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32.. (3) 42. (1) E=0+eV,, , 32.(3) Stopping potential is independent of intensity he | prev,, 33, (1), 34.(1) ‘The current depends on intensity Ve = 35.(2) Here intensity is doubled due to frequency and not by, , number of photons hence number of ejected electrons V,vis : graph:, , remains same. Therefore saturation current remains, same, Maximum K.E, will be more than double., , 36, (4) Wavelength of ultraviolet < 5000 A, , ye pF eat sheiing), e e, , BE! css, 37. (4) Fora point source, J « lie « number of electrons Slope = tan 6 = (iii) is OK., n, | al 1 Forplate 1: Plate 2: Plate 3:, mln) Ws 4 Soon %-0002 20.004, he he he, , 38, (4) ise Poe Li?, , : . G+ b): @=1:2:4, (i) is OK., 39, (4) Stopping potential is independent of intensity, , , , joclo Ud For plate 2, threshold wavelength, saa =500nm, . 2 2 % I 1, b. | 4 7 (Fi el For plate 3, threshold wavelength, nod 250nm, i \y, 06) 9 :, Mt Avioit = 400 nim, 60 Ayioie < Arnesnod for plate 1, Acie, > Arreshotg for plate 3, = Bam, he 9 ~ 9 ~ Therefore violet colour light will eject photo-electrons, from plate 2 and not plate 3., . E=0+K,, =Eso 9+ Kose > Kr $ 43, (2) Stopping potential is same for a and b, hence their, my 2m Kou frequencies are same. The maximum current values, = Bq Be are different for a and b so they, will have different, intensities, ee 44(4) 45, (4) 46.0). (3), e, 1240 48. (4) 49, (2), 41.(1) Ey, =550nm= —— eV =2.25 eV, 550 4), 50. (, 1240, Ey Olan ev Rea E=W+k, 1389 E=W+05 wa(i), £,,=350nm= ig eV 12E=W+08 sail), From plate p, electrons are ejected by wave lengths (ini), 550, 450 and 350 nm W+08, From plate q electrons are ejected by wave lengths 450 12=——— 5 12 W+0.6=W+08, and 350 om W+05, From plate r electrons are ejected by wave length W=leV, , 350 nm

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51., , 52., , 53,, , 4., 55., , 56., , 57,, , 58., 59., , 60., , pa JA _2x25x108 x15 x10, (l c 3x10°, =2.5x10°N, (1) de-Broglie wavelength, hk, Pp V2mK, By Bie |B el, a, VK, YieK 4, , % change = (-1}100-(4-1)x100=-158, , , , ie. 75% decrease, , (b) : The least energy required to eject an electron from the surface of metal is called work functional energy of that metal., , (d) + Knowledge based question., , (b): Work: function of a metal. decreases with increase in, temperature as well as if coated with an oxide layer. The, yalue of work function is less for lead. (= 4:25 eV) than for, platinum (= 5-65:eV)., , (0) : Work function of a metal $o'= hyo where vp is the threshold, frequency. So vp = d/h. If v < vg or v < bo/h, no photoelectric, emission takes place., , ():t Here 9 = 165 eV = 1465 x 16x 10-9], , “ _y _165x16x10=- = oa, , bh 66x10", , (6) : The photoelectric effect can be understood on the basis of, quantum nature of light only., , (b)s Max. KE = hy - 4) ie, maximum KE of the emitted, photoelectrons depends on the frequency v of the incident, radiation. i :, , (@ : The best material to be used for photoelectric emission is, caesium as its work function is lowest., , =4x 104 Hy, , 61,(a) : Knowledge based question., , 62., 63., , (@) : Knowledge based question., , (b): The phenomenon of photoelectric emission. was first, observed by Heinrich Hertz., , 64, (a) : Photocell is a device to measure. intensity of light as, photoelectric current is proportional to the intensity of light, falling on the photo cathode of photoelectric cell., , 65. (4) : Knowledge based question., , 66, (b) : Photoelectrons are ejected from a metal surface if its, frequency of the incident light is greater than the threshold, frequency of thé surface, whatever may be the intensity of, incident light, It is true for option (6)., , 1, , 67. @: a= or Bx v As the energy of photon of wavelength, 1800 A is more than that: of 2300 A, so photoelectric, emission takes place,, , 68. cue, Vat i, , 0, Wate My gy adhe ste, Ora ae Bk ae se, , From (i) and (i),, , 69.@, , 70.), , 71.@, , 2.0, , 73.0), , 74.0), , a: 4, dy Roig ag ae, 1 eVy=tmiba=h¥~ 69 or Vox ty, 2 a, comparing it, with the equation of straight line., yomi+e, we have, slope of the straight line, (m) = he, , + For photoelectric emission, E, = hv — oy where B, ev and, is the intercept cut by straight line on negative E, axis. Thus,, option (c) is true., , : From the graph, we note that the saturation current is same, for curves b and c but different for curve a. Therefore,, intensities of b and ¢ will be equal but different from that of, aie, I, # f, but I, = f.. As stopping potential is same for, curves a and b, hence f,, = fj. Thus answer (a) is correct., , : Photoelectric current + intensity of incident light., , stopping /potetial is independent ‘of the intensity of the, incident light. Hence graph between V, and J is'a st. line, parallel to I-axis., , + For a given metal surface, the number of ‘photoelectrons, emitted is directly proportional to the intensity of incident, radiation but maximum KE of emitted photoelectrons, depends on the frequency of incident radiation and is, independent of the intensity of the incident radiation. Thus, option (6) is correct., , , , 75. (c) + No. of electrons emitted depends upon the intensity (7) of the, , 76, @):, , 76.(0) +, , 71, @:, , 78, .(a), , 79. @:, , 80. :, , incident light where I =, r, , he Ixt, a eth ag,, , B iny, Thus no. of photoelectrons will increase by a factor 4,, , Since 1, 2h > A, and wavelength of red light is greater than, the threshold wavelength of photosensitive surface, hence no, photoelectron is emitted., , ‘As number of photoelectrons emitted is directly proportional, to intensity of light falling on photosensitive surface and, , ity « ——., Intensity Goune?, In photoelectric effect experi-ment, The photoelectric current, 1 increases linearly with the intensity of incident light as, shown in Fig. 11.4(H).1. I, , intensity, Photoelectric current depends on the number of, photoelectrons emitted per second which in tum depends on, the intensity of incident light., , As for different metals, the value of threshold frequency (7p), is different, s0 options (a) and (c) are wrong. Since threshold, frequency can not be negative, hence. option (6) is wrong., Since max. KE, (K)nex = hv ~ 0p = hv - hy = h(v~v)‘Therefore the variation of K,,,, and v is a straight line of slope, h after threshold frequency vp. Hence option (d) is true., Since Vyrays > Yaraviolt» $0 Photoelectric effect occurs with, X-rays.