CUET Physics

224 PHYSICS 215. A monochromatic visible light consists of 233. We have, m = c   \\ v = c light of single wavelength. v µ 216. All the colours meet in phase at the centre, Time taken = Thickness of slab hence the central fringe is white. velocity 217. In case the slits are of unequal widths, I1 ≠ I2 = c d = µd /µ c ⇒ a ≠ b 234. The speed of image towards you = Distance So, Imin = (a – b)2 ≠ 0 Time In other words, the position of minimum = (8.5 + 8.5) cm intensity will no longer be completely dark. 1 218. Path difference for all wavelengths is zero at = 17 cm s–1 the central fringe. So, the central fringe will be sharp. As one moves away from the centre, 235. Velocity of light in vacuum is an absolute the sharpness decreases. constant, hence does not depend on any factor. 219. b ∝ l ⇒ b’ = l’ b = 5000 Å × 0.036 236. The size of the particle is of the order of l 6000 Å wavelength. = 0.03 cm l = c = 3 × 108 = 1014 Hz v 3 × 10−6 lD 6 × 10−7 × 1 = 10–4 m 220. b = d = 6 × 10−3 However, for higher frequency, l is still smaller and resolution is better. Required number of fringes = 1 × 10−2 = 100 10−4 1 v n 237. Magnification, m = n = u  ⇒ v = u 221. For diffraction, l ≈ size of obstacle (1 cm), We have, 1+1 = 1 which is true for ultrasonic waves. vu f 223. In order to invisible in vacuum, m of medium \\ n − 1 = 1 must be equal to m of vacuum, which is 1. u µ f 224. Resolution of an optical instruments limited ⇒ (n – 1) = u  ⇒ m = (n – 1)f because of diffraction of light. f 227. Polaroid glass reduces the light intensity to 238. Angle of deviation = ∠BOA’ half on account of polarization. = 180° – 70° = 110° 228. In the electron microscope, electron beam is AB used because waves of small wavelengths are associated with electrons. So, the resolving 35° 35° power of electron microscope increases. 231. Interference at thin films causes colouring of soap bubbles. 232. Doppler’s shift, D l = v l O 110° Mirror c 106 × 5400 Å A¢ = 3 × 108 = 18 Å

ANSWERS   225 239. When viewed vertically above, the length of 259. gmw = g µa = 5 = 5 the fish will not change. Its apparent length wµa 4 alone would change. 3 4 240. A transparent material would become 3 invisible when its refractive index equals the refractive index of the surrounding medium So, sin C = 1 = 4 i.e. vacuum, for which m = 1. g µw 5 241. There is no change in frequency on reflection. ⇒ C = sin–1 (4/5) 242. The change of phase on reflection at the 260. i = 2r (Given) denser medium is p radian. sin i sin 2r 2 sin r cos r 245. m∝ 1 m = sin r = sin r = sin r T µ 246. m∝ 1 ⇒ cos r = 2 l2 247. For air or vacuum, the value of m (= 1) is ⇒  r = cos–1 µ  2  minimum. 248. According to Rayleigh’s scattering law But i = 2r I∝ 1 \\ i = 2 cos−1  µ  l4 2 249. m increases with increasing frequency of light. 261. We know, 251.        m = Real depth P = P1 + P2 = 1 −1 =1D Apparent depth 0.5 1 = Real radius of curvature 262. We know, m= 1 = c App. radius of curvature sin C v 252. There would not be any scattering of light. ⇒ v = c sin C 253. Because light bends towards the normal = 3 × 108 × sin 30° in going from air to glass, the point of = 1.5 × 108 m/s convergence will move closer to the glass slab. 1 4 sin C 3 254.       fw = 4fa = 4 × 2.4 cm = 9.6 cm 263. We have, m= = 255. Total internal reflection occurs only when 3 light goes from denser into rarer medium. 4 256. It must be noted that the parallel rays can be \\ sin C = reflected as parallel only from a plane mirror. 257. We have, P = P1 + P2 ⇒ tan C = 3 7 = 12000 + 100 25 If r is radius of the circle, then = (5 + 4) dioptres = 9 dioptres tan C = r h 258. Given,    f1 = f f2 = –f ⇒ r = h tan C Suppose F is focal length of the combination, = 12 × 3 cm we have, 1 = 1+ 1 = 1−1 =0 7 ⇒ F f1 f2 ff F = ∞

226 PHYSICS 264. We have, m = Real depth ⇒ 1 = 0 or f = ∞. Apparent depth f \\ Real depth = 4 × 24 274. The red colour has maximum wavelength. 3 So, scattering is minimum for red because it is inversely proportional to the fourth power = 32 cm of wavelength. 265. When deviation is minimum, refracted ray 275. Given, i1 = 0°, r1 = 0° inside the prism is parallel to base of the prism. ⇒ r2 = A – r1 So, (b) is correct. 266. Total apparent depth = (7 + 3) cm = 10 cm = 60° – 0° = 60° m = Real depth Also, r2 > C i.e., critical angle, hence total App. depth internal reflection occurs at the base. So, net deviation is 60°. 276. As per Cauchy’s formula, ⇒ 1.5 = Real depth m = A + b + C , App. depth l2 l4 ⇒ Apparent depth = (1.5 × 10) cm where A, B, C are constants. = 15 cm In case of different colours, l is different hence m is different, hence deviation (d) is different. 267. sin C = 1 This results in dispersion. µ But m = c c =2 277. The formation of a rainbow is because of /2 dispersion of light that takes place due to refraction of different colours through \\ sin C = 1 = sin 30° different angles. 2 278. Magnifying power, ⇒ C = 30° 268. P = P1 + P2 M = 1+ D = 1 + 25 = 11 P2 = P – P1 f 2.5 ⇒ = 100 − 100 ⇒ fe = 5 cm 80 20 So, f0 = 50 cm = (1.25 – 5) D = – 3.75 D 279. For large Magnifying power, 270. For minimum deviation, f0 > fe. r1 = r2 = r 280. As water sticks to glass, the shape of water and 2r = A drop is correctly shown by (a). ⇒ A = A = 60° = 30° 2 2 281. Given, f0 = 10 fe Length of the telescope, L = f0 + fe = 11 fe 271. In case of a concave mirror, when the object ⇒ 11 fe = 55 cm is moved from infinity to focus of the mirror 282. We know, i.e., u decreases from ∞ to f, the image moves 1 =  µ2 − 1  1 − 1  Given, f  µ1  R1 R2  from focus to ∞ i.e. v increases from f to ∞. The \\    correct variation is shown in (c). ⇒ m1 = m2 272. Given, m = m2 1 \\ 1  µ2 − 1  1 − 1  f = 0 f =  µ1   R1 R2    f = ∞ 

ANSWERS   227 Case Study–1 Case Study–4 1. Time taken for 360° shift = 24 h 2. f = 6.25 cm,  D = 25 cm,  u = ?,  v = – 25 cm Time taken for 1° shift = 24 h = 4 minutes. 1−1 = 1  or   1 =v1 − 1 =− 1 − 1 360 vu f u f 25 6.25 4. n = c = 3 × 108 = 1.5 Hence, u = – 5 cm v 2 × 108 Apparent depth = Real depth =6 cm 4. m = D = 1 + PD = 1 + 20 × 1 =6 n 1.5 f 4 = 4 cm Case Study–7 Shift = Real depth – Apparent depth = 6 cm – 4 cm = 2 cm 1. Wavefronts are perpendicular to rays of light. Case Study–2 2. d and c are in same phase because they are on 1. Mirage is an optical illusion of water observed the same wavefront. Similarly, e and f are also in the same phase. in hot deserts. \\ fd – ff = fc – fe Case Study–3 3. Separation between successive wavefronts in 1. Using Snell’s law, medium-1 is larger than that in medium-2. 1.72 sin qc = 1.50 sin 90° So, the speed of light in medium 1 is larger than in medium‑2. 90° Case Study–8 3. Given: d = 0.28 mm = 0.28 × 10–3 m,  D = 1.4 m x = 1.2 cm = 1.2 × 10–2 m,  n = 4,  l = ? θc For constructive interference, we have, x = nl D d 1.50 75 or sin qc = 1.72 = 86 or l = xd = 1.2 × 10−12 × 0.28 × 10−3 nD 4 × 1.4 or qc = sin–1  75  = 6 × 10–7 m  86  4. Thus, the wavelength of light is 6 × 10–7 m. = sin–1 (0.872) = 60.69° I1 a2 = 81 sin iC = µouter = 1.44 4. Given:  I2 = b2 1 µinner 1.68 a 9 Hence, iC = 59° \\     b = 1  or a = 9b r = 90° – iC = 31° In the interference pattern, we have, Hence, rmax = 31° Imax . = (a + b)2 = (9b + b)2 = 1.68 Imin . (a − b)2 (9b − b)2 Therefore, imax = 60° = 16040bb22 = 1265 Thus, total internal reflection will take place \\ Imax . = 25 for all angles for which i < 60°. Imin . 16

228 PHYSICS Case Study–9 3λ 3 × 5890 × 10−10 2a 2 × 0.25 × 10−3 λ q = = 2 4. a sin q = (2n + 1) = (2 × 1 + 1),  when q is small sin q ≈ q = 3.530 × 10–3 rad \\  Total angular spread aq = 3λ = ± 3.530 × 10–3 rad 2 UNIT 7:  DUAL NATURE OF MATTER AND RADIATION 1. (d) 2. (a) 3. (a) 4. (d) 5. (b) 6. (b) 7. (d) 8. (d) 9. (a) 10. (b) 11. (b) 12. (a) 13. (d) 14. (b) 15. (b) 16. (a) 17. (b) 18. (a) 19. (a) 20. (d) 21. (a) 22. (b) 23. (a) 24. (b) 25. (d) 26. (d) 27. (c) 28. (b) 29. (d) 30. (d) 31. (a) 32. (c) 33. (a) 34. (b) 35. (a) 36. (c) 37. (a) 38. (d) 39. (b) 40. (b) 41. (a) 42. (d) 43. (a) 44. (b) 45. (c) 46. (b) 47. (a) 48. (a) 49. (a) 50. (b) 51. (b) 52. (c) 53. (b) 54. (b) 55. (c) 56. (d) 57. (a) 58. (a) 59. (b) 60. (c) 61. (a) 62. (b) 63. (c) 64. (a) 65. (a) 66. (d) 67. (a) 68. (b) 69. (a) 70. (a) 71. (c) 72. (d) 73. (d) 74. (d) 75. (d) 76. (c) 77. (c) 78. (d) 79. (d) 80. (b) 81. (a) 82. (d) 83. (b) 84. (a) 85. (b) 86. (a) 87. (d) 88. (d) 89. (a) 90. (b) 91. (c) 92. (d) 93. (a) 94. (a) 95. (d) 96. (b) 97. (d) 98. (c) 99. (b) 100. (b) 101. (d) 102. (d) 103. (b) 104. (c) 105. (b) 106. (b) 107. (a) 108. (a) 109. (a) 110. (c) 111. (a) 112. (c) 113. (c) 114. (d) 115. (c) 116. (b) 117. (d) 118. (a) 119. (c) 120. (c) 121. (b) 122. (d) 123. (c) 124. (a) 125. (a) 126. (d) 127. (d) 128. (c) 129. (d) 130. (c) 4. (d) Case Study–1 4. (c) 1. (a) 2. (b) 3. (a) 4. (a) Case Study–2 4. (b) 1. (a) 2. (a) 3. (b) Case Study–3 1. (d) 2. (c) 3. (b) Case Study–4 1. (d) 2. (d) 3. (d)

ANSWERS   229 HINTS/SOLUTIONS 1. Radius of the electron is of the order of ⇒ nmax = eV 10–15 m. h 2. |e| = 1.6 × 10–19 C ⇒ nmax ∝ V 9. The intensity of X-rays increases with the me = 9.1 × 10–31 kg |e| 1.6 × 10−19 C number of electrons striking the filament, me 9.1 × 10−31 kg \\ = hence depends upon the filament current. = 1.76 × 1011 C kg–1 10. g-rays are produced by the nucleus of an atom. The ratio of charge to mass is also known as 11. The energy of X-rays of wavelength l is given specific charge. by, 3. X-rays are em waves of wavelength ranging E = hn from 0.1 Å to 100 Å. (a) l = 0.01 nm = hc l   = 0.01 × 10–9 m ⇒ E ∝ 1   = 10–11 m = 0.1 Å ( 1 Å = 10–10 m) l 4. The value of specific charge is same for all the 12.        E = mc2 = hn particles of cathode rays but they are different for different particles of positive rays. The or mc2 = h c specific charge (q/m) of positive rays is much l smaller than that of cathode rays and is not a universal constant. or mc = p = h l 5. When a charged particle is moving making \\ p = h = 6.6 × 10−34 → l .001nm an angle q with the direction of B , where q ≠ 0° or 90° or 180°, the path of the charged 6.6 × 10−34 = 6.6 × 10–22 kg ms–1 particle in the magnetic field is helical path. = 10−12 6. E = 7.2 × 106 NC–1 B = 1.8 T v = BE = 7.2 × 106 = 4 × 106 ms–1. 13. E = hn = hc 1.8 l 7. X-rays are produced when fast moving   = 6.6 × 10−34 × 3 × 108 = 19.8 × 10–13 J electrons are suddenly stopped on a metal 10−13 of high atomic number. When high energy 14. The penetrating power of X-rays is directly electrons penetrate the target atoms and knock proportional to potential difference between out the electrons from the inner completely the cathode and the anticathode. The greater filled orbits of the atoms, the valency gets the p.d., the larger the penetrating power (also created. The valency so created gets filled by called quality) of X-rays. The penetrating the jump of electrons from higher orbit to that power of X-rays is directly proportional to of lower orbit. This results in the production frequency (n) and is inversely proportional to of characteristic X-rays. wavelength (l) of the X-rays. 8. The maximum frequency of continuous 15. X-rays cannot be emitted from a hydrogen X-rays is given by, atom because its energy levels are very very hnmax = eV close to each other.

230 PHYSICS 16. X-rays can pass through flesh and blood and Putting values of m (= 9.1 × 10–31 kg) and not through bones. So, intestines won’t cast a e(= 1.6 × 10–19 C) good shadow. 6.6 × 10−34 17. According to Mosley’s law, the frequency (n) we get, l = of characteristic X-ray spectral line is directly proportional to the square of the atomic 2 × 9.1 × 10−31 × 1.6 × 10−19 × V number (Z) of the target element. 12.3 Å i.e., n ∝ Z2 Solving we have, l = V or ν ∝ Z 12.3 Å 24. We know,   le = V 18. X-rays are produced when high energy electrons get accelerated through a very high when V = 100 volt potential difference (≈ 10,000 V) in a highly evacuated glass tube, and strike a target le = 12.3 Å = 1.23 Å (≈ 0.001 cm of Hg). 100 19. Millikan’s oil drop experiment makes use 25. If a charged particle of charge q is accelerated of Stoke’s law. When the oil drop is falling by a potential difference V, then freely under the effect of gravity in a viscous medium with terminal speed v, then p2 qV = 2m F = mg = 6 phrv or p = 2mqV Here, the symbols have their usual meanings. De-Broglie wavelength, l = h = h p 2mqV 20. X-rays are not deflected by electric and magnetic fields, which shows that X-rays are Putting values for proton, unchanged. we have, lp = 0.286 Å 21. Alkali metals like lithium, sodium, potassium V etc. show photoelectric effect with visible light. So, in case of alkali metals, the threshold 26. 1 eV = 1.6 10–19 J frequency lies in the visible region. 1 J = 107 erg 22. Metals like zinc, magnesium, cadium, etc. are sensitive only to ultra-violet light for photo- So, J, eV and erg are units of energy and (d) is electric effect. Infra-red radiations cannot unit of power, as 1 W = 1 Js–1. eject photoelectrons from a metal surface, etc. whereas X-rays will always do it. 27. Work function (W0) varies from metal to metal. That material is better for photoelectric Energy (E) ∝ 1 , hence more the l, lesser the emission whose work function is least. l Moreover, with decrease in atomic number, work function will increase. energy of incident radiations. 28. Lesser the work function of a material, better it is for photoelectric emission. Because 23. eV = 1 mv2 or v = 2eV caesium has least work function, it is the best 2 m metal for photoelectric emission. 29. E = hn But l = h or h = E mv ν = h m = ML2T −2 = [ML2T–1] m × 2eV T −1 = h SI units of h kg m2s−1 2meV \\ cgs units of h = g cm2s−1 = 103 × 104 = 107

ANSWERS   231 30. 1 KWh = 3.6 × 106 J h= h p mv Among the given units of energy, it is the 40. l = largest unit of energy. ⇒ l ∝ v–1 (a) 1 eV = 1.6 × 10–19 J h (b) 1 Ws = 1 J 41. We have, l = p (c) 1 MeV = 1.6 × 10–13 J 31. E = number of photons × (energy of a photon) Taking log on both sides we get, log1l0 = log1h0 – log1p0 \\ Number of photons = E , where n is νh = – log1p0+ log1h0 frequency of the photon. It is of the form, y = –mx + c. The graph is a 32. Gases under normal pressure act as bad straight line, making an angle of 135° with conductors. However, when gases are the positive direction of x-axis. As tan q = subjected to low pressure and high voltage, m = –1 = tan (180° – 45°) they act as conductors. = tan 135° 33. A photocell converts light energy into electrical energy. ⇒ q = 135° 34. E = hn = hc Y l =  6.63 × 10−34 × 3 × 108  J  45 × 10−12  135° Solving we get, X E = 4.42 × 10–15 J O 35. 1 keV = 103 × 1.6 × 10–19 J 42. l = h = 1.6 × 10–16 J p \\ E = 4.42 × 10−15 l ∝ 1p 1.6 × 10−16 1 = 27.6 keV It is of the form, y ∝ x , which is a rectangular 36. Since energy associated with the continuous hyperbola. spectrum of radiation = 27.6 keV 43. Energy gained by an electron as it jumps = eVmin across a potential difference of 1 volt is given by, 1 eV = 1.6 ×10–19 J. ⇒ Vmin = 27.6 kV 44. l = h ≈ 30 kV mv 37.   Emax = 1.16 V = eV0 Taking log on both sides, \\ V0 = 1.16 Volt log l = log h – log (mv) 38. In photoelectric effect, the incident photon is or log (mv) = – log l + log h completely absorbed by the electrons of the photosensitive material. It is of the form, 39. l = h y = – mx + c p \\ slope (m) = – 1 It is of the form, y ∝ 1 , and represents 45. Refer to Answer 44 above, x intercept (c) = log h rectangular hyperbola as shown in (b).

232 PHYSICS 46. We have, For n > n0, V is positive and V ∝ v. log(1m0v) = – log1l0+ log1h0 The correct variation is shown by (a). It is of the form, 50. The maximum value of the photoelectric current is called the saturation current. The y = – mx + c, here m = 1 graph between photoelectric current and p.d between the electrodes (I Vs V) is non-linear. and tan q = – 1 = tan 135° With incident light of some frequency but higher intensity, the saturation current also ⇒ q = 135° increases. However, the stopping potential (V0) remain same. The correct graph is shown in (b). 47. Einstein’s photoelectric equation is given by, In the given graph, since the magnitude of saturation current increases, intensity of 1 mv2 = hn – W0 radiation increases. So, I2 > I1. 2 51. The retarding potential (which is given to = hn – hn0 the collector) for which the photoelectric current becomes zero is known as the cut-off = h(n – n0) or stopping potential for the corresponding frequency of the incident light. Also Ek = 1 mv2 = h(n – n0)...(i) 2 We have, ⇒ Ek ∝ (n – n0) 1 mvm2 ax = eV0 = Kmax where v is frequency of the incident photons, 2 n0 is threshold frequency of the metal, v is ⇒ Kmax = eV0 velocity of ejected photoelectron from the From the given graph, the cut-off potential metal surface and m is mass of photoelectron. From the above equation, photoelectric is – V, for which the photoelectric current is emission is possible only if n > n0. Equation (i) is of the form, y = mx – c, which is zero. a straight line with negative intercept on the Y-axis. It is correctly shown in (a). 52. The minimum value of the frequency of light below which the photoelectric emission stops completely, howsoever large the intensity of light may be, is called the threshold frequency. It is different for different metals, lesser the work function of the metal lesser the threshold frequency. 48. The graph between I and V is a non-linear W = hn0 curve i.e., does not obey Ohm’s law. When From the graph we have, n1 < n2 < n3 stopping potential i.e., negative potential is 53. W = hn0 = hn (given) applied, I becomes zero. Therefore, the correct As n1 < n2 < n3, we have, W ∝ n graph is shown by (a). ⇒ W1 < W2 < W3 54. The magnitude of saturation photoelectric 49. We have, current depends upon the intensity of eV = 1 mv2 = hn – hn0 incident radiation. 2 ⇒ 55. When a photon collides with an electron, it When V = hν − hν0 imparts some energy to the electron, hence n e e decreases, and l increases. n = n0, V = 0.

ANSWERS   233 56. 1 Vm–1 = 1 JC–1 m–1 has higher wavelength and lower energy than that of yellow light, therefore, no photoelectric = 1 NmC–1 m–1 emission will take place. = 1 NC–1 64. Photoelectric effect was explained by Einstein following quantum theory of light that was 57. The deflection of the electron in the electric predicted by Max Plank. Quantum theory of light gives the concept of photons. field is given by, y = 1 at2 = 1  eE  l2 , where 2 2  m  v2 l is distance through which electric field is So, photoelectric emission relates to applied. quantisation of energy. Also, v = E . So,  e = 2yE 65. E = hn = mc2  or  m = hν B m l2 B2 c2 66. Lasers can be of three kinds, viz solid, liquids ⇒  me ∝ E ∝ V and gaseous. Here, V is p.d. between the electrodes. 67. We know,  l = h  ⇒ l ∝ 1 2mE E 58. X-rays can pass through black paper (photographic film) whereas other given rays ∴ l2 = E2 = E=1 can’t do so. l1 E1 2E 2 59. The photoelectric current establishes the ⇒ l2 = l1 particle nature of electrons. 2 60. E = hn 68. The emission of electrons by heat is called thermionic emission. E or h = ν 69. Above the threshold frequency, maximum K.E. (or velocity) of the emitted photoelectrons = ML2T −2 depends upon the frequency (or wavelength) T −1 of the incident light but is independent of the intensity of the incident light. = [ML2T–1] 61. When an electron jumps across a p.d. of V 70. A photocell transforms light energy into volt, energy gained by the electrons is eV. electrical energy. So,when accelerated through 1.2 V, energy 71. E = hν = mc2 gained by the electron is 1.2 e.V. hν 1 mv2 or m = c2 2 62. We know, hn = W0 + or 1 mv2 = hv – W0 or p = mc = h ( c = νl) 2 l Also, eV = 1 mv2 72.    E= 1 mv2 2 2 ∴ eV = hn – W0 or v = 2E m or V = h ν − W0 e e But l = h mv It is of the form, y = mx – c, and represents a straight line with slope h . = h = h e m × 2E 2mE m 63. Yellow light has greater wavelength or lower energy than that of green light. But red light

234 PHYSICS 73. de-Broglie wavelength of electron accelerated 82. Electron volt is used to represent energy. through V is given by, 1 eV = 1.6 × 10–19 J l = 12.3 Å 83. A phototube converts light energy into V electrical energy. Given, V = 40 × 103 V 84. The maximum kinetic energy of 12.3 × 10−10 12.3 × 10−10 photoelectrons depends on frequency and \\ l = = 2 × 102 m nature of material but not on the intensity (of 40 × 103 incident light).  6.1 × 10–12 m 85. SI unit of l is m. hc l  6.1 pm ( 1 p = 10–12) 86. We have, eV = hn = 74. E = 1 mv2 ⇒   l = hc = 6.6 × 10−34 × 3 × 108 2 eV 1.6 × 10−19 × 20 × 103 Also, E = hν = hc  ⇒ E ∝ 1  i.e. deuteron ; 0.62 Å l l will have shortest wavelength. 87. Cathode rays give rise to X-rays when they strike a substance of high atomic number. 75. l= h mv 1 mv2 = hn – W0 2E 88. Since, eV = 2 m Also, v = \\ l = h ⇒ V = h ν − W0 2mE e e or l ∝ E–1/2 It is of the form, y = mx – c. 76. me = 9.1 × 10–31 kg (i) Slope of the graph is h , hence h can be 77. Discharge tube is used to produce and analyze e found. the passage of electricity through gases at low pressure (ii) Intercept on the Y-axis is W0 , hence e 78. E = hν work function can be found. or h = E (iii) W0 = hn0, so knowing W0 can enable us ν h ML2T −2 to find n0 T −2 = However, charge on an electron can’t be ∴ h = [ML2T–1]...(i) found from the given graph. Also, angular momentum (L) = mvr 89. We have, Ek = 1 mv2 = hn – hn0 2 or L = [ML2T–1]...(ii) Given, n = 4n0 From (i) and (ii), h has same dimensions as that of angular momentum. ⇒  (K.E.)max = 4 hn0 – hn0 = 3h n0 90. Work function, W0 = hn0 79. Photoelectric current is of the order of mA i.e. 10–6 A.   = hc l0 80. Gamma rays have highest frequency (or minimum wavelength) among the given rays.  6.6 × 10−34 ×3× 108   2000 × 10−10  81. A photocell is also known as an electronic   = J eye.

ANSWERS   235   =  6.6 × 10−34 ×3× 108 ×1  eV graph between the stopping potential and  2000 × 10−10 1.6 × 10−19  intensity of incident light will be constant i.e., it will be a straight line parallel to I-axis.    6.2 eV 101. Lead is the best absorber of X-ray. 92. We know, l = h = h 102. X-rays are electromagnetic waves which are mv p emitted in the form of photons. So, momentum will be same. 103. X-rays can pass through flesh and blood but not through bones. Also, X-ray cannot 92. We have, eV = hn be used in RADAR as they are not reflected by the target. X-rays can be used to produce = hc  ⇒ l ∝ 1 photoelectric effect. l V So, (b) is an incorrect statement. 93. Interference, diffraction and Doppler’s effect 104. The radius of electron is of the order of are associated with the wave nature of light. 10–15 m. 94. We know, l = h 105. When atomic number of elements increases, 2 mE the work function will decrease. ⇒ l ∝ 1 106. When the temperature of a metal increases, m its work function will decrease. So, lp = m=a =4 2 1 la mp 1 1 107. E ∝ l 95. E = hn So, the graph between E and l is a rectangular hyperbola as shown by (a). ⇒ h = E 108. The photoelectric cell is also called as electric ν eye. So, SI units of h are Js. 109. The graph between the frequency (n) of the h =h incident light and photoelectric current (I) is a 96. l= mv p non-linear relation i.e. it does not obey Ohm’s law. The graph meets X-axis at n = n0, which Also, p2 2mE defines threshold frequency for which I = 0. E = 2m  ⇒ p = \\ l = h I 2mE 97. Diffraction establishes the wave nature of O v0 v radiation. Also, photoelectric effect proves the particle nature of light. So, dual nature of 110. The de-Broglie hypothesis established the radiation is shown by (d). wave nature of the moving material particles. The de-Broglie waves are not electromagnetic 98. Momentum of photon = mc waves. So, l = h = h mc p 111. Momentum of the photon = mc 99. When a metal is heated, electrons are ejected = E ( E = mc2) out of it. There ejected electrons are commonly c known as thermions. = hν = h 100. Stopping potential (Vs) does not depend on c l the intensity of the incident light. So, the

236 PHYSICS 112. Photons have zero rest mass. A photon is a 122. We have the photoelectric equation given particle having zero rest mass but moves with by, eV0 = hn – hn0 the velocity of light.  h  h 113. Photoelectric effect involves the law of ⇒ V0 =  e  ν − e ν0 conservation of energy.   114. E = hν It is of the form, y = mx – c. ⇒ h = E = ML2T −2 So, the graph between V0 and n is a straight ν T −1 line. ⇒ [h] = [ML2 T–1] Slope of the line, m = h and intercept of the e 115. Energy of photon, E = hn line is  h ν0  on the Y-axis below the origin.  e  hc = l 123. When zine plate is irradiated with light, photoelectric emission results. So, E ∝ 1 i.e. E is less if l is longer. l 124. de-Broglie wavelength of an electron when it falls through a p.d. of V volts is given by, 116. [h] = [ML2 T–1] l = 12.3 Å So, the required ratio of SI to CGS units of h is V given by, = 12.3 Å kg m2s−1 103 × (102 )2 = 107 10 × 103 g cm2s−1 = 1×1 117. The number of ejected photoelectrons relates = 12.3 Å to the photoelectric current which is directly 100 proportional to the intensity of light. = 0.123 Å 118. l= h  ⇒ l ∝ 1 , for a fixed value of v. 125. The number of ejected photoelectrons (which mv m relates to the photoelectric current) is directly proportional to the intensity of incident light. Since mass of b-particle is minimum, l would be maximum. 126. l = h , l depends on mass, velocity, mv 119. Photoelectric emission is possible for n > n0. Because n for X-rays is greater than that momentum and frequency and not on the of ultra violet light, photoelectric emission will be possible using X-rays. charge of the moving particle. 120. Maximum kinetic energy = 4 eV 127. We have, l = h If V0 is stopping potential, then eV0 is equal to 2mE maximum kinetic energy of photoelectrons. or E = h2  i.e., E ∝ 1 2ml2 l2 So, eV0 = 4 eV ⇒ V0 = 4 V \\ E′ = l2 =  1 2 =4 121. Maximum energy of emitted photoelectrons = E (l′)2    0.5  2 eV ⇒ E′ = 4 E So, V0 = 2 V For no electron to reach the anode, the So, energy should be added to decrease the wavelength = E′ – E = 4E – E stopping potential (V0) should be – 2 V. = 3E.

ANSWERS   237 128. We know, W0 = hc Maximum kinetic energy of the emitted e l0 electrons is given by, 6.6 × 10−34 × 3 × 108 Emax. = hn – hn0 = hc  1 − 1 = (1.6 × 10−19 ) × (4100 × 10−10 )  l l0  Solving we get, = 1297.8×4 1×89 × 10−8 J W0 = 3.018 eV or Emax. = 19.84 × 10−18 eV 54 1.6 × 10−19 Because work function of A and B is less than 3.018 eV, so both these metals will emit = 2.296 eV photoelectrons. 129. K.E. (k) = 1 mv2 Case Study–2 or 2 4. Mass of one photoelectron = 9.1 × 10–31 kg Also, ⇒ mv = 2mk \\  Number of photoelectrons in two kg (say n) ∴ l = h = h = 9.1 ×210k−g31 = 2 × 1031 mv 2mk 9.1 ⇒ kg l ∝ 1 Change on one photoelectron k = 1.6 × 10–19 C l2 = k1 = k = 1 \\  Total charge on  2 × 1031  photoelectrons l1 k2 2k 2  9.1  l2 = l1 = 2 ×91.1031 × 1.6 × 10–19 = 3.52 × 1011 C 2 130. We have, Case Study–3 hn = W0 + K.E. (Ek) 4. l = h = 6.62 × 10−34 = 1.1 × 10–34 m mv 3 × 2 ⇒ Ek = hn – W0 Compare it with the equation of straight line, Case Study–4 y = mx – c. 2. l = h  also, v = 2E mv m \\ Slope of st. line = h, which is Planck’s constant. Case Study–1 \\  l = h  or l ∝ E–1/2 2. order of frequency 2mE X-rays > ultraviolet rays > visible light rays > 3. Momentum gained ion absorption of infrared rays photon p = h So infrared red cannot eject photoelectrons λ due to less energy. \\  Total momentum gained = nh = nh . 4. l = 1800 Å = 18 × 10–8 m, λ l0 = 2700 Å = 27 × 10–8 m 4. As photons are not deflected by electric and magnetic fields.

238 PHYSICS UNIT 8:  ATOMS AND NUCLEI 8. (a) 16. (a) 1. (d) 2. (a) 3. (c) 4. (d) 5. (b) 6. (a) 7. (b) 24. (b) 9. (b) 10. (b) 11. (d) 12. (c) 13. (c) 14. (c) 15. (a) 32. (d) 17. (a) 18. (a) 19. (b) 20. (a) 21. (a) 22. (c) 23. (a) 40. (a) 25. (c) 26. (d) 27. (a) 28. (a) 29. (b) 30. (a) 31. (d) 48. (a) 33. (b) 34. (c) 35. (c) 36. (a) 37. (b) 38. (a) 39. (a) 56. (b) 41. (c) 42. (a) 43. (a) 44. (b) 45. (a) 46. (d) 47. (a) 64. (c) 49. (b) 50. (a) 51. (b) 52. (c) 53. (b) 54. (d) 55. (c) 72. (b) 57. (d) 58. (b) 59. (c) 60. (c) 61. (a) 62. (c) 63. (b) 80. (d) 65. (a) 66. (c) 67. (b) 68. (c) 69. (c) 70. (d) 71. (d) 88. (d) 73. (d) 74. (a) 75. (a) 76. (c) 77. (b) 78. (c) 79. (c) 96. (c) 81. (d) 82. (a) 83. (c) 84. (a) 85. (d) 86. (b) 87. (a) 104. (b) 89. (b) 90. (b) 91. (a) 92. (a) 93. (a) 94. (b) 95. (c) 112. (a) 97. (a) 98. (a) 99. (d) 100. (d) 101. (b) 102. (a) 103. (d) 120. (a) 105. (d) 106. (c) 107. (b) 108. (a) 109. (b) 110. (d) 111. (d) 128. (d) 113. (d) 114. (d) 115. (b) 116. (c) 117. (d) 118. (d) 119. (c) 136. (d) 121. (b) 122. (d) 123. (b) 124. (b) 125. (c) 126. (c) 127. (c) 144. (d) 129. (b) 130. (d) 131. (b) 132. (a) 133. (c) 134. (a) 135. (b) 152. (d) 137. (a) 138. (c) 139. (a) 140. (b) 141. (b) 142. (c) 143. (d) 160. (b) 145. (d) 146. (a) 147. (a) 148. (d) 149. (c) 150. (c) 151. (c) 153. (a) 154. (d) 155. (d) 156. (d) 157. (a) 158. (a) 159. (c) 161. (c) 162. (d) 163. (c) 164. (d) Case Study–1 4. (b) 1. (a) 2. (d) 3. (c) 4. (b) Case Study–2 4. (c) 1. (a) 2. (c) 3. (a) 4. (d) Case Study–3 1. (c) 2. (d) 3. (c) Case Study–4 1. (b) 2. (a) 3. (c) HINTS/SOLUTIONS 3. Nuclear forces are the strongest forces in 1. Isotones are the nuclides which contain the nature. These are the forces of attraction same number of neutrons. amongst the nucleons and are 1038 times 2. Nuclear density is very large. It is of the order stronger than electrostatic forces. of 1017 kg/m3.

ANSWERS   239 4. V = 1 = 2p2mk2Z2e4  1 − 1  12. The potential energy (U) of an electron l  n22  revolving in the nth orbit is given by, ch3  n12  kZe2 U = − r , = RZ2  1 − 1  1 Nm2C–2 and Ze is the charge  n12 n22  4pε0   where k = where R = 2p2mk2e4 on the nucleus. ch3 n2h2 R is called Rydberg’s constant and has the Also, r = 4p2mkZe2 dimensions of wave number, i.e., [M0L–1T0]. 5. Isotopes of an element are the atoms of the Putting this value in U we get, element which have same atomic number but different atomic weights. So, isotopes have U ∝ 1 , we have ignored the negative sign. same Z. n2 6. Isobars are the atoms of different elements 13. The kinetic energy (K) of an orbital electron which have the same atomic weight but revolving in the nth orbit is given by, different atomic numbers. So, isobars have same A. = KZe2 2r 7. Isotopes have the same Z, as they have the Putting value of r we get, same atomic number. K.E ∝ n12 8. A b-particle carries unit negative charge It must be noted that U is negative and K is and has negligible mass. So, the emission of b-particle (or b-decay) produces isobars. positive. 9. r= n2h2  i .e. r ∝ n2 14. Total energy of electron is sum of U and K. 4p2mkZe2 \\ E = K.E. + P.E. 10. The velocity of electron in the nth orbit is = kZe2 − kZe2 = − kZe2 given by, 2r r 2r c  Z  Putting value of r we get: 137  n  vn = , kZe2 4p2mkZe2 2 n2h2 Total energy(E) = − × where Z is atomic number of the element and = − 2p2mk2Z2e4 c is velocity of light. n2h2 So, v ∝ 1 So, total energy of revolving electron in the n 11. The frequency (f) of an orbital electron in the nth orbit is negative i.e., nth orbit is related to n as, E ∝ − 1 n2 Z2m f ∝ n3 It implies that for large n, E is less negative So, i.e., E becomes more. Hence, in the outer f ∝ 1 , stationary orbits, total energy of revolving n3 electron is large. here Z is atomic number and m is mass of the 15. In a hydrogen atom, single electron revolves electron. around a single proton forming the nucleus.

240 PHYSICS So, the question of s nuclear force is not \\ E = 9.1 × 10−31 × 9 × 1016 significant in case of a hydrogen atom. 1.6 × 10−13 16. Rutherford’s experiments on scattering of a = 9.1 × 9 × 10−2 MeV particles proved that atoms are mostly empty. 1.6 17. Energy equivalent of mass defect is called = 0.511 MeV binding energy. 18. Binding energy of a nucleus is the energy with 28. Mass of a neutron is 1.00866 u, which nucleons are bound in the nucleus. It where u is atomic mass unit. is measured by the amount of work required to be done to separate the nucleons by some 1 u = 1.66 × 10–27 kg infinite distance such that they do not interact with each other. 29. The element with chemical symbol X, mass 19. B.E. of 24He = 4 × B.E. = 4 × 7.1 MeV number A and atomic number Z is represented nucleon by, AZX or ZXA. 30. In AZX, (A – Z) represents the number of = 28.4 MeV neutrons. So, in case of 29328U there are 92 protons. 20. B.E. of 21He = 2 × B.E. Since number of protons equals the number nucleon of electrons, it also has 92 electrons. = 2 × 1.1 MeV 31. 23982X has 146 neutrons. 32. The nucleus, does not contain any electrons. = 2.2 MeV In 29328U, the nucleus has 92 protons and 146 21. 1 atomic mass unit (1u) = 1.66 × 10–27 kg (= 238 – 92) neutrons but no electrons. In terms of energy, 1 amu  931 MeV 33. The decay or disintegration of a radioactive 22. Mass of a proton = 1.66 × 10–27 kg = 1 amu = 1u element follows an exponential law, 23. En = − 13.6 eV N = N0e–lt n2 Here, N0 = number of atoms in the sample at t = 0 − 13.6 eV For n = 2, E2 = 4 = – 3.4 eV N = number of atoms left under cayed at time t. 24. 1 amu = 1 u = 931 MeV l is decay constant of the element. \\ 1 MeV = 1u 34. N = N0e–lt 931 For t = 0, N = N0 Using E = mc2 For t = ∝, N = 0 = 1.66 × 10–27 × (3 × 108)2 J This decaying equation is shown by (c). 35. We know, N = N0e–lt 1.66 × 9 × 10−11  931 MeV = 1.6 × 10−13 Taking logs on both sides we get, (Q 1 MeV = 1.6 × 10–13 J) log N = log N0 – lt 25. The decay constant of a stable element is zero. or log N = –lt + log N0 It is of the form, y = – mx + c, and is correctly 26. In case of uranium ore, shown by (c). the ratio of 92U235 to 92U238 is 1 : 40. 38. We have, N = N0e–lt Differentiating with respect to t we get, 27. me = 9.1 × 10–31 kg Using E = mc2 dN d dt dt = 9.1 × 10–31 × (3 × 108)2 Joule = N0 e−lt 1 MeV = 1.6 × 10–13 J = N0e–lt (– l)

ANSWERS   241 − dN − dN Here, N is Avogadro’s number and A is atomic dt dt ⇒ ∝ N or = lN mass. 6.023 × 1023 \\ Number of atoms in gram = 238 So, the rate of disintegration is directly proportional to the number of atoms actually present. Also, the required number in 1 kg 39. Average life or Mean life of a radioactive = 6.023 × 1026 element is given by, 238 τ = 1 47. When one atom of 23952U undergoes fission, l about 200 MeV energy gets released. But T = 0.6931 , where T is half-life 48. 1 Rutherford = 106 disintegrations per second. l 49. Nuclear fuel can be fissioned by the thermal \\ τ = 1 = T = 1.44 T neutrons. The nuclei of 23952U (which is a l 0.6931 commonly used fissionable material) can be broken up by neutrons carrying low energy. 40. N =  1 n =  1 5 The energy of slow neutrons does not exceed N0  2   2  100 KeV. = 1 × 100  3% However, the slow neutrons with energies 32 upto 0.5 eV are called thermal neutrons. 41. 23982U → 29304Th + X 50. In case of fast breeder test reactor (FBTR) fuel Decrease in mass number = 238 – 234 = 4 used is a mixture of plutonium and natural Decrease is atomic number = 92 – 90 = 2 uranium. Sodium is used as a coolant in FBTR. 51. 1 curie = 3.7 × 1010 disintegrations per second So, mass number of the emitted particle is 4 or 1 Ci = 3.7 × 1010 decays/s (dps) and its atomic number is 2. This relates to the alpha particle. Hence, X is an alpha particle. 52. ZXA −b−particle→ ZA+1Y So, decay by b– –emission increases atomic 42. Total energy of electron (En) = K.E. + P.E. number by 1 and there is no change in mass − kZe2 = − 13.6Z2 eV = 2rn n2 number. This is also called the binding energy of 53. The reciprocal of the decay constant (l) of a electron. radioactive substance is called its mean life or average life. In case of ground state of hydrogen atom, 54. Radioactivitly is irreversible, spontaneous and self-disintegration process. Z = 1, n = 1 \\ E1 = − 13.6(1)2 eV 55. Binding energy per nucleon is maximum (1)2 for an element with A ≈ 56. Hence, it is most suitable for nuclear fission. So, the correct = – 13.6 eV choice is (c). 43. Neutrons are required for fission. 56. N =  1 4 = 1 × 100 N0  2  16 44. Protons are required for fusion. 45. In case of a thermal reactor, the fuel used is          6% natural uranium. 57. Energy released in a nuclear fission is about 46. Number of atoms present in 1 g of uranium 200 MeV. atom (23982U) is given by, N . A

242 PHYSICS 58. The antiparticle of electron is positron. The 67. As the isotope is stable, its half-life is infinite. symbol for electron is –01e and that of positron is +01e. A positron should not be confused with Also, l = 0.693 = 0.693 =0 a proton, as the positron has zero mass. T1/ 2 ∞ 59. Curie is a measurement of radioactivity. 68. Maximum number of spectral lines obtained 60. We have, N = N0e–lt on account of transition of electron present in nth orbit to various lower orbits = n(n − 1) . 2 If l = 1 , N = e–1 t N0 Also, nth excited state means (n + 1)th orbit. So, the required number of wavelengths = 1 = 1 ~ 0.37 e 2.718 4 × (4 − 1) = 2 =6 (Q e = 2.718) \\ N  37% N0 69. Given E3→2 is associated with the emission of So, the reciprocal of time during which the ultraviolet radiation. It must be remembered that with increasing quantum number, the number of atoms of a radioactive substance energy difference between adjacent energy gets reduced to 37% of its original value is levels in atoms decreases. In simpler words, called radioactive decay constant (l). as n increases, energy difference between adjacent energy levels decreases. 61. When the radioactive element emits +10b, there is no change in mass but a proton gets So, the transition from E4 to E3 i.e. E4→3 can released. So, the neutron – proton (n/p) ratio be associated with the emission of infrared increases. radiation i.e., decrease in energy. 62. We have, N = N0e–lt, which shows that the 70. As is clear from the given figure, graph is an exponential curve. 63. (i) AZX −b→ ZA+1Y C l1 (ii) ZA+1 l −a→ AZ––41K B (iii) AZ––41K −g→ AZ––41K l2 l3 Thus, the radioactive radiations given out A respectively are b, a, g. ECA = ECB + EBA 64. Average life or mean life of a radioactive ⇒ element is given by, ⇒ hn3 = hn1 + hn2 Also, τ = 1 = 1.44 T, ⇒ n3 = n1 + n2 l ⇒ E3 = E1 + E2 where T is half-life of the element. So, hc = hc + hc l3 l1 l2 l being constant for the element, its mean life remains unchanged. 1 1 1 l2 + l1 l3 l1 l2 l1l2 65. Alpha particles are helium nuclei (42He). = + = 66.  1 n = N = 1 , which gives n = 4.32 l3 = l1l2  2  N0 20 l1 + l2 t = nT1/2 71. The trajectory traced by an alpha particle = 4.32 × 2.8 days depends upon the impact parameter. Impact parameter (b) refers to the perpendicular = 16.5 days. distance of the velocity vector of an a-particle

ANSWERS   243 from the centre of the nucleus. Angle of 76. g-rays are electromagnetic radiations of very scattering (q) (which is the angle between short wavelength of the order of 10–12 m. the direction of approach and the direction of receding of the particle) increases as b Because of no mass, g-rays are highly decreases. An a-particle passing close to the penetrating, about 100 times more than b-rays nucleus suffers large angle deflection. and 10,000 times more than a-rays. 72. Refer to the answer of Q. 70 above. 77. Exponential law is obeyed by both the species of atoms, and is independent of their life time. l3 = l1l2 So, the correct option is (b). l1 + l2 78. When an electron jumps from the 1st orbit =  2.4 × 10−7 × 8 × 10−8  m to the ground state, energy (in the form of a  2.4 × 10−7 + 8 × 10−8  photon) given out = E1 – E0 = – 8 eV – (– 18 eV) =  2.4 × 8 × 10−15  m = 10 eV  (2.4 + 0.8) × 10−7    So, when an electron having 12 eV energy interacts, a photon of energy 10 eV and an = 19.2 × 10–8 m electron with 2 eV energy would emerge. 3.2 = 6 × 10–8 m 79. Radioactive substance left undecayed Note: It must be remembered that as per the = N0 – 3 N0 = N0 conservation of energy requirement, two 4 4 photons (4 → 2 and 2 → 1) have the same total energy as the single photon (4 → 1). \\ N = 1 =  1 n N0 4  2  73. Energy released is given by, E = (Dm)c2 It gives n = 2, i.e., two half-lives. = (2 × 10–9) × (3 × 108)2 = 1.8 × 108 J \\ t = nT1/2 = 2 × 4 = 8 months 74. Energy emitted in the first case, E1 = 2E – E 80. Half-life of a radioactive element depends = E...(i) only on the nature of the element. Energy emitted in the second case, 81. We know, E2 = 4E –E N = N0e–lt 3 when t = ∞, N = N0e–∞ = E  ...(ii) = N0 3 ∞ Also, E = hn = 0 = hlc 82. Yukawa proposed that adjacent nucleons (protons and neutrons) attract each other as ⇒ E ∝ 1 a result of constant exchange of particles, l called p-mesons (i.e. pions) between them. The p-meson may have charge 0, +e or –e. The From (i) and (ii), E2 is 1 rd of E1, therefore, l2 charged p-meson has a mass 272 times the 3 electron mass and a neutral p-meson has a mass 264 times the electron mass. must be 3 times i.e., 3l. proton ←pp+−((mmeessoonn))→ neutron 75. When n increases, the energy difference between adjacent energy levels decreases.

244 PHYSICS A proton changes into a neutron in the field 90. Positrons are produced during pair of p– and a neutron changes into a proton in production. the field of p+. However, the total number of n → p + e– + ν neutrons and protons in the nucleus remains p → n + e+ + n same. 91. The given reaction refers to the proton- 83. The fissionable material in the atom bomb was plutonium (23994Pu). When an atom bomb proton cycle, which takes place at very high is dropped, the pieces of 239Pu are brought temperature (~ 107 K). together by an explosive device, so that their mass and size become greater than the In fact the complete reaction is, corresponding critical values. 4(11H) → 42He + 2(+01e) + 2n + 2g + 24.7 MeV 84. Electron volt is unit used for energy. It is the Here, four protons combine to form one energy gained by an electron as it jumps across a potential difference of 1 volt. helium nucleus plus two positrons (+01e), two gamma rays (g) and two neutrinos with a 85. a-decay: AZX → AZ––24Y + 42He release of about 25 MeV. b-decay: AZX → AZ+1Y +–01e + ν This refers to nuclear fusion, as it is possible ν is antineutrino between light nuclei. g-decay: AZX* → AZY + g 92. Geiger Muller counter is used to detect X* denotes the nucleus in an excited state. a-particles and b-particles. From the above, mass of an alpha particle is 93. When an electron and a positron are four times that of a hydrogen atom. Mass of annihilated, two photons (moving opposite a b-particle is same as that of electron. g-rays to each other) each of energy about 0.53 MeV, possess no mass and no charge. are produced. So, mass of the end-product gets reduced by 94. Total energy, charge and momentum are 4 amu. conserved. 86. A positron is identical to the electron except \\ Minimum energy of photon that its charge is +e instead of –e. = 2 × 0.53 MeV The positron emission takes place when a nuclear proton changes into a neutron and in = 1.06 MeV the process a neutrino is also given out. p → n + e+ + n During pair production, an electron-positron pair is created when a photon passes close 87. Carbon–14 is used in determining the age of to a nucleus in the medium. The photon the organic samples. Carbon–14 is produced must have at least an energy of 1.02 MeV to continuously in the earth’s atmosphere on produce a pair. account of the following reaction. 147N + 10n → 146C + 11H + 0.6 MeV 95. In case of g-rays, l is of the order of 10–12 m. There exists a radioactive equilibrium 96. E = mc2 between the formation of Carbon–14 equilibrium in the atmosphere. E’ = m  2 c 2  3  88. Specific charge (Charge to mass ratio) remains the same for alpha particles. = 4 mc2 9 89. Annihilation of an electron and a positron establishes mass-energy equivalence. \\ E − E’ =  1 − 4  E  9  = 5 9 97. 73Li + 11H → 2(42He) + energy.

ANSWERS   245 98. In every nuclear reaction which represents The fission process results in the production transition of one nucleus into another, of several neutrons, generally 2 or 3. following conservation laws are obeyed: On an average, about 2.5 neutrons are (i) Conservation of charge number produced per fission. (ii) Conservation of nucleons (mass) 10n + 23952U → 15461Ba + 9326Kr + 3(11n) + Q Here, Q  200 MeV (iii) Conservation of energy 106. n = t = 9 =3 (iv) Conservation of linear momentum \\ T1/2 3 As per the question, N =  1 n =  1 3 10n + 150B → 42He + 37X N0  2   2  So, 73X is 73Li. 99. (a), (b) and (c) are units of energy or N = N0 8 1 Mev = 1.6 × 10–13 J  1  1 kWh = 3.6 × 106 J Thus, the activity reduces to  8  th. 100. A positron has the same mass as that of an 107. Average binding energy per nucleon is about electron. 7.6 MeV for nuclei with A ≈ 230 and about 8.5 MeV for nuclei with A ≈ 115. 101. In fusion reactions, about 1% of mass gets converted into energy. So, B.E. is about 8 MeV for heavier nucleon 102. Einstein’s mass-energy relation is, E = (Dm) c3, where Dm is mass changed into nuclei. energy and c is velocity of light. nuBc.lEeo. n is small for light nuclei and increases 103. E = (Dm)c2 Dm = 1 mg = 10–3 g = 10–6 kg to a maximum value for A ≈ 60. \\ E = 10–6 × (3 × 108)2 108. Half-life of 146C is 5600 years. = 10–6 × 9 × 1016 J 109. Pauli proposed that in b-decay, an additional particle is also emitted. The presence of this = 9 × 1010 J additional particle enables the electrons to have continuous distribution of energy. 104. The more readily fissionable isotope of Instead of two particles, now three particles uranium is 23952U. In fact, uranium has many (p, e– and ν ) share the energy. isotopes viz. n → p + e– + ν 92U233, 29325U, 29326U and 23982U 105. The fission of 23952U by a neutron is given by, 110. n= 6400 1600 01n + 29325U → 29326U* → X + Y + neutrons + Q = 4 23962U* is an intermediate excited state which lasts for a very-very small time (~ 10–12s) \\ N =  1 n =  1 4 = 1 before splitting into fission fragments X and N0  2   2  16 Y. There are many combinations of X and Y which satisfy the requirement of mass-energy, 111. Nuclear forces do not obey inverse square charge and nucleon number. Q is energy law. When r is more than 10 fm (10–14 m), the released, which is of the order of 200 MeV. nuclear force is negligible. When the nucleons 105. The process of fission is initiated by a neutron and is accompanied with the emission of are brought closer, attractive force comes into neutrons. play, which rapidly increases.

246 PHYSICS At a distance 0.5 fm (0.5 × 10–15 m), the force 119. γ-rays will be able to pass through 20 cm becomes strong repulsive in nature. This is thickness of steel because of their large correctly shown by (d). penetrating power. 112. N =  1 n =  1 4 120. Control rods are made of cadmium. N0  2   2  121. 105B + 42He → 137N + 10n So, the particle Y is neutron. N 1 × 100  6.25% or N0 = 16 122. Half-life can be expressed in units of time So, (d) is correct. 113. 6C12 + 0n1 → ZXA + –1e0 \\ A = 12 + 1 = 13 123. Mean life ar Average life (t) is related with half-life (T) of a radioactive element by the Z = 6 + 1 = 7 relation, 114. n = 72000 =3 τ = 0.6931 24000 T N  1 3 1 124. We have, N0  2  8 \\ = = N  1 n 1 N0  2  64 = = 115. When an a-particle approaches the nucleus, ⇒ n 6 ⇒   it experiences the electrostatic force of As  1 =  1  2  2 repulsion and its K.E. gets converted into ⇒ potential energy. At the distance (r0) of nearest n = 6 t approach, K.E. of a-particles gets completely n t = nT, we get T = converted into P.E. of the system. K.E. = P.E. Ze × 2e T = 60 = 10 s r0 6 = Ke × 125. Electrons occupy the orbitals in increasing 2Ze2 order of energy i.e., K.E. ⇒ r0 = Ke × Es < Ep < Ed < Ef 126. The reciprocal of wavelength (1/l) is called Putting values and solving we get, the wave number and it gives the number of r0 = 9 × 109 × 2 × 92 × (1.6 × 10−19 )2 wavelengths per unit length. 5 × 1.6 × 10−13 1 ≈ 10–14 m ν = l So, r0 ≈ 10–12 cm SI unit of ν is m–1. 116. A neutron cannot exist in free state 127. Maximum number of spectral lines obtained n → p + e– + ν on account of transition of electron present in So, the radioactive neutron decays into a proton, an electron and an antineutrino. the nth orbit to various lower orbits is given by, 117. 2131Na n(n − 1) 2 Z = 11, A = 23 128. A photon is associated with energy. So, the dimensions of photon are those of So, there are 11 protons, 11 electrons and 12 (i.e., 23 – 11) neutrons. energy. [Photon] = [ML2T–2] 118. The process of ionization is caused by the energy produced by electrostatic forces. 129. In fast breeder test reactor (FBTR) fuel used is a mixture of plutonium and natural uranium.

ANSWERS   247 130. Uranium in which fraction 23932U is increased 146. An isobar is produced in a b-emission. Isobars from 0.7% to 02.3% is called enriched uranium. have same mass number but different charge number. b-emission involves change in charge 131. Maximum number of electrons in a shell can number alone. So, isobars are produced with be 2n2. the emission of b-particles. 132. a-particles and γ-rays have line spectra. 147. Only charged a-particles are deviated by an 133. b-particles have a continuous spectra. electric field, others are not charged. 134. Activity of a radioactive element, A = − dN = lN 148. Half-life depends only on the nature of the dt radioactive element. ⇒ A = lN 149. The energy of a hydrogen atom in its ground It is of the form, y = mx state is – 13.6 eV. 135. Number of atoms left after n half-lives is We have, En = − 13.6 eV i.e., n2 given by, N = N0 E1 = − 13.6 eV ⇒ 2n (1)2 N = N02–n or N =  1 n Also, E∞ = 0 N0  2  \\ B.E. = E∞ – E1 = 13.6 eV 136. 1 millicurie = 37 Rutherford 150. The decay constant of a stable element is zero. 137. The decay constant of a stable element is zero. l = 0.693 = 0.693 = zero T1/2 ∞ 138. The source of stellar energy is nuclear fusion. 151. Critical mass of uranium is 10 kg. 139. Uncontrolled nuclear chain reaction is the basis of an atom bomb. 152. In uranium ore, the ratio of 23925U to 23982U is 1 : 40. 140. Controlled chain reaction is the basis of a nuclear reactor. 153. Neutrons are required for fission and protons are required for fusion. 141. Nuclear fusion is the basis of hydrogen bomb only. 154. Required number of emitted 142. In a nuclear reaction, atomic number, mass spectral lines = n(n − 1) = 4×3 =6 number and charge are conserved. 2 2 143. In a nuclear reaction, mass, energy, charge 155. P.E. of electron in nth orbit, and momentum are conserved. P.E. = −Ke2 = Un 144. Energy in the sun is generated mainly by the rn fussion of hydrogen atoms. Ke2 1 1 145. As per Bohr’s theory, radius of nth Bohr orbit rn 2 2 is given by, n2 K.E. of electron = − = − (P.E.) = − Un mZ rn ∝ Here, n is principal quantum number, Z  is \\ Total energy of electron = K.E. + P.E. charge number of element, m is mass of We have, K.E + P.E. = – 1.5 eV particle revolving around the nucleus. ⇒ −1 Un + Un = – 1.5 eV 2 1 So, rn ∝ m 1 2 r1 m2 27 ⇒ Un = – 1.5 eV r2 m1 64 ⇒ = = ⇒ Un = – 3.0 eV

248 PHYSICS 156. En = −  13.6 Z2  eV = 1  n2  2 ⇒ En ∝ 1 162. The emitted photon is with maximum energy n2 in case of transition III. We have, Z = 1 in case of H-atom E = E2 – E1 \\ E2 =  3 2 = 9 = − 13.6 − (− 13.6) E3  2  4 (2)2 (1)2 Note: 1st and 2nd excited states refer = (– 3.4 + 13.6) eV = 10.2 eV energy corresponding to 2nd and 3rd orbits 163. Given, T = 10 days respectively. \\ Number of half-lives, 157. We have, ν = 1 = RZ2  1 − 1  n = t l  n12 n22  T   We have, Z = 1 in case of hydrogen atom. = 30 =3 Lyman series (in ultraviolet region) is 10 obtained when an electron jumps to 1st orbit from any subsequent orbit. So, N =  1 n =  1 3 N0  2   2  ν = 1 = R(1)2  1 − 1  = 1 = 0.125 l  (1)2 (∞)2  8   = R(1 – 0) 164. In case of nuclear reactions, the number of fundamental particles is not conserved. 1 ⇒ l = R , where R is Rydberg constant. Case Study–1 158. With increasing quantum number, the energy 3. Energy of projectile difference between adjacent energy levels in atoms decreases. = Potential energy at closest approach = 4π1ε0 z1rz2 159. AZX −b→ ZA+1Y −a→ AZ––41K −g→ AZ––41K \\ Energy ∝ z1z2 b-particle is –01b, a-particle is 24a or 42He and γ-ray is 00γ. 4. The K.E. of the a-particles is equal to the potential energy of the a-particles, at the 160. Suppose there are x alpha particles and y beta distance of closest approach (r0). particles that are emitted. \\  Difference in mass number = 4x = 32 \\ K.E. = 1 × (Ze)(2e) 4πε0 r0 ⇒ x = 8 Also, difference in charge number or r0 = 1 × 2Ze2 4πε0 (K.E.) = 2x – y = 88 – 78 = 10 ⇒ y = 2x – 10 Putting 1 = 9 × 109 in SI units,  Z = 79 4πε0 = 2 × 8 – 10 = 6 (for gold),  e = 1.6 × 10–19 C  and 161. We have, K.E. = 7.9 × 10–13 J 2 × 79 × (1.6 × 10−19 )2 7.9 × 10−13 N  1 n 1 Then r0 = 9 × 109 × N0  2  = =  1 2  2    or   r0 = 4.61 × 10–14 m.

ANSWERS   249 Case Study–2 3. Because Cadmium rod absorbs neutrons 2. mv2 = 1 e2 .  Here  r = a0. without suffering disintegration and slow r 4πε0 r2 down the intensity of chain reaction. 4. Rate of production of energy, Case Study–4 P = 400 MW = 400 × 106 Js–1 2. Mass of U-235 converted into energy Energy produced in a day = 10 × 0.1 kg = 10 g E = P × 86400 J 100 = 400 × 86400 × 106 J Number of U235 atoms in 10 g = 3.456 × 1013 J = 6.02323×51023 × 10 = 25.63 × 1021 If m kg is the required mass, then \\ Energy released m = E   c2 = 25.63 × 1021 × 200 MeV 3.456 × 1013 = 51.26 × 1023 × 1.6 × 10–13 J or m = (3 × 108 )2 kg = 8.2 × 1011 J = 0.384 × 10–3 kg = 0.384 g UNIT 9:  ELECTRONIC DEVICES 1. (a) 2. (b) 3. (c) 4. (c) 5. (b) 6. (c) 7. (c) 8. (c) 9. (a) 10. (c) 11. (b) 12. (d) 13. (c) 14. (c) 15. (b) 16. (c) 17. (c) 18. (c) 19. (d) 20. (c) 21. (a) 22. (d) 23. (a) 24. (b) 25. (a) 26. (d) 27. (b) 28. (d) 29. (a) 30. (b) 31. (c) 32. (c) 33. (a) 34. (a) 35. (a) 36. (d) 37. (c) 38. (b) 39. (b) 40. (d) 41. (c) 42. (c) 43. (d) 44. (a) 45. (c) 46. (b) 47. (d) 48. (a) 49. (d) 50. (d) 51. (d) 52. (b) 53. (d) 54. (c) 55. (b) 56. (a) 57. (a) 58. (a) 59. (a) 60. (d) 61. (a) 62. (a) 63. (b) 64. (d) 65. (a) 66. (c) 67. (a) 68. (c) 69. (a) 70. (c) 71. (b) 72. (a) 73. (b) 74. (c) 75. (a) 76. (b) 77. (c) 78. (c) 79. (d) 80. (c) 81. (c) 82. (c) 83. (b) 84. (a) 85. (a) 86. (b) 87. (c) 88. (d) 89. (b) 90. (d) 91. (b) 92. (b) 93. (b) 94. (b) 95. (b) 96. (d) 97. (d) 98. (c) 99. (d) 100. (d) 101. (d) 102. (a) 103. (d) 104. (b) 105. (a) 106. (c) 107. (a) 108. (a) 109. (a) 110. (d) 111. (a) 112. (b) 113. (b) 114. (b) 115. (b) 116. (d) 117. (d) 118. (d) 119. (d) 120. (d) 121. (c) 4. (d) Case Study–1 4. (c) 1. (d) 2. (b) 3. (d) 2. (d) 3. (a) 4. (c) Case Study–2 2. (d) 3. (d) 1. (c) Case Study–3 1. (b)

250 PHYSICS HINTS/SOLUTIONS 1. Drift velocity per unit electric field is called It is called p-type because the conduction of mobility. It is denoted by m. electricity in such semiconductors is due to motion of holes. m = v E 11. With increase in temperature, more number of covalent bonds are broken, hence the 2. In case of a half wave rectifier, the output d.c. conductivity of semiconductivity increases. value of the wave is less than the r.m.s value of Here, the semiconductor is cooled from T1 to the a.c. component of the wave because of the T2, conductivity decreases. So, the resistance barrier voltage of diode. So, the r.m.s value of will increase on lowering the temperature of the a.c. component of the wave is greater than the semiconductor. the d.c. value. 3. Forbidden energy band gap is almost zero 12. b = 1 α  or  1−α = 1 in case of conductors, about 1 eV in case of −α α β semiconductors and about 6 eV in case of insulators. or 1 −1 = 1 α β So, EG1 < EG2 < EG3. \\ 1 = 1 + β 4. On increasing the reverse biased voltage to α β some large value in case of p-n junction, there or a = 1 β β occurs the break down of the p-n junction. So, + the reverse current suddenly increases. This means large current flows due to minority 13. A n-type semiconductor has excess of free carriers. electron for conduction. But, the total number of electrons in an atom equals the total 5. The given arrangement of p-n junctions number of protons inside the nucleus. Hence, represents a circuit of full wave rectifier. n-type semiconductor is bound to be neutral. 6. When the transistor is used as an amplifier, 14. The forbidden gap in case of a semiconductor the emitter-base junction is forward biased is of the order of 1 eV. and the base-collector junction is reverse biased. 15. Most commonly used material is Silicon. 7. OR, NOR and AND gates are respectively 16. Transistors are mainly used for amplification represented by 1, 3 and 2. of the input signals (voltage/power). In fact, a transistor can be used both as amplifier and So, (c) is correct. oscillator. 8. The thickness of depletion layer at 17. In common base circuit, the output resistance p-n junction is about 10–6 m. is moderate. 9. In a common base amplifier, the input 18. In half wave rectifier, only half of the wave voltage and output voltage are in phase with gets rectified. It is correctly shown by (c). each other. Therefore, the phase difference between output voltage and input voltage in 19. In Boolean algebra, a common base transistor circuit is zero. 1 + 0 = 1 10. When a pure semiconductor of Ge or Si is 0 + 1 = 1 doped with a controlled amount of trivalent atoms, like aluminium or boron, we get and 1 + 1 = 1, are all correct. p-type semiconductor. 20. (100)2 = 0 × 20 + 0 × 21 + 1 × 22 = 0 + 0 + 4 = 4

ANSWERS   251 21. (1011)2 + (100)2 2 15 33. The input to OR gate is A and (A ⋅ B). = (1×20+1×21+0×22+1×23) 2 7 − 1 Therefore, Y = A + A ⋅ B + 0 × 20 + 0 × 21 + 1 × 22 2 3 − 1 34. In a p-type semiconductor, silicon is doped with an impurity atom of valency 3. So, silicon = (1 + 2 + 0 + 8) + (0 + 0 + 4) 2 1 − 1 is doped with aluminium. = (15)10 0−1 35. The output of 2-input OR gates is zero only when its both inputs are 0. \\ (15)10 = (1111)2 22. In common collector circuit, the output resistance is very high. ABY 23. Boolean algebra is based on some logical 000 relationship. In fact, a logic gate is an electronic circuit which makes decisions A Y0 1 1 based on some logic. B Symbol of OR gate 1 0 1 24. In Boolean algebra, 111 1.1 = 0 is incorrect Truth table of OR gate as 1.1 = 1. 36. The forbidden energy gap in semiconductors 25. A = 1, B = 0 lies just below the valence band. A × A + B = 1 × 1 + 0 = 1 + 0 37. The input to AND gate will be A and B . = 1 = A Therefore, the output is given by, 26. To obtain n-type semiconductor from a Y = A ⋅ B germanium crystal, it must be doped with an impurity atom of valency 5. 38. Here, Y = A + B is for OR gate joined with 27. To obtain p-type semiconductor from a NOT gate. So, the final combination represents germanium crystal, it must be doped with an NOR gate. impurity atom of valency 3. 39. Here, Y = A ⋅ B is for AND gate joined with 28. In case of a reverse biased p-n junction, the NOT gate. So, the final combination represents reverse current is almost zero and the order NAND gate. of resistance is about 106 W. 40. For A = 1, A + A = 1 + 0 = 1 29. A photodiode is a semiconductor diode which is operated with reverse bias below the For A = 0, A + A = 0 + 1 = 1 breakdown voltage and whose conductivity gets modulated by the absorption of light in So, (d) is correct choice. or near the depletion layer at the p-n junction. 41. y = A ⋅ B + B ⋅ A = 1.0 + 1.0 30. In case of forward biasing of the p-n junction diode, the p-n junction region is caused by   = 0 + 0 = 0 diffusion of charge carriers. Also, the flow of current across the junction is mainly due to 42. A transistor can be used as an amplifier as diffusion of charges. well as an oscillator. However, it cannot be used as a rectifier. 31. The Boolean expression for the logic gate is y 43. A p-n junction diode cannot be used for = A ⋅ B = A ⋅ B , which represents AND gate. increasing the amplitude of an AC signal. 32. The Boolean expression of this gate is 44. In order to use a transistor as an amplifier, the Y = A ⋅ B = A + B emitter-base junction is forward biased and This represents OR gate. the collector-base junction is reverse biased. 45. In case of n-p-n transistor working as an amplifier, the electrons go from the collector region to the base region.

252 PHYSICS 46. For the given truth table, Boolean expression 58. For the given truth table, Boolean expression is Y = A ⋅ B, which is for AND gate. is Y = A ⋅ B , which is for NAND gate i.e. not 47. There is no forbidden gap or energy gap in case of a good conductor. AND gate. 48. A = 1, B = 0 59. In the transistor, emitter-base is forward biased. A × B + A = 1 × 0 + 1 = 1 + 1 = 1 = A 60. In a transistor, the emitter is heavily doped and its main function is to supply majority 49. A junction transistor is a transformer charge carriers when it is forward biased. The of resistance, which can be achieved by collector is doped moderately as compared to interchanging the biasing across the junction emitter. Its main purpose is to collect majority triode, hence the name junction transistor charge carriers. The base is very lightly is given. The parameters associated with doped as compared to emitter and collector. the transistor are current gain, resistance The main purpose of the base is to bring an gain and power gain. In simpler words, a interaction between the emitter and collector. transistor is essentially a current operating So, biasing the base has no meaning. device in which the emitter current controls the collector current. 61. The correctly marked p-n-p common base transistor is shown in (a). The electrons leave 50. In a common emitter transistor amplifier, the base and reach the emitter, hence the there occurs amplification in the current direction of current is from emitter to base. voltage and power of the given signal. So, the current gain is more than unity. 62. The correctly marked n-p-n common base transistor is shown in (a). The electrons leave 51. Since there is amplification in the current the emitter and reach the base, hence the voltage and power of the given signal, all the direction of current is from base to emitter. three parameters are greater than unity. 63. The correct symbol for p-n junction is shown 52. In case of a common base transistor amplifier, in (b). current gain is least. 64. The p-n junction acts as a half-wave rectifier D.C. current gain (a) = Ic , which is the ratio (changing a.c. to d.c.) and the output across Ie RL is shown in (d). of collector current to the emitter current. 65. A semiconductor diode comprises of one p-n junction. Also, A.C. current gain (aa.c.) = ∆Ic ∆Ie 66. Resistivity of a semiconductor depends on the atomic nature of the semiconductor. 53. We have the relation between a and b, which is given by, 67. An oscillator is basically an amplifier which maintains oscillations of the output signal, b = 1 α which is possible by providing positive −α feedback to the amplifier. 54. Zener diode is a voltage regulator, hence it is 68. The statements (a), (b) and (d) are correct but used for stabilisation. (c) is wrong. 55. A crystal lattice refers to an orderly The majority carriers in n-type semi- arrangement of atoms/molecules in a crystal. conductors are electrons, and not holes. 56. The length of the depletion layer gets reduced 69. The logic circuit is an AND gate with truth on increasing the forward voltage. table and logic gate as given below: 57. In order to operate the transistor in saturation ABC mode, the emitter-base and collector-base junctions are forward biased. A 000 C B 010 Symbol of AND gate 1 0 0 111

ANSWERS   253 70. Though the circuit acts as a full wave rectifier, are same. At room temperature, because of the contributions from diode D1 are B, D as thermal vibrations of atoms, few bonds of shown correctly by (c). intrinsic semiconductor are broken. This 71. Semiconductors have covalent bonding in results in the production of equal number of them. electrons and holes in semiconductor. 72. The p-n junction diode is a one way device. 85. Fermi energy refers to maximum energy of It offers low resistance when it is forward electrons in metals at 0°C. biased and high resistance when it is reverse 86. Two NAND gates are used to form AND gate. biased. Hence, it acts as a rectifier. 87. Holes are majority charge carriers in a p-type 73. Zener diode is like an ordinary diode except semiconductor. that it is heavily doped in order to have a sharp break down voltage (reverse bias). This 88. b= α , where a = Ic and b = Ic sharp break down voltage is called zener 1−α Ie Iβ voltage. 89. The forbidden gap in the energy bands of 74. At absolute zero, a pure semiconductor has germanium at room temperature is about filled valence band and empty conduction 0.67 eV. band. Hence, it will behave as an insulator. 90. The current gain (a or b) being the ratio of 75. In a good conductor, the energy gap between two similar quantities, has no units. the conduction band and the valence band is 91. A p-type semiconductor is electrically neutral. zero. 92. We know, 76. The temperature coefficient of resistance of a a = 1 β β = 99 = 99 = 0.99 semiconductor is negative, hence its resistance + 1 + 99 100 decreases with increase in temperature. 77. The p-n junction can be used as a detector 93. The width of the potential barrier in a p-n also. junction diode decreases, when it is forward biased. 78. When forward biased, the potential barrier of 94. p-n junction can be used as a rectifier. the p-n junction diode would decrease. 95. In a transistor, the emitter region is heavily 79. In a pure conductor, valence band and doped to produce a large number of majority conduction band partly overlap each other. carriers. Hence, there is no forbidden energy gap in 96. The depletion region consists of immobile case of a pure conductor. ions. 80. In a p-type semiconductor, conduction is due 97. In a junction diode, the holes represent to greater number of holes and lesser number missing of electrons. of electrons. 98. A rectifier changes a.c. to d.c. but the filter 81. Silicon is the most commonly used material ensures smooth d.c. output from a rectified for making transistor. signal. 82. When a pure semiconductor of germanium 99. The given truth table satisfies the relation: crystal is doped with a controlled amount of A · B = Y pentavalent atoms, such as phosphorous, we The corresponding logic gate is AND gate. get n-type (extrinsic) semiconductor. 100. At OK, intrinsic semiconductor behaves as a 83. When a pure semiconductor of germanium perfect insulator. crystal is doped with a controlled amount of 101. The temperature coefficient of resistance is trivalent atoms, such as indium, we get p-type negative for a semiconductor. So, when the semiconductor. temperature of a semiconductor increases, 84. The number of electrons and holes in an its resistance decreases hence its conductivity intrinsic semiconductor at room temperature increases.

254 PHYSICS 102. Phosphorous is a pentavalent material, hence 110. A zener diode is always connected in reverse when it is doped in a pure semiconductor of bias, and has a sharp break down voltage germanium, then in germanium we have, called zener voltage. So, the zener diode can act as a voltage regulator. ne >> nh 103. The resistance of a reverse bias p-n junction is 111. The barrier potential depends upon all the factors given in (b), (c) and (d). In other words, about 106 W. the barrier potential depends upon doping density, temperature and the forward bias. 104. When the resistance between the p and the n regions is high, then the p-n junction acts 112. The depletion region created at the junction as a capacitor. A p-n junction diode can be does not contain any free charge carriers. considered to be equivalent to a capacitor, with p and n regions acting as plates of a 113. Base region is least doped in a transistor. parallel plate capacitor and the depletion region as the dielectric medium. 114. The formation of energy bands in solids are a consequence of Pauli’s exclusion principle. 105. In case of forwarding biasing, the barrier 115. The p-n junction diode is reverse biased if potential decreases and the forward current p-side of p-n junction is at negative potential subsequently increases. w.r.t. n-side of p-n junction. This condition is satisfied in case of (b). 106. The p-n junction diode is said to be forward biased if p-side (region) of p-n junction diode 116. NAND gate is associated with ‘not AND’ is at positive potential w.r.t. n-side (region) output signal. This is correctly shown by the of p-n junction. It is correctly shown in case logic gate (d). of (c). 117. Given p-n junction is forward biased with 107. The p-n junction diode is said to be reverse voltage biased if p-side (region) of p-n junction diode is at negative potential w.r.t. n-side (region) = 6V – 4V = 2V of p-n junction. It is correctly shown in case of (a). \\ Current, I = 2 = 2 × 10–2 A 100 108. Form factor = Irms = Erms 118. During reverse biasing, when high voltage is Id.c. Ed.c. applied, electric break down of junction takes place. This results in large increase in reverse In case of a half wave rectifier, current. This high voltage applied is known as zener voltage. Irms = I0 2 120. We have, Id.c. = I0 I = R V Ri = 50 π + 5000 + 50 I0 π = 5 = 1  9.9 × 10–3 A 2 505 101 \\ Form factor = 2 = = 1.57 I0 ⇒ I = 9.9 mA π 121. Boolean expression of the given arrangement 109. The charge density near the p-n junction i.e. is given by, depletion region, varies with distance as shown by (a). The p-n junction is equivalent to Y = A + B = A ⋅ B = A · B a capacitor in which the depletion layer acts So, AND gate is produced. as a dielectric.

ANSWERS   255 Case Study–1 = 6.63 × 10−34 Js × 3 × 108 ms−1 4. Number of holes in the sample 1.9 × 1.6 × 10−19 J = Number of electrons in the sample = 6.5 × 10–7 m = ne × V = 6 × 1018 × 10–7 = 6 × 1011 Case Study–3 Case Study–2 4. Eg = hc = 1240 eV = 2.48 eV 4. λ 500 nm hc l= Eg UNIT 10:  COMMUNICATION SYSTEMS 1. (c) 2. (b) 3. (b) 4. (c) 5. (d) 6. (c) 7. (a) 8. (c) 9. (d) 10. (c) 11. (b) 12. (b) 13. (a) 14. (a) 15. (b) 16. (c) 17. (d) 18. (c) 19. (b) 20. (a) 21. (b) 22. (d) 23. (b) 24. (d) 25. (d) 26. (b) 27. (c) 28. (c) 29. (b) 30. (c) 31. (d) 32. (d) 33. (b) 34. (a) 35. (d) 36. (c) 37. (b) 38. (c) 39. (a) 40. (a) 41. (d) 42. (c) 43. (a) 44. (b) 45. (b) 46. (d) 47. (a) 48. (d) 49. (a) 50. (d) 51. (d) 52. (a) 53. (c) 54. (d) 55. (b) Case Study–1 3. (d) 4. (c) 5. (d) 1. (b) 2. (d) HINTS/SOLUTIONS 3. Waves with more than 30 MHz frequency 23. Area covered = pd2 = p (2Rh) will not be reflected back. 30. n = 1  or  LC = 1 7. Bandwidth = (nc + nm) – (nc – nm) = 2nm 2π LC 2πν 8. Power radiated by the antenna is proportional 1 = 1 2π × 2 × 106 4π × 106  l 2 =  λ  to . 9. Modulation index = Amax. − Amin . 31. In point-to-point communication mode, Amax. + Amin . communication takes place over a link between a single transmitter and a receiver. 11. Modulation does not change time lag between transmission and reception. 32. The maximum line-of-sight distance between the transmitting and receiving antennas is 12. h = d2 2R            dM = 2RhT + 2RhR where hT and hR are the heights of transmitting 14. Bandwidth = 2 × 4000 Hz = 8 kHz and receiving antennas respectively. 20. Amplitude of carrier wave = AC Here hT = hR = h Amplitude of side band = µAC 2 \\ dM = 2Rh + 2Rh Ratio = 2 = 2 2Rh = 8Rh µ

256 PHYSICS 34. The band pass filter retains the frequencies nc , nc = 2000 Hz, nc – nm and nc + nm and rejects the frequencies nm, 2nm and 2nc. nc – nm = (2000 – 5) kHz = 1995 kHz Here nc = 630 kHz, nm = 6 kHz 41. l = c = 3 × 108 m s−1 = 300 m The frequencies retained by the band pass ν 106 Hz filter are To transmit a signal, the length of an antenna nc = 630 kHz, is comparable to the wavelength of the signal nc – nm = (630 – 6) kHz = 624 kHz  atleast λ  . Thus the length of the antenna is  4  nc + nm = (630 + 6) kHz = 636 kHz 35. Any device that converts one form of energy 300 m. into another is called transducer. 43. Space waves are used for line-of-sight communication as well as satellite An electrical transducer may be defined communication. as a device that converts some physical variables (like pressure, displacement, 45. The gap between the frequency of the side force, temperature etc.) into corresponding bands (i.e., upper side band and lower side variables in the electrical signal. band) is called bandwidth and it is given by 36. Coverage range of TV tower is given by Bandwidth = (nc + nm) – (nc – nm) d = 2Rh = 2 nm d1 = h1 ,   d1 = 120  or h2 = 480 m = twice of the frequency of the d2 h2 2d1 h2 message signal Height to be added = 480 – 120 = 360 m 46. The maximum distance between the transmitter and receiver for satisfactory communication in LOS mode is 37. m= Am dM = 2RhT + 2RhR Ac where hT and hR are the heights of transmitting Side band frequencies are and receiving antennas respectively. Here, dM = 40 km = 40 × 103 m, USB = nc + nm, LSB = nc – nm hT = 20 m, R = 64 × 105 m \\ 40 × 103 = 2(64 × 105 )20 Here, Am = 10 V, Ac = 20 V + 2(64 × 105 )hR nc = 1000 kHz , nm = 10 kHz 40 × 103 = 16 × 103 + (128 × 105 )hR m = 10 = 0.5 20 USB = 1000 + 10 = 1010 kHz LSB = 1000 – 10 = 990 kHz 38. The wireless communication frequency bands (128 × 105 )hR = 24 × 103 for cellular mobile radio are in the range of 896 – 901 MHz (mobile to base station) and (24 × 103 )2 = 45 m 840 – 935 MHz (base station to mobile). 54. l = hR = 128 × 105 = 10 m = 1000 cm 39. nm = 5 kHz, nc = 2 MHz = 2000 kHz c = 3 × 108 The frequencies of the resultant signal are ν 30 × 106 nc + nm = (2000 + 5) kHz = 2005 kHz

CUET-UG (Mock Test Paper) 1. Which of the following statements is not true 6. Two identical capacitors are joined in parallel, about Gauss’s law? charged to a potential V, separated, and then connected in series, the positive plate of one (a) Gauss’s law is true for any closed surface. is connected to the negative of the other. (b) The term q on the right side of Gauss’s Which of the following is true? law includes the sum of all charges enclosed by the surface. (a) The charges on the free plated connected together are destroyed. (c) Gauss’s law is not much useful in calculating the electrostatic field when (b) The energy stored in this system the system has some symmetry. increases. (d) Gauss’s law is based on the inverse square (c) The potential difference between the dependence on distance contained in free plates is 2V. coulomb’s law (d) The potential difference remains 2. The SI unit of electric flux is constant. (a) N C–1 m–2 (b) N C m–2 7. Two spherical conductors each of capacity C are charged to potential V and –V. These are (c) N C–2 m2 (d) N C–1 m2 then connected by means of a fine wire. The loss of energy is 3. Consider a region inside which, there are various types of charges but the total charge (a) Zero (b) 12 CV2 is zero. At points outside the region, (a) the electric field is necessarily zero. (c) CV2 (d) 2 CV2 (b) the electric field is due to the dipole 8. In the series combination of two or more than moment of the charge distribution only. two resistances, (c) the dominant electric field is inversely (a) the current through each resistance is proportional to r3, for large r (distance the same. from the origin). (b) the voltage through each resistance is (d) the work done to move a charged the same. particle along a closed path, away from the region will not be zero. (c) neither current nor voltage through each resistance is the same. 4. The dielectric constant for metal is given by (d) both current and voltage through each resistance are the same. (a) Zero (b) Infinite (c) 1 (d) 10 9. In parallel combination of n cells, we obtain 5. When air is replaced by a dielectric medium (a) more voltage (b) more current of constant K, the maximum force of attraction between two charges separated by a distance (c) less voltage (d) less current 10. In a region of constant potential, (a) increases K times (a) the electric field is uniform. (b) remains unchanged (b) the electric field is zero. (c) decreases K times (c) there can be no charge inside the region. (d) increases K–1 times (d) both (b) and (c) are correct. M-1

M-2 PHYSICS  11. When a metal conductor connected to the left (d) the direction of field B does not depend gap of a meter bridge is heated, the balancing upon the direction of velocity v . point 16. The nature of parallel and anti-parallel (a) shifts towards right currents is (b) shifts towards left (a) parallel currents repel and anti-parallel currents attract. (c) remains unchanged (b) parallel currents attract and anti‑parallel (d) remains at zero. currents repel. 12. In a potentiometer of 10 wires, the balance (c) both currents attract. point is obtained on the 7th wire. To shift the balance point to the 9th wire, we should (d) both currents repel. (a) decrease resistance in the main circuit. 17. A moving coil galvanometer can be converted into an ammeter by (b) increase resistance in the main circuit. (a) introducing a shunt resistance of large (c) decrease resistance in series with the cell value in series. whose emf is to be measured. (b) introducing a shunt resistance of small (d) increase resistance in series with the cell value in parallel. whose emf is to be determined. (c) introducing a resistance of small value 13. The resistivity of alloy manganin is in series. (a) Nearly independent of temperature (d) introducing a resistance of large value in parallel. (b) Increases rapidly with an increase in temperature 18. The conversion of a moving coil galvanometer into a voltmeter is done by (c) Decreases with increase in temperature (a) introducing a resistance of large value in (d) Increases rapidly with a decrease in series. temperature. (b) introducing a resistance of small value 14. A charged particle is moving in a cyclotron, in parallel. what effect on the radius of the path of this charged particle will occur when the (c) introducing a resistance of large value in frequency of the radio frequency field is parallel. doubled? (d) introducing a resistance of small value (a) It will also be doubled. in series. (b) It will be halved. 19. A paramagnetic sample shows a net magnetization of 8 Am–1 when placed in (c) It will be increased by four times. an external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is (d) It will remain unchanged.  placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetization will 15. If an electron is movingwith velocity v be produces a magnetic field B, then (a) the direction of field B w ill be same as (a) 32/3 Am–1 (b) 2/3 Am–1 the direction of velocity v . (c) 6 Am–1 (d) 2.4 Am–1 (b) the direction of field B will be opposite to the direction of velocity v. 20. A magnetic needle is kept in a non-uniform (c) the direction of field B will be magnetic field. It experiences perpendicular to the direction of (a) a torque but not a force. velocity v . (b) neither a force nor a torque.

CUET-UG (MOCK TEST PAPER) M-3 (c) a force and torque. (a) 100 mH (d) a force but not torque. (b) 1 mH 21. The magnetic susceptibility of an ideal (c) 10 mH diamagnetic substance is (d) cannot be calculated unless R is known (a) +1 (b) 0 29. The line that draws the power supply to your house from the street has (c) – 1 (d) ∞ 22. At a certain place on earth, BH = 13 BV (a) zero average currents. angle of dip at this place is (b) 220 V average voltage. (a) 60° (b) 30° (c) voltage and current out of phase by 90°. (c) 45° (d) 90° (d) voltage and current are in phase. 23. Which of the following is correct about 30. In a purely capacitive circuit if the frequency magnetic monopole? of the ac source is doubled, then its capacitive reactance will be (a) Magnetic monopole exists. (a) remains same (b) doubled (b) Magnetic monopole does not exist. (c) halved (d) zero (c) Magnetic monopoles have a constant 31. Mirage is a phenomenon due to value of monopole momentum. (a) refraction of light (d) The monopole momentum increases (b) reflection of light due to an increase in its distance from the field. (c) total internal reflection of light 24. The earth behaves as a magnet with a (d) diffraction of light magnetic field pointing approximately from the geographic 32. Which of the following forms a virtual and erect image for all positions of the object? (a) North to South (b) South to North (a) Concave lens (b) Concave mirror (c) East to West (d) West to East (c) Convex mirror (d) Both (a) and (c) 25. The core of any transformer is laminated so as 33. Which of the following colours of white light to deviated most when passes through a prism? (a) reduce the energy loss due to eddy (a) Red light (b) Violet light currents. (c) Yellow light (d) Both (a) and (b) (b) make it lightweight. 34. A rod of length 10 cm lies along the principal axes of a concave mirror of a focal length of (c) make it robust and strong. 10 cm in such a way that its end is closer to the pole is 20 cm away from the mirror. The (d) increase the secondary voltage. length of the image is 26. If the coil is open, then L and R becomes (a) infinity, zero (b) zero, infinity (a) 10 cm (b) 15 cm (c) infinity, infinity (d) zero, zero (c) 2.5 cm (d) 5 cm 27. The peak value of ac voltage on 220 V mains 35. The best metal to be used for photoemission is is: (a) 200 2 V (b) 230 2 V (a) Potassium (b) Lithium (c) Sodium (d) Cesium (c) 220 2 V (d) 240 2 V 36. Which of the following is not the property of photons? 28. What is the value of inductance L for which the current is maximum in a series LCR‑circuit (a) charge (b) rest mass with C = 10 mF and w = 1000 s–1? (c) energy (d) momentum

M-4 PHYSICS 37. The term used “to collect the information photon. This is called spontaneous emission. This about an object and a place without physical may also so happen that the excited electron absorbs contact” is called: another photon, releases two photons, and returns to the lower energy state. This is known as stimulated (a) Modulation (b) Communication emission. Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and (c) Amplification (d) Remote sensing ablating the stone inside the human body organ. 38. The forbidden energy bandgap in conductors, Apart from medical usage, the laser is used for semiconductors and insulators are EG1, EG2 the optical disk drive, printer, barcode reader, etc. and EG3 respectively. The relation among them is 41. What is the full form of LASER? (a) EG1 = EG2 = EG3 (a) Light amplified by stimulated emission (b) EG1 < EG2 < EG3 of radiation (c) EG1 > EG2 > EG3 (d) EG1 < EG2 > EG3 (b) Light amplification by stimulated 39. A radioactive nucleus emits a beta particle. emission of radiation The parent and daughter nuclei are (c) Light amplification by simultaneous emission of radiation (a) Isotopes (b) Isotones (d) Light amplified by synchronous (c) Isomers (d) Isobars emission of radiation 40. The radius of a spherical nucleus as measured 42. The “stimulated emission” is the process of: by electron scattering is 3.6 FM. What is the mass number of the nucleus most likely to (a) release of a photon when the electron be? comes back from a higher to lower energy level. (a) 27 (b) 40 (b) release of two photons by absorbing one (c) 56 (d) 120 photon when the electron comes back from a higher to lower energy level. Read the following case/passage and answer the questions from 41 to 45: (c) absorption of a photon when the electron moves from lower to higher energy level. Electromagnetic radiation is a natural phenomenon found in almost all areas of daily life, from radio (d) absorption of a photon when an electron waves to sunlight to X-rays. Laser radiation – like moves from higher to lower energy level. all light – is also a form of electromagnetic radiation. Electromagnetic radiation that has a wavelength 43. What is the range of amplitude of LASER? between 380 nm and 780 nm is visible to the human eye and is commonly referred to as light. At (a) 150 nm – 400 nm wavelengths longer than 780 nm, optical radiation is termed infrared (IR) and is invisible to the eye. At (b) 700 nm – 11000 nm wavelengths shorter than 380 nm, optical radiation is termed ultraviolet (UV) and is also invisible to the eye. (c) 10 nm – 100 nm The term “laser light” refers to a much broader range of the electromagnetic spectrum than just the visible (d) Both (a) and (b) spectrum, anything between 1.50 nm up to 11000 nm (i.e., from the UV up to the far IR). The term laser is 44. Lithotripsy is: an acronym that stands for “light amplification by stimulated emission of radiation”. (a) an industrial application Einstein explained the stimulated emission. In an (b) a medical application atom, an electron may move to a higher energy level by absorbing a photon. When the electron comes (c) laboratory application back to the lower energy level it releases the same (d) process control application 45. LASER is used in which one of the following: (a) optical disk drive. (b) transmitting satellite signal. (c) radio communication. (d) ionization.

CUET-UG (MOCK TEST PAPER) M-5 Read the following case/passage and answer the (c) moderately doped p-n junction diode. questions from 46 to 50: (d) two back to back p-n junction diode. Light Emitting Diode (LED) is a heavily doped p-n junction that under forward bias emits spontaneous 47. LED emits light: radiation. The diode is encapsulated with a (a) when reversed biased. transparent cover so that emitted light can come out. When the diode is forward biased, electrons are sent (b) when forward biased. from n → p (where they are minority carriers) and holes are sent from p → n (where they are minority (c) when forward or reverse biased. carriers). At the junction boundary, the concentration of minority carriers increases as compared to the (d) when heated. equilibrium concentration (i.e., when there is no bias). 48. During recombination at the junction, emitted Thus at the junction boundary on either side of photons have: the junction, excess minority carriers are there which recombine with majority carriers near the junction. (a) energy equal to or slightly less than the On recombination, the energy is released in the form bandgap. of photons. Photons with energy equal to or slightly less than the bandgap are emitted. When the forward (b) energy greater than the bandgap. current of the diode is small, the intensity of light emitted is small. As the forward current increases, the (c) the energy which has no relation with intensity of light increases and reaches a maximum. the bandgap. Further, an increase in the forward current results in a decrease of light intensity. LED’s are biased such that (d) very low energy compared to the the light emitting efficiency is maximum. bandgap. The V-I characteristics of a LED are similar to that 49. A threshold voltage of LED is: of a SI junction diode. But, the threshold voltages are (a) lower compared to other p-n junction much higher and slightly different for each colour. The reverse breakdown voltages of LED’s are very diodes and slightly different for each low, typically around 5 V. So care should be taken colour. that high reverse voltages do not appear across them. LED’s that can emit red, yellow, orange, green and (b) lower compared to other p-n junction blue light are commercially available. diodes and the same for all colours. 46. LED is a: (c) higher compared to other p-n junction diodes and the same for all colours. (a) lightly doped p-n junction diode. (d) higher compared to other p-n junction (b) heavily doped p-n junction diode. diodes and slightly different for each colour. 50. The reverse breakdown voltages of LED’s are: (a) very low and typically around 0.5 V. (b) very high and typically around 50 V. (c) very low and typically around 5 V. (d) very low and typically around 0.05 V. ANSWER KEY 1. (c) 2. (d) 3. (c) 4. (b) 5. (c) 6. (c) 7. (c) 8. (a) 9. (b) 10. (d) 11. (a) 12. (b) 13. (a) 14. (d) 15. (c) 16. (d) 17. (b) 18. (b) 19. (b) 20. (c) 21. (c) 22. (a) 23. (b) 24. (b) 25. (a) 26. (a) 27. (c) 28. (a) 29. (c) 30. (c) 31. (c) 32. (c) 33. (b) 34. (d) 35. (b) 36. (a) 37. (d) 38. (a) 39. (a) 40. (a) 41. (b) 42. (b) 43. (d) 44. (b) 45. (a) 46. (b) 47. (b) 48. (a) 49. (d) 50. (c)

MOCK TEST PAPER–1 1. Quantization of charge implies: (d) The materials move from a region of strong magnetic field to weak magnetic field. (a) charge does not exits. (b) charge exists on particles. 6. The north pole of a long bar magnet was pushed slowly into a short solenoid connected to a (c) there is a minimum permissible magnitude galvanometer. The magnet was held stationary of charge. for a few second with the north pole in the middle of the solenoid and then withdrawn (d) charge can’t be created. rapidly. The maximum deflection of the galvanometer was observed when the magnet 2. Identify the false statement. was (a) Inside a charged or neutral conductor, electro-static field is zero. (b) The electrostatic field at the surface of the (a) moving towards the solenoid. charged conductor must be tangential to the surface at any point. (b) moving into the solenoid. (c) at rest inside the solenoid. (c) There is no net charge at any point inside (d) moving out of the solenoid. the conductor. 7. When an alternating current flows through a (d) Electrostatic potential is constant circuit consisting of a resistor in series with a throughout the volume of the conductor. capacitor, during the cycle at some instant it is possible to have 3. Four resistances 10 , 5 , 7  and 3  are connected so that they form the sides of a (a) voltage across the circuit zero but current rectangle AB, BC, CD and DA respectively. through it not zero. Another resistance of 10  is connected across the diagonal AC. The equivalent resistance (b) current through the circuit zero but the between A and B is voltage across it not zero. (a) 2  (b) 5  (c) current through the capacitor not zero but the voltage across it zero. (c) 7  (d) 10 . (d) all of these. 4. The coil of a ballistic galvanometer is wound 8. Consider the following statements about on a non-metallic frame because the moving electromagnetic waves and choose the correct system should have the least possible : ones. (1) moment of inertia, (2) electrical resistance, S1: em waves having wavelengths 1000 times (3) damping smaller than light waves are called X-rays. (a) 1, 2, 3 correct (b) 1, 2 only correct S2: Ultraviolet waves are used in the treatment of swollen joints. (c) 2, 3 only correct (d) 3 only correct. 5. Which one of the following is not a characteristic S3: Alpha and gamma rays are not electromagnetic waves. of diamagnetism ? (a) The diamagnetic materials are repelled by S4: De Broglie waves are not electromagnetic a bar magnet. in nature. (b) The magnetic susceptibility of the materials S5: Electromagnetic waves exhibit polarisation is small and negative. while sound waves do not. (c) The origin of diamagnetism is the spin of (a) S1, S4 and S5 (b) S3, S4 and S5 electrons. (c) S1, S3 and S5 (d) S2, S3 and S4. M-6