26.Electron & Photon

Electron and Photon

Chapter Overview Discovery of Cathode Rays (Eiedron) Positive Rays " Photoelectric Effect

• Planck's Quantum Theory • Compton Effect " Dual Character of Radiation

/

t>· Discovery of Cathode Rays (Electron) Sir William Crooks studied various gases in a gas discharge tube (a glass tube with a very high potential applied to its ends) at low pressures. If the pressure in the tube is lowered to about 10-4 atm, glass begins to fluoresce (glow) faintly. It was established that the glow was due to bombardment of the glass by a certain kind of rays emerging from cathode (negative electrode) which travel in a straight line until they strike the anode (positive electrode). These rays were called as cathode rays . Sir J J Thomson demonstrated that when cathode rays were deflected on to an electrometer, it acquired negative charge. He also showed that the rays were deflected on application of an electric field. The cathode ray beam was deflected away from the negatively charged plate. These results were found to be identical, irrespective of the gas taken in the discharge tube. He concluded that the cathode rays were a stream of fast moving negatively charged particles called electrons (named by stoney). He also calculated the velocity and specific charge for an electron. The specific charge is the ratio of charge to the mass of an electron, denoted as elm ratio . The el m ratio was found to be same for all gases. This led to the conclusion that the electron must be a fundamental or universal particle common to all kinds of the atoms. ·

988

Chapter 26 • Electron and Photon

Properties of Cathode Rays (i) Cathode rays travel in straight lines. (ii) The cathode rays are independent of the nature of the gas or electrodes employed to produce them in the

(iii) (iv) (v) (vi) (vii) (viii)

(ix)

(x) (xi) (xii) (xiii)

discharge tube. Therefore, .::_ for cathode rays is a m universal constant equal to 1. 7592 x 10 11 C kg-1 . They can be deflected by electric and magnetic fields . They have penetrating power and can penetrate through small thickness of matter. On striking the target of high atomic weight and high m~lting point, they produce X- rays. They produce fluorescence and phosphorescence in certain substances and hence affect photographic plate. They have small ionising power and ionise the gas through which they pass. They travel in straight line with high velocity, momentum and energy and cast shadow of objects placed in their path. They emerge normally from surface of cathode and can therefore be brought to focus by using a concave cathode. They heat up the material on which they fall. They exerts mechanical pressure, so they can rotate a small paddle wheel. They can produce physical and chemical change. They can exhibit interference and diffraction phenomena under suitable arrangements. Th~s, they may behave as waves.

r> Positive Rays Positive rays were discovered by Goldstein. Positive rays are moving positive ions of the gas filled in the discharge tube. The mass of these particles is nearly equal to the mass of the atoms of gas.

Properties of Positive Rays (i) (ii) (iii) (iv) . (v) (vi)

These consist of fast moving positively charged particles. These rays are deflected in magnetic field. These rays are deflected in electric field. These rays travel in straight line. Speed of positive rays is less than that of cathode rays. These rays can affect the photographic plate. (vii) These rays penetrate through the thin aluminium foil. (viii) These rays can produce fluorescence and phosphorescence.

tt> Photoelectric Effect The phenomenon of p ejection of electrons from a metal surface, when light of sufficient high frequency A falls on it, is known as the photoelectric effect. ~------~+~v~---------~ The figure given below shows the experimental set up to study photoelectric effect. When ·a suitable radiation is incident on the electrode P, electrons are ejer::ted from it. The electrons whi'cf.. 'ilave sufficient kinetic en~~gy, ·~~ach' 'e'lectr~de Q, (de~pite its negative polarity).

the

,

The potential difference between the two electrodes acts as the retarding potential. As electrons reach on electrode Q, so it becomes more and more negative, so fewer and fewer electrons reach on electrode Q and photo electric current recorded by ammeter falls . The particular potential difference V0 (say) at which no electron reach on electrode Q, is called stopping potential. In this case; the work done by stopping potential is equal to the maximum kinetic energy of the electrons. ie,

laws of Photoelectric Effect Following are the laws of photoelectric effect (i) For each emitting metal, there is a certain mirumum frequency v0 ( or maximum wavelength /..0 ) , called the threshold frequency of the incident radiation, below which no emission of photoelectron takes place, no matter how great is the intensity. The value of v0 (or /..0 ) is different for different emitting surfaces: (ii) The process of emission of photoelectrons is an instantaneous process. There is no time lag ( ejEcts electrons ~m Srrtain photosensitive surface, yellow light does not. Will (a) red light (a) violet light eject ~~o~ge~sc~~ from the same substance ? (iv) A photon and an electron have same wavelength. Which particle is moving faster?

~>

Dual Character of Radiation

In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be explained only on the basis of its particle nature. Thus light is said to have a dual character. Such studies on light wave were made by Einstein in 1905. Louis-de-Broglie, in 1942 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual character as wave and as particles .

Instance 4 Calculate the de-Broglie wavelength of neutrons whose energy is 1 eV (Given mass of electrons= 1.67 x 10-27 kg) Interpret

We know that h A= .J2mE 6.6 x 10-34

~2 X 1.67 X 10-27 X 1.6 X 10~ 19 =

0.2857

X

10-lQ

= 0.2857 A.

De-Broglie Relation

Davisson and Germer Experiment

According to de-Broglie a wave is associated with energy moving particle. These waves are called de-Broglie waves or matter waves. According to quantum theory, energy of photon E= h ... (i) If mass of the photon is taken as m, then as per Einstein's equation ••• (ii) E = mc 2 From Eqs. (i) and (ii) h\' = mc 2

The wave nature of the material particles as predicted by de-Broglie was confirmed by Davisson and Genner (1927) in United States and by G P Thomson (1928) in Scotland. Experimental arrangement used by Davisson and Germer is as shown in figure . Electrons from hot tungsten cathode (C) are accelerated by a potential difference V between the cathode and anode (A). A narrow hole in the anode renders the electrons into a fine beam of electrons and allows it to strike the nickel crystal. The electrons are scattered in all directions by the atoms in the crystal. The intensity of the electron beam scattered in a given direction is found by the use of a detector. By rotating the detector about an axis through the point 0, the intensity of the scattered beam can be measured for different values of~. the angle between incident and the scattered direction of electron beam.

h ~ = mc 2 , where A. = wavelength of photon. A. h A. =-me de-Broglie asserted that the above equation is completely a general function and applies to photon as well as all other moving particles. A-!!__ =-hSo, - mv .J2mE where m is mass of particle and v is its velocity. (i) -de-Broglie wavelength associated with charged particle

Electron gun

A

·· .. Detector

A-~- _h__ h - p - .JzmE - ~2mqV (ii) de-Broglie wavelength of a gas molecule

A = __ h_

where

absolute

T

.J3mkT k = Boltzmann's constant

= 1.38 x

temperature

OJ CD

w 3

10-23 J/K

(iii) Ratio of wavelength of photon and electron. The

)

wavelength of photon of energy E is given by A = he P

E

'while the wavelength of an electron of kinetic energy K is given by A. = c

ratio

,

b .Therefore for same energy, the

-v2mK

AP = .:._.JzmK = ~2mc Ae

E

2

. E2

K

Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. It is maximum for diffracting angle' 60° at 54 volt potential difference.

992

Chapter 26 • Electron and Photon

,,

(9

9}0 .

0 .....,:.___,__ _ _ _ _ __

\:~~.Yd·· ....

54V

If the de-Broglie waves exist· for electrons then these should be diffracted as X-rays. Using the Bragg's formula 2d sine = n A, we can determine the wavelength of these waves. Where d = distance between diffracting planes e = (ISO- ) = glancing angle for incident beam= Bragg's angle. 2 The distance between diffracting planes in Ni-crystal for this experiment is d = 0.91 Aand the Bragg's angle = 65 °. This gives for n = 1, A = 2 x 0.91 x 10-lo sin 65° = 1.65

···..,~tori1i.c planes

the

Now de-Broglie wavelength can also be determined by using l 1 _ 12.27 _ 12.27 _ Ao !OrillU a ".JS4 -- 1 67 Thus the de-Broglie C

rv -

0

0

hypothesis is verified.

lntext Questions 26.2 tic energy; which of the following has smallest de-Broglie }Vavelength, electron, proton, a-particle? ''nature of lnatter not apparent to our daily observations ? ... . wavelength pf a photon of an electromagnetic radiation elnl.~ltothewaV'elength of the radiatip:;?> ,, .'> pn needs to be therm~se~.;~th the environment, before it can be used for neutron diffraction 'exp€rip~nts.

,

.

W:4y?

Chapter 26 • Electron and Photon

993

kinetic energy with which an "'""''-w.vu einitted from a metal surface is independent of the f the light and depends upon its frequency. The number of photoelectrons emitted ,is mctep,en,aeJlt the frequency of th · cident llght · its intensity. : Einstem's equation of photoelectric effect . 1

2

2mvmax 1

2

2mvrriax

bove relation is the wave ve1enl'lll of particle. veH~n?;Inof

16. de-Broglie wavelength of

,

llluSlrative Example 1 What is frequency and energy of a photon of wavelength 6000 A? (Given h = 6.6 x 10-34 Js, c = 3 x 108 ms- 1) (a) 15 x 10 4 Hz and 3.3x 1o-10J (b) 6 x 10 10 Hz and 4.4X 10-19J (c) 5 x 10 12 Hz and 3.3 x 10-19J (d) 5 x 10 14 Hz and 3.3 x 10-19J . e 3x 10 8 SolutiOn Frequency v =-;;- = = 5 x 10 14 Hz 11, 6000xl0-10

Example 4 Find the ratio of de-Broglie wavelength of proton and a-partide which has been accelerated through the same potential difference. (a)

3J2

(b) 2J2

(c)

2.)3

(d)

A.h AP _ - ~2mqV 'A.a. -

Solution

and energy= hv = 6.6 x 10-34 x 5 x 1014 = 3.3 x 10-19 J

Example 2 The work function of sodium is 2.3 eV. Calculate the maximum wavelength for the light that will cause photoelectrons to be emitted from the sodium. (aJ 64oo A (bJ 5ooo A (c) 54oo A (d) 5200 A Solution

KEmax

= hv -

0 > 0

= he - q> 0 A.

(Kin~tic

energy is always +ve)

A-o A. max

8 34 = eh = 3x10 x 6.62 x 10- = 5400 q> 0 2.3 X 1.6 X 10- 19

A

Example 3 In an experiment on photoelectric effect it was observed that for incident light of wavelength 1. 98 X 10-7 m, stopping potential is 2.5 V. Wh~t is the energy of photoelectrons with maximum speed, workfunction ~ 0 and threshold frequency? (a) 6.25 eV, 3.75 eV, 9.10 x 1014 Hz (b) 3.75eV, 6.25 eV, 9.10x10 14 Hz (c) 4.75 eV, 6.25 eV,8.10x10 14 Hz (d) None of the above Solution I

-->

and its speed increases. The relations amongst E , B and -->

velocity v must be such that -->-->

-->

(a) E · B = 0, v is arbitrary -->-->

-->

(b) E , B and v are all parallel to each other. --7

22. An electric field of intensity 6 X 104 Vm-l is applied perpendicular to the direction of motion of the electron. A magnetic field of induction 8 x 10-2 wm-2 is applied perpendicular to both the electric field and direction of motion of the electron. What is the velocity of the electron if it passes undeflected? (b) 7.5 x 10-5 ms-1 (a) 7.5 x 10 5 ms- 1 (c) 48 X 10- 2 ms-1 (d) It is never possible

---?

--t --7

---? ---?

(c) E · v = 0; B · v = 0 but E · B ,t 0 -->

-->

-->

(d) v is parallel to E and perpendicular to B

16. An electron with (rest ma:;s m0 ) moves with a speed of 0.8 c. Its mass when it moves with this speed is (a) m 0 (b) m 0 / 6 (c) 5m 0 /3 (d) 3m 0 /5 17. A charged dust particle of radius 5 x 10-7 m is located in a horizontal electric field having an intensity of 6.28 x 105 vm- 1 . The surrounding medium in air with 2 coefficient of viscosity 11 = 1.6 x 10-15 Nsm- . Ifthis particle moves with a uniform horizontal speed of 0.01 ms-1, the number of electrons on it will be (a) 20 (b) 15 (c) 25 (d) 30 18. The momentum of a charged particle moving in a perpendicular magnetic field depends on (a) its charge (b) the strength of magnetic field (c) radius of its path (d) All of the above 19. If in a Thomson's mass spectrograph, the ratio of the electric fields and magnetic fields, in order to obtain coincident parabola of singly ionised and doubly ionised positive ions are 1 : 2 and 3 : 2 respectively, then the ratio of masses of particles will be (a) 3 : 1 (c) 9 : 4

(b) 2 : 1

(d) 9 : 2 20. The specific charge for positive rays is much less than that for cathode rays. This is because (a) masses of positive rays are much larger (b) charge on positive ray is less (c) pGJsitive rays are positively charged (d) experimental method is wrong

24. The mass of a particle is 400 times than that of an electron and charge is double . The particle is acce'lerated by SV. Initially the particle remained in rest, then its final kinetic energy will be (b) 10 eV (a) 5 eV · (c) 100 eV (d) 200 eV 25. A charged particle is moving in a uniform magnetic field in a circular path. The energy of the particle is tripled. If the initial radius of the circular path was R, the radius of the new circular path after the energy is tripled will be (a)

~

(b) ../3R

3 (c) 3 R

(d) R / ../3 26. An electron moving with a variable linear velocity v in a variable magnetic field B will remain rotating in a circle of constant radius r only when (a) B is held constant (b) vis held constant (c) Both v and Bare constant (d) None ofthe above 27. el m ratio of anode rays produced in a discharge tube , depends on the (a) nature of the gas filled in the tube (b) nature of the material of anode (c) nature of the material of cathode (d) All of the above 28. A positively charged particle enters a magnetic field of value with a velocity vk. The particle will move along (a) +X axis (b) -X axis (d) -Z axis (c) + Z axis

BJ

29. In a mass spectrograph, an ion X of mass number 24 and charge +e and another ion Y o[ mass number 22 and charge + 2 e enter in a perpendicular magnetic field with the same velocity. The ratio of the radii of the circular path in the field will be (a) 11/22 (b) 11/ 2 (c) 22/ 11 (d) 24/ 11

Chapter 26 • Electron and Photon 30. A beam of electrons of velocity 3 x 10 7 ms- 1 is deflected 1.5 mm is passing 10 em through an electric field of 1800 vm-1 perpendicular to their path. The value of elm for electron is (b) 2 X 10 11 Ckg-1 (a) 1.78 x 10 11 C kg-1 (d) 3.5 X 1011 Ckg-1 (c) 1.5 X 10 11 Ckg-1

Particle Nature' ofLight 31. Planck's constant has the dimensions of (a) energy (b) mass (c) frequency (d) angular momentum 32. The wavelength of a 1 keV photon is 1.24 x 10-9 m. What is the frequency of 1 MeV photon? (a) 2.4 X 10 15 Hz (b) 2.4 X 1020 Hz 15 (c) 1.24 x 10 Hz (d) 1.24 x 1020 Hz 33. A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is (a) 3.2 X 10- 8 N (b) 3.2 X 10-7 N 7 (d) 5.12 X 10- 8 N (c) 5.12 X 10- N 34. Calculate the energy of a photon with momentum 3.3 x 10-13 kg-ms-1, given Planck's constant to be 6.6 x 10- 34 Js (b) 9.9 X 10-5 J (a) 7.3 x 104 J (d) 8.1 X 10 3 J (c) 1.3 x 105 J 35. Momentum of a photon is p. The corresponding wavelength is (a) hi/P (c) ph

(b) plh • (d) hlp

36. Which of the following statement about photon 1s incorrect? (a) Photons exert no pressure (b) Momentum of photon is h\'lc (c) Photon's rest mass is zero (d) Photon's energy ish\' 37. A photon in motion has a mass equal to (a) clh\' (c}

h\'

(b) hi'A (d)

hvlc 2

38. Momentum of a photon of wavelength A. is (a) hi'A (b) h A;c2 (c) hAle (d) zero

39. A photon will have less energy, if its (a) amplitude is higher (b) frequency is higher (c) wavelength is longer (d) wavelength is shorter 40. An important spectral emission line has a wavelength

of 21 em. The corresponding photon energy is ( h= 6.62 x 10-34 Js and c = 3 x 108 ms- 1) (a) 5 .9 X 10-8 eV (b) 5.9 X 10-4 eV (c) 5.9 X 10-6 eV (d) 11.8 X 10-6 eV

(a) 0.267 x 10 18 (b) 0.267 X 10 19 (c) 0.267 x 10 20 (d) 0.267 X 10 17 43. If the energy of photons corresponding to the wavelength of 6000A is 3.2 x 10-19 J, the photon energy for a wavelength of 4000 A will be (b) 2.22 X 10-19J (a) 1.11 X 10-19 J (c) 4.40 x 10-19 J (d) 4.80 x 1o-19 J 44. A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of P;hotons emitted per second is 1 (a) 1.72 X 10 31 (b) 1.327 X 1025 (c) 1.327 x1o 37 (d) 1.327 x1o 45

Emission of Electrons and Photoelectric Effect 45. The photoelectric threshold of Tungsten is 2300 A. The

energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 A is (h = 6.6 x 10-34 J-s) (a) 0.15 eV (b) 1.5 eV (c) 15 eV (d) 150 eV 46. A light of wavelength 4000 A iS"allowed to fall on a metal

surface having work function 2 eV. The maximum velocity of the emitted electrons is (R = 6.6 x 10-34Js) (a) 1.35 X 10 5 ms-1 (b) 2.7 X 10 5 ms-1 5 1 (d) 8.1 x 105ms- 1 (c) 6.2 x 10 ms47. Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the fastest electron emitted is approximately (b) 3.2 X 10-19 (a) 3 X 10-21 (c) 3 X 10-17 (d) 3 X 10-15 48. Radiations of two photon's energy, twice and ten times

the work function of metal are incident on the metal surface successively. The ratio of maximum velocites of photoelectrons emitted in two cases is (a) 1 : 2

(b) 1 : 3

(c) 1 : 4 (d) 1 : 1 49. The variation of photoelectric current given by the photocell, with the intensity of light, is given by a graph, which is a straight line with (a) +ve slope with intercept on current axis (b) -ve slope with intercept of current axis (c) +ve slope passing through origin (d) - ve slope passing through origin 50. When the photons of energy hv fall on a photosensitive metallic surface (work function h\'0 ) electrons are emitted from the metallic surface. The electrons coming out of the surface have some kinetic energy. The most energetic ones have the kinetic energy equal to Vs

41. The energy of a photon of green light of wavelength

50000 A is (a) 3.459 X 10-19 J (c) 4.132 X 10-19 J

A

B

(b) 3.973 X 10-19 J (d) 8453 X 10-19 J

42. What will be the number of photons emitted per second by a 10 W sodium vapour lamp assuming that 90% of the consumed energy is convei,ted into light? Wavelength of sodium light is 590 nm, h = 6.63 x 10-34J-s.

997

(a) less (c) equal

(b) more (d) Nothing can be said

998

Chapter 26 • Electron and Photon

51. When the photons of energy hv fall on a photosensitive metallic surface (work function h\'0 ), electrons are emitted from the metallic surface. The electrons coming out of the surface have some kinetic energy. The most energetic ones have the kinetic energy equal to (a) hv0 (b) h (c) h - hv0 (d) hv + h'·o 52. For a certain metal ,. = 2 v0 and the electrons come out with a maximum velocity of 4 x 106 ms-1 . If the value of v = 5 v 0 , then maximum velocity of photoelectrons will be (a) 2 x 10 7 ms-1 (b) 8 x 106 ms- 1 6 1 (c) 2 x 10 ms(d) 8 x 10 5 ms- 1 53. The wavelength of the photoelectric threshold for silver is !..0 . The energy of the electron ejected from the surface of silver by an incident light of wavelength A(A < A0 ) will be (a) hc(A 0 - A)

(c)

!!._(.!_ __!___] c A A0

54. A metal surface is illuminated by a lig4t of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become (a) four times the original value (b) twice the original value (c) l/6th of the original valu~ (d) unchanged 55. The work function of a metal is leV. Light of wavelength 3000 A. is incident on this metal surface. The velocity of emitted photoelectons will be (b) 10 3 ms- 1 (a) 10 ms-1 1 4 (c) 10 ms(d) 10 6 ms- 1 56. In the photoelectric effect the velocity of ejected electrons depends upon the nature of the target and . (a) the ~equency of the incident light (b) the polarisation of the incident light (c) the time for which the light has been incident (d) the intensity of the incident light 57. Light of wavelength 4000 A. is incident on a metal plate whose work function is 2 eV. The maximum KE of the emitted photoelectron would be (a) 0.5 eV (b) 1.1 eV (c) 1.5 eV (d) 2.0 eV 58. A photon of energy 3.4 eV is incident on a metal having work function 2 eV. The maximum KE of photoelectrons is equal to (b) 1.7 eV (a) 1.4 eV (c) 5.4 eV (d) 6.8 eV 59. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is (a) double the earlier value (b) unchanged (c) more than doubled (d) less tham doubled

60. A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photoelectrons is (b) 2.66 eV (a) 1.07 eV (c) 3.7 eV (d) 4.81 eV 61. If the work function for a certain metal is 3.2 x 10- 19 J and it is illuminated with light of frequency v = 8 x 1014 Hz, the maximum kinetic energy of the photoelectron would be (b) 3.2 X 10- 19 J (a) 2.1 X 10-19 J (c) 5.3 x lo-19 J (d) 8 .5 x 10- 19 J 62. Ultraviolet radiations of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV) (a) 3.2 x 10-21 J (b) 3.2 X 10-19 J (c) 7 x 10-25 J (d) 9 x lo-32 J 63. Ultraviolet light of wavelength 300 nm and intensity 1.0 wm-2 falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly (a) 9.61 x 10 14 s- 1 (b) 4.12 x 10 13 s-1 12 1 (c) 1.51 x 10 s(d) 2.13 x 10 11 s- 1 64. The photoelectric threshold wavelength for a metal surface is 6600 A.. The work function for this metal is (a) 0.87 eV (b) 1.87 eV (c) 18.7 eV (d) 0.18 eV 65. Light of wavelength 4000 A. incident on a sodium surface for w~ich the threshold wavelength of photoelectrons is 5420 A. The work function of sodium is (a) 0.57 eV (b) 1.14 eV (c) 2.29 eV (d) 4.58 eV 66. The difference between kinetic energies of photoelectrons emitted from a surface by light of wavelengths 2500 A. and 5000 A. will be (b) 2.47 eV (a) 1.61 eV (c) 3.96 eV (d) 3.96 x 1o-19 eV 67. When a point source oflight is 1 m away from~ photoelectric cell, the photoelectric current is found to be I rnA. If the same source is placed at 4 m from the same photoelectric cells, the photoelectric current (in rnA) will be (a) l/16 (b) l/ 4 (c) 41 (d) 161

Wave Nature ofParticle 68. An electron and photon have same wavelength. If E is the energy of photon and p is the momentum of electron, then the magnitude of E/p in SI unit is (a) 3.33 x 10-9 (b) 3.0 x 108 (c) 1.1 X 10-19 (d) 9 X 10 16 69. The wavelength of de-Broglie wave associated with a thermal neutron of mass m at absolute temperature T is given by (Here, k is the Boltzmann constant)

h

(a) (c)

~2mkT h ~3kmT

h

(b) (d)

~mkT h

2~mkT

Chapter 26 • Electron and Photon (a) 10 3 ms- 1 (c) 1.4 x 10 3 ms- 1

70. The de-Broglie wavelength of a neutron at 927°C is A. What will be its wavelength at 27°C? (a) A./2 (c) 4 A

(b) A./4 (d) 2 A

(d) hvlm 0 c

74. An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is -h -eEt (b) E (a) eEt 2 (c)

-mh

(d)

eEt 2

(b) 1.2 x 103 ms- 1 (d) 2.8 x 103 ms- 1

Principle of Uncertainty

71. What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 A? (a) 700 ms- 1 (b) 1000 ms- 1 (c) 1400 ms- 1 (d) 2800 ms- 1 ·n. If the mass of neutral 1.7 x 10-27 kg, then the de-Broglie wavelength of neutral of energy 3eV is (h = 6.6 x 10-34 J-s) (b) 1.6 x 10- 11 m (a) 1.6 x 10-16 m (d) 1.4 x 10- 11 m (c) 1.4 x 10- 10 m 73. A particle with rest mass m0 is moving with speed of light c. The de-Broglie wavelength associated with it will be (a) infinite (b) zero (c) m0 cl h

999

-h eE

75. What should be the velocity of an electron so that its momentum becomes f'!qual to that of a photon of wavelength 5200 A?

76. The correctness of velocity of an electron moving with velocity 50 ms- 1 is 0.005%. The accuracy with which its position can be measured will be (a) 4634 x 10-3 m (b) 4634 x 10-5 m (c) 4634 x 10-6m (d) 4634 x 10-8 m 77. If a proton and an electron are confined to the same region, then uncertaint'y in momentum (a) for proton is more, as compared to the electron (b) for electron is more , as compared to the proton (c) same for both the particles (d) directly proportional to their masses 78. If the uncertainty in the position of an electron is 10-10 m, then the value of uncertainty in its momentum (in kg-ms- 1 ) will be (a) 3.33 X 10-24 (b) 1.03 X 10-24 (c) 6.6 X 10-24 (d) 6.6 X 10-2 4 79. The uncertainty in the position of a particle is equal to the de-Broglie wavelength. The uncertainty in its momentum will be

lilA

(b) 2hi 3A.

(c) 'Alii

(d) 3A/2h

(a)

80. If the uncertainty in the position of proton is 6 x 108 m, then tl1e minimum uncertainty in its speed will be (b) 1 ms- 1 (a) 1 cms- 1 1 (c) 1 mms(d) 100 ms- 1

Exercise II Only One Correct Option 1. When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 V. If elm for the electron is 1.8 x 10 11 Ckg-1, the maximum velocity of ejected electrons is (a) 6 x 10 5 ms- 1 (b) 8 x 10 5 ms- 1 (c) 106 ms- 1 (d) 1.8 x 10 6 ms- 1 2. The intensity of X-rays from a coolidge tube is plcitted against wavelength A as shown in figure. The minimum wavelength found is "-c and the wavelength of Ka like is "-K· As the accelerating voltage is increased

(a) A.K- Ac increases (c) A.J! increases

(b) · A.K- "-c decreases (d) A.K decreases

3. The stopping potential V for photoelectric emission for a metal surface is plotted along Y-axis and frequency v of incident light along X-axis. A straight line is obtained as shown. Planck's constant is given by (a) slope of the line (b) product of slope of the line and charge on the electron (c) intercept along Y-axis divided by charge on the electron (d) product of intercept along X-axis and mass of the electron (e) product of slope and mass of the electron 4. Mixed He+ and 0 2 + ions (mass of He+ = 4 amu and that of o2 += 16 amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then (a) He+ ions will be deflected more than those of o2 + (b) He+ ions will be deflected less than that of o2 + (c) all the ions will be deflected equally (d) no ions will be deflected 5. In Millikan's oil drop experiment a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 V. Then charge on other drop of radius 2 r which is kept constant with a potential difference of 3200 Vis (b) 2 Q (a) Q 12 (c) 4 Q (d) Ql4

1000

Chapter 26 • Electron and Photon

6. An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is (a) zero (b) Infinity (c) equal to kinetic energy of the proton (d) greater than the kinetic energy of proton 7. Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 Afalls on the surface, the velocity of fastest electrons ejected from the surface is (a) 2.5 X 10 18 ms- 1 (b) 2.5 x 10 13 ms-1 18 1 (c) 6. 7 x 10 ms(d) None of these

(a) Zero (b) 1 eV (c) 2 eV (d) 10 eV 14. The de-Broglie wavelengthL associated with an elementary particle of linear momentum p is best represented by the graph

(a)

(a) I (c) III

(b) II (d) IV

10. Electron with energy 80 keV are incident on the tungsten

target of an X-ray tube. K shell electrons of tungsten have -72.5 keV energy. X-rays emitted by the tube contain only (a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of- 0.155 A (b) a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-rays spectrum (Bremsstrahlung) with a minimum wavelength of -0.155 A and the characteristic X-ray spectrum of tungsten 11. What is the de-Broglie wavelength (in A) of the a-particle accelerated through a potential difference V ?

(a) 0.287

(b) 12.27

-JV ( ) c

0.101

.Jv

-JV '

(d) 0.22

-JV

12. An oil drop carrying a charge q has a mass m kg. It is

falling freely in air with terminal speed v. The electric field required to make the drop move upwards with the same speed is (a) mg

(b) 2mg

q (c) mgv

7

q (d)

2mgv

q

13. ·Photons of energy of 6 eV are incident on a metal surface

whose work function ·is.4 eV. The minimuin kinetic energy . ofthe emitted photoelectronn~ill be

(b)

p

p

8. The maximum wavelength of radiation that can produce

photoelectric effect in certain metal is 200 run. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 run will be (a) 12.4 eV (b) 6.2 eV (c) 100 eV (d) 200 eV 9. Consider the following statements concerning electrons : I. Electrons are universal constituents of matter. II. J J Thomson received the very first Nobel prize in Physics for discovering the electron. III . The mass of the electron is about 1/2000 of a neutron. IV. According to Bohr the linear momentum of the electron is quantised in the hydrogen atom. Which of the above statements are not correct?

L

L

L

L

(c)

~

(d)

p

p

15. Maximum velocity of photoelectron emitted is 4.8 ms-1 . The e/m ratio of electron is 1. 76 x 10 11 Ckg-1, then stopping potential is given by (b) 3x10- 7 Jc-1 (a) 5 x 10-10 Jc-1 (c) 7x10-11 Jc-1 (d) 2.5x10- 2 Jc-1 16. Two identical metal plates shown photoelectric effect by

a light of wavelength leA falls on plate A and 'A 8 on plate B('AA = 2'A 8 ) . The maximum kinetic energy is (a) 2 KA = K8 (b) KA < K8 /2 (c) KA= 2KB (d) KA = K8 / 2 17. During X-ray production from coolidge tube if the current is increased, then (a) the penetration power increases (b) the penetration power decreases (c) the intensity of X-rays increases (d) the intensity of X- rays decreases 18. Light of wavelength 5000 Afalls on a sensitive plate with photoelectric work fi.mctional of 1. 9 eV. Th€ kinetic energy of the photoelectron emitted will be (b) 2.48 eV (a) 0.58 eV (c) 1.24 eV (d) 1.16 eV 19. A proton and an a-particle are accelerated through the same potential difference . The ratio of their de-Broglie wavelength ('AP!'A 0 ) is (a) 1! 2.fi (b) 1 (c) 2

(d)

2.J2

20. The de-Broglie wavelength of a neutron at 27°C is A..

What will be its wavelength at 927°C? (a) A/4 (b) A/3 (c) A/2 (d) 3 A/2 21. A potential difference of 10 4 Vis applied across an X-ray tube. The ratio of the de-Broglie wavelength of X-rays produced is 1

ra)' 20 (c) 1

(!... m

for electron = 1.8 x 10 11 Ckg-1 ) 1 (b) 10

1 (d) 100

Chapter 26 • Electron and Phofon 22. The minimum light intensity that can be perceived by the eye is about 10-10 wm-2 . The number of photons of wavelength 5.6 x 10-7 m that must enter the pupil of area 10-4 m 3 s-1 , for vision is approximately equal to (h = 6.6 x 10-34 J-s) (b) 3 x 103 photons (a) 3 x 102 photons 4 (d) 3 x 10 5 photons (c) 3_x 10 photons 23. A charged oil drop falls with terminal velocity v0 in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v0 . The initial charge on the drop lS

(b) q (d) 2q

(a) q/2 (c) 3q/2

24. The filament current in the electron gun of a coolidge tube is increased while the potential difference used to accelerate the electrons is decreased. As a result, in the emitted radiation (a) the intensity increases while the minimum wavelength decreases (b) the intensity decreases while the minimum wavelength increases (c) the intensity as well as the winimum wavelength increases (d) the intensity as well as the minimum wavelength decreases 25. What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 em. When light of wavelength 3800 A is incident on a barium emitter? (Given that work function of barium is 2.5 eV; h = 6.63 x 10-34 J-s; e = 1.6 x 10-19 C ; m = 9.1 x 10-31 kg.) (a) 6.32 X 10- 4 T (b) 6.32 X 10-5 T (c) 6.32 X 10--6 T (d) 6.32 X 10--8 T 26. Given that a photon of light of wavelength 10,000 Ahas an energy equal to 1.23 eV. When light of wavelength 5000 A and intensity I 0 falls on a photoelectric cell, the surface current is 0.40 x 10--D A and the stopping · potential is 1.36 V, then the work function is (a) 0.43 eV (b) 0 .55 eV (c) 1.10 eV (d) 1.53 eV 2 7. The wavelength of characteristic X-ray Ka. line emitted by hydrogen like atom is 0.32 A. The wavelength of K p line emitted b;r the same element is o (a) 0.21 A (b) 0.27 A (c) o.33 A Cd) o.4o A. 28. A photon and electron have same de-Broglie wavelength. Give that v is the speed of electron and c is the velocity of light . Ee, EP are the kinetic energy of electron and photon respectively. Pe• Ph are the momentum of electron and photon respectively. Then which of the following relation is correct? v 2c (a) ~= (b) Ee = EP 2c EP v (c) ~=~

Ph

2v

2c v

(d) ~= -

Ph

29. Which one of the following . statements regarding photo-emission of electrons is correct?

1001

(a) Kinetic energy of electrons increases with the intensity of incident light. (b) Electrons are emitted when the wavelength of the incident light is above a certain threshold wavelength. (c) Photoelectric emission is instantaneous with the incidence of light. (d) Photoelectrons are emitted whenever a gas is irradiated with ultraviolet light. 30. A 100 W light bulb is placed at the centre of a spherical chamber of radius 0.10 m. Assume that 66% of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is (a) 0.87 X 10--6 Pa (b) 1.77 X 10--6 Pa (d) None of these (c) 3.50 x 10--6 Pa

May have More than One Correct Option 31. When photons of energy 4.25 eV strike the surface of a metal, the ejected photoelectrons have a maximum kinetic energy EA eV and de-Broglie wavelength A.A- The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is E3 = (EA -l.SO)eV. If the de-Broglie wavelength of these photoelectrons is 'AB = 2'AA, then (a) the work function of A is 2.25 eV (b) the work function of B is 4.20 eV (c) EA = 2.0 eV (d) E8 = 2.75 eV 32. In Thomson's experiment, if the velocity of electron is greater than the ratio of electric field (E) and magnetic field (ie, v > E!B), then (a) the electron will reach the undeflected spot (b) the electron will not reach the undeflected spot (c) the electron will move to a spot above the undeflected position (d) the electron will move to a spot below the undeflected position 33. When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons heyve maximum kinetic energy TA eV and de-Broglie wavelength A.A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is T8 = (TA - 150) eV. If the de-Broglie wavelength of these photoelectrons 'AB = 2'AA, then (a) the work function of A is 1.50 eV (b) the work function of B is 4.0 eV (c) TA = 2.00 eV (d) All of the above 34. Electric conduction takes place in a discharge tube due to movement of (a) positive ions (b) negative ions (c) electrons (d) photons 35. The maximum KE of photoelectrons ejected from a photometer when it is irradiated of wavelength 400 nm is 1 eV. If the threshold energy of the surface is 0.9 eV (a) the maximum KE of photoelectrons when it is irradiated with 500 nm photons will be 0.42 eV (b) the maximum KE in case (a) will be 1.425 eV (c) the longest wavelength which will eject the photoelectrons from the surface is nearly 650 nm (d) maximum KE will increase if the intensity of radiation is increased.

1002

Chapter 26 • Electron and Photon

Comprehension Based Questions Passage I According to Einstein, when a photon or light of frequency v or wavelength A is incident on photosensitive metal surface of work function 0 , where W1). If current before and after change are 11 and 12 all other conditions remaining unchanged (assuming hv > W2 ) then 11 < 12 . Reason In above case 11 = 12 .

(b)

Kmax

Kmax

(a) 2 (c) l/2

(d)

42. Assertion Work function of copper is greater than the work function of sodium, but both have same value of threshold frequency and threshold wavelength. Reason The frequency is inversely proportional to wavelength. 43. Assertion Photocells are used in cinematography. Reason A photocell converts electrical energy into light energy.

Passage II Acco-rding to de-Broglie, a moving material particle exhibits dual nature (ie, a particle as well as a wave). He also predicted that a wave is associated with every moving material particle (which controls the particle) called matter wave and its wavelength is called de-Broglie wavelength given by A= h!mv where h is Planck's constant, m is the mass of the particle moving with velocity v. The existence of matter waves was firstly experimentally verified by Davisson and Germer using slow moving electrons which were accelerated with moderate accelerating potential. 38. An electron is accelerated under a potential difference of 64 V, the de-Broglie wavelength associated with electron is (use charge of electron 1.6 x 10-19 C, mass of electron 9.1 x 10-31 kg; h = 6.623 x 10- 34 J-s) . (a) 1.53 A Cb) 2.53 A Cc) 3.35 A Cd) 4.54 A

39. If a-pa ide and proton have ·s ame momenta, the ratio of de-Broglie wavelength of a,particle and proton is

44. Assertion A tube light emits white light. Reason Emission of light in a tube takes place on a very high temperature. 45. Assertion The de-Broglie wavelength of a molecule varies inversely as the square root oftemperature . Reason The root mean square velocity of the molecule depends on the temperature. 46. Assertion Stopping potential ·is a measure of KE of

photoelectron. 1

Reason W = eV5 = -mv 2 = KE 2 47. Assertion The graph of stopping potential (VJ versus frequency (v) of incident radiation is a straight line nor passing through the origin. Reason According to Einstein's photoelectric equation the slope of the graph between V5 and ,. is !!:_. e 48. Assertion A photon has no rest mass, yet it carries definite momentum. Reason Momentum of photon is due to energy hence its equivalent mass.

Chapter 26 • Electron and Photon 49. Assertion The relative velocity of two photons travelling in opposite direction is the velocity of light. Reason The rest mass of photon is zero. 50. Assertion Photoelectric effect demonstrates the particle nature of light. Reason The number of photoelectrons is proportional to the frequency of light.

Previous Year's Questions . 51. Light of wavelength A. falls on a metal having work

function he . Photoelectric effect will take place only if Ao

(DCE 2009)

A0

(b) A.~ 2"- 0

(c) A $; A0

(d) A.=4A 0

(a) A~

52. The surface of the metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of metal is (AIEEE 2009)

(a) 1.42 eV (c) 1.68 eV

(b) 1.51 eV (d) 3.0 eV

53. A radiation is incident on a metal surface of work function 2.3 eV. The wavelength of incident radiation is 600 nm. If the total energy of incident radiation is 23 J, then the number of photoelectron is (UP SEE 2009) (a) zero (b) >10 4 (c) =10 4 , (d) None of these 54. In a cathode ray oscillograph, the focusing of beam on the screen is achieved by (UP SEE 2008) (a) convex lenses (b) magnetic field (c) electric potential (d) All of these 55. A wrong argument for the particle nature of cathode rays is that they (UP SEE 2008) (a) produce fluorescence (b) travel through vacuum (c) get deflected by electric and magnetic fields (d) cast shadow 56. An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of 45 x 10-2 A. The maximum energy of a photon in the radiation in eV is (h = 6 .62 x to-34 J-s, c = 3 x 10 8 ms-1 ) (Karnataka CET 2008)

(a) 27, 500 (c) 17, 500

(b) 22, 500 (d) 12, 500

57. Millikan's oil drop experiment estabilish that {Kerala CET 2008)

(a) (b) (c) (d) (e)

electric charge depends on velocity electron has wave nature electric charge is quantised electron has particle nature electron has wave nature

58. Which phenomenon best supports the theory that matter has a wave nature? {VIT EEE 2008) (a) Electron momentum (b) Electron diffraction (c) Photon momentum (d) Photon diffraction

1003

59. Which one of the following statements is wrong in the context of X- rays generated from a X- rays tube? {liT JEE 2008)

(a) Wavelength of characteristic X-rays decreases when the atomic number of the target increases (b) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target (c) Intensity of the characteristic X-rays depends on the electrical power given to the X-rays tube (d) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube. 60. Out of a photon and an electron the equation E = pc, is valid for (BVP Engg. 2008) (a) both (b) neither (c) photon only (d) electron only 61. A and B are two metals with threshold frequencies 1.8 x 10 14 Hz and 2.2 x 10 14 Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted in (Take h = 6.6 x to-34 J-s) (Karn~taka

CET 2007)

(a) B alone (b) A alone (c) Neither A nor B (d) Both A and B 62. When a point source of light is at a distance of 50 em from a photoelectric cell, the stopping voltage is found to be V0 . If the same source is placed at a distance of 1 m from the cell, the stopping voltage will be (Gujarat CET 2007) (a) 2 V0

(b) V 0 (d) V0 /4

(c) V0 !2 63. Electrons with de-Broglie wavelength A. fall on the target in anX-raytube. The cut-off wavelength A.0 of the emitted X-rays is {liT JEE 2007) (a)

= 2mcA.

, A

0

2

2h

(b) "-o = -

me

h

(d) A. 0 =A

64. Light of frequency v falls on material of threshold frequency ,·0 . Maximum kinetic energy of emitted electron is proportional to (UP SEE 2009) (a) v- ,.0

(b) ,.

(c) ~v-v 0

(d) ''o

65. A metallic surface is irradiated by a monochromatic light of frequency ,. 1 and stopping potential is found to be V 1 . If the light of frequency v2 irradiates the surface, the stopping potential will be {Kerala PET 2006) h h (a) V1 +-(V 1 +V 2) (b) V1 +-(Vz -V 1 )

e e·

(c) V1 + h(v 2

e

-

v1)

h

(d) V1 + -;Cv 1 - v 2 )

e

(e) V1 -h(v 2 -v 1 )

66. Maximum velocity of the photoelectrons emitted by a metal surface is 1.2 x 106 ms- 1 . Assuming the specific charge of the electron to be 1.8 x 10 11 Ckg-1, the value of the stopping potential in volt will be (Karnataka CET 2006)

(a) 2 (c) 4

(b) 3 (d) 6

I

1004

Chapter 26 • Electron and Photon

67. Monochromatic light incident on a metal surface emits electrons with kinetic energies from zero to 2.6 eV. What is the least energy of the incident photon, if the tightly bound electron needs 4.2 eV to remove?

(a)

(c) 2

(BHU 2005)

(b)

J2

(d)

..!.

(AIEEE 2005)

2 70. A photocell is illuminated by a small bright source placed

(b) From 1.6 eV to 6.8 eV (d) More than 6.8 eV

(a) 1.6 V (c) 6.8 eV

1

J2

1 m away. When the same source of light is placed

..!. m

2 away, the number of electrons emitted by photocathode would (AIEEE 20\15) (a) increase by a factor of 2 (b) decrease by a factor of 2 (c) increase by a factor of 4 (d) decrease by a factor of 4

68. An element with atomic number Z= 11 emits Ka X-ray of wavelength A.. The atomic number of element which ~mits Ka-X-ray of wavelength 4/c (liT Screening 2006) (a) 6

(b) 4 (c) 11 (d) 44 69. If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

Answers Exercise I 1. (a)

2. (d) (a) 22. (a) 32. (b) 42. (c) 52. (b) 62. (b) 72. (b)

12.

11. (c) (b)

21. 31. 41. 51. 61.

(d) (b) (c) (a) (c)

71.

3. (c) 13. (d) 23. (c) 33. (b) 43. (d) 53. (d) 63,:c%,6c:) 73." (b)

4. (c) 14. 24. 34. 44. 54. ' 64. 74.

(b) (b) (b) (a) (d) (b) (a)

5. 15. 25. 35. 45. 55. 65. 75.

(a) (b) (b) (d) (b) (d) (c) (c)

6. 16. 26. 36. 46. 56. 66. 76.

(b) (c) (c) (a) (c) (a) (b) (b)

7. 17. 27. 37. 47. 57. 67. 77.

8. (c)

(a) (b) (a) (d) (b) (b) (a) (c)

18. 28. 38. 48. 58. 68. 78.

9. (c) (c) (d) (c) (c) (c) (c) (a)

(d) (b) (a) (b) (a) (b) (b)

19. 29. 39. 49. 59. 69. 79.

10. 20. 30. 40. 50. 60. 70. 80.

(c) (a) (c) (c) (a) (b) (d) (b)

10. 20. 30. 40. 50. 60. 70.

(d) (c) (b) (c) (c) (c) (c)

Exercise II 1. (d)

11. (c) (b) (a,b,c)

21. 31. 41. 51. 61.

(d) (c) (b)

2. 12. 22. 32. 42. 52. 62.

'j

.

·· · \1

,

-_i\

).

(a) (b) (c) (b,d) (d) (a) (b)

.ll

3. 13. 23. 33. 43. 53. 63.

(b) (a) (c) (b,c) , (b) (d) (a)

4. 14. 24. 34. 44. 54. 64.

(c) (d) (c) (a,b,c) (c) (c) (a)

5. 15. 25. 35. 45. 55. 65.

(b) (c) (c) (a,c) (b) (c) (b)

6. 16. 26. 36. 46. 56. 66.

(d) (b) (c) (a) (c) (a) (c)

7. 17. 27. 37. 47. 57. 67.

(d) (c) (b) (c) (a) (c) (c)

8. 18. 28. 38. 48. 58. 68.

(b) (a) (a) (a) (a) (b) (a)

9. (d) (d) (c) (b) (b) (b) (b)

19. 29. 39. 49, 59. 69.

Chapter 26 • Electron and Photon

1005

Hints &Solutions Exercise I ~ qV

2 1. qV = .!mv2 or v = 2 m ie v oc {q . vHe = 'q-He_x_m_H_

'

'lj-; ..

VH

qH

mHe

-

~- 1

'lj--; "t;; -

7. Asq vB = mV21r 19 qBr (2xl.6xl0)xlxl or m- - - ..::.._ _ _ _ _...;,_ ____ - v 1.6 x l0 7

.fi

DEql mv !1._ = X v2 0.02 X(10 6)2 m DEl 0.21 x (2 x 104 ) x (5 x 10-2) = 9.52 X 10 7 Ckg- 1

=2 xlo-26 k = 2xl026 =12 g 1.66 x to- 27 Therefore, particle must be c + +.

2. Here, x = - -2-

or

Mv 2

.

3. In mass spectrograph, - - = q v B 1 r

E and qE = B q v or v = B

or r =

Mv M (E) ME wq = B q B = qBB I

qvB

m

m

2xl.6xl0- 19 x 6x105 x0.2 6.65 x10- 27 = 5.77 x 10 12 ms-2 2 9. eV = mv l r; so V oc v2; 2 2 1 2 = 500 (2v) ... V2 = V1 (v ~ --;- = 2000Vcm _ 10.

I

4. E=.!mv 2 orv2 = 2Eim;' 2 If E' is the intensity of electric field applied then E'q = m V21r or r = m v2 IE' q m(2E lm) 2 E or r = Eq qE ie, roc 1 I q. 1

so .

4

n. = 5 x 107cm-3 = 5 x 107 x 106m-3 No. of positive ions, np = 5 x 10 7 x 10 6 = 5 x 10 13 m-3 v = 0.4 ms- 1; J = 4 x 10-6 Am-2; vp = ? Use the relation J = nee ve + np e vp and solve it for vp 4Xl0-6= (5x10 13 X 1.6 X 10-19 X 0.4) + (5 X10 13 X 1.6 X 10-19 Xvp) 6 6 _4xlo- -3 .2 x lo- 0.8xl0-6 _ -1 01 vP-.ms 8.0xlo- 6 8xlo-6 B21D q 6. For similar parabola; y 2 = -· - - x, will be same for

E m

B2q two particles. It means - - = a constant for these two particles. m 2 m1 = B{q 1 = (0.8) x~ = ~ m2 B~q 2 1.2 2e 9

qV

mg=qE or -nr 3 pg=- or Vocr 3 3 d V2 = V1 (

~:

J

= 400 x (

fJ

= 3200 V

11. When pressure in a tube is reduced in the range 1 em and 10-3 em; the mean free path of moving electron in the discharge tube increases. As a result of which the electron gets higher KE while moving towards anode and then cause ionisation of the atoms with which it will collide on its ways causing excitation phenomenon. 12. Here, u = O,a = qEim;s = l and v =? qEl As v2 =u 2 +2as; so v 2 = 0+2-m

v=~2~El

or

5. Here, No. of electrons

,

F

J

so r is related with (qiM) .

1

.

8. Acce 1erat10n, a = - = - -

13. From the symmetry of figure, the angle 8 = 45°. The path

of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure, AC=2rcos45°= 2rx .,1=.fir mv

2

As B q V = - -

r

AC =

... (i)

mv or r = Bq

27 7 .fi mv =.fix 1.67 x 10- x 10 = O.l 4 m Bq l xl.6x lo-19

14. qE = mg or E = mg = (SOx 10-6) x 9.8 = 98 Nc-1 q S x lo- 6 Since the force due to electric field on charged particle should be opposite to the gravity pull and charge on the drop is negative, hence the electric field must act vertically downwards.

1006

Chapter 26 • Electron and Photon

15. A charged particle moves along a straight line with acceleration, hence electric field should be parallel to the direction of motion of charged particle and no force should act on charged particle due to magnetic field. It will be so if charged particle is moving parallel to the direction of magnetic field. Sm0 16. mmo mo 2 2 2 2 3 - ~1- v /e - ~1- (0.8e) /e 6n11rv - 17. neE= 6n11rv or n = - - - · eE X

0.01 =

15

34. Given E/e = 3.3 x 10-13 kg ms- 1 ;

82

. '1 ar parab o1a, - l Dq c b ot h t h e 19. For s1rru - must b e same 10r Em ions. :: = (

!: J ::J !: X (

X

=

(%

JnJ i J ~ X

X(

=

20. Specific charge = charge/mass. The positive rays are stream of positive ions. The mass of positive ion is much more than that of electrons, hence specific charge of positive ions is less.

= _!_mv 2 or v = ~2qV /m

ie, v oc .J1i 2 22. V = E/8 = 6 X 10 4/8 X 10-2 = 7.5 X 105 ms-1 23. The length of Crooke's dark space will be equal to the length of tube ie, 15 em. 24. Gained in KE = q V = 2 x 5 = 10 eV

21. qV

25. Ek =

1 q2 82 r2 . te, r

2

.JE,;

oc

28. Force on the charged particle in magnetic field is 1\

"

"

1\

F = q(vk x Bj) = qv B(k xj) = qvB(-i)

29. r = m v ie, qB

So,

roc

E = 3 .3 X 10-13 X e = 3.3 X 10-13 X 3 X 108 = 9.9 X 10-5 J

35. E = hv =he I A.= me 2 , hence A.= hI me= hlp 36. Photons move with velocity of light and have energy hv. Therefore, they also exert pressure . 37. E

= h\' = me 2 or m = h"!e2

he 38. Energy of photon E = - = me 2 ; A. momentum of photon = me = h!A. 39. Energy of a photon E = he; E is less if A. is longer. A. 40. E = hv !A.= he (ineV) eA. 6.6 X 10-34 X 3 X 10 8 -------oc, - - - - = 5.9 1.6 X 10-19 X 0.21

X

10-6 eV

42. Energy of photon

26. For a charged particle in magnetic field B, r = m v/qB. The radius can be fixed for a charged particle if v and B both are fixed. 27. elm of the anode rays depend on the nature of the gas filled in the discharge tube. "

So,

41. E =he/A.= 6.6 x 10-34 x 3 x 10 8 15000 x 10-19 = 3.973 X 10-19 J

m r---So, r2 = r1 ~Ek2 /Ek 1 = R-./3 = -J3 R

~

33. Momentum of incident light per second E 60 -7 P1 =- = - - = 2x10 e 3x 108 Momentum of reflected light per second. 60 E 60 -7 100 e 3x10 Force on the surface = change in momentum per second = P2 -(-p1) = P2 + P1 = (2+1.2)x10- 7 = 3.2 x 10- 7 N

= mV.;r or mv = q r B .

So,

... (i) ... (ii)

p2 = - X - = - - =1.2x10 8

6 X 3.14 X 1.6 X 10-5 X 5 X 10-7 1.6 X 10-19 X 6.28 X 105 18. q v B

32. he I A.= 10 3 eV h\- = 106 eV Dividing Eq. (ii) by Eq. (i) we get, v = 10 3e I A. = 10 3 x 3 x 108 I 1.24 x 10-9 = 2.4.x 10 20 Hz

m/q

r m q 24x2e 24 -1 = -1X2- = - - - = r2 m2 q1 22xe 11

1 1 eE x2 e 2 y v2 30. y=-at 2 = - - -2 o r - = -22 2 m v m Ex 2xl.Sx10-3 x(3x10 7 f

34 8 E = he = 6.63 X10- X3 X 10 = _6.63 X 3 X _18 10 9 A. 590x1o59 90 Light energy produced per second = x 10 = 9W 100 :. Number of photons emitted per sec 9x59 18 = 2.67 X1019 6.63x3x10 43. E = he I A. or E oc l I A.; so E2 = E1 x /... 1 I /... 2 = 3.2 X 10-19 X 6000/ 4000 = 4.8 X 10-19 J 44. Number of photons emitted per sec, Power P n=-----Energy of photon hv 10000 = 1. 72 X1031 6.6 X 10-34 X 880 X10 3

1800 X(0.1) 2 = 1.5 X 10 11 Ckg-1

31. Planck's constant, h = E/v = [ML2 r 2; r1 ] = [ML 2r 1 ] Angular momentum, L =I w,;, [ML2r 1]

45. Ek =he

(.!. __!_)cineV) /...

e A.

0

34

= 6.6 X10- X 3 X 10 1.6 x Io- 19

8

10

(

10 I8oo

_

10

J

10 = l.S eV 2300

Chapter 26 • Electron and Photon

46.

1

2

59. Let £ 1 and E2 be the KE of photoelectrons for incident light of frequency v and 2v respectively.

he . 2 mv =T-o (meV)

Then hv = £ 1 + o and h 2 v = £ 2 + 0

6.6 X 10-34 X 3 X 108 4000 X 10-10 X 1.6 X 10-19

--~--~--~~~-2

So, 2(£1 + 0) = E2 + 0 or E2 = 2£1+ 0 It means the KE of photoelectron becomes more than double.

= 3.1-2 = 1.1eV = 1.1 x l.6 x 1o- 19 J =1.76 x 10-19 J 1.76x1o-19 x 2 v 9 X 10-13 = 6.2 x 10 5 ms-1

60. Retarding potential,

v =~-!tL s

47. Ek =E- 0 =6.2-4.2=2.0eV = 2.0 X 1.6 X 10-19 = 3.2 X 10-19 J . 1

1

2

2

48. -mv1 =2 0 - 0 = 0 and -mv2 =10 0 - 0 =9 0. 2 2 . v1 ..

V

2

[k_

1 =~~=3

49. Photoelectric current oc intensity of incident light. Therefore , the graph is a straight line having positive slope passing through origin. 50. The value of threshold frequency v0 for A is less than that forB , hence A < 8 . 51. Maximum KE of the emitted photoelectrons = hv -h\-0 . 52.

1

2

2

m v1 = 2hv0 1

2 So ,

1

-

hv 0 = hv 0 and

2

= 5hv 0 -

2

1

mv 2

2

2

2

v2 = 2v1 = 2x4 x 106 = 8 x 10 6ms- 1

54. The maximum KE of the emitted photoelectrons is independent of the intensity of the incident light but depends upon the frequency of the incident light.

61. Maximum KE = hv - o = 6.63 X 10- 34 X 8 X 1014 - 3.2 X 10- 19 = 2.1 x 10- 19 J 62. KE of fastest electron = E - 0 = 6.2-4.2 = 2.0eV = 2 X 1.6 X 10- 19 = 3.2 X 10- 19 J 63. Energy incident over 1 cm2 = 1.0 x 10-4 J ; Energy required to produce photoelectrons = 1.0 X 10-4 X 10-2 = 10-6 J . Number of photoelectrons ejected = number of photons which can produce photoelectrons = energy required fo r producing electron/ energy of photon. 10-6 X 300 X 10- 9 10- 6 = --- = - - - - - - he / A 6.6 x 10- 34 x 3 x 108 =1.51 x 10 12 s- 1 64. o = ..!!:__ (in eV) = 6.6 x 10-34 x 3 x 108 = 1.87 eV eAo . 6600 x 10- 10 x1. 6 x 10- 19 65. 0 = he I A0 (in eV) 6.62x10-34 x 3 x 108 - -----,-. , . . - - - -= = 2.29 eV 5420 X 10- 10 X 1.6 X 10- 19 66.

till = he-~

A1

6.63 X 10-34 X 3 X 108 1 (3 X 10- 7 ) X 1.6 X 10- 19

or

V=

56. The velocity of photoelectrons depends upon the frequency of the incident light. 57. Maximum KE =he - o

A

6.6x10- 34 x3x10 8 1 _:__ _ ___:~.:....:.._ X - 2 = 1.1 eV 400 X 10- 10 1.6 X 10- 19 58. MaximumKE' = E- 0 =3.4-2 .= 1.4eV t

he(A 2 - A1) . A1A2 (m eV)

= 2.47 eV

67. Photoelectric current (J) oc Intensity of incident light and .

.

mtens1ty

= 4.14- 1 = 3.14 eV 2x3.14xl.6x10-19 106 -1 = ms 9.1 x 10-31

A2

_ 6.62 X 10- 34 X 3 X 10 8 X (5000- 2500) X 10- 10 2500 X 5000 X 10- 20 X 1.6 X 10- 19

1 he 2 -zmv = T-o

55.

e

Ae

9 1240 x 1o- -l.0 7 9 330 x 10-

= 3.73-1.07 = 2.66 v

hv 0 .= 4 hv 0

- mv2 =4 x -mv1

or

1007

So, I

oc

1 oc - - - -

(distance)2

1

(1] 2

Hence I' = I (distance) 2 4 •

= -I 16

h he 68. For electron, p = -; and for photon, E = -

A

A

:. §_=he / A. =e=3 x 108ms- 1 p h/ A 69. KE of thermal neutron, .!mv 2 = ~kT

2

or

mv=~;

So,

2

A=!!:_=~ p

3kmT \

1008

Chapter 26 • Electron and Photon Rate of change of de-Broglie wavelength

h

70. We know that, A= ~; -y2mkT

dA

Aoc-

So,

,[f

A27 = A927 or

A

h or v = mA

71. Momentum, p = m v = h/A · h 6.62 x w-34 or v - - - ----;:::--------;:;- mA- 9.1x10- 31 x5.2x1o- 7 = 1.4 X 10 3 ms- 1=1400 ms-1 72. E = 3eV = 3xl.6x1o-19 J

&=-hmf>v 1.034 x 10-34 9.1 X10-31 X0.0025

----,~--- = 4634 x 10-5 m

-6.6 X 10-34

r======::::======:== = 1.65 x w- 11 m 27

~2xl.7x10- x 3 x l.6x10-19

h h)1-v 2 / c2 73. A= - = =0. mv m0 v

6.6x 10-34 9.1 X10-31 X5200 X1Q-10 =1.4 X 10 3 ms-1

0 005 50 76. Here, L'>v = · x = 0.0025ms-1 100

A=-h-

(·: v =c)

78. t>p = !!___ = 1.034 x w-

34

w-w

&

= 1.034 x 10- 24 kg-ms-1 n. h 79. op=-=-

A

&

eE 74. Here, u = 0; a= - ; v = ?;t = t

h 80. L'>p=mf>v=-

m

&

eE v = u+at = 0+-t m de-Broglie wavelength,

A=!!_= h mv m(eEt!m)

-h = eEt 2

h 75. mv=-

927+273 = 2 27 +273

A27 = 2An7 = 2A

~2mE

1)

h (

dt = eE -t2

1

or

n. 1.034 x w-34 L'>v = - - = ------::c=-------cm& 1.67x10-27 x6x1o-s

h eEt

Exercise II 1 2 1.. eV =-mv 2

v=~ ;; 2

or

rm

roc-

or

q

=)2x(l.8x10 )x9 =1.8 x10 ms- 1 11

6

2. As accelerating voltage V across X-rays tube increases, the value of minimum wavelength of X- rays, A = ~; c eV decreases ; so the separation between A.K and 'Ac increases. 3. Ek = eV = hv- 0 or

V=

~v-.fu

e e Slope of straight line between V and v is ~ h = e x slope of straight line. e 4. E = .!.mv 2 2

So,

--=8

Q'V' QV

Q I = 8 QV = 8Q X 800 = 2Q V' 3200

2

r= mv =_!!:_X qB

~2mEK

As per question;

r

,

Q'V' 4 --=-n(2r) 3 pg l 3

6 . As EK = .!.mv 2 or mv =

mv 2 qvB=or

and

or

v=J¥

or

r1 = r2 4 3 5. ---nr pg l 3 QV

qB

{2E = ~2Em {;-

qB

or

mpvp = meve

or

~2 mp EKP = )r:-2-m_e_E_K_,

Chapter 26 • Electron and Photon

EK p

or 7.

1

h h 1 14. A=- or L =- ie, L oc -. The curve (d) is correct.

EK, =->1 mP __

or

p

me

EKe >EKP

2 mv

2

V

he

15.

.

V

1

he

16. KA = -

AA

. KE = he- -he - (.meV) Max1mum A.e A0 e =

:e(i- :J

1 1 = 1240(-- - - - ) 100 200 = 6.2 eV 1 9. The mass of electron is about - - times that of a neutron 1836 and angular momentum of electron is quantised in the hydrogen atoms but not the li11ear momentum of electron. 10. Since the energy of incident electron, E = 80 keV. The minimum wavelength of X- rays produced is

A.=he= 6.6x10- 34 x 3x108 E 80 X 1000 X 1.6 X 10- 19 = 0.155 x m = 0.155 A Since the energy of K-shell electron is -72.5 keV, so the incident electron of energy 80 keV will not only produce continuous spectrum of minimum wavelength 0.155 A but shell also knock electron of K shell out of atom, ~-esulting emission of characteristics X- rays.

or V = mv; = v;;, s 2e 2(e/m)

he

- W2 • The photoelectric current is proportional to the intensity of incident light. Since, there is no change in intensity of light. Hence, 11 =12 . Therefore, reason is true but assertion is false.

42. When work function of copper is greater than the work function of sodium, then

s

e

... (i)

c =-::;: 0

Hence Eq. (i) becomes

e

2 hY E =me = hv => m = c2 Therefore, momentum of photon hY ,Jzy =mc= -xc = c2 c Thus, photon possessed momentum due to its equivalent mass even its rest mass is zero .

49. Velocity of first photon = u = c Velocity of second photon = v = - c Now, relative velocity of first photon with respect to second photon

u-v 1- uv

Na

But we know that Y0

V = hY-