26.electron & photon

8/22/2019 26.Electron & Photon

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Electron and Photon

Chapter OverviewDiscovery 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 lowpressures. If the pressure in the tube is lowered to about 10-4 atm, glassbegins to fluoresce (glow) faintly. It was established that the glow wasdue to bombardment of the glass by a certain kind of rays emerging fromcathode (negative electrode) which travel in a straight line until theystrike the anode (positive electrode). These rays were called as cathode

rays .Sir J J Thomson demonstrated thatwhen cathode rays were deflectedon to an electrometer, it acquired negative charge. He also showed thatthe rays were deflected on application of an electric field. The cathode ray

beam was deflected away from the negatively charged plate. These resultswere found to be identical, irrespective of the gas taken in the dischargetube. He concluded that the cathode rays were a stream of fast movingnegatively charged particles called electrons (named by stoney). He alsocalculated the velocity and specific charge for an electron. The specificcharge is the ratio of charge to the mass of an electron, denoted as elmratio . The el m ratio was found to be same for all gases. This led to theconclusion that the electron must be a fundamental or universal particlecommon to all kinds of the atoms.


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988 Chapter 26 Electron and PhotonProperties of Cathode Rays

(i) Cathode rays travel in straight lines.(ii) The cathode rays are independent of the nature of thegas or electrodes employed to produce them in thedischarge tube. Therefore, .::_ for cathode rays is amuniversal constant equal to 1.7592 x 1011 C kg-1.(iii) They can be deflected by electric and magnetic fields.(iv) They have penetrating power and can penetrate throughsmall thickness of matter.(v) On striking the target of high atomic weight and high

m ~ l t i n g point, they produce X-rays.(vi) They produce fluorescence and phosphorescence incertain substances and hence affect photographic plate.(vii) They have small ionising power and ionise the gasthrough which they pass.

(viii) They travel in straight linewith high velocity, momentumand energy and cast shadow of objects placed in theirpath.(ix) They emerge normally from surface of cathode and cantherefore be brought to focus by using a concave cathode.(x) They heat up the material on which they fall.(xi) They exerts mechanical pressure, so they can rotate asmall paddle wheel.(xii) They can produce physical and chemical change.(xiii) They can exhibit interference and diffractionphenomena under suitable arrangements. T h ~ s theymay behave as waves.

r> Positive RaysPositive rays were discovered by Goldstein. Positive rays aremoving positive ions of he 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) These consist of fast moving positively charged particles.(ii) These rays are deflected in magnetic field.(iii) These rays are deflected in electric field.(iv) These rays travel in straight line.. (v) Speed of positive rays is less than that of cathode rays.(vi) 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 ofejection of electrons froma metal surface, when lightof sufficient high frequencyfalls on it, is known as thephotoelectric effect.The figure given belowshows the experimental set upto study photoelectric effect.When a suitableradiation is incident on the

p

A~ - - - - - - ~ + ~ v ~ - - - - - - - - - ~

electrode P, electrons areejer::ted from it. The electronswhi'cf.. ' ilave sufficient kinetice n ~ ~ g y , the e ' l e c t r ~ d e Q, ( e ~ p i t e its negative polarity).,

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

laws of Photoelectric EffectFollowing are the laws of photoelectric effect(i) For each emitting metal, there is a certain mirumumfrequency v0 ( or maximum wavelength /..0 ) , called thethreshold frequency of the incident radiation, belowwhich no emission of photoelectron takes place, nomatter 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 aninstantaneous process. There is no time lag (


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(iv) The efficiency of photoelectric effect is less than 1% ie,number of photons less than 1% are capable of ejectingphotoelectrons.Thus, energy imparted by the photon = maximum kineticenergy of the emitted electron + work function of the metal.

or1 2

hv = 2mvmax +or !m v2 = hv-"'2 max 'I' ... (i)I f he energy of the incident photcm is just equal to the workfunction of the metallic surface, the kinetic energy of the metalsurface is zero.Eq. (i) is referred as Einstein's photoelectric equation.

Substituting = hv0 in Eq . (i)1 22mvmax = hv- hv 0

he heor A. A. 0Illustration 1 A metal has a work unction of2.0 eV. It is illuminatedby monochromatic ligh t of wavelength 500 nm. Calculate (a) thethreshold wavelength, (b) the maximum energy of photoelectrons, (c)the stopping potential. (Given, Planck's constant, h = 6.6 x 10-34 Js,charge on electron, e = 1.6 x 10 -19 C and 1 eV = 1.6 x 10 -19 J) .Solution Here, h = 6.6 x 10-34 Js, e = 1.6 x 10-19 c,1 eV = 1.6 X 10 -19 J= 2.0 eV = 2.0 X 1.6 X 10-19 = 3.2 X 10-19 J,A.= 500 run= 500x 10-9 m(a) If A.0 is the threshold w a v e l ~ n g t h , then

= heAo

Ao =he= 6.6xl0-34 x3x108 3.2x 10-19

or= 618.75x 10-9 m= 618.75 run(b) The maximum energy of the photoelectrons,

1 2 he-zmvmax = hv - =T-6.6x10-34 x3xl08 _ 3.2 x 10_19SOOxl0\9= 3.96 X 10 -19 -3 .2 X 10-19 = 0.76 X 10-19 J(c)"The stopping potential is given by

1 2 -1 9eV0 =2mvmax =0.76xl0or v. = 0.76 x l0- 19 = 0.76xl0-19 =0.475 v0

e 1.6x10-19Characteristics of Photoelectric Effect

(i) Effect of intensity Figure shows graph of photocurrentas a function of potential difference for light of constant frequencyand two different intensities. When potential difference V issufficiently large and positive, the current becomes constant. Thestopping potential difference V0 needed to reduce the current tozero is shown. For a given frequency if the intensity of light isincreased (or we can say that number of photons incident per unitarea per unit time is increased) the.photoelectric current increasesor viee-vers ; but the stopping potential remains the same.

Current(1)

Chapter 26 Electron and Photon 989

21

v =constant

- - f . , - - ! : , - - - - - - - - -+Po ten t ia l0 difference (V )(ii) Effect of frequency Figure shows the current as afunction of potential difference for two different frequencieswith the same intensity in each case.

Current(1)

(+)VIf the frequency is decreased, the stopping potenti&ldecreases and at a particular frequency of incidentlight, the stopping potential becomes zero. This valueof frequency of incident light for which the stoppingpotential is zero is called threshold frequency v0 Ifthe frequency of incident light (v) is less than thethreshold frequency (v 0), no photoelectric emissiontakes place.

(iii) Effect of photo-metal When frequency and intensityof incident light are kept fixed and photo-metal is changed, weobserve that stopping potential (V5) versus frequency (v) graphsare parallel straight lines, cutting frequency axis at differenpoints. This shows that threshold frequency is different fordifferent metals, the slope (:) for all the metals is same andhence universal constant.

Metal1

' 'Figure shows that threshold freql,lency and workfunctiOJ.are greater for metal 2 as compared to meta l l . . ,>(-10 NoteThe negative potential applied to the c o l f e c t c i { ~ ~ d ~ r ~ f ; v ; ; ; i f f i ~electrons reaching the collector (ie, to red'uc e1he c f t o ' ~ l ~ i t r c l i r r ~ t \ f t bzero) is kirown as stop'pingpOteOti'al . l ;r r ~ i , ~ ' "1: ,;;r : {.1 '_)A:.X.- . s


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990 Chapter 26 Electron and PhotonImportant Features

Three major features of the photoelectric effect could not beexplained by the wave theory of light which were later explainedby Einstein's photon theory.(a) Wave theory suggests that the kinetic energy of the

photoelectrons should increase with the increase inintensity of light. However, Kmax = eV0 suggests that itis independent of the intensity of light.(b) According to wave theory, the photoelectric effectshould occur for any frequency of the light, providedthat the light is intense enough. However, E ;::: orv 2:: v0 or A. $ /..0 suggest that photo mission is possibleonly when frequency of incident light is either greaterthan or equal to the threshold frequency ,.0 .

The electrons will be emitted from metal surface only if the frequency ofincident light is greater than threshold frequency v0. The e l ; c t ~ c > n s are emitted instantaneously. The interaction betweenphotons and electrons is one to one. So, weak incident light very fewphotonsarrive per unit time, but each one has enough energy to eject anelectron instantaneously.

The work . f u ~ c t i o n and threshold frequency \'o varies from metal tometal.

The frequency\' and wavelength A, are related as ,. = /'A..The energy ofphoton =hv =heA,'

For a given intensity I =nhv greater the frequency lesser will be numberof photons. Hence, lesser will be photoelectric current ie, p_hotocurrent0


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llltext Questions 26.1(i) Db Xcrays show phenomenon of photoelectric effect?

(ii) It is harder to r e m o y ~ a free electron frgm copper than from sodium? Which metal has greater work function? Which has highert p r e h g ~ d v y a v e l ~ l l g t ~ ?

G ; r e ! j ~ ~ p ~ ~ ejEcts electrons ~ m aSrrtain photosensitive surface, yellow light does not. Will (a) red light (a) violet light eject~ ~ o ~ g e ~ s c ~ ~ from the same substance ?(iv) A photon and an electron have same wavelength. Which particle is moving faster?

Dual Character of RadiationIn case of light some phenomenon like diffraction andinterference can be explained on the basis of its wave character.However, the certain other phenomenon such as black bodyradiation and photoelectric effect can be explained only onthe basis of its particle nature. Thus light is said to have a dualcharacter. Such studies on light wave were made by Einstein in1905. Louis-de-Broglie, in 1942 extended the idea of photons tomaterial particles such as electron and he proposed that matter

also has a dual character as wave and as particles .De-Broglie Relation

According to de-Broglie a wave is associated with energymoving particle. These waves are called de-Broglie waves ormatter waves.According to quantum theory, energy of photonE = h ... (i)If mass of the photon is taken as m, then as per Einstein'sequation

E = mc2 (ii)From Eqs. (i) and (ii)h\' = mc 2

h = mc 2 , where A. = wavelength of photon.A. hA. = - -mede-Broglie asserted that the above equation is completelya general function and applies to photon as well as all othermoving particles. A-!!__ =-h-So, - mv .J2mEwhere m is mass of particle and v is its velocity.(i) -de-Broglie wavelength associated with charged particle

A - ~ - _ h _ _ h- p - .JzmE - ~ 2 m q V(ii) de-Broglie wavelength of a gas molecule

A=_ _ _ where T absolute temperature.J3mkTk = Boltzmann's constant = 1.38 x 10 -23 J/K(iii) Ratio of wavelength of photon and electron. The

) wavelength of photon of energy E is given by A = heP E'while the wavelength of an electron of kinetic energy Kis given by A. = b . herefore for same energy, thec -v2mKratio

AP = .:._.JzmK = ~ 2 m c, Ae E . E2

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

hA= .J2mE6.6 x 10-34

X 1.67 X 10-27 X 1.6 X= 0.2857 X 10-lQ= 0.2857 A.

Davisson and Germer ExperimentThe wave nature of the material particles as predicted byde-Broglie was confirmed by Davisson and Genner (1927) inUnited States and by G P Thomson (1928) in Scotland.Experimental arrangement used by Davisson and Germer isas shown in figure . Electrons from hot tungsten cathode (C) areaccelerated by a potential difference V between the cathode andanode (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 thecrystal. The intensity of the electron beam scattered in a givendirection is found by the use of a detector. By rotating the detectorabout an axis through the point 0, the intensity of the scatteredbeam can be measured for different values the angle betweenincident and the scattered direction of electron beam.

OJCDw3

A

Electrongun

.. Detector

Davisson and Germer, found that the intensity of scatteredbeam of electrons was not the same but different at differentangles of scattering. It is maximum for diffracting angle' 60 at54 volt potential difference.


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992 Chapter 26 Electron and Photon

54VIf the de-Broglie waves exist for electrons then theseshould 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 thisexperiment is d = 0.91 Aand the Bragg's angle = 65 . This givesfor n = 1, A = 2 x 0.91 x 10-lo sin 65 = 1.65

(9,,9}0

. 0 ....,:.___,_______\ : ~ ~ . Y d .... . . , ~ t o r i 1 i . c planes

theNow de-Broglie wavelength can also be determined by usingC l 1 _ 12.27 _ 12.27 _ 1 67Ao!OrillU a "- rv - .JS4 -- 0 0 Thus the de-Broglie

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 e l n l . ~ l t o t h e w a V ' e l e n g t h of the radiatip:;?> , . > .pn needs to be t h e r m ~ s e ~ the environment, before it can be used for neutron i f f r a c t i e x p r i p ~ n t s . W:4y?


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,


kinetic energy with which an "'""''-w.vueinitt ed from a metal surface is independent of thef the light and depends upon its frequency.The number of photoelectrons emitted ,s mctep,en,aeJltthe frequency of th cident llght its intensity.

: Einstem's equation of photoelectric effect .1 22mvmax

1 22mvrriax

bove relation isv e H ~ n ? ; I n o f the waveve1enl'lll of particle.

16. de-Broglie wavelength of


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llluSlrativeExample 1 What is frequency and energy ofa photon ofwavelength6000 A? (Given h = 6.6 x 10-34 Js, c = 3 x 108 ms-1)

(a) 15 x 104 Hz and 3.3x 1o-10J(b) 6 x 1010 Hz and 4.4X 10-19J(c) 5 x 1012 Hz and 3.3 x 10-19J(d) 5 x 1014 Hz and 3.3 x 10-19J. e 3x 108SolutiOn Frequency v =-;;- =

10= 5 x 1014 Hz

11, 6000xl0-and energy= hv = 6.6 x 10-34 x 5 x 1014 = 3.3 x 10-19 JExample 2 The work function of sodium is 2.3 eV. Calculate themaximum wavelength for the light tha t will cause photoelectrons to beemitted from the sodium.

(aJ 64oo A (bJ 5ooo A(c) 54ooA (d) 5200A

Solution KEmax =hv - 0A.heT- q>0> 0 ( K i n ~ t i c energy is always +ve)A-o

A. = eh = 3x108 x 6.62 x 10 -34 = 5400 Amax q>0 2.3 X 1.6 X 10-19Example 3 In an experiment on photoelectric effect it was observedthat for incident light of wavelength 1.98 X 10-7 m, stopping potentialis 2.5 V. W h ~ t is the energy of photoelectrons with maximum speed,workfunction and threshold frequency?

(a) 6.25 eV, 3.75 eV, 9.10 x 1014 Hz(b) 3.75eV, 6.25 eV, 9.10x1014 Hz(c) 4.75 eV, 6.25 eV,8.10x1014 Hz(d) None of the aboveSolution I


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Chapter PracticeExercise I

Cathode and Positive Rays1. Doubly ionised helium atom and hydrogen ions areaccelerated, from rest, through the same potentialdifference. The ratio of final velocities of helium andhydrogen is

(a) 1 : ../2(c) 1 : 2 (b)../2 : 1(d) 2 : 12. In Thomson's mass spectrographs, when an electric field

of 2 X 104 Vm-1 is applied then the deflection producedon the screen is 20 mm. If the length of the plates is 5 emand the distance of the screen from plates is 21 em andthe velocity of positive ions is 106 ms- 1, then their specificcharge will be(a) 107Ckg-1(c) 5.9 X 107 Ckg-1

(b) 2.59 X 107 Ckg-1(d) 9.52 X 107 Ckg-l3. The working principle the mass spectrograph is thatfor a given combination of accelerating potential andmagnetic field , the ion beam (with charge q and mass M)to be collected at different positions of ion collectors will

depend upon the value of(a) M(c) q/M

(b) (q/ M)2(d ) qM

4. The mass of a proton is 1836 times that of an electron.An electron and a proton are projected into a uniformelectric field in a direction perpendicular to the field withequal initial kinetic energies. Then(a) the electron trajectory is less curved than the protontrajectory(b) the proton trajectory is less curved than the electrontrajectory(c) Both trajectories are equally curved(d) Both trajectories will be straight

5. An ionisation chamber, with parallel conducting plates asanode and cathodes has 5 x 107 cm-3 electrons and thesame number of singly charged positive ions per cm3. Theelectrons are moving toward the ar:ode with velocity 0.4ms-1. The current density from anode to cathode is 4J,!Am-2.The velocity of positive ions moving towards cathode is(a) 0.1 ms-1 (b) 0.4 ms-1(c) zero (d) 1.6 ms-1

6. In Thomson mass spectrograph, singly and doubly ionisedparticles from similar parabola corresponding to magneticfields of 0.8 T and 1.2 T for a constant electric field. Theratio of masses of ionised particles will be,

(a) 3 : 8(c) 8 : 3

(b) 2: 9(d) 9 : 2

7. In an ionisation experiment it is found that a doublionised particle enters a magnetic field of 1 T anmoves in a circular path of radius 1 m with a speed o1.6 x 107 ms-1. The particle must be(a) c+ + (b) Be++(c) Li++ (d) He++

8. An a-particle of mass 6.65 x 10-27 kg travels at righangles to a magnetic field of 0.2 T with a speed o6 x 105 ms-1. The acceleration of a-particle will be(a) 5.77 x 1011 ms-2 (b) 7.55 x 1011 ms-2(c) 5.77x 1012 ms-2 (d) 7.55 x 1012 ms-29. Cathode rays of velocity 106 ms-1 describe aapproximate circular path of radius 1 m in an electrfield of 500 V cm-1 . If the velocity of cathode rays doubled, the value of electric field needed so that thrays describe the same circular path is(a) 1000 V cm-1 (b) 1500 V cm-1(c) 2000 V cm-1 (d) 500 V cm-1

10. An oil drop with charge q is held stationary between twplates with an external potential difference of 400 V.the size of the drop is doubled without any change ocharge, the potentia l difference required to keep the drostationary will be(a) 400 V(c) 3200 V (b) 1600 v(d) 4000 v

11. Air becomes conducting when the pressure rangebetween(a) 76 em and 10 em (b) 10 em and 1 em(c) 1 em and 10-3 em (d) 10--4 em and 10-7 em

12. An electron of mass m and charge q is accelerated frorest in a un iform electric field of strength E. The velociacquired by it as it travels a distance l is(a) ~ 2 E q l ! m(b) ~ 2 E q ! m l(c) ~ 2 E m ! ql(d) ~ E q ! m l

13. A proton of mass 1.67 x 10-27 kg enters a uniformagnetic field of 1 Tat point A as shown in figure, withspeed of 107 ms-1. The magnetic field is directed normto the plane of paper downwards. The proton emergout of the magnetic field at point C, then the distance Aand the value of angle 8 will respectively be


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996 Chapter 26 Electron and PhotonX X X X45 ---7BX X X XAX X X XcX X X X

X X X X

(a) 0.7 m, 45 (b) 0.7 m, 90(c) 0.14 m, 90 (d) 0.14 m, 4514. An oil drop of mass 50mg and ofcharge -5 J.!C is just balancedin air against the force of gravity. Calculate the strength ofthe electric field required to balance is, (g = 9.8 rns-2)(a) 98 Nc-1 upwards(b) 98 Nc-1 downwards(c) 9.8 Nc -1 towards north(d) 9.8 Nc - 1 towards south15. A charged particle is moving in the presence of electric~ ----t --7field E and magnetic field B. The directions of E and B

are such that the charged particle moves in a stra ight line--> -->and its speed increases. The relations amongst E , B and

-->velocity v must be such that-->--> -->(a) E B = 0, v is arbitrary(b )(c)(d)

-->--> -->E, B and v are all parallel to each other.--7 ---? --t --7 ---? ---?E v = 0;B v = 0 but E B ,t 0--> --> -->v is paral lel to E an d perpendicular to B

16. An electron with (rest ma:;s m0) moves with a speed of0.8 c. Its mass when it moves with this speed is(a) m0 (b) m0 / 6(c) 5m0 /3 (d) 3m0 /5

17. A charged dust particle of radius 5 x 10 -7 m is locatedin a horizontal electric field having an intensity of6.28 x 105 vm-1. The surrounding medium in air withcoefficien t of viscosity 11 =1.6 x 10-15 Nsm-2 . Ifthis particlemoves with a uniform horizontal speed of 0.01 ms-1, thenumber of electrons on it will be(a) 20 (b) 15(c) 25 (d) 30

18. The momentum of a charged particle moving in aperpendicular magnetic field depends on(a) its charge(b) the strength of magnetic field(c) radius of its path(d) All of the above19. If in a Thomson's mass spectrograph, the ratio of theelectric fields and magnetic fields, in order to obtaincoincident parabola of singly ionised and doubly ionisedpositive ions are 1 : 2 and 3 : 2 respectively, then the ratio

of masses of particles will be(a) 3 : 1 (b) 2 : 1(c) 9 : 4 (d) 9 : 2

20. The specific charge for positive rays is much less than thatfor 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

21 . If a cathode ray tube has a potential difference V voltbetween the cathode and anode, then the speed v ofcathode rays is given by(a) v oc V 2 (b) v oc JV(c) v oc v-1 (d) v v

22. An electric field of intensity 6 X 104 Vm-l is appliedperpendicular to the direction of motion of the electron.A magnetic field of induction 8 x 10-2 wm-2 is appliedperpendicular to both the electric field and direction ofmotion of the electron. What is the velocity of the electronif it passes undeflected?(a) 7.5 x 105 ms -1(c) 48 X 10- 2 ms-1 (b) 7.5 x 10-

5 ms-1(d) It is never possible

23. The mean free path of the electrons in a discharge tube is20 em . The length of the tube is 15 em only. Then lengthof Crooke's dark space is(a) 5 em (b) 20 em(c) 15 em (d ) 2S em

24. The mass of a particle is 400 times than that of an electronand charge is double . The particle is acce'lerated by SV.Initially the particle remained in rest, then its final kineticenergy will be(a) 5 eV (b) 10 eV(c) 100 eV (d) 200 eV

25. A charged particle is moving in a uniform magneticfield in a circular path . The energy of the particle istripled. If the initial radius of the circular path was R,the radius of the new circular path after the energy istripled will be(a) ~ (b) ../3R3(c) 3 R (d) R / ../3

26 . An electron moving with a variable linear velocity v in avariable magnetic field B will remain rotating in a circleof constant radius r only when(a) B is held constant(b) vi s 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 ofvalue BJ with a velocity vk . The particle will movealong(a) +X axis(c) + Z axis (b) -X axis(d) -Z axis29. In a mass spectrograph, an ion X of mass number 24 andcharge +e and another ion Y o [ mass number 22 andcharge + 2 e enter in a perpendicular magnet ic field with

the same velocity. The ratio of the radii of the circularpath in the field will be(a) 11/22 (b) 11/ 2(c) 22/ 11 (d) 24/ 11


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30. A beam of electrons of velocity 3 x 107 ms-1 is deflected1.5 mm is passing 10 em through an electric field of1800 vm-1 perpendicular to their path. The value of elmfor electron is(a) 1.78 x 10 11 C kg-1(c) 1.5 X 10 11 Ckg-1

Particle Nature'ofLight(b) 2 X 1011 Ckg-1(d) 3.5 X 1011 Ckg-1

31. Planck's constant has the dimensions of(a) energy (b) mass(c) frequency (d) angular momentum32. The wavelength of a 1 keV photon is 1.24 x 10 -9 m. Whatis the frequency of 1 MeV photon?(a) 2.4 X 1015 Hz (b) 2.4 X 1020 Hz(c) 1.24 x 1015 Hz (d) 1.24 x 1020 Hz33. A parallel beam of light is incident normally on a planesurface 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(c) 5.12 X 10 -7 N (d) 5.12 X 10-8 N34. Calculate the energy of a photon with momentum3.3 x 10 -13 kg-ms-1, given Planck's constant to be6.6 x 10-34 Js(a) 7.3 x 104 J(c) 1.3 x 105 J

35 . Momentum of awavelength is(a) hi /P(c) ph

(b) 9.9 X 10-5 J(d) 8.1 X 103 J

photon is p. The corresponding(b) plh

(d ) hlp36. Which of the following statement about photon 1sincorrect?(a) Photons exert no pressure

(b) Momentum of photon is h\'lc(c) Photon's res t mass is zero(d) Photon's energy ish\'37. A photon in motion has a mass equal to

(a) clh\' (b) hi'A(c } h\' (d ) hvlc2

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, i f ts(a) amplitude is higher (b) frequency is higher(c) wavelength is longer (d) wavelength is shorter40 . 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 eV41. The energy of a photon of green light of wavelength

50000 A is(a) 3.459 X 10 -19 J (b) 3.973 X 10 -19 J(c) 4.132 X 10 -19 J (d) 8453 X 10 -19 J42. What will be the number of photons emitted per secondby a 10 W sodium vapour lamp assuming that 90% of theconsumed energy is convei,ted into light? Wavelength ofsodium light is 590 nm, h = 6.63 x 10-34J-s .

Chapter 26 Electron and Photon 997(a) 0.267 x 1018(c) 0.267 x 1020 (b) 0.267 X 10

19(d) 0.267 X 1017

43. If the energy of photons corresponding to the wavelengthof 6000A is 3.2 x 10 -19 J, the photon energy for awavelength of 4000 A will be(a) 1.11 X 10-19 J (b) 2.22 X 10 -19J(c) 4.40 x 10-19 J (d) 4.80 x 1o-19 J44. A radio transmitter operates at a frequency 880 kHz anda power of 10 kW. The number of P;hotons emitted pesecond is 1(a) 1.72 X 1031 (b) 1.327 X 1025(c) 1.327 x1 o37 (d) 1.327 x1 o45

Emission ofElectrons and Photoelectric Effec45 . The photoelectric threshold of Tungsten is 2300 A. Theenergy of the electrons ejected from the surface bultraviolet light of wavelength 1800 A is

(h = 6.6 x 10-34 J-s)(a) 0.15 eV(c) 15 eV

(b) 1.5 eV(d) 150 eV46. A light of wavelength 4000 A iS"allowed to fall on a metasurface having work function 2 eV. The maximum velocit

of the emitted electrons is (R = 6.6 x 10-34Js)(a) 1.35 X 105 ms-1 (b) 2.7 X 105 ms-1(c) 6.2 x 105 ms-1 (d) 8.1 x 105ms-147. Ultraviolet radiation of 6.2eV falls on an aluminiumsurface (work function 4.2eV). The kinetic energy in joul

of the fastest electron emitted is approximately(a) 3 X 10 -21 (b) 3.2 X 10 -19(c) 3 X 10 -17 (d) 3 X 10 -15

48 . Radiations of two photon's energy, twice and ten timethe work function of metal are incident on the metasurface successively. The ratio of maximum velocites ophotoelectrons 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 thephotocell, with the intensity of light, is given by a graphwhich 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 origin50 . When the photons of energy hv fall on a photosensitivmetallic surface (work function h\'0 ) electrons are emittefrom the metallic surface. The electrons coming ou t o

the surface have some kinetic energy. The most energetiones have the kinetic energy equal to

(a) less(c) equal

VsA B

(b) more(d) Nothing can be said


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998 Chapter 26 Electron and Photon51. When the photons of energy hv fall on a photosensitivemetallic surface (work function h\'0), electrons areemitted from the metallic surface. The electrons coming

out of the surface have some kinetic energy. The mostenergetic 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 outwith a maximum velocity of 4 x 106 ms-1. If the value ofv = 5 v0, then maximum velocity of photoelectrons will be(a) 2 x 107 ms-1 (b) 8 x 106 ms-1(c) 2 x 106 ms-1 (d) 8 x 105 ms-1

53 . The wavelength of the photoelectric threshold for silver is!..0 . The energy of the electron ejected from the surface ofsilver by an incident light of wavelength A(A < A0 ) willbe(a) hc(A0 - A)

(c) !!._(.!_ _ !___]c A A054. A metal surface is illuminated by a lig4t of given intensity

and frequency to cause photoemission. If the intensityof illumination is reduced to one-fourth of its originalvalue, then the maximum kinetic energy of the emittedphotoelectrons would become(a) four times the original value(b) twice the original value(c) l/6th of the original a l u ~(d) unchanged

55. The work function of a metal is leV. Light of wavelength3000 A. is incident on this metal surface. The velocity ofemitted photoelectons will be(a) 10 ms-1 (b) 103 ms-1(c) 104 ms-1 (d) 106 ms- 1

56. In the photoelectric effect the velocity of ejected electronsdepends upon the nature of the target and. (a) the ~ e q u e n c y 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 light57. Light of wavelength 4000 A. is incident on a metal platewhose work function is 2 eV. The maximum KE of theemitted photoelectron would be(a) 0.5 eV (b) 1.1 eV(c) 1.5 eV (d) 2.0 eV58. A photon of energy 3.4 eV is incident on a metal havingwork function 2 eV. The maximum KE of photoelectronsis equal to(a) 1.4 eV(c) 5.4 eV (b) 1.7 eV(d ) 6.8 eV59. The frequency of the incident light falling on aphotosensitive 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 irradiatedwith light of wavelength 332 nm. The retarding potentialrequired to stop the escape of photoelectrons is(a) 1.07 eV (b) 2.66 eV(c) 3.7 eV (d) 4.81 eV61. If the work function for a certain meta l is 3.2 x 10- 19 J andit is illuminated with light of frequency v = 8 x 1014 Hz,the maximum kinetic energy of the photoelectron wouldbe(a) 2.1 X 10-19 J (b) 3.2 X 10 - 19 J(c) 5.3 x lo -19 J (d) 8.5 x 10- 19 J62. Ultraviolet radiations of 6.2 eV falls on analuminium surface. KE of fastest electron emitted is(work function = 4.2 eV)

(a) 3.2 x 10 -21 J(c) 7 x 10-25 J (b) 3.2 X 10-

19 J(d) 9 x lo -32 J

63. Ultraviolet light of wavelength 300 nm and intensity1.0 wm-2 falls on the surface of a photosensitivematerial. If one percent of the incident photons producephotoelectrons, then the number of photoelectronsemitted from an area of 1.0 cm2 of the surface is nearly(a) 9.61 x 1014 s- 1 (b) 4.12 x 1013 s-1(c) 1.51 x 10 12 s-1 (d) 2.13 x 1011 s- 1

64. The photoelectric threshold wavelength for a metalsurface 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 sod ium surfacefor w ~ i c h the threshold wavelength of photoelectrons is5420 A. The work function of sodium is(a) 0.57 eV (b) 1.14 eV(c) 2.29 eV (d) 4.58 eV66. The difference between kinetic energies of photoelectrons

emitted from a surface by light of wavelengths 2500 A.and 5000 A. will be(a) 1.61 eV (b) 2.47 eV(c) 3.96 eV (d) 3.96 x 1o-19 eV67. When a point source oflight is 1 m away r o m ~ photoelectriccell, the photoelectric current is found to be I rnA . If thesame source is placed at 4 m from the same photoelectriccells, the photoelectric current (in rnA) will be

(a) l/16 (b) l/ 4(c) 41 (d) 161

Wave Nature ofParticle68. 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 101669. The wavelength of de-Broglie wave associated with a

thermal neutron of mass m at absolute temperature T isgiven by (Here, k is the Boltzmann constant)h h

(a) ~ 2 m k T (b) ~ m k T(c) h~ 3 k m T (d) h2 ~ m k T


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

71. What should be the velocity of an electron so thatits momentum becomes equal to that of a photon ofwavelength 5200 A?(a) 700 ms-1 (b) 1000 ms-1(c) 1400 ms-1 (d) 2800 ms-1n. If the mass of neutral 1.7 x 10-27 kg, then thede-Broglie wavelength of neutral of energy 3eV is(h = 6.6 x 10-34 J-s)(a) 1.6 x 10-16 m(c) 1.4 x 10-10 m

(b) 1.6 x 10-11 m(d) 1.4 x 10 -11 m73. A particle with rest mass m0 is moving with speed of lightc. The de-Broglie wavelength associated with it will be(a) infinite (b) zero

(c) m0cl h (d) hvlm 0 c74. An electron of mass m and charge e initially at rest getsaccelerated by a constant electric field E. The rate ofchange of de-Broglie wavelength of this electron at time tignoring relativistic effects is

-h(a) eEt2-mh(c) eEt 2

-eEt(b) E(d ) -h

eE75. What should be the velocity of an electron so thatits momentum becomes f'!qual to that of a photon ofwavelength 5200 A?

Chapter 26 Electron and Photon 999(a) 103 ms-1(c) 1.4 x 103 ms-1

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

76 . The correctness of velocity of an electron moving withvelocity 50 ms-1 is 0.005%. The accuracy with which itsposition can be measured will be(a) 4634 x 10-3m (b) 4634 x 10-5m(c) 4634 x 10-6m (d) 4634 x 10-8m

77. I f a proton and an electron are confined to the sameregion, 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 masses78. 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 -24

79. The uncertainty in the position of a particle is equal to thede-Broglie wavelength. The uncert ainty in its momentumwill be(a) lilA (b) 2hi 3A.(c) 'Alii (d ) 3A/2h

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

Exercise IIOnly One Correct Option

1. When radiation is incident on a photoelectron emitter,the stopping potential is found to be 9 V. If elm for theelectron is 1.8 x 1011 Ckg-1, the maximum velocity ofejected electrons is(a) 6 x 105 ms-1(b) 8 x 105 ms-1(c) 106 ms-1(d) 1.8 x 106 ms-1

2. The intensity of X-rays from a coolidge tube is plcittedagainst wavelength A as shown in figure. The minimumwavelength 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 ametal surface is plotted along Y-axis and frequency v ofincident light along X-axis. A straight line is obtained asshown. Planck's const ant 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 an d mass of theelectron(e) product of slope and mass of the electron

4. Mixed He+ and 0 2+ ions (mass of He+ = 4 amu andthat of o2+= 16 amu) beam passes a region of constantperpendicular magnetic field. If kinetic energy of all theions 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 andradius r is kept constant between two plates of potentialdifference of 800 V. Then charge on other drop of radius2 r which is kept constant with a potential difference of3200 Vi s(a) Q 12(c) 4 Q (b) 2 Q(d ) Ql4


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1000 Chapter 26 Electron and Photon6. An electron and a proton have the same de-Brogliewavelength. 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 proton7. Energy required to remove an electron from an aluminiumsurface is 4.2 eV. If light of wavelength 2000 A alls on thesurface, the velocity of fastest electrons ejected from thesurface is

(a) 2.5 X 1018 ms-1(c) 6. 7 x 1018 ms-1

(b) 2.5 x 1013 ms-1(d) None of these8. The maximum wavelength of radiation that can producephotoelectric effect in certain metal is 200 run. Themaximum kinetic energy acquired by electron due toradiation of wavelength 100 run will be(a) 12.4 eV (b) 6.2 eV(c) 100 eV (d) 200 eV9. Consider the following statements concerning electrons :

I. Electrons are universal constituents of matter.II. J J Thomson received the very first Nobel prize inPhysics for discovering the electron.III. The mass of the electron is about 1/2000 of aneutron.IV. According to Bohr the linear momentum of theelectron is quantised in the hydrogen atom.Which of the above statements are not correct?(a) I (b) II(c) III (d) IV

10. Electron with energy 80 keV are incident on the tungstentarget 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 o f - 0.155 A(b) a continuous X-ray spectrum (Bremsstrahlung) withall wavelengths(c) the characteristic X-ray spectrum of tungsten(d) a continuous X-rays spectrum (Bremsstrahlung)with a minimum wavelength of -0 .155 Aand thecharacteristic X-ray spectrum of tungsten

11. What is the de-Broglie wavelength (in A) of the a-particleaccelerated through a potential difference V ?(a) 0.287

-JV( ) 0.101c .Jv

(b) 12.27-JV'(d) 0.22-JV

12. An oil drop carrying a charge q has a mass m kg. It isfalling freely in air with terminal speed v. The electricfield required to make the drop move upwards with thesame speed is(a) mgq(c) mgv7

(b) 2mg(d)

q2mgv

q13. Photons of energy of 6 eV are incident on a metal surfacewhose work function is.4 eV. The minimuin kinetic energy. ofthe emitted p h o t o e l e c t r o n n ~ i l l be

(a) Zero(c) 2 eV (b) 1 eV(d) 10 eV

14. The de-Broglie wavelengthL associated withan elementaryparticle of linear momentum p is best represented by thegraphL L

(a) (b)

p pL L

(c) (d) ~ p15. Maximum velocity of photoelectron emitted is 4.8 ms-1 .The e/m ratio of electron is 1. 76 x 1011 Ckg-1, then

stopping potential is given by(a) 5 x 10-10 Jc -1 (b) 3x10-7 Jc -1(c) 7x10-11 Jc -1 (d) 2.5x10-2 Jc-1

16. Two identical metal plates shown photoelectric effect bya light of wavelength leA falls on plate A and 'A8 on plateB('AA = 2'A8 ) . The maximum kinetic energy is(a) 2 KA = K8 (b) KA < K8 /2(c) KA= 2KB (d) KA = K8 / 217. 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 A alls on a sensitive plate withphotoelectricwork fi.mctional of 1. 9 eV. Th kinetic energyof the photoelectron emitted will be(a) 0.58 eV (b) 2.48 eV(c) 1.24 eV (d) 1.16 eV

19 . A proton and an a-particle are accelerated through thesame potential difference . The ratio of their de-Brogliewavelength ('AP!'A 0 ) is(a) 1 ! 2.fi (b) 1(c) 2 (d) 2.J2

20. The de-Broglie wavelength of a neutron at 27C is A..What will be its wavelength at 927C?(a) A/4 (b) A/3(c) A/2 (d) 3 A/221 . A potential difference of 104 Vi s applied across an X-raytube. The ratio of the de-Broglie wavelength of X-rays

produced is (!... for electron = 1.8 x 1011 Ckg-1)1ra)-' 20

(c) 1

m 1(b) 101

(d) 100


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22 . The minimum light intensity that can be perceived bythe eye is about 10-10 wm-2. The number of photonsof wavelength 5.6 x 10 -7 m that must enter the pupilof area 10-4 m3s-1, for vision is approximately equal to(h = 6.6 x 10-34 J-s)(a) 3 x 10

2photons(c) 3_x 104 photons (b) 3 x 10

3photons(d) 3 x 105 photons

23. A charged oil drop falls with terminal velocity v0 inthe absence of electric field. An electric field E keeps itstationary. The drop acquires charge 3q , it starts movingupwards with velocity v0 . The initial charge on the droplS(a) q/ 2(c) 3q/2

(b) q(d ) 2q

24 . The filament current in the electron gun of a coolidgetube is increased while the potential difference used toaccelerate the electrons is decreased. As a result, in theemitted radiation(a) the intensity increases while the minimum wavelengthdecreases(b) the intensity decreases while the minimumwavelengthincreases(c) the intensity as well as the winimum wavelengthincreases(d) the intensity as well as the minimum wavelengthdecreases

25 . What is the strength of transverse magnetic field requiredto bend all the photoelectrons within a circle of a radius50 em. When light of wavelength 3800 A s incident ona barium emitter? (Given that work function of bariumis 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 Ahasan energy equal to 1.23 eV. When light of wavelength5000 A and intensity I0 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

27. The wavelength of characteristic X-ray Ka. line emit ted byhydrogen like atom is 0.32 A. The wavelength of Kp lineemitted 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 oflight . Ee, EP are the kinetic energy of electron and photonrespectively. Pe Ph are the momentum of electron andphoton respectively. Then which of the following relationis correct?(a) ~ = v (b) Ee = 2c- -

EP 2c EP v~ = ~ (d ) ~ = 2c(c) -Ph 2v Ph v

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

Chapter 26 Electron and Phofon 1001(a) Kinetic energy of electrons increases with the intensity

of incident light.(b) Electrons are emitted when the wavelength of theincident light is above a certain threshold wavelength.(c) Photoelectric emission is instantaneous with theincidence of light.(d) Photoelectrons are emitted whenever a gas isirradiated with ultraviolet light.30 . A 100 W light bulb is placed at the centre of a sphericalchamber of radius 0.10 m. Assume that 66% of the energysupplied to the bulb is converted into light and that thesurface of chamber is perfectly absorbing. The pressureexerted by the light on the surface of the chamber is(a) 0.87 X 10--6 Pa (b) 1.77 X 10--6 Pa(c) 3.50 x 10--6 Pa (d) None of these

May have More than One Correct Option31. When photons of energy 4.25 eV strike the surface ofa metal, the ejected photoelectrons have a maximumkinetic energy EA eV and de-Broglie wavelength A.A- The

maximum kinetic energy of photoelectrons liberatedfrom another metal B by photons of energy 4.70 eV isE3 = (EA -l.SO)eV. If the de-Broglie wavelength ofthese photoelectrons is 'AB = 2'AA, then(a) the work function ofA is 2.25 eV(b) the work function of B is 4.20 eV(c) EA = 2.0 eV(d) E8 = 2.75 eV32. In Thomson's experiment, if the velocity of electron isgreater than the ratio of electric field (E) and magneticfield (ie, v > E!B), then(a) the electron will reach the undeflected spot(b) the electron will not reach the undeflected spot(c) the elec tron will move to a spot above the undeflected

position(d) the electron will move to a spot below the undeflectedposition33. When photon of energy 4.0 eV strikes the surface of ametal A, the ejected photoelectrons heyve maximumkinetic energy TA eV and de-Broglie wavelength A.A.The maximum kinetic energy of photoelectrons liberatedfrom 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 ofA 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 tomovement of(a) positive ions (b) negative ions(c) electrons (d) photons35. The maximum KE of photoelectrons ejected from aphotometer when it is irradiated of wavelength 400 nm is1 eV. If the threshold energy of the surface is 0.9 eV(a) the maximum KE of photoelectrons when it isirradiated with 50 0 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 thephotoelectrons from the surface is nearly 650 nm(d) maximum KE will increase if the intensity of radiationis increased.


16/27

1002 Chapter 26 Electron and PhotonComprehension Based Questions

Passage IAccording to Einstein, when a photon or light of frequencyv or wavelength A is incident on photosensitive metalsurface of work function 0 , where


17/27

49. Assertion The relative velocity of two photons travellingin opposite direction is the velocity of light.Reason The rest mass of photon is zero.

50. Assertion Photoelectric effect demonstrates the particlenature of light.Reason The number of photoelectrons is proportional tothe 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 ifAo (DCE 2009)

(a) A ~ A0(c) A $; A0

(b) 2"-0(d) A.=4A0

52 . The surface of the metal is illuminated with the light of400 nm. The kinetic energy of the ejected photoelectronswas 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 inciden t on a metal surface of work function2.3 eV. The wavelength of incident radiation is 60 0 nm.If the total energy of incident radiation is 23 J, then thenumber of photoelectron is (UP SEE 2009)(a) zero (b) >104(c) =1 04 , (d) None of these

54 . In a cathode ray oscillograph, the focusing of beam on thescreen is achieved by (UP SEE 2008)(a) convex lenses (b) magnetic field(c) electric poten tial (d) All of these

55 . A wrong argument for the particle nature of cathode raysis 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 p roduces a continuous spectrum of radiationwith its shortest wavelength of 45 x 10-2 A. The maximumenergy of a photon in the radiation in eV is(h = 6.62 x to -34 J-s , c = 3 x 108 ms-1)(Karnataka CET 2008)(a) 27, 500(c) 17 , 500

(b) 22, 500(d) 12, 50057 . Millikan's oil drop experiment estabilish that{Kerala CET 2008)

(a) electric charge depends on velocity(b) electron has wave nature(c) electric charge is quantised(d) electron has particle nature(e) electron has wave nature

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

Chapter 26 Electron and Photon 100359 . 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 whenthe atomic number of the target increases(b) Cut off wavelength of the conti nuous X-rays dependson the atomic number of the target

(c) Intensity of the characteristic X-rays depends on theelectrical 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, isvalid for (BVP Engg. 2008)(a) both (b) neither

(c) photon only (d) electron only61 . A and B are two metals with threshold frequencies

1.8 x 1014 Hz and 2.2 x 1014 Hz. Two identical photonsof energy 0.825 eV each are incident on them. Thenphotoelectrons are emitted in (Take h = 6.6 x to -34 J-s)

( K a r n ~ t a k a CET 2007)(a) B alone (b) A alone(c) NeitherA nor B (d) Both A and B

62. When a point source of light is at a distance of 50 em froma photoelectric cell, the stopping voltage is found to beV0 . If the same source is placed at a distance of 1 m fromthe cell, the stopping voltage will be

63 .(a) 2 V0(c) V0 !2

(b) V0(d) V0 /4(Gujarat CET 2007)

Electrons with de-Broglie wavelength A. fall on the targetin anX-raytube. The cut-off wavelength A.0 of the emittedX-rays is {liT JEE 2007), 0 = 2mcA.

2 2h(a) A (b) "-o = -h me

(d) A.0 =A64. Light of frequency v falls on material of threshold

frequency ,0 . Maximum kinetic energy of emitted electronis proportional to (UP SEE 2009)(a) v - ,.0 (b) ,.(c) ~ v - v (d) ''o

65. A metallic surface is irradiated by a monochromatic lightof frequency ,.1 and stopping potential is found to be V1.If the light of frequency v2 irradiates the surface, thestopping potential will be {Kerala PET 2006)

h h(a) V1 +-(V1 +V2) (b) V1 +-(Vz -V 1 )e ee h(c) V1 +h(v2 - v1) (d) V1 + -;Cv1 - v 2 )e(e) V1 - h ( v 2 - v 1)

66 . Maximum velocity of the photoelectrons emitted by ametal surface is 1.2 x 106 ms-1. Assuming the specificcharge of the electron to be 1.8 x 1011 Ckg-1, the valueof the stopping potential in volt will be(a) 2(c) 4 (b) 3(d) 6

(Karnataka CET 2006)


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1 04 I Chapter 26 Electron and Photon67. Monochromatic light incident on a metal surface emits

electrons with kinetic energies from zero to 2.6 eV. Whatis th e least energy of th e incident photon, if th e tightlybound electron needs 4.2 eV to remove?(a) 1.6 V(c) 6.8 eV

(BHU 2005)(b) From 1.6 eV to 6.8 eV(d) More than 6.8 eV

68. An element with atomic number Z= 11 emits Ka X-rayof wavelength A.. The atomic number of element which

~ m i t s Ka-X-ray of wavelength 4/c (liT Screening 2006)(a) 6 (b) 4(c) 11 (d) 44

69. If th e kinetic energy of a free electron doubles, itsde-Broglie wavelength changes by the factor

Exercise I1. (a) 2. (d) 3. (c) 4. (c) 5. (a)11. (c) 12. (a) 13. (d) 14. (b) 15. (b)

21 . (b) 22 . (a) 23. (c) 24. (b) 25 . (b)31. (d) 32 . (b) 33. (b) 34 . (b) 35. (d)41. (b) 42 . (c) 43. (d) 44. (a) 45. (b)51 . (c) 52 . (b) 53 . (d) 54. (d) 55 . (d)

(b) 63,:c%,6c:) ' (b)1. (a) 62. 64 . 65 . (c)71. (c) 72. (b) 73." (b) 74. (a) 75. (c)Exercise II

1. (d) 2. (a) 3. (b) 4. (c) 5.11. (c) 12. (b) 13. (a) 14. (d) 15.21 . (b) 22 . (c) 23. (c) 24 . (c) 25 .31. (a,b,c) 32 . (b,d) 33. (b,c) , 34 . (a,b,c) 35 .41. (d) 42 . (d) 43. (b) 44 . (c) 45 .51 . (c) 52. (a) 53 . (d) 54. (c) 55 .61 . (b) 62. (b) 63. (a) 64 . (a) 65 .

' j ll. \1 ) .-_i\,

(a) J2 1(b) J2(c) 2 (d) ..!.2

(AIEEE 2005

70. A photocell is illuminated by a small bright source place

6.16.26 .36 .46 .56.66.76.

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

1 m away. When the same source of light is placed ..!. m2away, the number of electrons emitted by photocathodwould (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

Answers(b) 7. (a) 8. (c) 9. (c) 10. (c)(c) 17. (b) 18. (d) 19. (c) 20. (a)(c) 27. (a) 28. (b) 29 . (d) 30 . (c)(a) 37. (d) 38. (a) 39. (c) 40. (c)(c) 47. (b) 48. (b) 49 . (c) 50 . (a)(a) 57 . (b) 58 . (a) 59 . (c) 60. (b)(b) 67 . (a) 68 . (b) 69. (c) 70. (d)(b) 77. (c) 78. (b) 79. (a) 80 . (b)

6. (d) 7. (d) 8. (b) 9. (d) 10. (d)16 . (b) 17. (c) 18. (a) 19. (d) 20 . (c)26. (c) 27. (b) 28 . (a) 29. (c) 30. (b)36. (a) 37. (c) 38 . (a) 39. (b) 40 . (c)46. (c) 47. (a) 48 . (a) 49, (b) 50 . (c)56 . (a) 57 . (c) 58. (b) 59 . (b) 60. (c)66. (c) 67. (c) 68. (a) 69. (b) 70. (c)


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Hints &SolutionsExercise I

1. qV = .!mv2 or v =2 mie v oc {q . vHe = 'q-Hex_m_H_ - ~ - 1' 'lj-; . VH qH mHe - 'lj--; " t ; ; - . f i

DEql2. Here, x =- -2mvor !1._ = X v2 0.02 X(106)2m DEl 0.21 x (2 x 104 ) x (5 x 10-2)

= 9.52 X 107 Ckg-1. Mv 23. In mass spectrograph, - - = qv B 1r

Eand qE = B q v or v = -BMv M (E) MEor r =wq = B q B = qBB I

so r is related with (qiM) .4. E=.!mv2 orv2 = 2Eim;'2If E' is the intensity of electric field applied then

E'q = m V21ror r = m v2 IE' qm(2E lm) 2 Eor r = E 1q q E 1ie, roc 1 I q.so .

5. Here, No. of electronsn. = 5 x 107cm-3 = 5 x 107 x 106m-3No. of positive ions, np = 5 x 107 x 106= 5 x 1013 m-3v = 0.4 ms-1; J = 4 x 10-6 Am-2; vp = ? Use the relationJ = nee ve + np e vp and solve it for vp4Xl0-6= (5x1013 X 1.6 X 10-19 X 0.4)+ (5 X1013 X 1.6 X 10-19 Xvp)

_4xlo -6-3 .2 x lo-6 0.8xl0-6 _ 01 -1vP- 6 6 -.ms8.0xlo - 8xlo-B21D q6. For similar parabola; y 2 = --- x, will be same forE mB2qtwo particles. It means - - = a constant for these twoparticles. m

, m1 = B{q 1 = (0.8)2 x ~ =m2 1.2 2e 9

7. Asq vB = mV21rqBr (2xl.6xl0-19 ) x l x lr m - -- - ..::.._____ ..;,_____- v - 1.6 x l07

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

1 . F qvB8. Acce erat10n, a = - =- m m2xl.6xl0-19 x 6x105 x0.2

6.65 x10-27= 5.77 x 1012 ms-2

9. eV = mv2l r; so V oc v2;. (v 2J2 (2v)2 _1.. V2 = V1 ~ = 500 --;- = 2000Vcm

4 qV10 . mg=qE or - n r 3pg=- or Vocr33 dV2 = V1 ( J 400 x (fJ 3200 V

11. When pressure in a tube is reduced in the range 1 em and10-3 em; the mean free path of moving electron in thedischarge tube increases. As a result of which the electrongets higher KE while moving towards anode and thencause ionisation of the atoms with which it will collide on

12.its ways causing excitation phenomenon.Here, u = O,a = qEim;s = l and v =?As v2 =u 2 +2as;or v = ~ 2 ~ E l

qE lso v2 = 0+ 2 - m

13. From the symmetry of figure, the angle 8 = 45. The pathof moving proton in a normal magnetic field is circular. Ir is the radius of the circular path, then from the figure,AC=2rcos45= 2rx . ,1=.fir ...(i)

mv2 mvAs B qV = -- or r = -r BqAC = . f imv =.f ix 1.67 x 10-27 x 107 = O.l4 mBq l xl.6x lo -19

14. qE = mgor E = mg = (SOx 10-6) x 9.8 = 98 Nc-1q Sx lo-6Since the force due to electric field on charged particleshould be opposite to the gravity pull and charge on thedrop is negative, hence the electric field must act verticallydownwards.


20/27

1006 Chapter 26 Electron and Photon15. A charged particle moves along a straight line withacceleration, hence electric field should be parallel tothe direction of motion of charged particle and no forceshould act on charged particle due to magnetic field. It

will be so if charged particle is moving parallel to thedirection of magnetic field.16. m - mo - mo- v2 /e 2 - (0.8e) 2 /e 2

Sm03

- 17. 6n11rvneE= 6n11rv or n = - - - eE6 X 3.14 X 1.6 X 10-5 X 5 X 10-7 X 0.01 = 151.6 X 10-19 X 6.28 X 105

18. q v B = mV.;r or mv = q r B.. '1 b 1 82 lDq b c b h h9. For s1rru ar para o a, - - - must e same 10r ot t eEmions.

So, :: = ( !:J ( ::J X !: = (%Jn (iJ ~20. Specific charge = charge/mass. The positive rays arestream of positive ions. The mass of positive ion is muchmore than that of electrons, hence specific charge ofpositive ions is less.21. qV = _!_mv 2 or v = 2 q V /m ie, v oc .J1i222. V = E/8 = 6 X 104/8 X 10-2 = 7.5 X 105 ms-123. The length of Crooke's dark space will be equal to thelength of tube ie, 15 em.24 . Gained in KE = q V = 2 x 5 = 10 eV

1 q2 82 r2 .25 . Ek = te, r oc .JE,;2 mr----So, r2 = = R-./3 = -J3 R26 . 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 gasfilled in the discharge tube.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, ro c m/qqB

30.

So, r1 m1 q2 24x2e 24-=-X-=-- -=-r2 m 2 q1 22xe 111 2 1 eE x 2 e 2 y v2y=-a t =-- -or -=- -2 2 m v2 m Ex 22xl.Sx10-3 x(3x107f

1800 X (0.1)2= 1.5 X 1011 Ckg-1

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

32. he I A.= 103 eV ... (i)h\- = 106 eV ... (ii)Dividing Eq. (ii) by Eq. (i) we get, v = 103e I A.= 103 x 3 x 108 I 1.24 x 10-9 = 2.4.x 1020 Hz

33. Momentum of incident light per secondE 60 -7P1 =- =- - - = 2x10e 3x 108

Momentum of reflected light per second.60 E 60 -7p2 =-X-=---8 =1.2x10100 e 3x10Force on the surface = change in momentum per second

= P2 -(-p1) = P2 + P1 = (2+1.2)x10-7 = 3.2 x 10-7 N34. Given E/e = 3.3 x 10-13 kg ms-1;

So, 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= hl p36. 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 238. Energy of photon E = - = me ;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 108-------oc, - - - - = 5.9 X 10-6 eV1.6 X 10-19 X 0.21

41. E =he/A.= 6.6 x 10-34 x 3 x 108 15000 x 10-19= 3.973 X 10-19 J42. Energy of photon

E = he = 6.63 X10-34 X 3 X 108 = _6 .63 X3 X 10_18A. 590x1o-9 59Light energy produced per second = 90 x 10 = 9W100:. Number of photons emitted per sec

9x59 18 = 2.67 X 10196.63x3x1043. 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 J44. Number of photons emitted per sec,

Power Pn=- - - - - -Energy of photon hv10000 = 1. 72 X10316.6 X 10-34 X880 X 103

45. Ek =he (.!. _ !_)cineV)e A. /...0= 6.6 X10-34 X3 X 108 ( 1010 _ 1010 J= l.S eV1.6 x Io -19 I8oo 2300


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1 2 he .46. 2mv =T-o (meV)

v

6.6 X 10-34 X 3 X 108- - ~ - - ~ - - ~ ~ ~ - 24000 X 10-10 X 1.6 X 10-19= 3.1-2 = 1.1eV = 1.1 x l. 6 x 1o-19 J=1.76 x 10-19 J

1.76x1o-19 x 29 X 10-13= 6.2 x 105 ms-1

47. Ek =E-0 =6 .2-4.2=2 .0eV= 2 .0 X 1.6 X 10-19 = 3 .2 X 10-19 J .

48. 1 2 1 2-mv1 =20 - 0 =0 and -mv2 =100 -0 =90.2 2. v1 [k_ 1.. V2 = ~ ~ = 3

49. Photoelectric current oc intensity of incident light.Therefore, the graph is a straight line having positiveslope 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 .

1 252. 2m v1 = 2hv0 - hv0 = hv0 and

So ,

1 22mv 2 =5hv0 - hv0 .=4 hv01 2 1 2- mv2 =4 x -mv12 2

or v2 = 2v1 = 2x4 x 106 = 8 x 106ms-1

54 . The maximum KE of the emitted photoelectrons isindependent of the intensity of the incident light butdepends upon the frequency of the incident light.55. 1 2 he-zmv = T-o

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

= 4.14- 1 = 3.14 eV1

2x3.14xl.6x10- 19 106 -1= ms9.1 x 10-31or V=56. The velocity of photoelectrons depends upon thefrequency of the incident light.57 . Maximum KE =he - oA

6.6x10-34 x3x108 1_ :____ _ : ~ . : . . . . : . . _ - 2 = 1.1 eV400 X 10-10 1.6 X 10-1958. MaximumKE' = E- 0 =3.4-2 .=1.4eV

t

Chapter 26 Eectron and Photon 100759. Let 1 and E2 be the KE of photoelectrons for incidentlight of frequency v and 2v respectively.

Then hv = 1+ o and h2v= 2+ j>0So, 2(1 +0) = E2 + 0 or E2 = 21+ 0It means the KE of photoelectron becomes more thandouble.60. Retarding potential,

v = ~ - ! t L 1240 x 1o-9 - l .0 7s Ae e 330 x 10-9

= 3.73-1.07 = 2.66 v61. Maximum KE = hv - o

= 6.63 X 10-34 X 8 X 1014 - 3.2 X 10- 19=2.1 x 10- 19 J62. KE of fastest electron= E - 0 = 6.2-4.2 = 2.0eV

= 2 X 1.6 X 10-19 = 3 .2 X 10-19 J63. 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 photonswhich can produce photoelectrons = energy required fo rproducing electron/ energy of photon.

10-6 10-6 X 300 X 10-9= - - - =-------he / A 6.6 x 10-34 x 3 x 108=1.51 x 1012 s-1

64. o = ..!!:__ (in eV) = 6.6 x 10-34 x 3 x108 = 1.87 eVeAo . 6600 x 10-10 x1. 6 x 10-19

65. 0 = he I A0 (in eV)

66.

67.

6.62x10-34 x 3 x 108- -----,-.,..----= = 2 .29 eV5420 X 10-10 X 1.6 X 10-19t i l l = h e - ~ he(A2- A1) .A1 A2 A1A2 (m eV)

_ 6.62 X 10-34 X 3 X 108 X (5000- 2500) X 10-10- 2500 X 5000 X 10-20 X 1.6 X 10-19= 2.47 eV

Photoelectric current (J)oc Intensity of incident light and. . 1mtens1ty oc- - - (distance)2

1 (1] Io, I oc Hence I ' = I - = -(distance)2 4 16h he68. For electron, p = -; and for photon, E = -A A

:. _=he / A. =e=3 x 108ms-1p h/ A69. KE of thermal neutron, .!mv2 =2 2

or m v = ~ So , A = ! ! : _ = ~p 3kmT \


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1 08 Chapter 26 Electron and Photonh70. We know that, A= ~ ;-y2mkT

1So, Aoc -, [ fA27 = 927+273 = 2A927 27 +273

or A27 = 2An7 = 2A71. Momentum, p = m v = h/A

h 6.62 x w-34or v--- ---- ; : : : -------- ; : ; -- m A- 9.1x10-31 x5.2x1o-7= 1.4 X 103 ms-1=1400 ms-1

72. E = 3eV = 3xl.6x1o-19 JA= - h -~ 2 m E - -6.6 X 10-34r======::::======:== = 1.65 x w- 11 m~ 2 x l . 7 x 1 0 x 3 x l.6x10-19

h h)1 -v 2 / c273. A=-= =0 .mv m0 veE74. Here, u = 0; a= - ;v = ?;t = tm eEv = u+at = 0 + - tmde-Broglie wavelength,

A=!!_= h hmv m(eEt!m) eE t

(: v =c )

75.

76.

78.

79.

Rate of change of de-Broglie wavelengthdA h ( 1) -hdt = eE -t2 = eEt 2hmv=-A

hor v=mA6.6x 10-34

9.1 X10-31 X5200 X1Q -10=1.4 X 103 ms-1

Here, L'>v = 0005 x 50 = 0.0025ms-1100&=-h -mf>v

1.034 x 10 -34- - - - , ~ - - - = 4634 x 10-5m9.1 X10 -31 X0.0025t>p = !!___ = 1.034 x w-34

& w-w= 1.034 x 10-24 kg-ms-1

n. ho p = - = & Ah80. L'>p=mf>v=-&

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

Exercise II1 1 2. eV =-mv. 2

or =)2x(l .8x10 11 )x9 =1.8 x1 06 ms-12. As accelerating voltage V across X-rays tube increases,

the value of minimum wavelength of X- rays, A = ~ ;c eVdecreases ; so the separation between A.K and 'Acincreases.3. Ek = eV = hv- 0

or V = ~ v - . f ue eSlope of straight line between V and v is ~h = e x slope of straight line. e

4. E = .!.mv22or v=J

or ,mv 2qvB=r

r= mv =_!!:_X {2E = ~ 2 E mqB qB {;- qB

rmor roc-

r1 = r2QV 4 35. ---nr pgl 3

q

Q'V' 4and --=-n(2r)

3pgl 3Q'V'So, --=8QV

Q = 8QV = 8Q X 80 0 = 2QV' 3200or6. As EK = .!.mv2 or mv = ~ 2 m E K2As per question;

or mpvp = meveor mp EKP = )r:-2-m_e_E_K_,


23/27

EK mP__=->1EK meporEK >EKe Por1 2 he .7. 2mv = ~ - < ! J ( m e V )6.6 X 10-34 X 3 X 108

2000 X 10-lO X 1.6 X 10-19= 2eV = 2x1.6x10-19 J

4.2

V = X 2 X 1.6 X 10-19 /9.1 X 10-31= ..}6.4/9.1 = 106 ms-1.

8. Here , A0 = 200nm; A= 100nm;he/e = 1240eVnmM . KE he he (. V)ax1mum =-- - meA.e A0e

= :e(i- :J= 1240(-1- - - 1-)100 200= 6.2 eV

9. The mass of electron is about - 1- times that ofa neutron1836and angular momentum of electron is quantised in thehydrogen atoms bu t not the li11ear momentum of electron.10. Since the energy of incident electron, E = 80 keV. The

minimum wavelength of X- rays produced isA.=he= 6.6x10 -34 x 3x108

E 80 X 1000 X 1.6 X 10-19= 0.155 x 10 -10 m = 0.155 ASince the energy of K-shell electron is -72.5 keV, sothe incident electron of energy 80 keV will not onlyproduce continuous spectrum of minimum wavelength0.155 Abut shell also knock electron of K shell out ofatom, ~ - e s u l t i n g emission of characteristics X- rays.

h h11. A . = - = - r = ==mv ~ 2 m e V6.6 X 10'--34

X (4 X 1.66 X 10-27 ) X (2 X 1.6 X 10-19 ) XV0.101=..;v

12. When the oil drop is falling freely under the effect ofgravity is a viscous medium with terminal speed v, thenmg = 6nytr v ... (i)To move the oil drop upward with terminal velocity v if Eis the electric field intensity applied, the

Eq = mg +6n11rv = mg +mg = 2mgSo E = 2 mg/q1 213. -mv = hv-


24/27

1010 Chapter 26 Electron and Photon23. When drop is stationary, thenq1E'=6n:11rv0 or q1 =6n:11rv0 / EWhen drop moves upwards, then

3q = ! ? _ ~ l ] _ I _ ' ~ Q + v o ) = 2x 6 n : ~ r v 2q l3q] =2.q

24 . When filament current is increased, ::-;:nre electrons areemitted from electron gun. pne to which the intensity ofelectrons increases.heAs Amin = - , so as V decreases, Amin increases.eV

25. ~ m v ~ a x = l J or v ~ , , ~ -

27.

Now,

29: ix 10.::31

x[ 6.63 X 10-34 X ; X 108 _ 2.5 X 1.6 X lO 19]3.8xlo-= 27. 12 x 1010 m2s 2

B = !_71Vmax = ! ~ _ ! Q ~ ~ ~ _ . 2 1 ~ 1 0 _eRmax 1.6xl0-19 x0.5

'-' 6.32 X 10-0 T

1 LNow, hv - =2 mvmax =eV5or = hv2 - eV5 =- E2 - eV5= 2.46 - 1.36 = 1.10 eV- 1- = R(Z - b 2 [_!_- ]__]A 12 22K"

- 1- =R(Z - b)2 (I_ __)A 12 32(3/4 ) 3 9 27- -= - -= -X-=-AKa (8/9) 4 8 3227 27 0or A-K =- x AK =- x 0.32 = 0.27 A.32 " 32

28. E = ..!. mv2 =..!_(mv)v=.:!.(I1 )ve 2 2 2 Aand heE- = - .,,, A '

s._ = _:!__EP 2e

hPe = mv = h! A- and Ph= I..&= 1.p,.

29. KE of photoelectrons increases with increase m frequencyof the incident light and is independent of the intensity ofincid1nt light.

I'llOtDelectror.s are em itted if the wavelength of ti n n d e ~ t t light is less than threshold wavelength,he= - - .},{)Phc,LOelectric erruss10n is an instantaneous procphotoelectrons may not be emitted from a gas \ITu:traviolet light if the work function of tha t gas is larthan the energy W light.

30 . I ight falling pe r second on tb.e surf'l.ce of sphere. 66 6 .E '"'- )

,_ _, p ~ ~ ~ l i =(1- rI j )2l2 . t.S1- - - -J/,

On s"lving, T1 2.0 eV.p 11 = 4 - 'f/, = 4 - 2 = 2.0 eV 11 ""6-'1 11 ~ b - - 2 = 4 . 0 eV

.


25/27

34. In a discharge tube electric conduction takes place due tomovement of positive ions, negative ions and electrons.E -1eV_hY - - he -max- - A 0 - 400x10-9 0

As W2 The photoelectric current isproportional to the intensity of inciden t light. Since, thereis 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 workfunction of sodium, then Na

cBut we know that Y0 =-::;:0Hence Eq. (i) becomes

... (i)

Chap,er 26 Electron and Photon 101143. It is true that photocells are utilised to reproduce sound

in cinematography and also in camera l'nd television forscanning and telecasting the scene. Now, the photocellis such a device in which light energy is convetted intoelectrical energy.44. A tube light is a gas discharge tube which can emit lightof different colours. This colour mainly depends upon the

nature of the gas inside the tube and the nature of theglass. The light emitted is due to fluorescence emissionof light when argon is filled in tube. It takes place at lowpressure but not at high temperature.

45 . de-Broglie wavelength associated with gas raolecules

46.

47.

varies as 1Aoc -.JTAlso root mean square velocity of gas molecules isVrms = ~ ~ T

It is fact that, greater is KE of photoelectron, greater is thepotential required to stop it. Hence, stopping potent ial isa measure of KE of photoelectron. It ce.n be understoodfrom the relation eV5 = KEor V, = KE(ineV)Einstein's photoelectric equation is

or

1 2-mv = eV5 = hY - 02V = hY-


26/27

1012 Chapter 26 : Electron and Photon50. Photoelectric effect is based upon quantum theory of lightor particle nature of light.The number of photoelectrons emitted is proportional tointensity of incident light. It does no t depend on frequency

of light.he he51. Ek =----A. Ao

Ek is positive, ie, photoelectric emission will take place ifhe he- ~ - o r A,::; A-0 .A. A. 0

52 "' =he -E = 1240eVnm 1_6SeV 'I'O A. k 400nm= 3.1-1.68 = 1.42eV

53. The energy of each incident photon isE = he = 1242 eV = 2.07eV0 A. 600

54. In a cathode ray oscillograph the focusing of beam onthe screen is achieved by electric potential. There are twoplates X andY. X plates consists two plates X1 and X2 invertical plane while Y plates also consist two plates Y1 andY2 in a horizontal plane. An electric is applied betwee nthe X and Y plates by an external source.55. A charge pmticle is deflected by electric and magnetic fields.If the cathode rays is deflected by electric and magneticfields then this is the strong argument for the particlenature of cathode rays.

he (6.62x10-34 )x(3x108)56. Energy= hv = - eV =_ _ _ : _ _ - = - - - - - - = ~ - ' - - - . . . . : _ ~Ae (45 X10-2 X10-10 ) X(1.6 X10-19 )= 27500 eV

57. Milhkan's oil-drop experiment established that electriccharge is quantised.58. Matter has a wave nature that is best supported by the phenomenon of electron diffraction.59. In X-ray tube, the cut-off wavelength is given by

heA. n = - . The cut-off wavelength depends on them' eVenergy eV of the accelerated electrons and is independentof the atomic number of target.

60. Relativistic energy is given by2E= moe

orJ1--v2 /e 2

2 4Ez = moe1 - v2/e 2m2c6or E2 =-0- -ez- vz

Momentum is given by

or ,

... (i)

... (ii)

61 .

E2 _ pze2 = m ~ c 4 /or Ez = p2c2 +For photon, rest mass m0 = 0 so E = pcFor electron, m0 '* 0, so E '* pe = hv eV = (6.6x10-34 )x(l.8x10 14 ) =0.?4 eVOA e 1.6 X 10-19

(6.6 X10 - 34) X(2.2 X1014)os= 19 =0 .91eV1.6 X 10-Since, the incident energy 0.825 eV is greater than0. 74 eV and less than 0.91 eV, so photoelectrons areemitted from metal A only.

62. By changing the position of source of light fromphotoelectric cell, there will be a change in the intensityof light falling on photoelectric cell. As stopping potentialis independent rf the intensity of the incident light, hencestopping potential remains same, ie, V0 .63. Let Ek be the KE of the incident electron. Its linearmomentum, p = J2m Ek 'de-Broglie wavelength related to electron is

h h h2'A= -p= J 2 m E k or Ek = 2mA.2The cut-off wavelength of the emitted X- rays is related tothe KE of the incident electron as

he h2 2meA.2- = Ek =-- or A. 0 =A-0 2mA.2 h64. When light falls on a metallic surface, ejection ofphotoelectron results. In this process, conservation ofenergy holds.

Thus, from law of conservation of energy, the energyimparted by the photonmaximum kinetic energy of the emittedelectron + work function of the metal.or hY = (KE)max +but e(V1 - V2 ) = h(Y1 -Y 2 )or hv1- Vz ' -=-(v1-v2)eor

1 266. eV5 = zmvmaxmv2or V = ~ =' 2e

V2 =V1+ !:_(v2 - Y1 )e2vmax2(e/m)


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67 . Maximum KE = 2.6 eV.Work function,

26.electron & photon

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