Title 24. Atomic Physics -3 Easy Ans.pdf ✅

1. (d) 2. (b) . 2 1 1 / 2 0 T t N N       = Hence fraction of atoms decayed = 8 7 2 1 1 2 1 1 1 60 3 60 0 1 / 2  =       = −      − = −  T t N N In percentage it is 100 87.5% 8 7  = 3. (a) C-14 is carbon dating substance. 4. (b) t T N N / 0 2 1       = 2 / T 4 2 / T 2 1 2 1 2 1 16 1        =              =        T = 0.5 hour = 30 minutes. 5. (c) N n N dt dN = −  = − (Given n dt dN = )  = −  N n  Half life sec 0.693 0.693 0.693 n N = = =   6. (a) 231 91 231 90 235 92 0 1 X X Y ⎯⎯ → ⎯⎯− e⎯→ 7. (d) 0 1 13 6 13 7 N C e → + + 8. (c) 1 / 2 2 1 0 T t A A       =  1 / 2 8 2 1 100 1600 T       =  T 2sec 1 / 2 = Again at t = 6 sec, 200 2 1 1600 2 6  =      A = counts/sec 9. (d) 4 2 234 92 238 92U → Th + He 10. (b) 11. (d) 12. (d) 13. (b) By using t N N e − = 0  1 / 2 0 0 2 T N e N − =  1 / 2 2 T e  = By taking loge both the side 2 1 / 2 loge = T  T1 / 2 = 0.693 14. (a) Number of half lives in 20 min = n 4 5 20 = = Fraction of material remains after four half lives 16 1 = Hence fraction that decays 93.75% 16 15 16 1 = 1 − = = 15. (d) In the given case, 12 days = 3 half lives Number of atoms left after 3 half lives. 10 3 10 0.8 10 2 1 = 6.4 10  =  16. (a) Decay constant remains unchanged in a chemical reaction. 17. (d) 8 4 238 206 4 ' = − = − = A A n 18. (b) 4 2 4 2 − − ⎯⎯→ − ⎯ ⎯→ A Z A Z A Z X Y X   19. (a) Both the -rays and the cathode rays are made up of electrons. -rays are EM waves, -particles are doubly ionized helium atoms and protons and neutrons have approximately the same mass. 20. (b) ; 14 6 4 2 4 2 22 10 Ne → He + He + X hence X is carbon. 21. (c) For 80 minutes, number of half lives of sample 4 20 80 A = nA = = and number of half lives of sample 2. 40 80 B = nB = = Also by using n N N       = 2 1 0  n N 2 1   4 1 2 2 2 2 4 2 = = = A B n n B A N N 22. (d) 4 1 4 2 − − − − ⎯⎯→ − ⎯ ⎯→ m n m n m n X X X   23. (c) Half-life T 6476 years 1.07 10 0.693 0.693 1 / 2 4 =  = = −  24. (d) Number of nuclei decreases exponentially t N N e − = 0 and Rate of decay N dt dN  =       − Therefore, decay process losts up to t = . Therefore, a given nucleus may decay at any time after t = 0. 25. (a) To becomes 4 1 th, it requires time of two half lives i.e., t = 2(T1 / 2 ) = 25800 = 258 centuries 26. (a) Carbon dating 27. (d) 18 8 1 1 4 2 15 7 X + He → P + Y
28. (c) 29. (d) 234 91 234 90 238 92U ⎯⎯→ X ⎯ ⎯→ Y −   30. (d) 31. (c) After three half lives (i.e., 30 days) it remains , 8 1 2 1 3  =      so it will remain th 10 1 , approximately in 33 days. 32. (a) 234 91 234 90 238 92U ⎯⎯→ Th ⎯⎯→ Pa   234 92 ( ) 1 U o ⎯⎯E −⎯⎯→ 33. (d) By using 1 / 2 2 1 0 T t A A       =  8 1 2 1 9 / 3 0  =      = A A 34. (d) 115 50 115 ) 48 2( 1 Cd Sn o ⎯⎯−⎯⎯→  35. (b) In two half lives, the activity becomes one fourth. 36. (a)  decay decreases the mass number by 4 and atomic number by 2,  decay increases the atomic number by 1. Here atomic number of C is same as that of A. 37. (a) Number of half lives in two days four substance 1 and 2 respectively are n1 = 4 12 2 24 =  and 3 1.6 2 24 2 =  n = By using n N N       = 2 1 0  2 1 2 1 2 1 ( ) ( ) 0 2 0 1 2 1 n n N N N N             =  1 1 2 1 2 1 1 2 3 4 =             =  38. (b) 39. (c) 0.3 2.3 0.693 0.693 1 / 2 = = = T  40. (d) Number of  − particles emitted 4 4 238 222 = − = This decreases atomic number to 90 − 4 2 = 82 Since atomic number of 222 83Y is 83, this is possible if one  − particle is emitted. 41. (d) Number of half lives in 150 years 2 75 150 n = = Fraction of the atom of decayed n       = − 2 1 1 0.75 4 3 2 1 1 2  = =      = −  Percentage decay = 75% 42. (b) 5 0 975 9750 − −  =  =   A A e e t 10 5  =  e  5 = loge 10 = 2.3026 log1 0 10 = 2.3026   =  43. (a) Mass number decreases by 8 4 = 32 Atomic number decreases by 8  2 − 5 = 11 44. (d) / 3 6 5 / 0 2 1 5 10 64 10 2 1 1 / 2 t T t A A          =       = − − 21 2 1 128 1 / 3   =       = t t days 45. (b) 46. (c) Decayed fraction 4 3 = , so undecayed fraction 4 1 = Now 2 2 1 4 1 2 1 0   =        =      = n N N n n  t n T 2 3.8 7.6 days =  1 / 2 =  = 47. (c) 4 2 198 82 202 decay 84 X ⎯⎯⎯⎯→ Y + He  − and 0 1 198 83 198 decay 82   − − Y ⎯⎯⎯⎯→ Z + 48. (a) 3 2 1 8 1 2 1 0   =        =      = n N N n n Now t n T 3 3.8 11.4 days =  1 / 2 =  = 49. (d) 3; 24000 72000 n = = Now 8 1 2 1 0  =      = n N N 50. (d) 51. (a) 52. (b) 6 4 232 208 4 ' = − = − = A A n
n = 2n − Z + Z' = 2  6 − 90 + 82 = 4 53. (c) 54. (a) Number of half lives 5 1 5 n = = Now 32 1 2 1 2 1 5 0 0  =        =      = N N N N n 55. (d) 56. (b) Number of half lives 2, 5 10 n = = now 4 1 2 1 2 0  =      = N N Fraction decayed 4 3 4 1 1 1 0 = − = − = N N  In percentage 100 75% 4 3 =  = 57. (b) Number of half lives 5; 3.8 19 n = = Now n N N       = 2 1 0 5 5 2 1 10.38 2 1 10.38         =        = N N = 0.32 gm 58. (b) 5 0 2 1 2 1        =      = n N N 59. (d)   log 2 2.303 log10 2 1 / 2 = = e T 60. (a) X Y He Z B A Z A Z A Z 0 1 4 3 4 2 4 → 2 + → + − − − − 61. (d) 4 1 2 1 , 2 2 1 2 0 0  =       =  =      = N N n N N n So disintegrated part 4 3 4 1 1 1 0 = − = − = N N 62. (c) Number of half lives 4 2.5 10 n = =  A curie N N A A n 0.1 2 1 1.6 2 1 4 0 0  =        =       = = 63. (b) By using t N N e − = 0 and   1 t = = Substance remains 3 0.37 ~– 0 0 0 N N e N = N = =  Substance disintegrated 3 2 3 0 0 0 N N = N − = 64. (c) th 4 3 active decay takes place in time 2 ( ) T1 / 2 t =  2( ) 4 3 = T1 / 2  sec 8 3 T1 / 2 = 65. (c) By using t N N e − = 0 and average life time  1 t = So e e N N N N e N e 1 1 0 1 0 1 / = 0 =  = = −  − − Now disintegrated fraction e e N e N 1 1 1 1 0 − = − = − = 66. (d) Complete reaction will be as follows 0 1 4 2 207 82 235 92 X Y 7 He 4 e → + + − i.e., seven  − particles and four  − particles will be emitted. 67. (d) 7 4 235 207 4 ' = − = − = A A n n = (2n − Z + Z') = (2 7 − 92 + 82) = 4 68. (c) During -decay, a neutron is transformed into a proton and an electron. 69. (b) 4 2 − ⎯⎯→ − A Z A Z X X  70. (a) 71. (a) 72. (d) After emitting -particle (–1e 0 ) mass of nucleus doesn't change. 73. (a) n N N       = 2 1 0  n        =      = 2 1 2 1 16 1 4  n = 4. Also T1 / 2 t n =  T 10 days 4 40 1 / 2 = = 74. (c) As the  − particle has no charge and mass. 75. (d) With emission of an  particle ( ) 4 2 He mass number decreases by 4 unit and atomic number decrease by 2 units and with emission of 1 2 −  particle atomic number increases by 2 units. So Z will remain same and N will become N − 4. 76. (a) n n N N N N         =      = 2 1 2 1 0 0 2 100 2 1 100 1   =       = n n n comes out in between 6 and 7.
77. (d) 1 / 2 / 0 0 2 1 2 1 n t T N N N N         =      = 8 1 2 1 2 1 1 3 2.7 8.1  =       =       N =  N mg 0.125 mg 8 1  = = 78. (d) T T t N N N N 16 0 0 0 2 1 2 4 1         =      =  T = 8 days. 79. (d) 234 91 0 1 234 90 4 2 238 92U → He + X →− e + U Hence, A = 234, Z = 91 80. (c) Mean life  1 = 6.67 10 sec. 8 =  81. (d) N dt T dN N dt dN = −  =  1 / 2 0.693  4 10 2.3 10 /sec 1.2 10 0.693 1 5 8 7   =  atoms  = 82. (c) Remaining material t T N N / 0 2 = N 3.96 gm 2.15 10 (2) 10 20 / 15  = = = So decayed material = 10 − 3.96 = 6.04 gm 83. (b) Number of atoms remains undecayed t N N e − = 0 Number of atoms decayed = (1 ) 0 t N e − −       = −         = − −  e N e N 1 1 0 1 1 0   = 0.63 N0 = 63% of N0 . 84. (d) 85. (b) 4 2 1 16 1 2 1 0   =       =             = n N N n n also T days T t n 30 4 120 1 / 2 1 / 2 =  = = 86. (c) / / 10 0 t T1 / 2 t N N e e − − =   =  t years e t 20 1 2 1 / 10 2  =  =       87. (a) 2 8 1 2 1 0 5 1 5 0 0 1 / 2 N N N N T t  =       =      = 88. (a) 89. (b) 1 / 2 1 / 2 6 0 0 2 1 64 1 2 1 T T t A A         =      = 10 sec 2 1 2 1 1 / 2 6 0 6 1 / 2   =       =       T T 90. (a, c) 91. (b) A Z A Z U X   = ( =92) ( 238 ) ⎯⎯(8⎯, 6⎯)→ so A = A − 4n  = 238 – 4  8 = 206 and Z = n − n + z  2  = 6 – 2  8 + 92 = 82. 92. (c) ; / 0 0 t t  A A e A e − − = = where  = mean life So t T A A e / 1 0 − 1 =  t T t T A e e A A / / 1 1 0 1 1 = = − t T t T t T t t T A A e A e e A A e ( )/ 2 1 / / 1 / 2 0 1 2 1 2 ( ) − − −  = =  = 93. (c) 94. (d) 4 4 228 212 = − n = and n = 2  4 − 90 + 83 = 1 95. (c) In a gamma decay process. There is no change in either A or Z. 96. (a) The radioactivity of a sample decays to th 16 1 of its initial value in four half lives. 97. (d) / / 48 0 2 1 16 1 2 1 t T t N N         =      = 192 . 2 1 2 1 4 / 48 t hour t   =       =       98. (c) If  is the decay constant of a radioactive substance than average life =  1 Also half life 0.693 (Average life) 0.693 = =   in single average life, more than 63% of radioactive nuclei decay 99. (b) 100.(d) ; 0 t M M e − = given       =  1 t 2 M gm e M e 1.35 1 10 10 2 2   =       = =       −  