488 Optics and Modern Physics 9. E µ Z2 16. Q E = hc - W …(i) n2 l …(ii) \\ (- 13.2) ´ (Z )2 = - 13.6 4E = hc -W …(i) (2)2 l/3 …(ii) \\ Z=2 Solving these equations, we get 10. lmin (in Å) = V 12375 W = hc (in volts) 3l For lmin = 1Å 17. f µ (Z - 1) for K-series V = 12375 V » 12.4 kV \\ f µ (Z - 1)2 h f2 = (Z2 - 1)2 l f1 (Z1 - 1)2 11. P = 12. l = 12375 = 7734 Å \\ f2 = æçèç Z2 - 11÷ø÷ö 2 f1 Z1 - 1.6 2 12375 çæçè 51 - 11ö÷÷ø 13. DE = 1085 = 11.4 eV = 31 - f Third Balmer line is corresponding to the = 25 f 9 transition, n = 5 to n = 2. 2 E5 - E2 = 11.4 18. f µ (Z - b) æèçç 1 - 1 ö÷÷ø n12 n2 \\ E1 - E1 = 11.4 (5)2 (2)2 1 1 E1 = - 54.28 eV \\ Slope µ n12 - n22 \\ | E1 | = 54.28 eV \\ Slope1 = 1- 1/9 = 32 14. OtherwiseU 1 = - 27.2 eV. Therefore, we have Slope2 1- 1 27 increased it by 27.2 eV. It implies that we have 4 increased it by 27.2 eV in all states. For Kb, n1 = 1, n2 = 3 U 2 = - 6.8 eV For Ka, n1 = 1, n2 = 2 \\ U 2¢ = (- 6.8 + 27.2) eV = + 20.4 eV 19. 1 = Rçæçè 1 - 1 ÷÷öø = R çæè 1 - 91 ø÷ö = 5R E2¢ = U 2¢ + K2 = (20.4 + 3.4) eV l n12 n22 4 36 = 23.8 eV \\ l = 36 15. eV0 = hc -W …(i) 5R l1 20. Energy of electrons = 10000 eV = hc - Similarly , e(2V0) l2 W …(ii) 150 = 0.122 10000 Subtracting Eq. (i) from Eq. (ii), we get \\ l1 (in Å) = eV0 = hc(l1 - l2) \\ l2 (in Å) = 12375 » 1.2 l1l2 10000 21. \\ V0 = hc(l1 - l2) l1 » 0.1 el1l2 l2 = (6.63 ´ 10-34) (3 ´ 108 ) (110) e (5V0) = hc - W 1.6 ´ 10-19 ´ 330 ´ 220 ´ 10-9 l » 1.8 volt eV0 = hc - W 3l
Chapter 33 Modern Physics-I 489 Solving these equations, we get 30. l = h W = hc 2Em 6l where, l = constant 22. eV0 = 2hn0 - hn0 \\ Em = constant \\ hn0 - eV0 or Eµ 1 In second condition, m eV = 3hn0 - hn0 = 2hn0 - 2eV0 31. U 1 = - 27.2 eV or V = 2V0 U 1 is assumed zero. Therefore, it is increased by 23. Frequency µ energy and energy µ Z 2 27.2 eV at all points. 24. E µ Z2 Eµ = 0 under normal conditions. n2 Hence, in charged conditions it is 27.2 eV. Z2 32. Number of photons emitted per second n2 \\ =1 = Energy radiated per second = P Energy of one photon hf or n = Z = 3 1000 25. 1 mv12 = hn1 - W = 6.63 ´ 10-34 ´ 880 ´ 103 2 1 mv22 = hv2 - W = 1.7 ´ 1030 2 33. E = 12375 = 4.125 eV From these two equations, we can see that 3000 v12 - v22 = 2h (n1 - n2) Since, E < W no photoelectric effect will be m observed. 26. Longest wavelength of Lyman series means, 34. E µ 1 n2 minimum energy corresponding n = 2 to n = 1. \\ (E2 - E1)H = (En - E2)He+ \\ E5 = - 13.6 (5)2 \\ - 13.6 + 13.6 = - 13.6(Z )2 + 13.6(Z )2 (2)2 (1)2 n2 (2)2 = - 0.544 eV Putting Z = 2, we get n = 4. 35. Kmax = E - W = 2 eV = 3.2 ´ 10-19 J 27. 1 = R (Z - 1)2 çèçæ 1 - 1 ÷÷øö 36. meve = h l n12 n22 l For Ka-line, n1 = 1 and n2 = 2 \\ ve = h lme 28. = 6.63 ´ 10-34 ´ 10- 10) ´ 9.1 ´ 10-31 La (5200 Ka Kb » 1400 m/s 37. K max = 1 mvm2 ax = E -W 2 \\ EKb = EKa + ELa \\ U max = 2(E - W ) \\ hg2 = hg1 + hg3 Here, m 29. Q l = g2 = g1 + g3 E = 12375 = 4.125 eV h or l µ 1 3000 \\ 2qVm qm = 2 ´ 3.125 ´ 1.6 ´ 10-19 9.1 ´ 10-31 lP = ( qm )a = 2´4 =2 2 la ( qm )P 1´1 » 1.09 ´ 106 m/s
490 Optics and Modern Physics Subjective Questions 4. (a) p = E Þ E = pc 1. c (b) l = 12375 La E (in eV) Ka Kb = 12375 = 8035 Å 1.54 » 804 nm ELa = EKb - EKa So, this wavelength lies in ultraviolet region. \\ hc = hc - hc 5. (a) f = c lLa lKb lKa l = lKa lKb (b) Q Number of photons emitted per second lKa - lKb or lLa N1 = Power of source = P Energy of one photon hc/ l = 0.71 ´ 0.63 = 5.59 nm N 1 = Pl 0.71 - 0.63 hc 2. lKa = 0.71nm = 7.1 Å = (75) (600 ´ 10-9 ) (6.63 ´ 10-34) (3 ´ 108 ) - = 127375 = 1743 eV E2 E1 7.1 » 2.3 ´ 1020 photons/s \\ E1 = E2 - 1743 6. (a) E = hf Þ f =E = - 2870 - 1743 h (b) = - 4613 eV l= c f lKb = 0.63 nm = 6.3 Å 7. (a) K = p2 or K µ p2 E3 - E1 = 12375 2m 6.3 If momentum is doubled, kinetic energy = 1964 eV becomes four times. \\ E3 = E1 + 1964 (b) p = E or E µ p (for a photon) = - 4613 + 1964 c = - 2649 eV If p is doubled, E will also become two times. 3. Number of photons incident per second 8. (a) Number of photons incident per second = Power = number of photons absorbed per second Energy of one photon = Power = P = Pl Energy of one photon (hc/ l) hc N = P = Pl hc/ l hc Number of electrons emitted per second (b) Force = Rate of change of momentum = = 0.1% of Pl = Pl (Number of photons absorbed per second) ´ hc 1000 hc (momentum of one photon) \\ Current = Charge (on photoelectrons per second) = æèç Pl øö÷ çæè h ø÷ö = P (Q P = power) = Ple hc l c 1000 hc 9. See the hint of above example. = (1.5 ´ 10-3) (400 ´ 10-9 ) (1.6 ´ 10-19 ) (1000) (6.63 ´ 10-34) (3 ´ 108 ) If surface is perfectly absorbing, force is P. If c = 0.48 ´ 10-6 A = 0.48 mA surface is perfectly reflecting, then force will be 2P. c
Chapter 33 Modern Physics-I 491 In this case, l4 = 150 as KE = | E4 | = 0.85 eV 0.85 F = 0.7P + 0.3 (2P) = 1.3P c cc » 13.3 Å 10. Force = Rate of change of momentum 2pr4 = 2p(n)2r1 (as r µ n2) = 53.15 Å » 4l4 = 2 [N ] é h ù × cos 60° ëê l ûú N = number of photons striking per second 17. Energy required to remove first electron is h = momentum of one photon. 24.6 eV. After removing first electrons from this l atom, it will become He+ . 11. l = h = h p mv E1 = - (13.6)(2)2 (as E µ Z 2 and Z = 2) Since, wavelength is too short, it does not behave = - 54.4 eV wave like property. \\ Energy required to remove this second electron 12. l = h will be 54.4 eV. mv h \\ Total energy required to remove both electrons l = 24.6 + 54.4 13. (a) l = h Þ p = = 79 eV p = = 12375 (b) l = l Þ h2 18. En E1 1023 2Km 2ml2 K = - 13.6 (n)2 3RT = v (say) or + 13.6 = 12.1 M 14. vrms = Solving this equation, we get n = 3 \\Total possible emission lines = n (n - 1) = 3 and l = h = h M mv m 3RT 2 Substituting the values, we get Longest wavelength means, minimum energy, 6.63 ´ 10-34 2 ´ 10-3 which is corresponding to n = 3 to n = Z (2/6.02 ´ 1023) ´ 10-3 3(8.31) (20 + 273) l = \\ lmax = 12375 E3 - E2 = 1.04 ´ 10-4 m = 1.04 Å Here, E3 - E2 = - 13.6 + 13.6 = 1.9 eV 15. KE = | Total energy | = 3.4 eV (3)2 (4 )2 l (in Å) = 150 for an electron \\ lmax = 12375 = 6513 Å 1.9 KE (in eV) = 150 19. In hydrogen atom, 3.4 DE1 = E3 - E1 = 12.1 eV = 6.6Å E µ Z2 16. (a) E1 = - 13.6 eV \\ For Z = 3, DE2 = (3)2 DE1 = (3)2 (12.1) \\ KE = | E1 | = 13.6 eV = 108.9Å l1 (in Å) = 150 for an electron KE (in eV) \\ l = 12375 » 113 Å 108.9 = 150 = 3.32Å 13.6 20. En - E1 = DE1 + DE2 2pr1 = 2p(0.529Å) » 3.32 Å \\ é - 13.6 + 13.6ùúû (Z )2 = 12375 + 12375 ëê n2 1085 304 E1 - 13.6 (b) E4 = (4 )2 = (4)2 = - 0.85 eV Putting Z = 2 for He+ , we get n = 5
492 Optics and Modern Physics 21. (a) Let these two levels are n1 and n2. Then, 25. F = - dU = - m2w2r 13.6 (Z )2 dr n12 - 0.85 = - …(i) Now, mv2 = | F | = m2w2r …(ii) r 13.6 (Z )2 - 0.544 = - n22 or v µ r Total number of lines between n2 and n1are or v = ar given by Now, m vr = n h 2p (n2 - n1) (n2 - n1 + 1) = 10 2 …(iii) or m (ar)r = n h 2p Solving these equations, we get \\ rµ n Z = 4, n1 = 16 and n2 = 20 (b) Smallest wavelength means maximum energy. 26. 1 µ çèæç 1 - 1 ÷÷öø l n12 n22 DEmax = En2 - En1 = - 0.544 + 0.85 = 0.306 eV lKb (1/ n12 - 1/n22)Ka lKa (1/ n12 - 1/n22)Kb lmin = 12375 = 40441Å \\ = 0.306 22. (a) Ionization energy = 15.6 eV = (1- 1/4) = 27 (1- 1/9) 32 \\ Ionization potential = 15.6 V (b) Eµ - E2 = 5.3 eV \\ lKb = 27 lKa = 27 l0 32 32 \\ l = 12375 = 2335Å 5.3 27. DE1 = 50% of 50 keV = 25 keV (c) E3 - E1 = 12.52 eV \\ l1 = 12375 = 0.495 Å \\ Excitation potential is 12.52 V. 25 ´ 103 (d) E3 - E1 = 12.52 eV = 49.5 pm \\ l = 12375 Å DE2 = 50% of 25 keV 12.52 = 12.5 keV = 988.41 Å \\ Wave number = 1 = 1 m-1 \\ l2 = 12375 = 0.99Å l 988.41´ 10-10 12.5 ´ 103 = 1.01 ´ 107 m-1 = 99 pm 23. (a) Energy of photon = 12375 eV = 1.44 eV 28. 1 = R (Z - 1)2 çæçè 1 - 1 ÷øö÷ lKa n12 n22 8600 \\ Internal energy after absorption or 0.76 1 = 1.097 ´ 107 (Z - 1)2 æèç1 - 1 ø÷ö ´ 10-10 4 = - 6.52 + 1.44 = - 5.08 eV (b) Energy of emitted photon = 12375 = 2.946 eV Solving this equation, Z = 41 4200 29. 26 pm = 0.26 Å Internal energy after emission = - 2.68 - 2.946 Now, 0.26 = 12375 - 12375 = - 5.626 eV V 1.5 V 24. Kmax (in eV) = E - W = 12375 - 4.3 30. Similar type of example is given in the theory. 2000 31. Transition if from L-shell to K-shell. = 1.9 eV 1 = R (Z - 1)2 ççæè 1 - 1 ÷÷öø l n12 n22 Therefore, stopping potential is 1.9 V.
Chapter 33 Modern Physics-I 493 or f =R (Z - 1)2 æèçç 1 - 1 ÷÷öø 38. E = 12375 = 6.875 eV c n12 n22 1800 or 4.2 ´ 108 = 1.1 ´ 107 (Z - 1)2 èæç1 - 1 ÷øö Kmax or K = E - W 3 ´ 108 4 = 4.875 eV Solving we get, Z » 42 r = 2Km Bq 32. (a) P = Vi = (40 ´ 103) (10 ´ 10-3) 2 ´ 4.875 ´ 1.6 ´ 10-19 ´ 9.1 ´ 10-31 = 400 W = 5 ´ 10-5 ´ 1.6 ´ 10-19 1% of 400 W is 4 W. (b) Heat generated = 400 W - 4 W = 0.148 m = 396 W = 396 J/s 39. fhigher = 8 ´ 1015 s-1 2p 33. Stopping potential = 10.4 V hf \\ Kmax = 10.4 eV E = 1.6 ´ 10-19 eV E = W + Kmax = 1.7 + 10.4 = 12.1 eV l = 12375 = 1022 Å = (6.63 ´ 10-34) (8 ´ 1015) 12.1 1.6 ´ 10-19 ´ 2p 12.1 eV is the energy gap between n = 3 and n = 1 = 5.27 eV in hydrogen atom. Kmax = E - W 34. E = hf = (6.63 ´ 10-34) (1.5 ´ 1015) eV = 6.21 eV = 3.27 eV 1.6 ´ 10-19 40. f = w = (1.57 ´ 107 ) c Kmax = E - W = 6.21 - 3.7 = 2.51 eV 2p 2p 35. l0 = 5000 Å E = 1.6 hf eV ´ 10-19 = 12375 = 2.475 eV W 5000 (6.63 ´ 10-34) (1.57 ´ 107 ) (3 ´ 108 ) 1.6 ´ 10-19 ´ 2´ p Stopping potential is 3V. Therefore, Kmax = 3 eV = eV E = W + Kmax = 5.475 eV l = 12375 = 2260Å = 3.1 eV 5.475 Kmax = E - W = 1.2 eV 36. (a) f0 = fA = 10 ´ 1014 Hz = 1015 Hz LEVEL 2 (b) W = | Kmax |c = 4 eV Single Correct Option (c) W = hf0 Þ h=W 1. mv2 = GMm …(i) f0 r r2 37. E1 = 12357 = 4.125 eV mvr = h (for n = 1) …(ii) 3000 2p E2 = 12357 = 2.0625 eV Solving these two equations, we can find v and r. 6000 Then, Maximum speed ratio is 3 : 1. Therefore, E = 1 mv2 - GMM maximum kinetic ratio is 9 : 1. 2r Now, 9 Kmax = 4.125 - W …(i) 2. M = q = constant …(ii) Kmax = 2.0625 - W L 2m Solving these two equations, we get M µL W » 1.81eV. and Kmax = 0.26 eV and = nh or L µ n 1 2p Putting K max = 2 mvm2 ax we can find vmax . Here, m L is the mass of electron. \\ M µn
494 Optics and Modern Physics 3. E2 - E1 = 40.8 eV Momentum of photon = Momentum of hydrogen \\ E1 - = atom (2)2 E1 40.8 eV \\ 3Rh = Mv Þ v = 3Rh 4 4M 3 or - 4 E1 = 40.8 eV 9. wmax = 1.5 ´ 1015 rad /s = 2pfmax or E1 = - 54.4 eV \\ fmax = 1.5 ´ 1015 Hz 2p | E1 | = 54.4 eV hfmax 4. i = qf = q æèç v øö÷ E = 1.6 ´ 10-19 eV 2pr or i µ v µ (1/ n) or i µ 1 = 6.63 ´ 10-34 ´ 1.5 ´ 1015 eV r (n)2 n3 2p ´ 1.6 ´ 10-19 i1 ççæè n2 ø÷÷ö 3 » 1.0 eV i2 n1 = = (2)3 = 8 Since, E < W , no photoemission can take place. 5. nf = 6 10. nf (nf - 1) = 6 Þ nf = 4 Total emission lines = nf (nf - 1) = 15 2 2 nf = 4 6. A = pr2 A µ r2 or A µ n4 (as r µ n2) ni = 2 \\ An = (n)4 A1 ln çèæç An ÷ø÷ö = 4 ln (n) From ni = 2, energy of six emission lines is either A1 greater than less or equal to the energy of absorption line. Therefore, ln æçèç An ÷öø÷ ln n graph is a straight A1 versus 11. T = 2pr or T µ r or T µ (n2/Z ) line of slope 4. v v (Z /n) 7. B = m 0i \\ T µ n3 2p Z2 or B µ i çæèç T1 ö÷÷ø çæçè n1 ÷öø÷ 3 ççèæ Z2 ö÷÷ø 2 r T2 n2 Z1 = = See the hint of Q.No. 4 of same section. 2 = æç 1÷ö 3 çæ Z ÷ö 2 è 2ø è 1ø i µ 1 and r µ n2 n3 \\ B = 1 \\ Z=4 n5 12. lKa will depend on atomic number Z and values Z B1 æèçç n2 öø÷÷ 5 B2 n1 is same of all three isotopes. \\ = = (2)5 = 32 13. 8. 1 = R æèç1 - 1 ÷öø = 3R n=2 l 4 4 \\ p = h 10.2 eV l n=1 = 3Rh = momentum of photon 4 l2 ® 1 = 12375 » 1213 Å 10.2 From conservation of linear momentum,
Chapter 33 Modern Physics-I 495 14. For K ³ 10.2 eV electrons can excite the hydrogen Now, K = minimum excitation energy required. 2 atom (as E2 - E1 = 10.2 eV). So, collision may be = 10.2 eV inelastic. Þ K = 20.4 eV 15. E3 - E1 = - 13/6 + 13.6 = 12.1 eV (3)3 22. KE = | E | = 3.4 eV From momentum conservation, l (in Å) = 150 Momentum of photon = Momentum of hydrogen KE (in eV) atom = 150 \\ E = mv 3.4 c = 6.6 Å or = E = 12.1 ´ 1.6 ´ 10-19 v mc 1.67 ´ 10-27 ´ 3 ´ 108 = 3.86 m/s More than One Correct Options 16. P = Vi = 150 ´ 103 ´ 10 ´ 10-3 = 1500 W 1. lKa and lKb will remain unchanged. But, l0 will 99% of this = 1500 ´ 99 J/s = 1485 J/s decrease çæè as l0 µ 1 ÷öø. 100 V = 1485 cal/s \\ Dl1 = lKa - l0 or Dl2 = lKb - l0 4.18 will increase. » 355cal/s 2. R µ n2, v µ 1 and E µ 1 n n2 17. E4 - E3 = 32.4 3. L µ n, r µ n2 - 13.6 (Z )2 13.6 (Z )2 \\ (4 )2 + (3)2 = 32.4 and T = 2pr or T µ r µ n2 v v (1/n) Solving, we get or T µ n3 Z =7 18. Q l = h 4. l = h = h = h 2qVm p mv 2Km V = h2 lµ 1 (if v is same) 2qml2 m (if K is same) 19. En µ 1 and Ln µ n lµ 1 n2 m 20. M = q (always) If change in potential energy is same, then change in kinetic energy is also same. But, this does not L 2m mean that kinetic energy is same. \\ M µL 5. nf (nf - 1) = 6 L = nh Þ L µn 2 2p Þ nf = 4 Hence, M µn nf = 4 1 Third excited state means n = 4. 21. K = p2 l0 25 ni = 2 2m 34 6 In collision, momentum p remains constant. \\ Kµ 1 mass After collision, mass has doubled. So, kinetic l1 and l4 are longer than l0. energy will remain K . Hence, loss is also K . l3, l5 and l6 belong to Lyman series. 22
496 Optics and Modern Physics 6. f µ (Z - b) …(i) Number of photons incident on sphere per second, If f versus Z is a straight line. èçæ N1 ÷øö r c N2 = 4 pR (pr2) f = l 2 1 = N1 r2 R l 4R2 Hence, versus Z is also a straight line. Comprehension Based Questions = (4 ´ 1015) (8 ´ 10-3)2 4 (0.8)2 12375 1. E1 = 4950 = 2.5 eV = 1011 K1 = maximum kinetic energy = 0.6 eV Number of photoelectrons emitted per second, W = E1 - K2 = 1.9 eV N3 = N2 = 105 2. K2 = 1.1eV 106 E2 = W + K2 = 3.0 eV Now, N3t = n 12375 \\ t = n = 111 s \\ l2 = 3.0 = 4125 Å N3 3. Magnetic field cannot change the kinetic energy of Match the Columns E µ Z2 2. For He+ , Z = 2 n2 charged particle. ————— E2 = - 3.4 eV 4. Kmax = E - W = 2 eV ————— l = 150 , for an electron H-atom KE (in eV) = 150 » 8.6 Å E1 = - 13.6 eV 2 Ionisation energy from first, excited state of 5. Kmax is 2 eV. Hence, stopping potential is 2V. H-atom Photoemission stops when potential of sphere = | E2 | = 3.4 eV = E (given) becomes 2V. (a) | E1 | = (13.6 eV) (2)2 \\ = q 2 4 pe0r = 16 (3.4 eV) = 16E \\ q = 8pe0 r (b) U2 = 2E2 = 2 (- 13.6) (2)2 6. Q (2)2 q = ne = 8pe0r \\ = - 8 (3.4 eV) = - 8E n = 8pe0r e (c) K1 = | E1 |= 16E = 2 ´ 8 ´10-3 (d) | E2 |= (13.6) (2)2 ´ 109 ´ 1.6 ´ 10- (2)2 9 19 = 1.11 ´ 107 = 4 (3.4 eV) = 4E So, this much number of electrons are required to 3. Kmax = hf - W be ejected from the sphere. \\ Kmax versus f graph is a straight line of slope h and intercept - W . Number of photons emitted per second, eV0 = hf = W N1 = Power of source \\ V0 = çæè heø÷ö f -W Energy of one photon e = 5 3.2 ´ 10-3 \\ V0 versus f graph is again a straight line of slope ´ 1.6 ´ 10-19 h and intercept W . ee = 4 ´ 1015
Chapter 33 Modern Physics-I 497 4. T = 2pr or T µ r µ n2 /Z m has become half, so l will become two times v Z/n v or 1312 nm or 1.31mm. \\ T µ n3 (b) E µ Z 2 \\ l µ 1 Z2 Z2 L = nh or L µ n For singly ionized helium atom Z = 2 2p \\ l is 1 th or 164 nm v µ Z and r µ n2 4 nZ 2. (a) f = v 2pr èççæ ÷öø÷ 5. 1 µ 1 - 1 f1 = (2.2 ´ 106) l n12 n22 (0.529 ´ 10-10) (2p) or l2 = (1/ n12 - 1/n22)i » 6.6 ´ 1015 Hz l1 (1/ n12 - 1/n22)f v µ 1 and r µ n2 or l2 = (1/4 - 1/16)l n (1/n12 - 1/n22)f f µv Þ f µ 1 r n3 æçè 3 l÷öø é 1 ù = 16 ê (1/ n12 - 1/n22)f ú \\ f2 = f1 ëê úû 8 (a) For first line of Balmer series, (b) DE = E2 - E1 = 10.2 eV = hf 10.2 ´ 1.6 ´ 10-19 n1 = 2, n2 = 3 \\ l2 = èçæ 27 ö÷ø l \\ f = 6.6 ´ 10-34 20 (b) For third line of Balmer series, » 2.46 ´ 1015 Hz n1 = 2, n2 = 5 \\ l2 = çæè 25 ÷øö l (c) In option (a), we have found that 28 f2 = 0.823 ´ 1015 Hz (c) For first line of Lyman series, l T2 = 1 n1 = 1, n2 = 2 \\ l2 = 4 f2 (d) For second line of Lyman series, N =t = tf2 = (10-8 ) (0.823 ´ 1015) n1 = 2, n2 = 3 \\ l2 = çèæ 12278ö÷ø l T2 = 8.23 ´ 106 revolutions 7. See the hint of Q.No. 3 of section Assertion and 3. (a) r µ 1 Reason. m \\ Stopping potential increases with increase in r = (r1H ) = 0.529 ´ 10-10 maximum kinetic energy of photoelectrons of m 207 frequency of incident light. = 2.55 ´ 10-13 m With increase in distance between cathode and (b) E µ m anode, f remains unchanged. \\ Ionization energy of given atom Kmax = hf - W = eV0 If W is decreased, Kmax and V0 both will increase. = (m) (ionization energy of hydrogen atom) Subjective Questions = (207) (13.6 eV) 1. (a) Reduced mass of positronium and electron is m , = 2815.2 eV = 2.81 keV 2 where, m = mass of electron 4. (a) En - E1 = 12.5 \\ - 13.6 + 13.6 = 12.5 n2 E µm \\ lµ 1 m Solving we get, n = 3.51
498 Optics and Modern Physics Hence, electron jumps to n = 3. So, possible lines \\ DE1 - W =5 …(i) are between n = 3 to n = 2, n = 3 to n = 1and DE2 - W Ans. between n = 2 to n = 1. Here, DE1 = E4 - E1 = 12.75 eV For n = 3 to n = 2, and DE2 = E3 - E1 = 12.09 eV DE = E3 - E2 = - 13.6 + 13.6 Substituting in Eq. (i) and solving, we get 9 4 W = 11.925 eV = 1.9 eV \\ l = 12375 Å = 6513 Å 7. For shorter wavelength 1.9 DE = - = (-13.6)(3)2 - é (-13.6)(3)2 ù E4 E3 (4 )2 ê ú » 651 nm ë (3)2 û Similarly, other wavelengths can also be = 5.95 eV obtained. W = E - Kmax = (5.95 - 3.95) eV = 2 eV (b) n = 3.51(in option-a) For longer wavelength A photon always transfers its energy completely. So, it cannot excite the ground state electrons to DE = - = (-13.6)(3)2 - é (-13.6)(3)2 ù n = 3 (like and electrons excited it in part-a). E5 E4 (5)2 ê ú ë (4 )2 û 5. (a) E = hf = (6.6 ´ 10-34)(5.5 ´ 1014) = 2.754 eV = 36.3 ´ 10-20 J \\ Kmax = E - W = 0.754 eV or stopping potential is 0.754 V. = 2.27 eV Ans. Ans. çèæ e ö÷ø …(i) (b) Number of photons leaving the source per 8. Magnetic moment, m = NiA = T (pr2) second, æèçç e ø÷÷ö evr 2pr/ 2 n = P = 0.1 or m = v (pr2) = E 36.3 ´ 10-20 = 2.75 ´ 1017 Ans. We know that mvr = nh …(ii) 2p (c) Number of photons falling on cathode per second, Solving Eqs. (i) and (ii), we get = 0.15 ´ 2.75 ´ 1017 m = neh Ans. 100 4 pm …(iii) n1 Magnetic induction, B = m 0i = m 0e …(iv) = 4.125 ´ 1014 m 0ev 2r 2rT (2r)(2pr) m 0ev Number of photoelectrons emitting per second, or B = = 4 pr2 n2 = 6 ´ 10-6 = 3.75 ´ 1013 From Newton's second law, e2 = mv2 1.6 ´ 10-19 4 pe0r2 r \\ % = n2 ´ 100 = 3.75 ´ 1013 ´ 100 or v2 = e2 n1 4.125 ´ 1014 4 pe0mr = 9% Ans. Solving all these equations, we get 6. K1 = 5 B = m 0pm2e7 Ans. 8e0h5n5 K2 E4 = –0.85 eV 9. Energy of electron in ground state of hydrogen E3 = –1.51 eV E2 = –3.4 eV atom is – 13.6 eV. Earlier it had a kinetic energy of 2 eV. Therefore, energy of photon released during formation of hydrogen atom, DE = 2 – (– 13.6) = 15.6 eV \\ l = 12375 = 12375 DE 15.6 E1 = –13.6 eV = 793.3 Å Ans.
Chapter 33 Modern Physics-I 499 10. U = – 1.7 eV = 12375 = 28.43 Å Ans. (13.6) (6)2 æç1 – \\ E =U = – 0.85 eV = –13.6 1 ÷ö 2 n2 è 9ø \\ n=4 (d) Ionization energy = (13.6) (6)2 = 489.6 eV Ans. Ejected photoelectron will have minimum l = 12375 = 25.3 Å Ans. 489.6 de-Broglie wavelength corresponding to transition from n = 4 to n = 1. 12. Pitch of helical path, p = (v cos q) T = vT . 2 DE = E4 – E1 = – 0.85 – (– 13.6) = 12.75 eV (as q = 60°) Kmax = DE – W = 10.45 eV (for an electron) T = 2pm = 2p æèç a = q ÷öø \\ l = 150 Å Bq Ba m 10.45 pv Ba = 3.8 Å Ans. \\ p= 11. (a) n (n – 1) = 3 or v = Bap …(i) p 2 \\ n=3 KE = 1 mv2 = E – W 2 i.e after excitation atom jumps to second excited state. Hence, nf = 3. So, ni can be 1 or 2. \\ W = E – 1 mv2 …(ii) If ni = 1, then energy emitted is either equal to 2 or less than the energy absorbed. Hence, the Substituting value of v from Eq. (i) in Eq. (ii), emitted wavelength is either equal to or greater we get than the absorbed wavelength. Hence, ni ¹ 1. = 4.9 – 1 W 2 9.1 ´ 10–31 ´ (2.5 ´ 10–3)2 (1.76 ´ 1011)2 ´ (2.7 ´ 10–3)2 p2 ´ 1.6 ´ 10–19 ni = 1 = (4.9 – 0.4) eV = 4.5 eV Ans. 13. (a) l = 1500 çæèç 1 p2 ÷ö÷ø 1/ 1 – lmax corresponds to least energetic photon with p = 2. \\ lmax = 1500 æççè 1 1 4 ÷÷øö = 2000 Å Ans. – 1/ ni = 2 lmin corresponds to most energetic photon with p = ¥ If ni = 2, then Ee ³ Ea. Hence, le £ lb \\ lmin = 1500 Å Ans. \\ ni = 2 Ans. (b) l¥ – 1 = 1500 Å (b) E3 – E2 = 68 eV Ans. \\ (13.6) (Z 2) æç 1 – 1ö÷ = 68 E3 = – 0.95 eV è 4 9ø E2 = – 2.05 eV \\ Z=6 (c) lmin = 12375 E1 = – 8.25 eV E3 – E1
500 Optics and Modern Physics \\ E¥ – E1 = 12375 eV (c) Energy of photon in first case, 1500 = 12375 = 8.25 eV 3000 \\ E1 = – 8.25 eV (as E¥ = 0) = 4.125 eV l2 – 1 = 2000 Å or E1 = 6.6 ´ 10–19 J \\ E2 – E1 = 12375 eV Rate of incident photons 2000 = P1 = 10–3 E1 6.6 ´ 10–19 = 6.2 eV \\ E2 = – 2.05 eV = 1.52 ´ 1015 per second Similarly, l31 = 1500 æççè 1 1 9 ÷öø÷ Number of electrons ejected – 1/ = 4.8 ´ 10–3 per second = 1687.5 Å 1.6 ´ 10–19 \\ E3 – E1 = 12375 eV = 7.3 eV = 3.0 ´ 1016 per second 1687.5 \\ Efficiency of photoelectrons generation \\ E3 = – 0.95 eV (c) Ionization potential = 8.25 V Ans. = 1.52 ´ 1015 ´ 100 3.0 ´ 1016 12375 14. (a) K1 = 3000 – W …(i) = 5.1% Ans. …(ii) K2 = 12375 – W …(iii) 15. Balmer Series 1650 l32 = 12375 = 12375 E3 - E2 v2 = 2v1 \\ K2 = 4K1 (13.6) çèæ 1 - 1 ÷öø 4 9 Solving these equations, we get = 6551 Å W = 3 eV = 655.1 nm \\ Threshold wavelength, 12375 12375 l0 = 12375 l42 = E4 - E2 = (13.6) æç 1 - 1 ö÷ 3 è 4 16ø = 4125 Å Ans. = 4853 Å (b) E2 = 12375 = 7.5 eV = 12 ´ 10–19 J = 485.3 nm 1650 12375 12375 Therefore, number of photons incident per l52 = E5 - E2 = second èæç 1 215 öø÷ (13.6) 4 - n2 = P2 = 5.0 ´ 10–3 = 4333 Å E2 12 ´ 10–19 = 433.3 nm = 4.17 ´ 1015 per second First two lie in the given range. Of these l42 Number of electrons emitted per second corresponds to more energy. (h = 5.1%) E = E4 - E2 = (13.6) èçæ 1 - 1 øö÷ = 5.1 ´ 4.17 ´ 1015 4 16 100 = 2.55 eV = 2.13 ´ 1014 per second \\ Kmax = E - W = (2.55 - 2.0) eV \\ Saturation current in second case = 0.55 eV Ans. i = (2.13 ´ 1014) (1.6=´31.04–1´91)0A–5 A 16. From the theory of standing wave, we can say that = 34 mA Ans. l = (2.5 - 2.0) = 0.5 Å 2
Chapter 33 Modern Physics-I 501 or l = 1 Å \\ mv2 = k …(i) rr …(ii) 2Å According to Bohr’s assumption, mvr = n h 2p Solving Eqs. (i) and (ii), we get r = nh 2p mk 2.5 Å Therefore, least value of d required will and v = k m correspond to a single loop. \\ E = U + 1 mv2 = k ln r – k + k = k ln r \\ dmin = l = 0.5 Å Ans. 2 22 2 Further for de-Broglie wavelength of an electron, Thus, rn = nh 2p mk l = 150 Å K (in eV) and En = k ln íìî nh üýþ Ans. 2p mk l =1Å \\ K = 150 eV Ans. 19. (a) r = mv …(i) Be 17. (a) Reduced mass nh m = m1m2 = (1837me) (207me ) mvr = 2p …(ii) m1 + m2 1837me + 207me Solving these two equations, we get = 186me = 186 ´ 9.1 ´ 10-31 r= nh 2pBe = 1.69 ´ 10-28 kg Ans. nhBe 2pm2 (b) En µ m and v= Here, reduced mass is 186 times mass of electron. Hence, ground state energy will also (b) K = 1 mv2= nhBe Ans. be 186 times that of hydrogen atom. 2 4 pm \\ E1 = 186(- 13.6) eV (c) M = iA = çæè e ø÷ö (pr2) = - 2529.6 eV » - 2.53 keV T Ans. e evr (c) E2 = 186 (- 3.4) eV = - 632.4 eV = (pr2) = 2 èçæ 2pr ÷øö \\ DE21 = 1897.2 eV v \\ l21 = 12375 Å =e nh nhBe DE21 2 2pBe 2pm2 = 12375 Å = nhe 1897.2 4 pm = 6.53 Å » 0.653 nm Ans. U = – MB cos 180° 18. Force of interaction between electron and = nheB 4 pm proton is F = – dU = –k Note Angle between M and B will be 180°. Think why? dr r (d) E =U + K = nheB Force is negative. It means there is an attraction 2pm between the particles and they are bound to each other. This force provides the necessary centripetal (e) | f | = Bpr2 = nh force for the electron. 2e
502 Optics and Modern Physics 20. (a) and (b) When hydrogen atom is excited, then Solving above two equations, we get W = 1.9 eV eV = E0 æç 1 – 1 ÷ö …(i) Ans. è1 n2 ø and l = 4125 Å When ion is excited, 22. E = 12375 = 3.1 eV 4000 eV = E0Z 2 é1 – 1ù …(ii) ê ú Number of photoelectrons emitted per second, ë 22 n12 û çèæ 1 ÷øö çèçæ 5 ÷øö÷ Wavelength of emitted light, n = 106 3.1 ´ 1.6 ´ 10–19 hc = E0 çæè 1 – 1 ÷öø …(iii) = 1.0 ´ 1013 per second l1 1 n2 …(iv) hc = E0 Z 2çæçè 1 – 1 ø÷÷ö \\ i = ne l2 1 n12 = 1.0 ´ 1013 ´ 1.6 ´ 10–19 Further it is given that = 1.6 ´ 10–6 A l1 = 5 …(v) = 1.6 mA Ans. l2 1 Ans. 23. (a) E = 12375 = 3.1 eV 4000 Solving the above equations, we get Z = 2, n = 2, n1 = 4 Energy of electron after first collision and V = 10.2 V E1 = 90% of E = 2.79 eV (as 10% is lost) (c) Energy of emitted photon by the hydrogen Energy of electron after second collision atom = E2 – E1 E2 = 90% of E1 = 2.51 eV KE of this electron after emitting from the = 10.2 eV Ans. metal surface and by the ion = E4 – E1 = (13.6) (2)2 æç1 – 1 ö÷ = (2.51 – 2.2) eV = 0.31 eV Ans. è 16ø (b) Energy after third collision, = 51 eV Ans. E3 = 90% of E2 = 2.26 eV 21. 0.6 = 12375 – W …(i) Similarly, 4950 …(ii) E4 = 90% of E3 = 2.03 eV 1.1 = 12375 – W So, after four collisions it becomes unable for l the electrons to come out of the metal.
34. Modern Physics-II INTRODUCTORY EXERCISE 34.1 Hence, the probability that a nucleus decays in two half-lives is 3 . 2. Penetrating power is maximum for g -rays, then of 4 b-particles and then a -particles because basically it depends on the velocity. However, ionization 9. Activity reduces from 6000 dps to 3000 dps in power is in reverse order. 140 days. It implies that half-life of the radioactive 3. Both the beta rays and the cathode rays are made sample is 140 days. In 280 days (or two half-lives) activity will remain 1 th of the initial activity. up of electrons. So, only option (a) is correct. 4 (b) Gamma rays are electromagnetic waves. Hence, the initial activity of the sample is (c) Alpha particles are doubly ionized helium 4 ´ 6000 dps = 24000 dps atoms and Therefore, the correct option is (d). (d) Protons and neutrons have approximately the 10. R = R0 æèç 1÷ö n …(i) same mass. 2ø Therefore, (b), (c) and (d) are wrong options. Here, R = activity of radioactive substance after n 4. During b-decay, a neutron is transformed into a half-lives = R0 (given) 16 proton and an electron. This is why atomic number Substituting in Eq. (i), we get n = 4 (Z = number of protons) increases by one and mass number (A = number of protons + neutrons) \\ t = (n) t1/ 2 = (4) (100 ms) = 400 ms remains unchanged during b-decay. 11. Using N = N 0e- lt 5. Following nuclear reaction takes place l = ln 2 = ln 2 t1/ 2 3.8 0n1 ¾¾® 1H1 + -1e0 + n where, n is antineutrino. N0 - ln 2 20 3.8 t R0çèæ 1 ÷øö n \\ = N 0e 2 6. From R = we have, 1 = 64 æç 1÷ö n Solving this equation with the help of given data è 2ø we find t = 16.5 days or n = 6 = number of half-lives \\ Correct option is (b). \\ t = n ´ tt1/2 = 6 ´ 2 = 12 h 12. 1000 = çæè 1 ö÷ø n 1 2 2 8000 7. Activity of S1 = (activity of S2) 1 \\ n = 3 = number of half-lives 2 or l1 N 1 = (l2 N2) These half-lives are equivalent to 9 days. Hence, l1 = N2 one half-life is 3 days. l2 2N 1 or tav = 1.44 t1/ 2 = 1.44 ´ 3 = 4.32 days or T1 = 2N 1 èçæT = half - life = ln 2 ö÷ø 13. R0 = lN 0 Þ N 0 = R0 T2 N 2 l l Given, N 1 = 2N 2 where, l = ln 2 \\ T1 = 4 t1/ 2 T2 N = N 0 e- lt \\ Correct option is (a). Find N 1 = N 0e- lt1 8. After two half-lives 1 th fraction of nuclei will and N 2 = N 0e- lt2 4 \\ Number of nuclei decayed in given remain un-decayed. Or, 3 th fraction will decay. time = N 1 - N 2 4
504 Optics and Modern Physics 14. (a) R = R0e-lt 2. Heavy water is used as moderators in nuclear R0 = 20 MCI reactors to slow down the neutrons. R = 8 MCI 4. During fusion process two or more lighter nuclei t = 4.0 h combine to form a heavy nucleus. Find l. Hence, the correct option is (c). (b) R0 = lN 0. Find N0 = R0 5. (a) m(c)2 = P ´ t l \\ = P´ t (c) Find R = R0 e-lt m c2 15. R0 = lN 0 = (109 ) (24) (3600) (3 ´ 108 )2 6.0 ´ 1011 = l (1015) \\ l = 6.0 ´ 10-4 s = 9.6 ´ 10-4 kg t1/ 2 = ln 2 = 0.693 (b) Number of fissions required per second l 6.0 ´ 10-4 = Energy required per second Energy released in one fission = 1.16 ´ 103 s 109 16. In 200 minute time, = 200 ´ 106 ´ 1.6 ´ 10-19 n1 = number of half-lives of X = 3.125 ´ 1019 = 200 = 4 50 6. Mass defect Smi - Smf = Dm n2 = number of half-lives of Y = (238.050784) - (234.043593 + 4.002602) = 200 = 2 = 4.589 ´ 10-3 u 100 Energy released = Dm ´ 931.48 MeV NX = N 0(1/2)4 = 1 = 4.27 MeV NY N 0(1/2)2 4 8. Q = (Dm in atomic mass unit) ´ 931.4 MeV INTRODUCTORY EXERCISE 34.2 = (2 ´ mass of 1H2 - mass of 2He4) ´ 931.4 MeV 1. 4 (2He4) = 8O16 = (2 ´ 2.0141 - 4.0024) ´ 931.4 MeV Q » 24 MeV Mass defect, Dm = {4 (4.0026) - 15.9994} = 0.011 amu 9. 2 1H2 ¾® 2He4 \\ Energy released per oxygen nuclei Binding energy of two deuterons, = (0.011) (931.48) MeV = 10.24 MeV E1 = 2 [2 ´ 1.1] = 4.4 MeV Binding energy of helium nucleus, \\ Correct answer is (c). E2 = 4 (7.0) = 28.0 MeV \\ Energy released DE = E2 - E1 = (28 - 4.4) MeV = 23.6 MeV Exercises LEVEL 1 3. By emission of one a -praticle, atomic number Assertion and Reason decreases by 2 and mass number by 4. But by the 1. Huge amount of energy is involved in any nuclear emission of one b-particle, atomic number increases by 1 and mass number remains process, which cannot be increased or decreased unchanged. by pressure or temperature. 4. In moving from lower energy state to higher 2. Some lighter nuclei are also radioactive. energy state electromagnetic waves are absorbed.
Chapter 34 Modern Physics - II 505 5. Neutrino or antineutrino is also produced during 9. Number of nuclei left = 1th b-decay. 8 6. Total binding energy per nucleon is more Now, 1 = æç 1÷ö n 8 è 2ø important for stability. \\ n = number of half-lives 9. (1 amu)(c2) = 931.48 MeV =3 10. a -particles are heaviest. Hence, its ionizing power 3 half-lives = 8 s is maximum. \\ 1 half-life = 8 s 3 11. In binding energy per nucleon versus mass number 10. Number of atoms disintegrated, graph binding energy per nucleon of daughter nuclei should increase (for release of energy) or the N = N 0 (1- e-lt ) daughter nuclei should lie towards the peak of the graph. \\ N = 1 - e-lt Objective Questions N0 1. Nuclear density is constant hence, mass µ volume At t = one mean life = 1 l or m µ V N = 1- e-1 = 1- 1 2. Radius of a nucleus is given by N0 e R = R0 A1/ 3 (where, R0 = 1.25 ´ 10-15 m) 11. Decayed fraction is 3 th. Therefore, left fraction is = 1.25 A1/ 3 ´ 10-15 m 4 1 th. Here, A is the mass number and mass of the 4 n n uranium nucleus will be N = N æèç 1 ÷øö or N = æçè 1 ø÷ö = 1 m » Amp , where mp = mass of proton 0 2 N0 2 4 = A (1.67 ´ 10-27 kg) \\ n = number of half-lives = 2 \\ Density, r = mass = 4 m Two half-lives are equivalent to 15 min. volume pR3 Therefore, one half-life is 7.5 min. 3 12. l = ln 2 = ln 2 day -1 = A (1.67 ´ 10-27 kg) t1/ 2 4 A (1.25 ´ 10-15m)3 Now, apply or r » 2.0 ´1017 kg/m3 R = R0e-lt Þ 5 = 100e-lt 5. Let n - a particles and m - b particles are emitted. Substituting value of l, we can find t. Then, 13. Let N nuclei decay per second. Then, 90 - 2n + m = 80 …(i) N (200 ´ 1.6 ´ 10-13) = 1.6 ´ 106 200 - 4n = 168 …(ii) Solving we get N = 5 ´ 1016 per sec Solving Eqs. (i) and (ii), we get 14. Nuclear density is independent of A. It is of the n = 8 and m = 6 order of 107 kg/m3. 7. By emitting one a -particle, atomic number 15. l = ln 2 = 0.693 = 1 sec-1 t1/ 2 6.93 10 decreases by 2. By emitting two b-particles, \\ atomic number increases by 2. Hence, N = N 0 e-lt æç 1 ö÷ N = è 10 ø ZA = ZC e-lt = - (10) or A and C are isotopes. N0 e 8. Binding energy = (Dm) ´ 931.5 MeV = e-1 » 0.63 = [2 ´ 1.00785 + 2 ´ 1.00866 - 4.00274] 16. Activity of atoms is 6.25% after four half-lives. ´ 931.5 \\ Four half-lives » 2 h = 120 min = 28.2 MeV \\ One half-life is 30 min.
506 Optics and Modern Physics 17. Probability of survival, = 1- e-lt = 1- e-0.1 ´ 5 P = Number of nuclei left = N 0e-lt = 0.39 Initial number of nuclei N 0 6. U238 = 3 1 Pb206 1 At t = one mean life = l P = e-1 = 1 N0 = 3+ 1= 4 …(i) e N =3 …(ii) N = N 0e-lt Subjective Questions l = ln 2 t1/ 2 1. (a) R = R0e-lt R = 2700 per minute, R0 = 4750 per minute From Eqs. (i) and (ii), we get t = 5 min t = 1.88 ´ 109 yr Find l. 7. l1 = ln 2 (T = half-life) T1 (b) t1/ 2 = ln 2 l l2 = ln 2 2. R = lN T2 6 ´ 1011 = 1.0 ´ 1015l R01 + R02 = 8 mCi (given) \\ l1(4N 0) + l2(N 0) = 8 mCi \\ l = 6 ´ 10-4 s From here we can find number t1/ 2 = ln 2 = 0.693 s l 6 ´ 10-4 after t = 60 yr = 1155 s = 19.25 min R = R1 + R2 = (4l1N 0)e-l1t + (l2N 0)e-l2t 3. Q R = lN \\ N = R = R= Rt1/ 2 9. (a) 82 + 10 = 92 , 206 + 10 + 20 = 236 l (ln 2)/t1/ 2 ln 2 So, this reaction is possible. = (8 ´ 3.7 ´ 1010) (5.3) (365) (24) (3600) (b) 82 + 16 - 6 = 92 , 206 + 32 = 238 0.693 But antineutrino is also emitted with b-1 = 7.14 ´ 1019 (or electron) decay. m = 7.14 ´ 109 ´ 60 g 6.02 ´ 1023 10. Binding energy = Dm ´ 931.5 MeV = 7.11 ´ 10-3 g = (7 ´ 1.00783 + 7 ´ 1.00867 - 14.00307) 931.5 ççèæ ö÷÷ø = 104.72 MeV 4. R = lN = ln 2 N 11. Energy released = binding energy t1/ 2 = Dm ´ 931.5 MeV = 4.5 ´ 109 0.693 ´ 3600 æçè 1 øö÷ (6.02 ´ 1023) = (8 ´ 1.007825 + 8 ´1.008665 - 15.994915) ´ 365 ´ 24 238 ´ 931.5 = 1.23 ´104dps = 127.62 MeV 1 = 10 days 12. (a) Number of nuclei in 1 kg of U235, l 5. N = æç 1 ÷ö (6.02 ´ 1026) è 235ø \\ l = 0.1 day -1 Probability of decay \\ Total energy released = Number of atoms decayed Initial number of atoms = (N ´ 200) MeV = N 0(1- e-lt ) N0 = æèç 1 ÷øö (6.02 ´ 1026) (200) (1.6 ´ 10-13) 235 = 8.19 ´ 1013 J
Chapter 34 Modern Physics - II 507 (b) m = 8.19 ´ 1013 g 2. l = l1 + l2 30 ´ 103 \\ ln 2 = ln 2 + ln 2 (T = Half-life) = 2.73 ´ 109 g T T1 T1 = 2.73 ´ 106 kg or T = T1T2 = 20 y 13. From momentum of conservation, T1 + T2 p1 = p2 1 th sample remains after 2 half-lives or 40 y. \\ 2K1m1 = 2K2m2 4 \\ K2 = K 1m1 = (6.802) (4) 3. Q-value = Final binding energy m2 208 Initial binding energy = 0.1308 MeV = E2N 2 + E3N 3 - E1N 1 14. Number of nuclei in 1 kg of uranium, 4. Energy released = Final binding energy N = æç 1 ÷ö (6.02 ´ 1026) - initial binding energy è 235ø = 110 ´ 8.2 + 90 ´ 8.2 - 200 ´ 7.4 Now, çæ 1 ÷ö (6.02 ´ 1026) (185 ´ 1.6 ´ 10-13) è 235ø = 160 MeV = (100 ´ 106)t 5. It means we are getting only 100 MeV of energy \\ t = 7.58 ´ 105 s by the fission of one uranium nucleus. = 7.58 ´ 105 day 60 ´ 60 ´ 24 Number of nuclei per second = Energy required per second Energy obtained by one fission = 8.78 days = 100 16 ´ 106 = 1018 ´ 1.6 ´ 10-13 15. Q-value = (Dm) (931.5) MeV 6. When the rate production = rate of disintegration, (a) Q-value = (2 ´ 2.014102 - 3.016049 number of nuclei or maximum. - 1.007825) ´ 931.5 \\ lN = A = 4.05 MeV ln 2 AT Similarly, Q-value of other parts can also be or N = A or N = ln 2 = maximum obtained. T 16. Q-value = (Dm) ´ 931.5 MeV 7. R0 = 15 ´ 200 = 3000 decay/min from 200 g = (2 ´ 4.0026 - 8.0053) ´ 931.5 carbon. = - 0.0931 MeV = - 93.1 keV èçæ 1 ÷øö n 2 Using R = R0 LEVEL 2 \\ 375 = 3000 èæç 21öø÷ n Single Correct Option \\ n = number of half-lives = 3 \\ t = 5730 ´ 3 = 17190 y 1. N0 = N0 æç 1÷ö n \\ n=4 16 è 2ø 8. AP = A0e-lt1 So, 3t times is equivalent to four half-lives. Hence, AQ = A0e-lt2 one half-life is equal to 3t. lt1 = ln (A0 / AP ) 4 \\ = 1 ln (A0 / AP ) =T ln (A0 / AP ) The given time 11 - t = 9 t is equivalent to 6 t1 l 2t 2 Similarly, t2 = T ln (A0 / AQ ) half-lives. \\ N = N 0 èæç 1 ÷ö6 = N0 \\ t1 - t2 =T ln çèæç AQ ö÷ø÷ 2ø 64 AP
508 Optics and Modern Physics 9. Combining two given equations, \\ t1 = 1 ln èççæ A0 ÷öø÷ = T ln A0 l A1 ln 2 A1 we have, 31H2 = 2He4 + p + n Dm = 3 ´ 2.014 - 4.001- 1.007 - 1.008 A2 = 2A0 e-lt2 = 0.026 u t2 = T 2 ln èæçç 2 A0 ÷öø÷ Energy released by 3 deuterons ln A2 = 0.026 ´ 931.5 ´ 1.6 ´ 10-13J t1 - t2 = T èççæ A0 ´ A2 ÷øö÷ = 3.9 ´ 10-12J ln 2 A1 2 A0 Now, (1016 ´ t) = çæèç 10 40 ö÷ø÷ (3.9 ´ 10-12) = T ln ççèæ A2 ÷÷øö 3 ln 2 A1 2 Solving we get, t » 1.3 ´ 1012 s More than One Correct Options 10. R1 = R2 1. y = lx = ln 2 × x R01e-l1t = R02e-lt T \\ l1N 0 e-l1t = l2N 0 e-l2t …(i) x = 1 = constant y l l1 = ln 2 = 0.693 x = T or x > T (as ln 2 = 0.693) t1 t1 y ln 2 y l2 = 0.693 Further, t2 xy = x (lx) = lx2 Substituting these values in Eq. (i), we can get After one half-life, x remains half. Hence, x2 the t. remains 1 th. 11. A = 232 - 4 = 228 4 From conservation of momentum, 3. By the emission of an a-particle, atomic number pa = pg = 2Kama = 2Kg mg decreases by 2 and by the emission of two or Ka = mg = 228 particles atomic number increases by 2. Hence, net Kg ma 4 atomic number remains unchanged. 4. At t = 4T \\ Ka = æçèç 228 4 ÷÷öø KTotal Number of half-lives of first n1 = 4 and number of 228 + half-lives of second n2 = 2 = æèç 228 ö÷ø KTotal N1 = x = N0 (1/ 2)4 = 1 232 N2 N0 (1/ 2)2 4 12. From momentum of conservation, y = R1 = l1 N0 (1/ 2)4 R2 l2 N0 (1/ 2)2 momentum of photon = photon of nucleus \\ E = 2Km = l1 = T2 c 4l2 4T1 \\ K = E2 = 2T = 1 2mc2 4T 2 = ´24 (7 ´ 1.6 ´ 10-13)2 )2 keV 6. R = R0A1/ 3 or R µ A1/ 3 ´1.67 ´ 10-27 ´ (3 ´ 108 2 Comprehension Based Questions 1. Energy released = (Dm) (931.48) MeV ´ 1.6 ´ 10-16 = [2 ´ 2.01102 - 3.0160 - 1.007825] ´ 931.5 = 1.1 keV = 4.03 MeV » 4 MeV 13. Activity A µ Number of atoms A1 = A0e-lt1
Chapter 34 Modern Physics - II 509 2. Let N number of fusion reactions are required, (c) l to X-rays is of the order of 1Å - 100 Å then E = é 12375 ù in eV êë l (in Å)ûú N ´ 4 ´ 1.6 ´ 10-13 = 103 ´ 3600 N = 5.625 ´ 1018 = é 12.375ù in keV ëê l ûú 3. In one fusion reaction two 2 H nuclei are used. 1 where, l is in Å. Hence, total number of 2 H nuclei are 2N . 1 (d) l is of the order of 4000 Å - 7000 Å or 1.125 ´ 1019 Mass in kg 12375 = çèæç 1.125 ´ 1019 øö÷÷ (2) kg Now, E (in eV) = l (in Å) 6.02 ´ 1026 = 3.7 ´ 10-8 kg Subjective Questions Match the Columns 1. N = R = 109 = 7.43 ´ 1013 l 0.693 çèæ - ÷öø 1. dN = lN 14.3 ´ 3600 dt dN = q - lN N dN t dt 0 q - lN \\ y = lx or l = y Now, ò òor = dt x 0 t1/ 2 = ln 2 = (ln 2) (x / y) \\ N = q (1 - e-lt ) l l R = R0e-lt Substituting the values, = ye-l(1/ l) = y/e 7.43 ´ 1013 = 2 ´ 109 [1 - e-(0.693/ 14.3 ´ 3600) t ] 0.693 R = y = lN e 14.3 ´ 3600 \\ N = y = y = x Solving this equation we get, Ans. el e( y/x) e t = 14.3 h 2. Energy is released when daughter nuclei lie 2. (a) AB C 0 towards peak of this graph, so that binding energy At t = 0 N0 0 per nucleon or total binding energy in the nuclear At t N1 N2 N3 process increases. Here, N 1 = N 0e-lt …(i) 3. A will continuously decrease, but C will increase. dN 2 = l (N 1 - N 2) (A + B) will continuously decrease as C is formed dt only from A and B. or dN 2 = lN 0e-lt - lN 2 (C + B) = Total - A dt A is continuously decreasing. Hence, (C + B) will continuously increase. or dN 2 + lN 2dt = lN 0e-lt 4. (a) Z ¢ = Z - Z + 1 = Z - 1 \\ eltdN 2 + lN 2eltdt = lN 0dt A¢ = A - 4 or d(N 2elt ) = lN 0dt (b) Z ¢ = Z - 2 ´ 2 + 1 = Z - 3 \\ N 2elt = lN 0t + C A¢ = A - 2 ´ 4 = A - 8 (c) Z ¢ = Z - 2 + 2 ´ 1 = Z At t = 0, N 2 = 0, \\ C =0 A¢ = A - 4 \\ N 2 = lN 0(te-lt ) (d) Z ¢ = Z - 2 ´ 2 + 2 ´ 2 = Z - 2 (b) Activity of B is A¢ = A - 2 ´ 4 = A - 8 R2 = lN 2 = l2N 0(te-lt ) 5. (a) E is of the order of kT , where k = Boltzmann constant and T » 300 K
510 Optics and Modern Physics For maximum activity, dR2 = 0 \\ N A0 = çæçè RA0 ÷öø÷ (t1/ 2)A = èæç 00..7237 øö÷ çèæ 10 ø÷ö dt N B0 RB0 (t1/ 2)B 5 \\ t = 1 Ans. = 5.4 l Ans. \\ Rmax = lN 0 5. Let N be the number of radio nuclei at any time t. e Then, net rate of formation of nuclei at time t is 3. (a) Let at time t, number of radioactive nuclei dN = a - lN dt are N . N dN t 0 a - lN Net rate of formation of nuclei of A ò òor = dt dN = a - lN dN 0 dt a - lN or = dt a e-lt l or N = (1 - ) N dN t N0 a - N ò òor = dt 0 Solving this equation, we get N N = 1 [a - (a - l N 0)e-lt ] …(i) Rate of formation = a Rate of decay = lN l Number of nuclei formed in time t = at ln 2 (b) (i) Substituting a = 2lN 0 and t = t1/ 2 = l and number of nuclei left after time in Eq. (i), we get N = 3 t = a (1 - e-lt ) 2 N0 l (ii) Substituting a = 2lN 0 and t ® ¥ in Eq. (i), Therefore, number of nuclei disintegrated in time we get t = at - a (1 - e-lt ) l a N = l = 2N 0 \\ Energy released till time, or N = 2N 0 t = E0 é at - a (1 - e-lt )ûùú ëê l 4. (a) LBe. tTRhAe0n,and RB0 be the initial activities of A and But only 20% of it is used in raising the RA0 + RB0 = 1010 dps …(i) temperature of water. Activity of A after time t = 20 days So, 0.2 E0 é at - a (1 - e-lt )úùû = Q êë l (two half-lives of A) is where, Q = msDq RA = çæ 1÷ö 2 RA0 = 0.25 RA0 \\ Dq = increase in temperature of water = Q è 2ø ms Similarly, activity of B after t = 20 days 0.2 E0 é at - a ( 1 - e-lt )ûúù (four half-lives of B) is ëê l \\ Dq = 4 ms èçæ 1 ö÷ø RB = 2 RB0 = 0.0625 RB0 6. We have for B dN B = P - l2N B dt Now, it is given that RA + RB = 20% of 1010 NB dN B t 0 P - l2N B or 0.25RA0 + 0.0625RB0 = 0.2 ´ 1010 dps …(ii) ò òÞ = dt 0 Solving Eqs. (i) and (ii), we get Þ ln èæç P - l2N B ÷öø = - l2t P RA0 = 0.73 ´ 1010 dps P(1 - e-l2t ) and RB0 = 0.27 ´ 1010 dps Þ NB = l2 (b) RA0 = lAN A0 = (t1/ 2)B × N A0 The number of nuclei of A after time t is RB0 lBN B0 (t1/ 2) A N B0 N A = N 0e-l1t
Chapter 34 Modern Physics - II 511 Thus, dN c = l1N A + l2N B = R1 = 1 Þ dt R9 Þ dN c = l1N 0e-l1t + P(1 - e-l2t ) and probability of getting b -particles dt = R2 = 8 Ans. çèæç t e-l2t - 1÷øö÷ R9 Nc = N 0(1 - e-l1t ) + P + l2 (ii) R01 = R02 7. 210 Po ¾® 28026Pb + 24He \\ N 01 = N 02 84 T1 T2 Dm = 0.00564 amu \\ N 01 = 1 N 02 4 Energy liberated per reaction = (Dm)931 MeV = 8.4 ´ 10-13 J Let N 0 be the total number of nuclei at t = 0. Electrical energy produced = 8.4 ´ 10-14 J N0 4N 5 5 Let m g of 210Po is required to produce the desired Then, N 01 = and N 02 = 0 energy. + = N0 N = m ´ 6 ´ 1023 Given that N1 N2 2 210 çèæ 1 ÷öø t/ 405 4N 0 èçæ 1 ÷øö t / 1620 l = 0.693 = 0.005 per day N0 2 5 2 N0 t1/ 2 or 5 + = 2 çæ - dN ö÷ = lN = (0.005)(6 ´ 1023 m) per day Let çæ 1ö÷ t/ 1620 = x è dt ø 210 è 2ø \\ Electrical energy produced per day Then, above equation becomes x4 + 4x - 2.5 = 0 = (0.005)(6 ´ 1023 m) ´ 8.4 ´ 10-14 J èæç 1 ÷øö t/ 1620 210 2 \\ x = 0.594 or = 0.594 This is equal to 1.2 ´ 107 J (given) Solving, it we get t = 1215 s. \\ m = 10 g Ans. Ans. Activity at the end of 693 days is 9. N = 10-3 ´ 6.02 ´ 1023= 2.87 ´ 1018 0.005 ´ 6 ´ 1023 ´ 10 1021 çæè 1 ø÷ö n 210 210 7 2 R = = per day = R0 During one mean life period 63.8% nuclei are decayed. Hence, energy released Here, n = number of half-lives = 693 = 5 138.6 E = 0.638 ´ 2.87 ´ 1018 ´ 5.3 ´ 1.6 ´ 10-13 J \\ = R(2)5 = 32 ´ 1021 = 4.57 ´ 1021 per day = 1.55 ´ 106 J Ans. 7 R0 R0 = lN = 0.693 ´ 6.02 ´ 1023per sec ´ 3600 Ans. 10. 14.3 ´ 24 8. (i) At t = 0, probabilities of getting a and b = 3.37 ´ 1017 per sec particles are same. This implies that initial activity After 70 hours activity, R = R0e-lt = (3.37 ´ 1017 ) e-(0.693/ 14.3 ´ 24)(70) of both is equal. Say it is R0. Activity after t = 1620 s, = 2.92 ´ 1017 per sec çæè 1 ÷öø 1620/ 405 R0 2 16 R1 = R0 = In fruits activity was observed 1mCi or 3.7 ´ 104 per sec. Therefore, percentage of activity = æç 1ö÷ 1620/ 1620 = R0 and R2 R0 è 2ø 2 transmitted from root to the fruit. Total activity, R = R1 + R2 =9 R0 = 3.7 ´ 104 ´ 100 16 2.92 ´ 1017 Probability of getting a-particles, = 1.26 ´ 10-11 % Ans.
35. Semiconductors INTRODUCTORY EXERCISE 35.1 1. Eg Carbon = 5.4 eV A B Y1 Y2 Y Eg Silicon = 1.1 eV 01101 Eg Germanium = 0.7 eV \\ (Eg)C > (Eg)Si > (Eg)Ge 11001 INTRODUCTORY EXERCISE 35.2 Clearly output resembles an ‘OR’ gate. 1. Hole diffusion from p to n side can be viewed as 3. For given amplifier, “electron diffusion” from n to p side. V0 = 2V, R0 = 2 kW Diffusion occurs due to difference in concentrations in different regions. bac = 100, Ri = 1 kW An electron (or hole) diffuses where its concentration is less. We have, output voltage, V0 = ICR0 2. Due to forward biasing depletion layer thickness Þ IC = collector current decreases, potential barrier is reduced and = V0 = 2 2 = 10-3A diffusion of electrons from n to p side occurs. R0 ´ 103 INTRODUCTORY EXERCISE 35.3 = 1 mA 1. In a transistor, base must be very thin and lightly Also, current amplication b = iC iB doped so that all of the charge carriers are not combined in base and majority of them passes the Þ iB = iC = 10-3 A reverse bias layer to collector side. b 100 2. Voltage gain is maximum and constant for mid = 10-5A = 10 ´ 10-6A frequency range but is less for both low and high = 10 mA. frequencies. and voltage amplification AV = bR0 = V0 Ri Vi Av Þ Vi = V0Ri = 2 ´ 1 ´ 103 (Voltage bR0 100 ´ 2 ´ 103 gain) = 10-2 volts = 10 mV w INTRODUCTORY EXERCISE 35.4 (Input frequency) (b) We construct truth table to see the logical 1. Input waveforms are as shown t5 t6 operation. A¢ t1 t2 t3 t4 A NAND Y1 O B¢ NAND Y O B NAND Y2 These are fed to an NAND gate, A B Y1 Y2 Y A Y 00110 10011 B Output waveform is as shown
t1 t2 t3 t4 t5 t6 Chapter 35 Semiconductors 513 1 Y Y: A B Y1 Y 0 0010 2. Truth table for the circuit given is 1010 0110 A NAND 1101 NAND As the output resembles output of a AND gate so, given circuit behaves like an ‘AND’ gate. B Y1 Exercises 1. With increase in temperature, number of electrons As IC = 95 IE Þ IE = 100 ´ IC 100 95 reaching conduction band increases but, mean relation time decreases. Effect of decrease in = 100 ´ 10 mA = 10.53 mA relation time is much less than that of increase in 95 number density of charge carriers. So, IB = 0.53 mA 2. For diode D1, 8. Depletion zone is formed due to diffusion of e- Vp - Vn = - 10 - 0 = - 10V So, D1 is in reverse bias. from n to p side. Due to diffusion positive For diode D2, immobile ions exists on n side and negative ions Vp - Vn = 0 - (-10) = 10 V So, D2 is in forward bias. on p-side. 3. Hole is a vacancy created by an electron moving Recombination of e- and holes takes place on from valance band to conduction band. p-side. This results in (n-side more positive than 4. Capacitor is charged to maximum potential p-side) formation of a junction field and potential difference barrier across the depletion zone. \\ V = Vmax = Vrms ´ 2 9. When applied voltage approaches zener potential = 220 2 volts diode breakdown to conduct excess current. This 5. Electrons take the energy and move up to neat causes a change of current in series resistance. higher energy level both in conduction and valance 10. Reverse breakdown may be “Avlanche break band. So, a hole move downwards due to movement of electron. down” (breakdown due to high velocity collision of minority carrier) or it may be a zener breakdown 6. In an n- p-n transistor, (breakdown of bonds due to strong field). IE E B C 11. No, it cannot be measured by using a voltmeter. Barrier potential is less than 0.2 V for germanium diode and is less than 0.7 V for silicon. Also there are no free charge carriers which provide ‘current’ for working of voltmeter. 12. An “OR” gate may be made to operate motor relays for garage gates. -+ IB Pick up -+ signal to relay Electrons move from emitter to base and electrons + which are not combined with holes in base region, – crosses to collector side. Sensor output 7. In an n- p-n transistor, from gate IE = IB + IC
514 Optics and Modern Physics 13. Such combination is called “Cascade” 19. Supply voltage may be 7 V and zener breakdown combination. at 5 V. AV = AV1 ´ AV2 So, voltage drop across resistance R = 7 - 5 = 2V. Now power rating of zener = 1W. (net ) = 10 ´ 20 = 200 So, current through zener must not exceed \\ V0 = output voltage I = P = 1A = AV ´ Vi V5 (net ) Hence, series resistance = 200 ´ 0.01 = 2V R=V = 2 = 10 W. i 1 14. We use, Eg = hc Þ Eg = 1240 eV× nm l 600 nm 5 » 2 eV 20. Current flows only in branches AB and EF as So, photon energy is less than that required diode of CD branch is in reverse bias. (\\ I3 = 0) (2.8 eV). So, detector is not able to detect this wavelength. So, given circuit is equivalent to 15. (i) Diode is a rectifier diode. I4 25 W 125 W A (ii) Point P represents zener breakdown potential. B 16. Given is a p - n - p CE configuration, when R1 is 25 W 125 W E increased, base current decreases as a result I1 I2 F current also decreases. G 25 W H Hence, both ammeter and voltmeter readings 5V decreases. 17. A NOT gate can be formed by proper biasing of a Current through cell is transistor as shown. i1 = V = 5 = 0.05 A RTotal 100 VCC=V(1) Also by symmetry, C Y I2 = I4 = 0.05 = 0.025 A B 2 A 21. Given circuit is equivalent to 10 V 3 kW VB=V(1) 400 W 10 V If input (A) is low (0), then the transistor is in cut off stage and output (Y) is same as VCC = V (1). If In base emitter loop, input (A) is high, then transistor current is in saturation and the net voltage at output (Y) is V (0) VB = IBRB or is in state 0. Þ IB = VB = 10 18. Given is an AND gate, its truth table is RB 400 ´ 103 ABY = 25 ´ 10-6 A 000 = 25 mA 100 In emitter-collector loop, VC = ICRC 010 111
Chapter 35 Semiconductors 515 Þ IC = 3 10 23. Consider the figure to solve this question, ´ 103 IE = IC + IB and IC = bIB K(i) = 3.33 mA ICRC + VCE + IERE = VCC K(ii) RIB + VBE + IERE = VCC K(iii) and bDC = IC = 3.33 mA IB 25 mA IE = IC = bIB = 133 22. Consider the figure given here to solve this IB + I IC VCC = 12 V A problem RC =100 kW RC IC = IE [As base current is very small] C RC = 7.8 kW H VCE = 3 V From the figure, IC (RC + RE ) + VCE = 12 (RE + RC ) ´ 1 ´ 10-3 + 3 = 12 I IB E IE RE + RC = 9 ´ 103 = 9 kW VBE = 0.5V RE = 1 kW R = 20 kW RE = 9 - 7.8 = 1.2 kW VE = IE ´ RE JD = 1 ´ 10-3 ´ 1.2 ´ 103 = 1.2 V Voltage, VB = VE + VBE = 1.2 + 0.5 = 1.7 V From Eq. (iii), Current, I = VB = 1.7 (R + bRE ) IB = VCC - VBE 20 ´ 103 20 ´ 103 Þ IB = VCC - VBE = 0.085 mA R + bRE Resistance, RB = 12 - 1.7 = 10.3 = 12 - 0.5 0.01 + 0.085 80 + 1.2 ´ 100 IC + 0.085 b = 11.5 mA [Given, b = 100] 200 = 108 kW From Eq. (ii), VCC = 12 V (RC + RE ) = VCE - VBE IC IC + I IC A VCC - VCE RB RC = 7.8 kW bI B C = (Q IC = bIB ) H VCE = 3 V (RC + RE ) = 2 (12 - 3) kW 11.5 I IB VBE = 0.5V E IE = 1.56 kW R = 200 kW RE JD RC + RE = 1.56 RC = 1.56 - 1 = 0.56 kW
36. Communication System Exercises Single Correct Option = 2 ´ 6.4 ´ 106 ´ 240 1. Range of frequencies is as follows : » 55 ´ 103 m Ground wave : 300 Hz to 300 kHz = 55 km (it may go upto 3 MHz) 10. Wavelength of signal is Sky wave : 300 kHz to 3 MHz Space wave : 3 MHz to 300 GHz. l= c = 3 ´ 108 = 20 ´ 103 m F 15 ´ 103 2. Length of antenna ³ l So, size of antenna required (at least) 4 =l Þ l³l 4 4 = 5 ´ 103 m = 5 km Þ l£4 ´l or l £ 400 m Also, effective power radiated by antenna is very less. 3. Sideband frequencies are : 11. LSB = wc = wm Lower sideband (LSB) = wc - wm = 1.5 ´ 106 - 3 ´ 103 = 1 ´ 106 Hz - 3 ´ 103 Hz = 1497 kHz = (1000 - 3) ´ 103 Hz and USB = wc + wm = 2997 kHz = 1503 kHz and = 2.997 MHz Bandwidth required Upper sideband (USB) = wc + wm = USB - LSB = 2wm = 1 ´ 106 Hz + 3 ´ 103 Hz = 2 ´ 3 kHz = 6 kHz = 1003 kHz 12. Due to over modulation m > 1, sidebands overlaps = 1.003 MHz and fading (loss of information) may occur. Also 4. Frequency of amplitude modulated wave is same when m » 1, distortion in carrier waveform occurs. as that of carrier wave. 13. Frequency of signal obtained = 2p 1 = 1 MHz LC 5. Basic communication system is Þ LC = (2p 1 ´ 106)2 Transmitter Channel Receiver Þ LC = 0.25 ´ 10-13 s2 From source To user Þ LC = 2.5 ´ 10-14 s2 6. Beyond horizon, a signal can reach via ionospheric 14. m = modulation index reflection or sky wave mode. Frequency range = Am suitable is 3 MHz to 30 MHz. Ac 7. UHF band is in range of 150 to 900 MHz. Þ Am = m ´ Ac = 0.75 ´ 12 = 9 volts So, suitable mode of communication is “space wave mode”. 15. Vibrating tuning fork and string produces all 8. Digital signals employs a discrete values of possible values of displacement hence, these signals are analog. amplitudes which are coded using binary system. A light pulse and output of NAND gate gives only 9. In LOS communication maximum distance upto discrete values of output. So, signals are digital these. which a signals can be received from tower is d = 2Rh
Chapter 36 Communication System 517 16. Penetrating power of signal a frequency of signal. (iii) Percentage increase in area covered = A2 - A1 ´ 100 So, 3 MHz signal travels longer distance in A1 ionosphere. = 3608 - 804 ´ 100 804 17. Modulation index m, » 349 % = Amax - Amin 21. That can be done by using six antennas, Amax + Amin (A = amplitude of modulated wave) \\ m = 15 - 3 = 12 = 2 1 15 + 3 18 3 18. Man made noises and atmospheric interferences 6R 2 affect only amplitude of a signal. R So, an AM signal is more noisy than a FM signal. 53 19. In LOS communication, it is not necessary that 4 transmitting antenna and receiving antenna are at 1 same height. Only requirement is that there must 6 R2 not be any obstacle in between. R T 53 Given, h = 81 m 4 \\ Distance upto which transmission can be made From above figure, side of triangle = R + h = d = 2Rh and, area covered in broadcast Also, altitude of triangle = pd2 = 3 ´ side = R = 2pRh 2 = 2 ´ p ´ 6.4 ´ 106 ´ 81 (m2) = 3258 km2 Þ 3 (R + h) = R 2 20. (i) When receiver is at ground level, then Þ 3 (R + h) = 2R service area covered = pd2 = p( 2Rh)2 Þ R+ h= 2 = 2pRh R = 2 ´ p ´ 6.4 ´ 106 ´ 20 3 » 804 km2 Þ h = 2 R - R » 0.15 R (ii) When receiver is at height of 25 m, area 3 covered = pd12 + pd22 1 = p ( 2RhT )2 + p ( 2RhR )2 = 2pR(hT + hR ) 22. From fmax = 9(N max )2 = 2 ´ p ´ 6.4 ´ 106 ´ 45(m2) » 3608 km2 Þ N max = fm2ax 81 Þ N max = (5 ´ 106)2 81 F1 = 0.3086 ´ 1012 m-3 = 3.086 ´ 1011 m-3 and N max F2 = (8 ´ 106)2 81 = 7.9 ´ 1011 m-3
518 Optics and Modern Physics or x = 3 ´ 108 ´ 4.04 ´ 10-3 2 23. Let the receiver is at point A and source is at B. = 6.06 ´ 105 = 606 km x h3 x Using Pythagoras theorem, d2 = x2 - hs2 = (606)2 - (600)2 = 7236 A B d d or d = 85.06 km So, the distance between source and receiver = 2d Velocity of waves = 3 ´ 108 m/s = 2 ´ 85.06 Time to reach a receiver = 4.04 ms = 4.04 ´ 10-3s = 170 km Let the height of satellite is The maximum distance covered on ground from the transmitter by emitted EM waves hs = 600 km Radius of earth = 6400 km d = 2RhT Size of transmitting antenna = hT or d2 = hT We know that 2R Distance travelled by wave = Velocity of waves or size of antenna hT = 2 7236 Time ´ 6400 2x = 3 ´ 108 = 0.565 km 4.04 ´ 10-3 = 565m
JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. An EM wave from air enters a medium. 4. An electron from various excited states of The electric fields are hydrogen atom emit radiation to come to the ground state. Let ln , lg be the E1 = E01x$ coséêë2pnçæè z - t øö÷ûúù in air and de-Broglie wavelength of the electron in c E2 = E02x$ cos[k(2z - ct)] in medium, where the nth state and the ground state, the wave number k and frequency n refer respectively. Let Ln be the wavelength of the emitted photon in the transition from to their values in air. The medium is the nth state to the ground state. For non-magnetic. large n, (A, B are constants) (2018) If er1 and er2 refer to relative permittivities of air and medium respectively, which of (a) Ln »A+ B (b) Ln » A + Bl2n l2n (d) L2n » l the following options is correct? (2018) (c) L2n » A + Bl2n (a) er1 =4 (b) er1 =2 (c) er1 = 1 (d) er1 = 1 er2 er2 er2 4 er2 2 5. If the series limit frequency of the Lyman 2. Unpolarised light of intensity I passes series is nL, then the series limit frequency of the Pfund series is (2018) through an ideal polariser A. Another (a) 25 nL (b) 16 nL (c) nL (d) nL 16 25 identical polariser B is placed behind A. The intensity of light beyond B is found to 6. It is found that, if a neutron suffers an be I . Now, another identical polariser C is elastic collinear collision with deuterium at 2 placed between A and B. The intensity rest, fractional loss of its energy is Pd ; while beyond B is now found to be 1. The angle for its similar collision with carbon nucleus 8 at rest, fractional loss of energy is Pc. The between polariser A and C is (2018) values of Pd and Pc are respectively (2018) (a) 0° (b) 30° (c) 45° (d) 60° (a) (.89, .28) (b) (.28, .89) 3. The angular width of the central maximum (c) (0, 0) (d) (0, 1) in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is 7. The reading of the ammeter for a silicon illuminated by monochromatic plane waves. If another slit of same width is diode in the given circuit is (2018) made near it, Young’s fringes can be observed on a screen placed at a distance 200W 50 cm from the slits. If the observed fringe width is 1 cm, what is slit 3V separation distance? (i.e. distance between the centres of each slit.) (2018) (a) 0 (b) 15 mA (c) 11.5 mA (d) 13.5 mA (a) 25 mm (b) 50 mm (c) 75 mm (d) 100 mm
2 Optics & Modern Physics 8. A telephonic communication service is 12. An electron beam is accelerated by a working at carrier frequency of 10 GHz. potential difference V to hit a metallic Only 10% of it is utilised for transmission. target to produce X-rays. It produces How many telephonic channels can be continuous as well as characteristic transmitted simultaneously, if each X-rays. If lmin is the smallest possible channel requires a bandwidth of 5 kHz? wavelength of X-rays in the spectrum, the (2018) variation of log lmin with log V is correctly (a) 2 ´ 103 (b) 2 ´ 104 represented in (2017) (c) 2 ´ 105 (d) 2 ´ 106 9. A diverging lens with magnitude of focal (a) (b)loglmin loglmin log V length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude log V of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final (c) (d)loglmin loglmin image formed is (2017) (a) virtual and at a distance of 40 cm from log V log V convergent lens 13. A radioactive nucleus A with a half-life T , (b) real and at a distance of 40 cm from the decays into a nucleus B. At t = 0, there is divergent lens (c) real and at a distance of 6 cm from the no nucleus B. After sometime t, the ratio convergent lens of the number of B to that of A is 0.3. (d) real and at a distance of 40 cm from Then, t is given by (2017) convergent lens (a) t =T log 1.3 (b) t = T log 1.3 log 2 10. In a Young’s double slit experiment, slits T T log 2 are separated by 0.5 mm and the screen is (c) t = log 1.3 (d) t = 2log 1.3 placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm 14. A particle A of mass m and initial velocity and 520 nm, is used to obtain interference v collides with a particle B of mass m/ 2 fringes on the screen. The least distance which is at rest. The collision is held on, from the common central maximum to the and elastic. The ratio of the de-Broglie point where the bright fringes due to both wavelengths lA to lB after the collision is the wavelengths coincide, is (2017) (2017) (a) 7.8 mm (b) 9.75 mm(c) 15.6 mm(d) 1.56 mm (a) lA = 2 (b) lA = 2 lB lB 3 11. Some energy levels of a molecule are (c) lA = 1 (d) lA = 1 shown in the figure. The ratio of the lB 2 lB 3 wavelengths r = l1 / l2 is given by 15. In a common emitter amplifier circuit (2017) –E using an n-p-n transistor, the phase –4/3E l2 difference between the input and the –2E l1 output voltages will be (2017) (a) 90° (b) 135° (c) 180° (d) 45° –3E 16. A transparent slab of thickness d has a refractive index n (z) that increases with z. (a) r = 2 (b) r = 3 (c) r = 1 (d) r = 4 Here, z is the vertical distance inside the 3 4 3 3 slab, measured from the top. The slab is placed between two media with uniform
Previous Years’ Questions (2018-13) 3 refractive indices n1 and n2(> n1), as shown (b) a = lL and bmin = æèçç 2l2 øö÷÷ L in the figure. A ray of light is incident with angle qi from medium 1 and emerges in (c) a = lL and bmin = 4lL medium 2 with refraction angle q f with a l2 (d) a = L and bmin = 4lL lateral displacement l. (2016) n1=constant qi 1 20. Radiation of wavelength l, is incident on z n (z) d a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3l, the speed of the fastest emitted 4 n2=constant l 2 electron will be (2016) qf æèç 4 ÷øö1/ 2 æçè 4 ÷öø1/ 2 3 3 (a) >v (b) < v Which of the following statement(s) is (c) = v æçè 4 ÷øö1/ 2 (d) = v èçæ 3 ø÷ö1/ 2 (are) true? 3 4 (a) l is independent on n (z) 21. Half-lives of two radioactive elements A (b) n1 sinqi = (n2 - n1) sinqf and B are 20 min and 40 min, respectively. (c) n1 sinqi = n2 sinqf Initially, the samples have equal number (d) l is independent of n2 of nuclei. After 80 min, the ratio of decayed numbers of A and B nuclei will be (2016) 17. In an experiment for determination of (a) 1 : 16 (b) 4 : 1 (c) 1 : 4 (d) 5 : 4 refractive index of glass of a prism by i-d plot, it was found that a ray incident at 22. The temperature dependence of resistances an angle 35° suffers a deviation of 40° and of Cu and undoped Si in the temperature that it emerges at an angle 79°. In that range 300-400 K, is best described by (2016) case, which of the following is closest to (a) linear increase for Cu, linear increase for Si (b) linear increase for Cu, exponential increase the maximum possible value of the for Si refractive index? (2016) (c) linear increase for Cu, exponential decrease (a) 1.5 (b) 1.6 (c) 1.7 (d) 1.8 for Si (d) linear decrease for Cu, linear decrease for Si 18. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To observer the 23. Identify the semiconductor devices whose tree appears (2016) characteristics are as given below, in the (a) 10 times taller (b) 10 times nearer order (a),(b),(c),(d). (2016) (c) 20 times taller (d) 20 times nearer II 19. The box of a pin hole camera of length L, has a hole of radius a. It is assumed that (a) V (b) V when the hole is illuminated by a parallel beam of light of wavelength l the spread of the spot (obtained on the opposite wall of the camera) is the sum of its I I Dark Resistance geometrical spread and the spread due to (c) V (d) V diffraction. The spot would then have its Intensity Illuminated of light minimum size (say bmin ) when (2016) (a) a = l2 and bmin = çæèç 2l2 ö÷ø÷ L L
4 Optics & Modern Physics (a) Simple diode, Zener diode, Solar cell, Light (a) 2 MHz only dependent resistance (b) 2005 kHz 2000 kHz and 1995 kHz (c) 2005 kHz and 1995 kHz (b) Zener diode, Simple diode, Light dependent (d) 2000 kHz and 1995 kHz resistance, Solar cell 28. A red LED emits light at 0.1 W uniformly (c) Solar cell, Light dependent resistance, Zener diode, Simple diode around it. The amplitude of the electric (d) Zener diode, Solar cell, Simple diode, Light field of the light at a distance of 1 m from dependent resistance 24. Monochromatic light is incident on a glass the diode is (2015) prism of angle A. If the refractive index of (a) 2.45 V/m (b) 1.73 V/m the material of the prism is m, a ray incident at an angle q, on the face AB (c) 5.48 V/m (d) 7.75 V/m 29. As an electron makes a transition from an would get transmitted through the face excited state to the ground state of a AC of the prism provided (2015) hydrogen like atom/ion (2015) (a) q< cos- 1 é sin ì + sin- 1 çèæç 1 ø÷ö÷ýþüûùúú (a) kinetic energy, potential energy and total ëêêm íA m energy decrease î (b) kinetic energy decreases, potential energy (b) q< sin- 1 êëéêm sinìíA - sin- 1çæçè 1 ÷öø÷ üù increases but total energy remains same î m þýúúû (c) kinetic energy and total energy decrease but (c) q > cos- 1 é siníìA + sin- 1èæçç 1 öø÷÷þýüúûùú potential energy increases ëêêm î m (d) its kinetic energy increases but potential é siníìA çæçè 1 ÷ø÷öþýüúûúù energy and total energy decrease êm î m (d) q> sin- 1 - sin- 1 30. Match Column I (fundamental experiment) ëê with Column II (its conclusion) and select 25. Assuming human pupil to have a radius the correct option from the choices given of 0.25 cm and a comfortable viewing below the list. (2015) distance of 25 cm, the minimum Column I Column II separation between two objects that A Franck-Hertz 1. Particle nature of experiment light human eye can resolve at 500 nm wavelength is (2015) B Photo-electric 2. Discrete energy experiment levels of atom (a) 30 mm (b) 1 mm (c) 100 mm (d) 300 mm 26. On a hot summer night, the refractive C Davisson-Germe 3. Wave nature of index of air is smallest near the ground r experiment electron and increases with height from the 4. Structure of atom ground. When a light beam is directed horizontally, the Huygens principle leads ABC ABC (a) 1 4 3 (b) 2 1 3 us to conclude that as it travels, the light (c) 2 4 3 (d) 4 3 2 beam (2015) (a) becomes narrower 31. A green light is incident from the water to (b) goes horizontally without any deflection the air-water interface at the critical angle (q). (c) bends upwards Select the correct statement. (2014) (d) bends downwards (a) The entire spectrum of visible light will come out of the water at an angle of 90° to the 27. A signal of 5 kHz frequency is amplitude normal modulated on a carrier wave of frequency (b) The spectrum of visible light whose frequency is less than that of green light will 2MHz. The frequencies of the resultant come out of the air medium signal is/are (2015)
Previous Years’ Questions (2018-13) 5 (c) The spectrum of visible light whose magnetic field of 3 ´ 10-4 T. If the radius frequency is more than that of green light will come out to the air medium of the largest circular path followed by (d) The entire spectrum of visible light will come these electrons is 10.0 mm, the work out of the water at various angles to the normal function of the metal is close to (2014) (a) 1.8 eV (b) 1.1 eV (c) 0.8 eV (d) 1.6 eV 32. A thin convex lens made from crown glass 36. The forward biased diode connection is (m = 3/ 2) has focal length f. When it is (a) +2V (2014) –2V measured in two different liquids having (b) –3V –3V refractive indices 4/3 and 5/3. It has the (c) 2V 4V focal lengths f1 and f2, respectively. The (d) –2V +2V correct relation between the focal length 37. During the propagation of is (2014) electromagnetic waves in a medium, (2014) (a) f1 = f2 < f (b) f1 >f and f2 becomes negative (a) electric energy density is double of the (c) f2 >f and f1 becomes negative magnetic energy density (d) f1 and f2 both become negative (b) electric energy density is half of the 33. Two beams, A and B, of plane polarised magnetic energy density light with mutually perpendicular planes (c) electric energy density is equal to the magnetic energy density of polarisation are seen through a (d) Both electric and magnetic energy densities polaroid. From the position when the are zero beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB 38. Match List I (Electromagnetic wave type) respectively, then IA / IB equals (2014) with List II (Its association/application) (a) 3 (b) 3/2 (c) 1 (d) 1/3 and select the correct option from the 34. Hydrogen (1H1), deuterium (1H2), singly choices given below the lists. (2014) ionised helium (2He4)+ and doubly ionised lithium (3Li8)++ all have one electron List I List II around the nucleus. Consider an electron A. Infrared waves 1. To treat muscular transition from n = 2 to n = 1. If the strain wavelengths of emitted radiation are B. Radio waves 2. For broadcasting l1, l2, l3 and l4, respectively for four elements, then approximately which one C. X-rays 3. To detect fracture of bones of the following is correct? (2014) D. Ultraviolet 4. Absorbed by the ozone layer of the (a) 4 l1 = 2l2 = 2l3 = l4 atmosphere (b) l1 = 2l2 = 2l3 = l4 (c) l1 = l2 = 4l3 = 9l4 Codes CD (d) l1 = 2l2 = 3l3 = 4l4 AB 21 43 35. The radiation corresponding to 3 ® 2 (a) 4 3 14 (b) 1 2 transition of hydrogen atom falls on a (c) 3 2 34 metal surface to produce photoelectrons. (d) 1 2 These electrons are made to enter a
6 Optics & Modern Physics 39. The graph between angle of deviation (d) frequency of radiation emitted is and angle of incidence (i) for a triangular proportional to (2013) prism is represented by (2013) (a) 1 (b) 1 (c) 1 (d) 1 n n2 n4 n3 dd (a) (b) 44. The I-V characteristic of an LED is (2013) R YGB R Oi Oi (a) I (b) G d d Y R (c) (d) O VO V (c) I VO O - Red Oi Oi (d) - Yellow R I - Green 40. Two coherent point sources S1 and S2 are Y - Blue separated by a small distance d as shown. G VB The fringes obtained on the screen will be 45. The anode voltage of a photocell is kept (2013) fixed. The wavelength l of the light Screen falling on the cathode is gradually d changed. The plate current I of photocell varies as follows (2013) S1 S2 II D (a) (b) (a) points (b) straight lines l l (c) semi-circle (d) concentric circles 41. A beam of unpolarised light of intensity I0 I I (c) (d) is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes ll an angle of 45° relative to that of A. The intensity of the emergent light is (2013) 46. A diode detector is used to detect an (a) I0 (b) I0 / 2 amplitude modulated wave of 60% (c) I0 / 4 (d) I0 / 8 modulation by using a condenser of 42. Diameter of a plano-convex lens is 6 cm capacity 250 pico farad in parallel with a load resistance 100 kW. Find the and thickness at the centre is 3 mm. If speed of light in material of lens is 2 ´ 108 maximum modulated frequency which could be detected by it. (2013) m/s, the focal length of the lens is (2013) (a) 10.62 MHz (b) 10.62 kHz (a) 15 cm (b) 20 cm (c) 5.31 MHz (d) 5.31 kHz (c) 30 cm (d) 10 cm 47. The magnetic field in a travelling 43. In a hydrogen like atom electron makes electromagnetic wave has a peak value of transition from an energy level with 20 nT. The peak value of electric field quantum number n to another with strength is (2013) quantum number (n - 1). If n >> 1, the (a) 3 V/m (b) 6 V/m (c) 9 V/m (d) 12 V/m
Answer with Explanations 1. (c) Speed of progressive wave is given by, v = w For first minima, k a sin q = 2l, [here, a = 10-6 m, q = 30°] 2 As electric field in air is, E1 = E01x cos èçæ 2 pnz - 2 pnt ø÷ö Þ l = 10-6 ´ sin 30° Þ l = 10-6 m c 2 \\ Speed in air = 2 pn = c Now, in case of interference caused by bringing èçæ 2 pn ÷øö second slit, b = lD c \\ Fringe width, 1 d Also, c= m 0er1 e0 …(i) [here, l = 10-6 m, b = 1 cm = 1 m, 2 100 In medium, 50 =? and D = 50 cm = 100 m] E2 = E02xcos (2 kz - kct ) d \\ Speed in medium = kc = c So, d = lD = 10-6 ´ 50 = 25 ´ 10-6 m 2k 2 b 1 2 ´ 1 ´ 100 Also, c = m 0er2 e0 …(ii) 100 2 or d = 25 mm As medium is non-magnetic medium, m medium = mair 4. (a) If wavelength of emitted photon in On dividing Eq. (i) by Eq. (ii), we have de-excitation is Ln; er2 er1 1 2= er1 Þ er2 = 4 n 2. (c) A CB < n I/2 I 1 I 8 ab Then, hc = En - Eg hc = pn2 - 2pmg2 , é = p2 ù Ln Ln 2m êQ E ú ë 2 m û Using Malus law, intensity available after C As energies are negative, we get = I ´ cos2 a hc = pg2 - pn2 2 Ln 2m 2m and intensity available after B = I cos 2 a ´ cos2b 2 2 Þ = pg2 çæ 1 - ççèæ pn ÷ø÷ö ö÷ = h2 ççèæ1 - l2g ÷öø÷ = I (given) 2m ç pg ÷ 2 ml2g l2n 8 è ø 1 So, I ´ cos2 a × cos2b = I Þ cos2 a × cos2b = 4 [Q p µ l-1, p = hl] 2 8 This is satisfied with a = 45° and b = 45° 2 ml2g c ççèæ1 l2g ö÷÷ø -1 h l2n So, angle between A and C is 45°. Þ Ln = - 3. (a) Angular width of diffraction pattern = 60° 2 ml2gc èçæç1 l2g ÷øö÷ h l2n Þ Ln = + [Q(1 - x)-n = 1 + nx] Þ Ln ~- A + B l2n 30° mcl2g where, A = é 2 ù and B = é2 mcl4g ù are ê ú ê ú êë h úû ëhû constants.
8 Optics & Modern Physics 5. (d) Series limit occurs in the transition n2 = ¥ to n1 = 1 - 1 mv12 + 1 mv 2 - v 2 + v 2 2 2 0 0 0 in Lyman series and n2 = ¥ to n1 = 5 in pfund series. For Lyman series, = 1 = 9 2 2 0 mv 2 v 0 n2=¥ = çèæ - 1 + 1ø÷ö = 8 = 0.88 = 0.89 9 9 hnL = Eg= E0 1 1 12 ¥ = 13.6 eV m 12m m v1 + 12m v2 …(i) v0 n1=1 hnL = 13.6 Similarly, for neutron-carbon atom collision; In Pfund series Momentum conservation gives; n2= ¥ v 0 = v1 + 12 v 2 and e = 1 Þ v0 = v2 - v1 11 hnp = E0 1 1 = 13.6 So, v1 = 13 v0 52 ¥ 52 æèç 121 1ö÷ø n2 = 5 \\Loss of energy = - 169 + = 0.28 hnp = 13.6 …(ii) So, Pd = 0.89 and Pc = 0.28 52 7. (c) Potential drop in a silicon diode in forward bias is From Eqs. (i) and (ii), we get around 0.7 V. 25hnp = hnL nL In given circuit, potential drop across 200 W resistor is 25 3 - 0.7 \\ np = I= DVnet = 200 R 6. (a) Neutron-Deuterium collision; Þ I = 0.0115A Þ I = 11.5 mA m 2m m 2m 8. (c) Only 10% of 10 GHz is utilised for transmission. v0 v = 0 v1 v2 \\Band available for transmission 10 Momentum conservation gives; = 100 ´ 10 ´ 109 Hz = 109 Hz mv0 = mv1 + 2 mv2 …(i) Now, if there are n channels each using 5 kHz then, Þ v 0 = v1 + 2 v 2 …(ii) n ´ 5 ´ 103 = 109 Þ n = 2 ´ 105 Collision given is elastic . 9. (d) So, coefficient of restitution, e = 1 15 cm \\ e = 1 = Velocity of separation Velocity of approach Þ 1= v 2 - v1 v0 - 0 Þ v 0 = v 2 - v1 On adding Eqs. (i) and (ii), we get 2 v0 = 3v2 Here, f1 = - 25 cm, f2 = 20 cm 2v0 For diverging lens, v = - 25 cm Þ 3 = v2 For converging lens, u = - (15 + 25) = - 40 cm So, from Eq. (i), we get 1 1 1 \\ v - - 40 = + 20 v1 = v0 - 24vv20 = v0 - 3 Þ 1 = 1 - 1 = 1 Þ v = 40 cm v 20 40 40 Þ v1 = - v0 3 10. (a) Let nth bright fringes coincides, then Fractional loss of energy of neutron yn1 = yn2 = æçèç -Kf + K i ø÷ö÷ for neutron Þ n1l1D = n2l2D Þ n1 = l1 = 520 = 4 Ki dd n2 l2 650 5
Previous Years’ Questions (2018-13) 9 Hence, distance from the central maxima is 15. (c) y = n1l1D = 4 ´ 650 ´ 10-9 ´ 1.5 = 7.8 mm vi vo d 0.5 ´ 10-3 t t 11. (c) We have, l = hc DE (Input) (Output) èçæ 4 E öø÷ l1 hc / DE1 DE2 3E - E =1 In a CE n-p-n transistor amplifier, output is 180° out of l2 hc / DE2 DE1 3 phase with input. \\ = = = 2E - lmin = hc 16. (a, c, d) From Snell's law, eV 12. (b) nsin q = constant log (lmin ) = log èçæ hc öø÷ - log V Þ y=c - x \\ n1 sin qi = n2 sin qf e Further, l will depend on n1 and n( z). But it will be independent of n2. 13. (a) Decay scheme is , N atoms of B 17. (a) d = (i1 + i2) - A Þ 40° = (35° + 79°) - A Þ A = 74° A A, B Now, we know that Let N atoms decays sinçæè A + dm ö÷ø 2 No into B in time t No – N at t=0 atoms of A m = sinèçæ A ÷øö 2 Given, NB = 0.3 = 3 Þ NB = 30 NA 10 NA 100 So, N0 = 100 + 30 = 130 atoms It we take the given deviation as the minimum By using N = N0e -lt deviation then, We have, 100 = 130e -lt sin èæç 74° + 40° öø÷ 2 1 m = = 1.51 Þ 1.3 = e - lt Þ log 1.3 = lt èæç 74° øö÷ sin 2 Þ log 1.3 = log 2 × t The given deviation may or may not be the minimum T deviation. Rather it will be less than this value. log (1.3) Therefore, m will be less than 1.51. \\ t = T × log 2 14. (a) For elastic collision, Hence, maximum possible value of refractive index is 1.51. pbefore collision = pafter collision . 18. (d) Telescope resolves and brings objects closer. mv = mv A + m Þ 2v = 2vA + vB K (i) 2 vB Hence, telescope with magnifying power of 20, the tree appears 20 times nearer. Now, coefficient of restitution, e = vB - vA 19. (c) uA - vB Here, uB = 0 (Particle at rest) and for elastic collision a e =1 q \\ 1 = vB - vA Þ v = vB - vA K (ii) q v From Eq. (i) and Eq. (ii) 4v 3 vA = v and v B = L 3 lA ççèæ h öø÷÷ VB 4/3 asin q » l Þ a çæè a öø÷ » l lB mVA 2 VA 2/3 L Hence, = = = =2 h Þ a = lL Spread = 2 a = 4 lL m . VB 2
10 Optics & Modern Physics 20. (a) According to the law of conservation of energy, i.e. where, qc = sin-1 æç 1 ö÷ è m ø Energy of a photon (hn) = Work function (f) + Kinetic 1 energy of the photoelectron æçè 2 mvm2 ax ÷öø \\ sin-1 æç sin q ö÷ > A - qc or sin q > sin( A - qc ) è m ø m According to Einstein’s photoelectric emission of light \\ q > sin-1[m sin( A - qc )] i.e. E = (KE)max + f é siníì æç 1 ÷ö üù As, hc = (KE)max + f or q > sin-1 êm î A - sin-1 è m ø ýþûú l ë If the wavelength of radiation is changed to 34l, then 25. (a) Retina Þ 4 hc = æèç 4 (KE )max + f öø÷ + f qy 3l 3 3 (KE)max. for fastest emitted electron = 1 mv ¢2 + f 25 cm 2 Þ 1 mv ¢2 = 4 æçè 1 mv 2 öø÷ + f Resolving angle of naked eye is given by 2 3 2 3 q = 1.22 l çæè 4 ö÷ø1/ 2 D 3 i.e. v¢ >v 1.22 ´ 500 ´ 10-9 0.25 ´ 2 ´ 10-2 N N N N \\ 25 y = 2 4 8 16 ´ 10-2 21. (d) A : Numbers left : N ® ® ® ® \\ y = 30 ´ 10-6 m = 30 mm N 15 \\ Number decayed, NA = N - 16 = 16 N 26. (c) B: Numbers left : N ® N ® N 2 4 3 m2 1 \\ Numbers decayed, NB = N - N = 4N m1 4 (15/16)N Ratio : NA = (3/ 4)N = 5 2 NB 4 m 2 >m1 22. (c) As, we know Cu is conductor, so increase in Dotted line is the normal. temperature, resistance will increase. Then, Si is According to Huygen’s principle, each point on semiconductor, so with increase in temperature, wavefront behaves as a point source of light. resistance will decrease. 23. (a) Theoretical question. Therefore, no solution is required. Ray 2 will travel faster than 1 as m 2 > m1. So, beam will bend upwards. 24. (d) q M A N 27. (b) fc = 2 MHz = 2000 kHz, fm = 5 kHz r1 r2 Resultant frequencies are, fc + fm, fc and fc - fm or, 2005 kHz, 2000 kHz and 1995 kHz Applying Snell’s law at M, 28. (a) Intensity at a distance r from a point source of sin q m = sin r1 power P is given by çæ sin q ö÷ sin q I = P ...(i) è m ø m 4pr 2 ...(ii) \\ r1 = sin-1 or sin r1 = 1 Also, = 2 e0E02c I Now, r2 = A - r1 = A - sin-1 æç sin q ÷ö è m ø where, E0 is amplitude of electric field and c the speed Ray of light would get transmitted form face AC if of light. Eqs. (i) and (ii) we get r2 < qc E0 = 2P = 2 ´ 9 ´ 109 ´ 0.1 sin 4pe0r 2c (1)2 ´ 3 ´ 108 or A - sin-1 çæ m q ö÷ < qc è ø = 6 = 2.45 V/m
Previous Years’ Questions (2018-13) 11 29. (d) The expressions of kinetic energy, potential energy Focal length in second liquid and total energy are 1 = èæçç m s - 1ö÷÷ø çæçè 1 - 1 ø÷÷ö me 4 f2 m l2 R1 R2 Kn = 8e20n2h2 Þ Kn µ 1 n2 1 ççèæ 3/2 1÷÷øö èæç 1 øö÷ Un = - me 4 Þ Un µ - 1 and Þ f2 = 5/3 - x 4e02n2h2 n2 Þ f2 is negative. - me 4 1 En = 8e20n2h2 Þ En µ- n2 33. (d) By law of Malus i.e. I = I0cos2 q In the transition from some excited state to ground Now, IA¢ = IA cos2 30° state value of n decreases, therefore kinetic energy IB¢ = IB cos2 60° increases, but potential and total energy decreases. As, IA¢ = IB¢ IA cos2 30° = IB cos2 60° 30. (b) No Solution is required 31. (d) For total internal Air Initially Finally IA IA reflection of light take Water IB place, following Polaroid conditions must be C obeyed. Green IB Polaroid (i) The ray must travel from denser to rarer medium. (ii) Angle of incidence (q) must be greater than or equal to Transmission axis Transmission axis critical angle(C ) 3 1 IA 1 i.e. C = sin-1 é m rarer ù Þ IA 4 = IB 4 Þ IB = 3 ëê m denser ûú Here, sin C = 1 and nwater = a+ b 34. (c) For hydrogen atom, we get nwater l2 1 = R Z 2 èæç 1 - 1 ø÷ö l 12 22 If frequency is less Þ l is greater and hence, RI n(water) is less and therefore, critical angle increases. So, they do Þ 1 = R(1)2 æçè 3 ÷øö Þ 1 = R(1)2 çèæ 3 ÷öø not suffer reflection and come out at angle less than 90°. l1 4 l2 4 32. (b) It is based on lens maker's formula and its Þ 1 = R(2 )2 èæç 3 ö÷ø Þ 1 = R(3)2 æçè 3 ÷øö l3 4 l4 4 magnification. i.e. 1 = (m - 1) èæçç 1 - 1 ø÷ö÷ 1 1 1 1 f R1 R2 l1 4l3 9l4 l2 Þ = = = According to lens maker’s formula, when the lens in the air. 35. (b) Thinking Process The problem is based on 1 = çèæ 3 - 1÷øö çæèç 1 - 1 ÷öø÷ frequency dependence of photoelectric emission. f 2 R1 R2 When incident light with certain frequency (greater 1 = 1 Þ f =2x than on the threshold frequency is focus on a metal f 2x surface) then some electrons are emitted from the ççèæ 1 1 1 ÷÷öø metal with substantial initial speed. x R1 R2 Here, = - When an electron moves in a circular path, then In case of liquid, where refractive index is 4 and 53, we get r = mv Þ r 2e 2B2 = m2v 2 3 eB 2 2 Focal length in first liquid KE max = (mv )2 Þ r 2e 2B2 = (KE)max 2m 2m 1 çèçæ m - 1ö÷ø÷ èçæç 1 1 ø÷ö÷ 1 çèçæ 3/2 1÷÷öø 1 f1 = m s R1 - R2 Þ f1 = 4/3 - x Work function of the metal (W), i.e. W = hn - KEmax l1 1.89 - f = r 2e 2B2 1 eV = r 2eB2 eV 2 m 2 2m Þ f1 is positive. 1 1 1 1 f1 = 8x = 4(2 x) = 4f Þ f1 = 4f [hv ® 1.89 eV, for the transition on from third to second orbit of H-atom]
12 Optics & Modern Physics = 100 ´ 10-6 ´ 1.6 ´ 10-19 ´ 9 ´ 10-8 R 2 = (3)2 + (R - 0.3)2 Þ R » 15 cm 2 ´ 9.1 ´ 10-31 Refractive index of material of lens m = c f = 1.89 - 1.6 ´ 9 = 1.89 - 0.79 = 1.1 eV v 2 ´ 9.1 Here, c = speed of light in vacuum = 3 ´ 108 m/s 36. (a) For forward bias, p-side must be a higher potential v = speed of light in material of lens = 2 ´ 108 m/s than n-side. = 3 ´ 108 = 3 2 ´ 108 2 +2 V –2 V So, (a) is forward biased. From lens maker’s formula 37. (c) Both the energy densities are equal i.e. energy is 1 = (m - 1) æççè 1 - 1 ÷öø÷ f R1 R2 equally divided between electric and magnetic fields. Here, R1 = R and R2 = ¥ (For plane surface) 38. (d) (a) Infrared rays are used to treat muscular strain. 1 3 1 f = èçæ 2 - 1øö÷ çæè 15 ÷øö (b) Radiowaves are used for broadcasting purposes. (c) X-rays are used to detect fracture of bones. Þ f = 30 cm (d) Ultraviolet rays are absorbed by ozone. 43. (d) DE = hn 39. (c) We know that, the Y n = DE = é 1 - 1ù h ëê(n - 1)2 angle of deviation k n2 ú û depends upon the angle d d k2 n 2k 1 of incidence. If we = n2(n - 1)2 » n3 µ n3 determine experimentally, dm the angles of deviation 44. (a) For same value of current higher value of voltage is corresponding to different X required for higher frequency. angles of incidence and i1 i i i2 then plot i (on-X-axis) and 45. (d) As l is increased, there will be a value of l above d (on-Y-axis), we get a curve as shown in figure which photoelectron will cease to come out. So, photocurrent will be zero. Clearly if angle of incidence is gradually increased, from a small value, the angle of deviation first 46. (b) D decreases, becomes minimum for a particular angle of incidence and then begins to increase. 40. (d) A fringe is a locus of points having constant path Signal CR difference from the two coherent sources S1 and S 2. It will be concentric circle. 41. (c) Relation between intensities is t = RC = 100 ´ 103 ´ 250 ´ 10-12 s = 2.5 ´ 10-5 s B I0 (I0/2) 45° The higher frequency which can be detected with (Unpolarised) IR tolerable distortion is 1 A f = 2 pmaRC IR = æèç I0 öø÷ cos 2 (45°) = I0 ´ 1 = I0 = 1 Hz 2 2 2 4 0.6 ´ 2.5 2p ´ ´ 10-5 42. (c) By Pythagoras theorem 100 ´ 104 4 ´ 104 = 25 ´1.2 p Hz = 1.2 p Hz = 10.61 kHz 3 cm 3 mm 47. (b) Peak value of electric field R E0 = B0c = 20 ´ 10-9 ´ 3 ´ 108 = 6V/m
JEE Advanced 1. Sunlight of intensity 1.3 kWm-2 is incident threshold frequency, so that the photoelectrons are emitted with negligible normally on a thin convex lens of focal kinetic energy. Assume that the length 20 cm. Ignore the energy loss of photoelectron emission efficiency is 100%. light due to the lens and assume that the A potential difference of 500 V is applied lens aperture size is much smaller than between the cathode and the anode. All its focal length. The average intensity of the emitted electrons are incident light, in kW m-2, at a distance 22 cm from normally on the anode and are absorbed. the lens on the other side is ............ . The anode experiences a force F = n ´ 10-4N due to the impact of the (Numerical Values, 2018) electrons. The value of n is ........... . 2. A wire is bent in (Take, mass of the electron, me = 9 ´ 10-31kg and eV = 1.6 ´ 10-19J) the shape of a (Numerical Value, 2018) right angled 45° triangle and is —2f f placed in front of a concave mirror 5. Consider a hydrogen-like ionised atom of focal length f with atomic number Z with a single as shown in the figure. Which of the figures shown in the electron. In the emission spectrum of this four options qualitatively represent(s) the atom, the photon emitted in the n = 2 to shape of the image of the bent wire? n = 1 transition has energy 74.8 eV higher (These figures are not to scale.) than the photon emitted in the n = 3 to (Single Correct Option, 2018) n = 2 transition. The ionisation energy of ¥ the hydrogen atom is 13.6 eV. The value of Z is ............ . (Numerical Value, 2018) (a) a a>45° (b) 6. For an isosceles prism of angle A and refractive index m , it is found that the (c) 0<a<45° (d) ¥ angle of minimum deviation dm = A. a Which of the following option(s) is/are correct? (More than One Correct Option, 2017) 3. In a radioactive decay chain, 232 Th (a) For the angle of incidence i1 = A, the ray 90 inside the prism is parallel to the base of the prism nucleus decays to 212 Pb nucleus. Let N a 82 and N b be the number of a and b- (b) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first particles respectively, emitted in this æç ÷ö è ø decay process. Which of the following refracting surface are related by r1 = ii 2 statements is (are) true? (c) For this prism, the emergent ray at the (More than One Correct Option, 2018) (a) Na = 5 (b) Na = 6 second surface will be tangential to the (c) Nb = 2 (d) Nb = 4 surface when the angle of incidence at the first surface is ù sin-1êésinA Aú 4. In a photoelectric experiment, a parallel i1 = ë 4 cos2 A - 1 - cos û beam of monochromatic light with power 2 of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. (d) For this prism, the refractive index m and the The frequency of light is just above the 1 cos-1çæè m ø÷ö angle prism A are related as A = 2 2
14 Optics & Modern Physics 7. A monochromatic light is travelling in a (b) A dark spot will be formed at the point P2 medium of refractive index n = 1.6. It (c) The total number of fringes produced enters a stack of glass layers from the between P1 and P2 in the first quadrant is bottom side at an angle q = 30°. The close to 3000 (d) At P2 the order of the fringe will be maximum interfaces of the glass layers are parallel 9. 131I is an isotope of Iodine that b decays to to each other. an isotope of Xenon with a half-life of 8 The refractive indices of different glass days. A small amount of a serum labelled with 131I is injected into the blood of a layers are monotonically decreasing as person. The activity of the amount of 131I injected was 2.4 ´ 105 Becquerel (Bq). It is nm = n - m Dn, where nm is the refractive index of the mth slab and Dn = 0.1 (see known that the injected serum will get distributed uniformly in the blood stream the figure). The ray is refracted out in less than half an hour. After 11.5 h, 2.5 ml of blood is drawn from the person’s parallel to the interface between the body, and gives an activity of 115 Bq. The total volume of blood in the person’s body, (m - 1) th and mth slabs from the right in litres is approximately (you may use e2 » 1 + x for|x| << 1 and ln 2 » 0.7). side of the stack. What is the value (Single Integer Type, 2017) of m ? (Single Integer Type, 2017) m n–mDn m–1 n–(m–1)Dn 3 n–3 Dn 10. A photoelectric material having 2 n–2 Dn 1 n– Dn work-function lf0ççæèils illuminated with light of wavelength ÷÷öø. n < hc The fastest q f0 8. Two coherent monochromatic point photoelectron has a de-Broglie sources S1 and S2 of wavelength l = 600 nm are placed symmetrically on either side of wavelength ld . A change in wavelength of the centre of the circle as shown. The the incident light by Dl results in a sources are separated by a distance Dld d = 1.8 mm. This arrangement produces change Dld in ld . Then, the ratio Dl is interference fringes visible as alternate bright and dark spots on the proportional to (Single Correct Option, 2017) circumference of the circle. The angular separation between two consecutive (a) l2d (b) ld (c) l3d (d) l3d bright spots is Dq. Which of the following l2 l l l2 option(s) is/are correct? 11. A parallel beam of P (More than One Correct Option, 2017) q light is incident P1 P1 a from air at an angle Dq Dq a on the side PQ of Q n=Ö2 R a right angled S1 d S2 P2 S1 d S2 P2 triangular prism of refractive index n = 2. Light undergoes total internal reflection in the prism at the face PR when a has a minimum value of 45°. The angle q of the (a) The angular separation between two prism is (Single Correct Option, 2016) consecutive bright spots decreases as we move from P1 to P2 along the first quadrant (a) 15° (b) 22.5° (c) 30° (d) 45°
Previous Years’ Questions (2018-13) 15 12. A plano-convex lens is made of material of mirror is at an angle q = 30° to the axis of the lens, as shown in the figure. refractive index n. When a small object is placed 30 cm away in front of the curved (Single Correct Option, 2016) surface of the lens, an image of double the f=30 cm size of the object is produced. Due to (– 50, 0) q X reflection from the convex surface of the (0, 0) R=100 cm lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is (are) true? (More than One Correct Option, 2016) 50 cm (a) The refractive index of the lens is 2.5 (50 + 50Ö3, –50) (b) The radius of curvature of the convex surface If the origin of the coordinate system is is 45 cm taken to be at the centre of the lens, the (c) The faint image is erect and real coordinates (in cm) of the point (x, y) at (d) The focal length of the lens is 20 cm which the image is formed are 13. A transparent slab of thickness d has a (a) (125/ 3, 25/ 3) (b) (50 - 25 3, 25) refractive index n (z) that increases with (c) (0, 0) (d) (25, 25 3) z. Here, z is the vertical distance inside 15. While conducting the Young’s double slit the slab, measured from the top. The slab experiment, a student replaced the two is placed between two media with uniform slits with a large opaque plate in the x-y refractive indices n1 and n2(> n1), as shown in the figure. A ray of light is plane containing two small holes that act incident with angle qi from medium 1 and emerges in medium 2 with refraction as two coherent point sources (S1, S2) angle q f with a lateral displacement l. emitting light of wavelength 600 mm. The n1=constant qi 1 student mistakenly placed the screen n (z) d parallel to the XZ plane (for z > 0) at a distance D = 3 m from the mid-point of z S1S2, as shown schematically in the n2=constant l qf 2 figure. The distance between the source d = 0.6003 mm . The origin O is at the intersection of the screen and the line joining S1S2. (More than One Correct Option, 2016) Which of the following statement(s) is (are) Z Screen true? (More than One Correct Option, 2016) (a) l is independent on n (z) O y (b) n1 sinqi = (n2 - n1) sinqf x (c) n1 sinqi = n2 sinqf S1 S2 (d) l is independent of n2 d D 14. A small object is placed 50 cm to the left Which of the following is (are) true of the of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of intensity pattern on the screen? curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The (a) Semi circular bright and dark bands mirror is tilted such that the axis of the centered at point O (b) The region very close to the point O will be dark (c) Straight bright and dark bands parallel to the x-axis (d) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
16 Optics & Modern Physics 16. In a historical experiment to determine the radiation was 64 times more than the Planck's constant, a metal surface was permissible level required for safe irradiated with light of different operation of the laboratory. What is the wavelengths. The emitted photoelectron minimum number of days after which the energies were measured by applying a laboratory can be considered safe for use? stopping potential. The relevant data for (Single Correct Option, 2016) the wavelength (l) of incident light and the corresponding stopping potential (V0) (a) 64 (b) 90 are given below: (Single Correct Option, 2016) (c) 108 (d) 120 l(mm ) 20. The electrostatic energy of Z protons 0.3 V0(Volt) uniformly distributed throughout a 0.4 2 1 spherical nucleus of radius R is given by E = 3 Z (Z - 1)e2 The measured masses of 5 4pe0R Given that c = 3´ 108ms-1 and e =1.6´ 10-19 C, the neutron, 11H, 15 N and 15 O are 7 8 Planck's constant (in units of J-s) found 1.008665 u, 1.007825 u, 15.000109 u and from such an experiment is) 15.003065 u, respectively. Given that the (a) 6.0 ´10-34 (b) 6.4 ´10-34 radii of both the 15 N and 15 O nuclei are 7 8 (c) 6.6 ´10-34 (d) 6.8 ´10-34 17. Highly excited states for hydrogen-like same, 1 u = 931.5 MeV/c2 (c is the speed of atoms (also called Rydberg states) with light) and e2/(4pe0) = 1.44 MeV fm. nuclear charge Ze are defined by their Assuming that the difference between principle quantum number n, where the binding energies of 15 N and 15 O is n >>1. Which of the following statement(s) 7 8 is (are) true? purely due to the electrostatic energy, (More than One Correct Option, 2016) the radius of either of the nuclei is (1fm = 10- 15 m) (Single Correct Option, 2016) (a) Relative change in the radii of two consecutive orbitals does not depend on Z (a) 2.85 fm (b) 3.03 fm (c) 3.42 fm (d) 3.80 fm (b) Relative change in the radii of two consecutive orbitals varies as 1/ n 21. A Young’s double slit interference arrangement with slits S1 and S2 is (c) Relative change in the energy of two immersed in water (refractive index = 4/ 3) consecutive orbitals varies as 1/ n3 as shown in the figure. The positions of maxima on the surface of water are given (d) Relative change in the angular momenta of by x2 = p2m2l2 - d 2, where l is the two consecutive orbitals varies as 1/n wavelength of light in air (refractive 18. A hydrogen atom in its ground state is index = 1 ), 2d is the separation between irradiated by light of wavelength 970Å. the slits and m is an integer. The value of Taking hc/ e =1.237´ 10-6 eVm and the p is (Single Integer Type, 2015) ground state energy of hydrogen atom as S1 - 13.6 eV, the number of lines present in dx the emission spectrum is d (Single Integer Type, 2016) S2 19. An accident in a nuclear laboratory Air resulted in deposition of a certain amount Water of radioactive material of half-life 18 days inside the laboratory. Tests revealed that
Previous Years’ Questions (2018-13) 17 22. Consider a concave mirror and a convex Passage (Q. Nos. 25-26) lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of Light guidance in an optical fibre can be 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a understood by considering a structure distance of 15 cm from the mirror. Its erect image formed by this combination comprising of thin solid glass cylinder of has magnification M1.When the set-up is kept in a medium of refractive index 7, refractive index n1 surrounded by a medium of 6 lower refractive index n2. The light guidance in the magnification becomes M2. The the structure takes place due to successive total magnitude½½M 2½½ is ½M1½ (Single Integer Type, 2015) internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im. (Passage Type, 2015) 25. For two structures namely S1 with n1 = 45 and n2 = 3, and S2 with n1 = 8 4 2 5 15 cm and n2 = 7 and taking the refractive index 50 cm 5 23. Two identical glass rods S1 and S2 of water to be 4 and that to air to be 1, (refractive index = 1.5) have one convex 3 end of radius of curvature 10 cm. They are placed with the curved surfaces at a the correct option(s) is/are n1>n2 distance d as shown in the figure, with their axes (shown by the dashed line) Air q Cladding n2 aligned. When a point source of light P is i Core placed inside rod S1 on its axis at a n1 distance of 50 cm from the curved face, the light rays emanating from it are found (a) NA of S1 immersed in water is the same as to be parallel to the axis inside S2. The that of S2 immersed in a liquid of refractive distance d is (Single Correct Option, 2015) index 16 3 15 (b) NA165ofisS1thime msaemrseedasinthliaqtuoidf So2f refractive index immersed in S1 P S2 50 cm d water (a) 60 cm (b) 70 cm (c) NA of S1 placed in air irseftrhaecstiavemiendaesxtha14t 5S2 (c) 80 cm (d) 90 cm immersed in liquid of 24. A monochromatic (d) NA of S1 placed in air is the same as that of S2 placed in water beam of light is incident at 60° on 60° q 26. If two structures of same cross-sectional one face of an area, but different numerical apertures NA1 and NA2(NA2 < NA1) are joined equilateral prism longitudinally, the numerical aperture of of refractive index n and emerges from the opposite face the combined structure is making an angle q with the normal (see figure). For n = 3 the value of q is 60° (a) NA1NA2 (b) NA1 + NA2 NA1 + NA2 and dq = m. The value of m is dn (Single Integer Type, 2015) (c) NA1 (d) NA2
18 Optics & Modern Physics 27. A nuclear power plant supplying 31. A fission reaction is given by 92326U ® 51440Xe + 3984Sr + x + y, electrical power to a village uses a radioactive material of half life T years as where x and y are two particles. the fuel. The amount of fuel at the Considering 236 U to be at rest, the kinetic 92 beginning is such that the total power energies of the products are denoted by requirement of the village is 12.5% of the K Xe, KSr, K x (2 MeV) and K y(2 MeV), electrical power available from the plant respectively. Let the binding energies per at that time. If the plant is able to meet nucleon of 92236U, 140 Xe and 3984Sr be 7.5 the total power needs of the village for a 54 maximum period of nT years, then the MeV, 8.5 MeV and 8.5 MeV, respectively. value of n is (Single Integer Type, 2015) Considering different conservation laws, the correct options is/are 28. Match the nuclear processes given in (Single Correct Option, 2015) Column I with the appropriate option(s) in Column II. (a) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV (b) x = p, y = e-, KSr = 129 MeV, KXe = 86 MeV Column I Column II (c) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV A. Nuclear fusion (d) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV P. absorption of thermal B. Fission in a 235 nuclear reactor neutrons by 92 U 32. A light source, which emits two wavelengths l1 = 400 nm and l2 = 600 nm, C. b-decay Q. 60 Co nucleus is used in a Young’s double-slit 27 D. g-ray emission experiment. If recorded fringe widths for R. Energy production in l1 and l2 are b1 and b2 and the number of stars via hydrogen fringes for them within a distance y on conversion to helium one side of the central maximum are m1 S. Heavy water and m2, respectively, then T. Neutrino emission (More than One Correct Option, 2014) (Matching Type, 2015) (a) b2 >b1 (b) m1 >m2 29. For a radioactive material, its activity A (c) from the central maximum, 3rd maximum of and rate of change of its activity R are l2 overlaps with 5th minimum of l1 defined as A = - dN and R = - dA, where (d) the angular separation of fringes of l1 is dt dt greater than l2 N (t) is the number of nuclei at time t. Two radioactive source P(mean life t) and Q (mean life 2t) have the same activity at 33. A transparent thin film of uniform thickness and refractive index n1 = 1.4 is t = 0. Their rate of change of activities at coated on the convex spherical surface of t = 2t are RP and RQ, respectively. If radius R at one end of a long solid glass RP = n , then the value of n is cylinder of refractive index n2 = 1.5, as RQ e shown in the figure. (Single Integer Type, 2015) 30. An electron in an excited state of Li2+ ion n1 has angular momentum 3h . The de 2p Broglie wavelength of the electron in this Air n2 state is ppa0 (where a0 is the Bohr radius). The value of p is (Single Integer Type, 2015) Rays of light parallel to the axis of the cylinder traversing through the film from
Previous Years’ Questions (2018-13) 19 air to glass get focused at distance f1 from 36. If lCu is the wavelength of K a , X-ray line the film, while rays of light traversing of copper (atomic number 29) and lMO is from glass to air get focused at distance f2 the wavelength of the K a , X-ray line of from the film. Then molybdenum (atomic number 42), then the ratio lCu/ lMo is close to (More than One Correct Option, 2014) (Single Correct Option, 2014) (a) | f1 | = 3R (b) | f1 | = 2.8R (c) | f2 | = 2R (d) | f2 | = 1.4R (a) 1.99 (b) 2.14 (c) 0.50 (d) 0.48 34. Four combinations of two thin lenses are 37. The image of an object, formed by a given in Column I.The radius of curvature plano-convex lens at a distance of 8 m of all curved surfaces is r and the behind the lens, is real and is one-third refractive index of all the lenses is 1.5. the size of the object. The wavelength of Match lens combinations in Column I light inside the lens is 2 times the with their focal length in Column II and 3 select the correct answer using the code wavelength in free space. The radius of given below the columns. the curved surface of the lens is Column I Column II (Single Correct Option, 2013) (a) 1 m (b) 2 m (c) 3 m (d) 6 m P. 1. 2r 38. A ray of light travelling in the direction 1 (i + 3j) is incident on a plane mirror. 2 Q. 2. r/2 After reflection, it travels along the direction 1 (i - 3j). The angle of 2 incidence is (Single Correct Option, 2013) R. 3. -r (a) 30°a (b) 45° (c) 60° (d) 75° 39. In the Young’s double slit experiment using a monochromatic light of wavelength l the path difference (in S. 4. r terms of an integer n) corresponding to any point having half the peak intensity is (Single Correct Option, 2013) Codes (Matching Type, 2014) (a) (2n + 1) l (b) (2n + 1) l PQ 2 4 RS PQ RS (a) 1 2 34 (b) 2 4 3 1 (c) (2n + 1) l (d) (2n + 1) l (c) 4 1 23 (d) 2 1 3 4 8 16 35. A metal surface is illuminated by light of 40. A right angled prism of refractive index m1 is placed in a rectangular block of two different wavelengths 248 nm and refractive index m 2, which is surrounded by a medium of refractive index m3 , as 310 nm. The maximum speeds of the shown in the figure, A ray of light ‘e’ photoelectrons corresponding to these enters the rectangular block at normal wavelengths are u1 and u2, respectively. If incidence. Depending upon the the ratio u1 : u2 = 2 : 1 and hc = 1240 eV relationships between m1, m 2 and m3, it nm, the work function of the metal is takes one of the four possible paths ‘ef’, nearly (Single Correct Option, 2014) ‘eg’, ‘eh’ or ‘ei’. (a) 3.7 eV (b) 3.2 eV (c) 2.8 eV (d) 2.5 eV
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