 10–5

(C) 4 10–6 (D) 6 10–5

Miscellaneous Davisson and Germer Experiment11.4

+

Diffracted electron beam

Nickel target

Movable collector

To galvanometer

LT

HT

Electron gun Vacuum chamber

Electron beamA

The key to crack this question lies in comprehending that, electrons moving with large velocities are used in electron microscope in the same manner as beam of light is used in optical microscope.

Mind over Matter

SAMPLE C

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422


422

Practice Problems

Photoelectric effect 11.1

62. A metallic sphere of radius 2 cm and work function 3.6 eV is suspended from an insulating thread in free-space. It is continuously irradiated with light of wavelength 100 nm. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A 10n (where 1 < A < 10). The value of n is

(A) 6 (B) 7 (C) 8 (D) 9 63. A metallic surface of work function 1.2 eV is

illuminated with a light of wavelength 500 nm. Assume an electron makes three collisions before being emitted and in each collision 12% of initial energy is lost. Find the K.E. of this electron as it comes out of the metal.

(A) 2.2 eV (B) 1.2 eV (C) 1.5 eV (D) 0.5 eV 64. A saturation current of I A is shown by a

photometal when source is at a distance of 9 m from it. If source is moving towards the photometal at a rate of 0.25 m for every 4 s, then what will be the relation between the saturation current I and saturation current I after 24 s?

(A) I = 1.63 I (B) I = 1.2 I (C) I = 1.44 I (D) I = I 1. In a photon-electron collision, _______ are

conserved. (i) number of photons (ii) total energy (iii) total momentum (A) (i) and (ii) (B) (ii) and (iii) (C) (i) and (iii) (D) (i), (ii) and (iii) 2. The following graph shows the variation of

stopping potential V0 as a function of frequency for two different metallic surfaces A and B. The work function of A, as compared to that of B is

(A) equal (B) less (C) more (D) cannot be judged 3. A photon in motion has a mass

(A) ch

(B) h

(C) h (D) 2

hc

4. If mean wavelength of light radiated by 200 W

lamp is 3500 Å, then number of photons radiated per second are

(A) 3 1023 (B) 3.5 1020 (C) 2.5 1020 (D) 5 1023

5. The momentum of a photon of energy 103 eV

in kg m/s will be (A) 5 1025 (B) 0.33 106 (C) 7 1024 (D) 1022

6. An important spectral emission line has a

wavelength of 24 cm. The corresponding photon energy is (h = 6.62 1034Js, c = 3 108 m/s)

(A) 5.2 104 eV (B) 5.2 106 eV (C) 5.2 108 eV (D) 11.8 106 eV 7. Threshold frequency for a metal is 8 1014 Hz.

Light of = 5000 Å falls on its surface. Which of the following statements is correct?

(A) No photoelectric emission takes place. (B) Photoelectrons come out with zero speed. (C) Photoelectrons come out with 103 m/s

speed. (D) Photoelectrons come out with 105 m/s

speed. 8. Assertion: Number of photoelectrons emitted

always increases linearly with the intensity of light. Reason: When incident light has frequency lesser than threshold frequency of metal then no photo electrons are emitted.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True.

V0

A B

SAMPLE C

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11.2 9. If certain metals (Cs, Al, Cu, Pb and Ni)

undergo photoelectric emission, then which type of radiations give photoemission for all metals?

( Cs = 2.14 eV, A l = 4.28 eV, Cu = 4.65 eV,

Pb = 4.25 eV and Ni = 5.15 eV) (i) X-ray (10–11 m to 10–8 m ) (ii) Infrared (7 10–7 m to 10–3 m) (iii) Visible (3.9 10–7 m to 7.5 10–7 m) (iv) Ultraviolet (10–8 m to 3.9 10–7 m) (A) (i), (iii), (iv) (B) (i), (iv) (C) (iii), (iv) (D) (i), (ii), (iii), (iv) 10. The work function of a substance is 2.5 eV.

The longest wavelength of light that can cause the photoelectron emission from this substance is approximately

(A) 500 nm (B) 400 nm (C) 310 nm (D) 220 nm 11. Light of wavelength strikes a photo-sensitive

surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to where

(A) = 2 (B) = 2

(C) 2 < < (D) >

12. A metal surface is illuminated by light of

two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2 respectively. If the ratio u1 : u2 = 3 : 1 and hc = 1240 eV nm, the work function of the metal is nearly

(A) 2.625 eV (B) 3.875 eV (C) 4.125 eV (D) 6.215 eV 13. Light of wavelength 5 107 m is incident on

a metal surface. The stopping potential for emitted photoelectrons is 1.4 V. The wavelength of incident light is reduced to 2.5 107 m, then stopping potential for emitted photoelectrons is

(A) 1.90 volt (B) 2.52 volt (C) 3.87 volt (D) 4.36 volt

14. From the following graph determine the work function (in eV) of the photosensitive metal?

(A) 1 (B) 1.5 (C) 1.25 (D) 2 15. Choose the correct statement/s (i) Matter waves are associated with

material particles only if they are in motion.

(ii) Larger the momentum, larger is the wavelength.

(iii) Wavelength associated with material particle is independent of the charge.

(A) (i), (ii), (iii) (B) (i), (ii) (C) (ii), (iii) (D) (i), (iii) 16. Three particles A, B and C of masses m, 2m

and 3m having charges 2q, 3q and q respectively are accelerated from rest through a variable potential difference of V. Which of the following graph will represent de-Broglie wavelength () associated with these particles for different values of V?

(A) (B) (C) (D)

A BC

V

A CB

V

A B C

V

A C B

V


5 4 32 1

( 1015) in Hz

(V0) (in eV)1 2 3

SAMPLE C

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424

17. Choose the correct option. (i) Electromagnetic radiation as well as

particles of matter may exhibit wave and particle properties in different experiments.

(ii) Electromagnetic radiation as well as particles of matter may exhibit wave and particle properties in different parts of the same experiment.

(A) only (i) is correct (B) only (ii) is correct (C) Both (i) and (ii) are correct (D) Both (i) and (ii) are incorrect 18. If the kinetic energy of the particle is

increased to 4 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

(A) 25 % (B) 75 % (C) 60 % (D) 50 % 19. A proton when accelerated through a potential

difference of V volts has a wavelength associated with it. If an alpha particle is to have the same wavelength , it must be accelerated through a potential difference of

(A) V8

volts (B) V4

volts

(C) 4 V volts (D) 8 V volts 20. The de-Broglie wavelength of an electron is

the same as that of a 100 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 0.5 MeV)

(A) 1 : 50 (B) 1 : 10 (C) 10 : 1 (D) 50 : 1 21. When the momentum of a proton is changed

by an amount p0, the corresponding change in the de Broglie wavelength is found to be 0.40%. Then, the original momentum of the proton was

(A) p0 (B) 100 p0 (C) 250 p0 (D) 400 p0 22. In Davisson and Germer’s experiment, an

electron beam of 64 eV is incident normally on a crystal surface. The maximum intensity is obtained at an angle of 45 from direction of incident beam. Then inter-atomic distance in the crystal lattice is

(A) 2.17 Å (B) 4.4 Å (C) 3.1 Å (D) 1.42 Å 23. The resolving power of an electron

microscope operated at 27 kV is R. The resolving power of electron microscope when operated at 3 kV is

(A) 2R (B) R/2 (C) R/3 (D) 3R 24. A small metal plate is placed at a distance of

3 m from a monochromatic light source of wavelength 5.4 107 m and power 50 watt. The light falls normally on the plate. The number of photons striking the metal plate per second per sq metre area will be

(A) 1.2 1018 (B) 1.2 1017

(C) 2.7 1016 (D) 2.7 1015

25. The energy flux of sunlight reaching the

surface of the earth is 1.388 103 watt/m2. If 4 1021 photons are incident on the earth per second, then calculate the wavelength of the incident photon.

(A) 800 nm

(B) 421 nm

(C) 573 nm

(D) 475 nm 26. If a photometal gives saturation current of

3.2 A when kept at certain distance from a point source (S), then what will be the value of its saturation current when the distance between the source and metal surface is doubled?

(A) 0.8 A (B) 3.2 A (C) 1.6 A (D) 6.4 A

Davisson and Germer Experiment11.4

Miscellaneous

The key to crack this question lies incomprehending that, energy flux is the totalenergy incident per area per second.

Mind over Matter

SAMPLE C

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425


1. (C) 2. (A) 3. (C) 4. (C) 5. (A) 6. (D) 7. (B) 8. (D) 9. (B) 10. (A) 11. (D) 12. (B) 13. (B) 14. (C) 15. (C) 16. (A) 17. (D) 18. (B) 19. (D) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) 25. (A) 26. (A) 27. (D) 28. (D) 29. (C) 30. (B) 31. (B) 32. (B) 33. (A) 34. (D) 35. (B) 36. (A) 37. (A) 38. (C) 39. (B) 40. (D) 41. (B) 42. (B) 43. (A) 44. (D) 45. (C) 46. (A) 47. (C) 48. (C) 49. (C) 50. (A) 51. (D) 52. (D) 53. (D) 54. (C) 55. (D) 56. (B) 57. (B) 58. (D) 59. (B) 60. (D) 61. (A) 62. (C) 63. (D) 64. (C) 1. (B) 2. (B) 3. (D) 4. (B) 5. (A) 6. (B) 7. (A) 8. (D) 9. (B) 10. (A) 11. (C) 12. (B) 13. (C) 14. (C) 15. (D) 16. (B) 17. (C) 18. (D) 19. (A) 20. (C) 21. (C) 22. (A) 23. (C) 24. (A) 25. (C) 26. (A) 1. The value of saturation current is independent

of the frequency of incident light but it increases with the increase in intensity of incident light.

2. r > y> g. Here, threshold wavelength < y. Since red light has maximum wavelength, so

frequency will be less than frequency of yellow light. Hence, no electrons are emitted.

3. Assertion is true because all the incident photons will not cause the emission of photoelectrons. i.e., in photoelectric emission, the value of photoelectric efficiency is less than 1.

4. The value of stopping potential is more negative for radiation of higher incident frequency.

i.e., V0 Hence, 1 < 2 < 3 5. Above threshold frequency (0), the stopping

potential increases with the increase in frequency.

6. The plate current reduces with increasing wavelength. When wavelength exceeds certain value, photoelectric effect ceases thus reducing current to zero.

7. Kmax = eV0 3 eV = eV0 V0 = 3V

8. Number of ejected electrons increases with the increase in intensity,

Intensity 2

1distance

9. P = h

, E = hc

Thus, if decreases, both P and E will increase. 10. = hc

E =

34 8

3 19

6.6 10 3 1035 10 1.6 10

= 3.5 10–11

= 35 10–12 m 11. E = hc

1

2

EE

= 2

1

E2 = 1 1

2

E

= 19 10

10

3.2 10 6000 104000 10

= 4.8 10–19 J

12. Condition of photoemission : 0 and UV < 5000 Å < IR

Hence, the radiation should be ultraviolet. Also, given n

t = 1.25 1020

Power (P) = nhct

P = nhct

= 20 34 8

101.25 10 6.63 10 3 10

5000 10

= 49.7 50 W


Practice Problems

Answer Key

Hints


SAMPLE C

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426


426

13. = h0. Thus, greater the work function, the higher is the threshold frequency and hence lower is the threshold wavelength.

14. (i) As eV0 = h – 0; If 0 increases, keeping the frequency

constant, then V0 decreases. (ii) Stopping potential is independent of

intensity. (iii) As eV0 = h – 0;

If frequency increases, keeping 0 constant, then V0 increases.

15. There will not be any photoemission possible for < 0. Ultraviolet radiations have < 0.

16. From Einstein’s Photoelectric equation,

2max

hc 1 = mvλ 2

+

0hc = eVλ + ...(

2max

1 mv2 = eV0)

V0 1

Thus, if incident wavelength is decreased, then stopping potential will increase.

17. Cut-off frequency is given as Work function = h Now, E = K.E. + 4h = 1

2mv2 + h

12

mv2 = 4h h

12

mv2 = 3h

v = 6hνm

18. Work function is given by, 0 = hc

For work function of 5 eV,

min = 15 84 10 3 10

5

= 240 nm,

For work function of 2 eV,

max = 15 84 10 3 10

2

= 600 nm

This means wavelength of 650 nm cannot be used.

19. K.Emax = hce

0 (eV)

K.Emax = 19hc

1.6 10

eV 2.2 eV

= 124316200

eV – 2.2 eV

K.Emax = 2eV 2.2 eV = 0.2 eV As kinetic energy can never be negative,

photo-emission doesn’t occur.

20. We know, (K.E.)max = h 0 2eV = 5eV 0 0 = 3eV Hence, when h = 6eV, (K.E.)max = 6eV 3eV = 3eV Also, (K.E.)max = eV0 = 3eV V0 = 3V As, stopping potential is a retarding potential,

potential of A relative to C = 3V 21. According to Einstein’s photoelectric equation, h = K.E. + 0 Initially, h = 0.4 + 0 ….(i) When frequency is increased by 30%, 1.3 h = 0.9 + 0 ….(ii) Subtracting equation (ii) by (i), 1.3 h h = 0.9 0.4 0.3 h = 0.5 h = 1.667 ….(iii) From equation (i) and (iii), 1.667 = 0.4 + 0 0 = 1.667 0.4 = 1.267 eV 22. E = 0 + K.E.; E = 12431

4000 = 3.10 eV

K.E. = E 0 = 3.10 eV 1 eV = 2.10 eV

2max

1 mv2

= 2.10 1.6 1019 J

vmax = 19

31

2 2.10 1.6 109.1 10

0.86 106 m/s

23. According to Einstein’s photoelectric equation,

12

mv2 = hc 0

1 1

Here, h = 4.14 1015 eVs = 4.14 1015 1.6 1019 h = 6.63 1034 Js = 2536 Å 0 = 3250 Å c = 3 108 m/s

12

mv2 = 6.63 1034 3 108

10 10

1 12536 10 3250 10

12

mv2 = 1.7 1019

v = 19

31

2 1.7 109.1 10

….( m = 9.1 1031 kg)

v = 0.6 106 m/s

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24. eV0 = h 0 = hc 0

eV0 = 34 8

9 196.63 10 3 10 1.7

332 10 1.6 10

eV

V0 = 2.04 V 25. E = 0 + K.E.max 1 = 0.5 + (K.E.max)1 (K.E.max)1 = 0.5 2.5 = 0.5 + (K.E.max)2 (K.E.max)2 = 2

max 1

max 2

(K.E. ) 0.5 1(K.E. ) 2 4

1 max 1

2 max 2

v (K.E. ) 1v (K.E. ) 2

26. When, 1 = 20, (K.E.1)max = h(20) – h0 = h0 ….(i) When, 2 = 50 (K.E.2)max = h(50) – h0 = 4h0 ….(ii) Dividing equation (i) by equation (ii),

1 max

2 max

K.E. 1K.E. 4

As, (K.E.)max = 2max

1 mv2

2122

v 14v

1

2

v 1v 2

27. 21 mv2

= hc

0 or

hc

= 21 mv2

+ 0 and

21

1 mv2

= hc34

0

= 43

20

1 mv2

0

2 21

4v v3

+ constant

So, v1 > 124v

3

i.e., v1 > v

2/ 443

28. Kmax = 2max

1mv2

= h 0 ….(i)

Now, when is doubled, then

12

m (2 vmax)2 = 2h 0

4 2max

1mv2

= 2h 0

4(h 0) = 2h 0 ….[from equation (i)]

30 = 2 h

0 = 2h

3

29. From Einstein’s photoelectric equation, h1 = + eV1 h2 = + eV2

1 1

2 2

eVeV

1 + eV21 = 2 + eV12

e = 2 1

1 2 1 2V V

30. K.E.1 = 0hc

K.E.2 = 0hc/ 2

= 02hc

K.E.2 = 3K.E.1

02hc

= 0hc3

20 = hc

0 = hc2

31. Using Einstein’s photoelectric equation, Case I:

eV = 0

hc hc

= hc

0

1 1

....(i)

Case II:

0

eV hc hc4 2

eV = 0

4hc 4hc2

= 4hc0

1 12

....(ii)

Equating (i) and (ii),

hc0

1 1

= 4 hc

0

1 12

0 0

1 1 2 4

0 = 3 32. Using photoelectric equation,

hc

= 0

hc

+ 3V0 ....(i)

hc2

= 0

hc

+ V0 ....(ii)

Subtracting (ii) from (i), hc

2= 2V0

V0 = hc4

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428

Substituting in equation (ii),

0

hc

= hc2

V0 = hc

2 hc

4

0

hc

= hc4

0 = 4 33. According to de-Broglie, permitted orbits can

exist only when the circumference of orbit equals an integral multiple of the wavelength of the electron.

i.e., 2r = n

But = h

mv

2r = nhmv

mvr = nh2

The above expression gives Bohr’s concept of stationary orbits.

34. de-Broglie wavelength,

= hp

But p = 2mE

1E

i.e., E 2

1

2 2

2 1

1 2

E 1 4E 0.5

As, E2 = 4E1 E = E2 – E1 = 4E1 – E1 = 3E1 35. = h

p

1p

36. de- Broglie wavelength of electron,

0 = 0

hmv

Force on electron in electric field,

F

= eE

= eE0 i

Acceleration of electron a

= 0F eE im m

Velocity of electron after time t,

v

= v0 i

+ 0eE i tm

= v0 0

0

eE1 t imv

de- Broglie wavelength at time t is = h

mv

= 0

00

heEm v 1 tmv

= 0

0

0

eE1 tmv

.... 00

hmv

37. Given,

v

= v0 i

B

= B0 j

Force on moving electron due to magnetic

field is,

F

= e v B

= e[v0 i

B0 j

] = ev0B0 k

As this force is perpendicular to v

and B

the

magnitude of v

will not change. Hence,

de-Broglie wavelength, = hmv

remains constant. 38. In an X-ray tube,

minhceV

min(hc / e)

V

Taking log both sides,

ln(min) = hcn n Ve

l l

i.e., y = C – mx Slope is negative and y-intercept is positive. Hence, 39. K.E. of electrons =

2p2m

here, p = h

....(de-Broglie hypothesis)

K.E. =

2h

2m

....(i)

Also, if 0 is cutoff wavelength, maximum

K.E. of X- ray photons = 0

hc

....(ii)

p

log min

log V

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Maximum K.E. of X- ray will be equal to that of electrons.

0

hc

= 2

2

h2 m

....[from (i) and (ii)]

0 = 22 mch

40. We know that de-Broglie wavelength,

= hmv

Velocity of a body falling from height H is given by

v = 2gH

= hm 2gH

= hm 2g H

Here, hm 2g

is a constant say ‘K’.

1H

41. For photon,

E = hc

For electron,

= h2mE

= hhc2m

= h2mc

42. E = 2 2

21 m vmv2 2m

21 p2m

….( momentum p = mv)

= 2

21 h

2m

…. hp

= 2

2h

2m 43. E = 120 eV = 120 1.6 1019J = 1.92 1017J

= 34

31 17

h 6.63 102mE 2 9.1 10 1.92 10

= 1.12 Å = 112 1012m = 112 pm

44. = hp

K.E. = 2p

2m =

2

2

h2m

= 34 2

31 7 2

(6.6 10 )2 9.1 10 (5.5 10 )

= 7.91 1025 8 1025 J 45. Let P be initial momentum of electron, Given, mP P P

as 1P

increase in P, decreases

0.5100

= 0.51100

= PP

0.51100

= m

PP P

0.995 P + 0.995 Pm = P

0.995 Pm = P200

P 200 Pm 46. Wavelength associated with photon, p = hc

E ….(i)

Wavelength associated with electron,

e = hp

But p = Ec

e = hcE

….(ii)

From equation (i) and (ii), p e 47. For electron, e

e

h2mE

For photon, pp

hcE

pe

p e

Ehhc2mE

1/2

e

p

1 Ec 2m

….( Ep = Ee)

48. h

p ....(de-Broglie formula)

For alpha particle,

h hp m v

As m = 4m, v = v

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430


430

h

4mv ....(i)

For deuteron,

dd d d

h hp m v

As md = 2m, vd = 2v

dh

4mv ....(ii)

Dividing equation (i) by (ii),

d

11

49. db 1m V

e e

1m V

p p

1m 9V

p e

e p

λ m V = λ m 9V

p

e

λ m 1 = λ M 3

....( me = m ; mp = M)

p = eλ m 3 M

p = λ m 3 M ....( e = )

50. The de-Broglie’s formula is

= h2m(K.E.)

=B

h2mK T

34

27 23

6.63 102 1.67 10 1.38 10 (27 273)

….(KB = Boltzmann constant = 1.38 1023)

= 1.78 1010 m = 1.78 Å 51. h

2mE

1

E

1 2

2 1

EE

10

210

1

4.5 10 E1.5 10 E

E2 = 9E1 Hence, added energy = E2 E1 = 8E1 52. = 12.27

VÅ

= 91.227 10

400

= 0.061 109 m = 0.06 nm

53. 1V

To decrease wavelength, potential difference between anode and filament is increased.

54. In Davisson – Germer experiment,

= 12.27V

Å

V 1

i.e., will decrease with increase in V. If there is a maxima of the diffracted electrons

at an angle , then 2dsin = Hence, sin i.e., with decrease in , will decrease. Thus when the voltage applied to A is increased,

the value of will be less than the earlier value. 55. Given d = 3 Å, = 14.7 for n = 1 2dsin = 2 3 1010 sin (14.7) = = 2 3 1010 0.253 = 1.5 Å 56. P = n hc

t

But P = Fc

Fc = n hct

n Ft h

=

5 7

346.62 10 5 10

6.62 10

= 5 1022

57. K.E. of a neutron in thermal equilibrium is

given by, E = 3

2kT

de-Broglie wavelength is given by

= h2mE

= h32m kT2

= h3mkT

58. EPh = 1240eV = 2.48 eV500

K.Emax = EPh – 0 = 2.48 – 2.28 = 0.2 eV For electron,

min = h2mE

=

34

31 19

6.63 10

2 9.1 10 0.2 1.6 10

= 2.747 109m λ ≥ λmin

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59. According to Bohr’s second postulate, mvrn = nh

2

2rn = nhmv

But, de-Broglie wavelength = hmv

2rn = n

No. of waves contained =in the orbit

Circumferenceoftheorbit

wavelength

= n2 r

= n = 2 ....( n = 2) 60. According to law of conservation of energy,

A Bm mmv (0) mv v2 2

A Bmmv mv v2

BA

vv v2

....(i)

As Collision is elastic,

B Av ve 1v

A Bv v v ....(ii) Equating (i) and (ii),

BA A B

vv v v2

BA

v2v2

BA

vv4

....(iii)

De-Broglie Wavelength is given by,

h

mv

A

B

= B

A

m v2mv

= B

A

v2v

= B

B

vv24

….[from (iii)]

A

B

2

61. Limit of resolution is given by d = 2sin

d ....(other factors are roughly same)

electron e

optical red

dd

=

1.227 nmV

646nm .... e

1.227 nmV

= 3

1.227100 10 646

6 10–6 62. Given: Radius of sphere, R = 2 cm = 2 10–2 m work function, 0 = 3.6 eV

Energy of incident radiation = hc

= 1240eV100

= 12.4 eV

Using Einstein’s photoelectric equation,

hc

= 0 + eV0

12.4 = 3.6 + eV0 V0 = 8.8 volt As soon as the surface potential of sphere

becomes 8.8 volt, it will stop emitting photoelectrons.

0

1 Q4 R

= 8.8

0

1 Ne4 R

= 8.8

where, N = number of photoelectrons emitted.

N = 2

9 19

0

8.8 R 8.8 2 109 10 1.6 101 e

4

N = 1.22 108 n = 8 63. E =

34 8

9hc 6.63 10 3 10

500 10

= 2.5 eV

Remaining energy after 1st collision, E1 = E – 12 E

100 = 2.5 – 12 2.5

100

E1 = 2.2 eV Remaining energy after 2nd collision, E2 = E1 – 1

12 E100

= 2.2 – 12 2.2

100

E2 = 1.936 eV Remaining energy after 3rd collision, E3 = E2 – 2

12 E100

= 1.936 – 12 1.936

100

E3 = 1.7 eV

B A v

Before collision

vA vB

After collision

A B

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432

Now, E3 = 0 + K.E. 1.7 = 1.2 + K.E. K.E. = 1.7 – 1.2 = 0.5 eV 64. Since saturation current is inversely

proportional to square of the distance between source and photometal.

I 2

1d

After 24 seconds, d = d – 0.25 6 = 9 – 1.5 ….( d = 9 m) = 7.5 m

I 1d

2I d

I d

2I 9

I 7.5

I = 1.44 I 1. During photon-electron collision, total energy

and momentum remain conserved but number of photons may not be conserved as all the photons would not cause emission of photo-electrons.

2. From the graph we observe that, (0)B > (0)A (0)B > (0)A 3. E = h = mc2 m = 2

hc

4. P = nhc

t

nt

= Phc

=

10

34 8

200 3500 106.63 10 3 10

= 3.5 1020 5. E = 103 eV, p = E

c =

3 19

8

10 1.6 103 10

= 0.53 1024

= 5 1025 kg m/s 6. E =

8 34

19

hc 3 10 6.62 100.24 1.6 10

= 5.2 106 eV

7. Frequency of light of wavelength ( = 5000 Å)

is = c

=8

10

3 105000 10

= 6 1014 Hz

which is less than the given threshold frequency. Hence no photoelectric emission takes place.

8. If frequency of incident light is less compared to threshold frequency of metal, then no photoelectron is emitted even if we increase the intensity.

9. Maximum wavelength required for photoemission with Cs metal,

9

CsCs

hc 1240 10 580nm2.14

Maximum wavelength required for photoemission with Ni metal,

9

NiNi

hc 1240 10 240nm5.15

For Al, Cu and Pb, the wavelength range will be between 240 nm to 580 nm. In order to cause photoemission in all experiments, the

wavelength of radiation must be below 240 nm, which is possible for X-ray and ultraviolet rays only.

10. max = h c

= 34 8

19

6.63 10 3 102.5 1.6 10

= 4.97 107 m 500 nm. 11. E = hc

– and 2E = hc

'

' E2E

1

E'2

E

Since 1

1E ,22

E

> 2

2 < <

12. 2

11

1 hcmu2

22

2

1 hcmu2

2 2

1 1

2

2

hcu 3

hcu 1

…. 1

2

u 3u

2 1

9hc hc9

2 1

9hc hc 8

9 1240 1240 8310 248

8 = 31 = 3.875 eV

Practice Problems

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433


13. = 1

hc 01

eV ....(i)

= 2

hc

02eV ....(ii)

Subtracting equation (ii) from equation (i), we get

1

hc

2

hc

= + 01eV 02

eV

hc 1 2

1 1

= e( 01V 02

V )

( 02V 01

V ) = hce 2 1

1 1

= 26

19

19.8 101.6 10

7 7

1 12.5 10 5 10

= 12.37 107 7

7 7

10 2.52.5 10 5 10

02V 01

V = 2.47 02

V = 01V + 2.47

= 1.4 + 2.47 = 3.87 V 14. From Einstein photo electric equation, eV0 = h h0 Now, when V0 = 0 0 = h h0 h = h0 = 0 From the above graph, 0 = 3 1015 Hz Work function is given by, 0 = h0 = 6.63 10–34 3 1014 0 = 2 10–19 J

0 (in eV) =

19

192 10

1.6 10

= 1.25 eV

15. Since, = hp

1p

Hence, larger the momentum, shorter is the wavelength.

16. Let EA, EB and EC be the energy with charge

2q, 3q and q respectively. Energy = charge potential difference EA = 2qV EB = 3qV Ec = qV

According to de-Broglie’s hypothesis,

AA A

h2m E

=h

2(m)(2qV) ....( mA = m)

=h 1

4qm V

Bh

2 2m (3qV) ....( mB = 2m)

=h 1

12qm V

Ch

2(3m)(qV) ....( mC = 3m)

=h 1

6qm V

Hence, A > C > B and 1V

Option (B) is the correct graph. 18. = h h

p 2mE

Now, = h2m(4E)

= 1 h2 2mE

= 0.502

% change = = 0.50 = 0.50 50% 19. The wavelength associated with a particle of

charge q, mass m and accelerated through a potential difference V is given by

hλ =2 m q V

or V = 2

2h

2 m q λ

For proton: V = 2

2p p

h2 m q

For -particle: V = 2

2h

2 m q

p pm qV 1 1 1V m q 4 2 8

….( m = 4mp and q = 2 qp)

Thus V = V8

. 20. Energy equivalent of electron mass = mc2 = 0.5 MeV

Using = h2mE

, we get

Eelectron = 2

2

h2m

and Ephoton = hc

photon

electron

EE

= 2

2

hc 2mh

= 2 5

3

2mc 2 5 10 10hc (100 10 ) 1

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434

Problems To Ponder

21. 1p

pp

pp

0p 0.40 1p 100 250

p = 250 p0 22. = d sin = 12.27

V Å

E = eV 64 eV =eV V = 64 volt

d = 12.27sin 45 ( 64)

= 2.17 Å

23. Resolving power, R 1

....(i)

and 1p

....(ii)

from equations (i) and (ii), R p

R p 2mER p 2mE

R E 3kVR E 27kV

R 1R 3

R = R3

24. Energy of photon emitted = hc

The energy of light falling per second per unit

area of a metal surface = 2

P4 r

Number of photons falling per second per unit area of metal surface = 2

P4 r hc

= 7

2 34 8

50 5.4 104 3.14 3 6.6 10 3 10

= 1.2 1018

25. Energy of each photon =

EnergyfluxNumberof photon

=

3

21

1.388 104 10

= 3.47 10–19 J

Now, E =

hc

=

34 8

19hc 6.63 10 3 10E 3.47 10

= 573 10–9 m

= 573 nm 26. Saturation current is inversely proportional to

the square of distance of metal surface from point source.

Let d be the initial distance between source and metal surface and I be the initial saturation current.

I 2

1d

….(i)

Now, when the distance between two metal and source in doubled.

d = 2d

I 2

1d

….(ii)

Dividing equation (ii) by (i),

2I d

I d

But d = 2d

2I d

I 2d

I = 1 I4 = 61 3.2 10

4

I = 0.8 A

1. A sphere S1 with mass m1 is moving with a velocity v collides (elastic collision) with a sphere S2 at rest with mass m2 (one dimensional motion). Find the change in the de-Broglie wavelength of sphere S1.

Ans: In elastic collision, According to law of conservation of momentum, m1u1 + m2u2 = m1v1 + m2v2 Here, u1 = v; u2 = 0 m1v = m1v1 + m2v2 m1 (v – v1) = m2v2 ....(i) According to law of conservation of energy,

2 2 2 21 1 2 2 1 1 2 2

1 1 1 1m u m u m v m v2 2 2 2

But u1 = v; u2 = 0

2 2 21 1 1 2 2

1 1 1m v m v m v2 2 2

2 2

1 11 m v v2

= 22 2

1 m v2

m1 (v – v1) (v + v1) = 22 2m v ....(ii)

Dividing equation (ii) by (i), v + v1 = v2 v1 = v2 – v ....(iii)

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From equation (i) and (iii), for v1, m1 (v – v1) = m2 (v + v1) m1v – m1v1 = m2v + m2v1

m2v1 + m1v1 = m1v – m2v v1 (m1 + m2) = v (m1 – m2)

1 21

1 2

m mv vm m

....(iv)

According to de-Broglie hypothesis, For sphere S1,

initial = 1

hm v

final =1 1

hm v

= final – initial

=

11 21

1 2

h hm vm m v

mm m

….[From (iv)]

=

1 2

1 1 2

m mh 1m v m m

= 1 2 1 2

1 1 2

h m m m mm v m m

= 2

1 1 2

h 2mm v m m

2. Explain the working of electron microscope. Ans: Principle: i. An electron of mass m moving with a

velocity v has a wave-packet associated with it, its de-Broglie wavelength is

given byh

mv

ii. High speed moving electrons can be focused by electric and magnetic fields in the same way as beam of light is focused by glass lenses.

Diagram:

Working: A beam of electron is emitted, when a

tungsten filament surrounded by a cylinder is heated. The cylinder has a small opening which is maintained at negative potential. The

beam of electron coming out of the cylinder is accelerated to a high potential of 105 V. Electrons at such high voltage possess small de-Broglie wavelength, hence they will have high magnifying and resolving power. The object used for scattering of electrons is kept outside the focus of objective lens. The electrons are directed towards the object using magnetic condensing lens. The electrons scattered by the object after passing through the objective form the first magnified image I1. This image lies just above the focal plane of the magnetic projection lens. This image later forms the final enlarged image I on the screen. In electron microscope, when electrons are accelerated through a potential difference of 60,000 V de-Broglie wavelength associated with electrons is given by,

4

12.27 12.27V 6 10

= 0.05 Å

Condensing Lens

ObjectiveLens

Projection Lens

Final image

O Object

I1

I

Magnetic Lens

Electron gun

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NEET-UG / JEE (Main) Challenger Physics Vol. - 2

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Transcript of NEET-UG / JEE (Main) Challenger Physics Vol. - 2