Title 24. Atomic Physics - Medium-2 aNS.pdf ✅

1. (c)Weight of positron is equal to weigth of electron. 2. (b) 3. (b)m = 4 3 r 3d r  (m)1/3 1 2 r r = 1/ 3 135 5       = (27)1/3 = 3. 4. (a)If R is the radius of the nucleus, the corresponding volume 4 3 R3 has been found to be proportional to A. This relationship is expressed in inverse from as R = R0 A1/3 The value of R0 is 1.2 × 10–15 m, e.e., 1.2 fm Therefore, AI Te R R = 1/ 3 0 AI 1/ 3 0 Te R (A ) R (A ) AI Te R R = 1/ 3 AI 1/ 3 Te (A ) (A ) = 1/ 3 1/ 3 (27) (125) = 3 5 or RTe = 5 3 × RAI = 5 3 × 3.6 = 6 fm 5. (d)Density of nuclear matter is independent of mass number, so the required ratio is 1 : 1, Alternative : A1 : A2 = 1 : 3 Their radi will be in the ratio R0 A1 1/3 : R0 A2 1/3 = 1 : 31/3 Density = 3 A 4 R 3   A1 : A2 = 3 3 3 1/ 3 3 0 0 1 3 : 4 4 R .1 R (3 ) 3 3   = 1 : 1 Their nuclear densities will be the same. 6. (d)Law of conservation of momentum gives m1 v1 = m2 v2  1 2 m m = 2 1 v v But m = 4 3 r 3 or m  r 3  1 2 m m = 3 1 3 2 r r = 2 1 v v  1 2 r r = 1/ 3 1 2        r 1 : r2 = 1 : 21/3 7. (d) 8. (b)  = p p 3 1/ 3 3 0 Am Am 4 4 R (R A ) 3 3 =   = 27 p 3 15 3 0 3m 3 1.67 10 4 R 4 3.14 (1.1 10 ) − −   =     = 3 × 1017 kg/m3 9. (c) 10. (c) Nucleus is stable but nuetrons and protons cannot be stable when seperated. So binding energy of nucleus is greater. So mass of nucleus is smaller. 11. (d) the binding energy per nucleon in a nucleus varies in a way that depends on the actual value of A. 12. (d) (a), (b) & (c) are correct descrition of binding energy of a nucleus. 13. (a)Q = (2BEHe – BELi) = (2 × 7.06 × 4 – 5.60 × 7) Mev = 17.28 Mev. 14. (c)Energy released = 2n Q E – 2 n P E = y – 2x = –(2x – y) 15. (c) 16. (c) 17. (a) 18. (b)Binding energy BE = ( M nucleus – M nucleous ) c2 = ( M0 – 8Mp – 9 Mn ) c2 19. (b) 20. (c) 21. (b) 22. (c)EP = (8 × 7.06 – 7 × 5.60) MeV = 17.28 MeV 23. (b) Nuclear binding energy = (mass of nucleus – mass of nucleons) C2 = (Mo – 8MP – 9MN )C2
24. (c) It is order of MeV 25. (c)As a proton is lighter than a neutron, proton can not be converted into neutron without providing energy from outside. Reverse is possible. The weak interaction force is responsible in both the processes (i) conversion of p to n and (ii) conversion of n to p. 26. (b)We know that whenever there is fusion or fission or nucleoids and nuclei. Some mass is lest (mass defect) which converts into energy. So, net mass of products is slightly less than that of initial substances. 27. (c)Process in which resultant nuclei with greater BEPN will release energy. so. R → 2s will consume energy P→ Q + S will consume energy P → 2R will release energy Q → R + S will consume energy 28. (a)Energy of each - ray photon = E = mc2 = 0.0016 × 931.5 MeV = 1.5 MeV 29. (a) 30. (a)1 amu = 931 Me V 31. (c)Mass energy equivalence relation E = mc2 was given by Einstein. 32. (b)In the case of formation of a nucleus the evolution of energy to the binding energy of the nucleus takes place due to disappearance of a fraction of the total mass, If the quantity of mass disappearing is M, then the binding energy is BE = MC2 From the above discussion, it is clear that the mass of the nucleuses must be less than the sum of the masses of the consituent neutrons and protons. We can then write. M = ZMp + NMn – M (A, Z) Where M (A, Z) is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is BE = [ZMP + NMn – M (A, Z)] C2 BE = [ZMP + (A – Z) Mn – M (A,Z)]C2 Where N = A – Z number of neutrons, 33. (a) In a nuclear reaction, atomic mass and charge number remain conserved, For a nuclear reaction to be completed, the mass number and charge number on both sides should be same. If we complete the equation by choice (a), then the complete reaction is Total atomic number on on LHS = 92 + 0 = 92 92U235 + o n 1 → 38Sr90 + 54Xe143 + 30 n 1 Total atomic number on RHS = 38 + 54 + 0 = 92 Total atomic number on RHS = 235 + 1 = 236 Total atomic number on RHS = 90 + 143 + 3 × 1 = 236 Thus, choice (a) is correct, 34. (d)If binding energy of product nuclei is greater then energy is relea 35. (b) 4 2He + 14 7 N → 17 8 O + 1 1H 36. (b)In  – emission, An antineutrino is produced n → p + e– +  37. (a) When a  − -particle is emitted from a nucleus, no. of proton increases and number of neutron decreases. Hence the neutron-proton ratio is decreased 38. (d) 39. (a) 40. (d) 41. (a) (a) beta rays are electron beam i.e., cathode rays (b) gamma rays are e.m. wave not neutrons. (c) alpha particle are doubly ionized helium atoms (d) neutrons are slightly heavier than protons so (b), (c) and (d) are wrong options 42. (d) 43. (b)According to question reaction may be expressed as 2He4 + 7N14 ⎯⎯→ 8O17 + 1X1 (proton) So, particle X is proton (1H1 ) 44. (d)When -particle is emitted, mass number decreases by 4 units and atomic number by 2 units. When -particle is is emitted, mass number remains same while atomic number increases by 1 unit. Let a – -particle and b – -particles be emitted. Then from conservation of mass unmber
234 = 222 + 4a  a = 234 – 222 4 = 3 From conservation of atomic number (charge) 92 = 87 + 2a – b × 1 or 92 – 87 = 2 × 3 – b  b = 6 – 5 = 1  a = 3, b = 1 45. (a) 46. (c)By conservation of linear momentum 0 = 234 v + 4 u v = 4u 234 −  speed v = 4u 234 47. (b)Since, 8 -particles and 2- particles are emitted so, new atomic number Z = Z – 8 × 2 + 2 × 1 = 92 –16 + 2 = 78 48. (a) 49. (b)Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in  - decay do not exist inside the nucleus. They are only created at the time of emission, They are only created at the time of emission Just as photons are created when an atom makes a transition from higher to a lower energy state. In negative  decay a neutron in the nucleus is transformed into a proton, an antineutrino Hence in radioactive decay process the negatively charged, emitted b - particles are the electrons produced as a result of the decay of neutrons present inside the nucleus. 50. (d) 51. (a) 52. (b) A 1 7 4 Z 0 3 2 X n Li He + → + It implies that A + 1 = 7 + 4  A = 10 and Z + 0 = 3 + 2  Z = 5 Thus, it is Boron 10 5 B 53. (d)Gamma-photon. 54. (d) 55. (b)Gamma ray is electromagnetic radiation which does not involve any change in proton number or neutron number 56. (c)For ZXA , Z = 0 + 5 – 2 = 3 and A = 1 + 10 – 4 = 7 ZXA , Z = 0 + 5 – 2 = 3 A = 1 + 10 – 4 = 7 57. (b) 22 10Ne → 14 6 X + 2  58. (d) (a) The emitted - particles have varying energy. (b) e– or e+ does not exists inside the nucleus. (c) v does carry momentum. (d) In -decay mass number does not change. 59. (b) 60. (d)When an -particle is emitted, mass number of nuclide X is reduced to 4, and its charge number is reduced to 2, But when a -particle is emitted, mass number of remains the same and its charge number is increased by 1. Hence, the resulting nuclide has alomic mass A – 4 and atomic number Z – 1. 61. (c)In any nuclear reaction mass number and atomic number should remain conserved. Reaction (c) satisfies this condition. Also, for 239 93 NP, neutron to proton ratio is greater than 1.52, which makes it unstable. 62. (b)We have K = y y m m m+  .Q  K = A 4 A − .Q  48 = A 4 A − .50  A = 100 63. (a)R = mv qB RP = m vP eB U R235 = U m .v 235 eB , R238U = m . V 238U eB  X = 2(R) = 2(m m ) V 238U 235U eB − = 2 3m VP eB  = 2 × 3 × 10 mm = 60 mm 64. (c)During -decay atomic number (Z) and mass number (A) does not change. So the correct option is (C) because in all other options either Z, A or both is/are changing.
65. (b)The magnitude of momentum of the daughter nucleus and -particles will be equal Q = KE of daughter nucleus + KE of -particle = 2 d p 2m + 2 p 2m = 2 d d p m m 2 m . m     +     KE of -particle = 2 p 2m = 1 m × d d m .m m m   + . Q = 216 220 × 5.5 Mev. = 5.4 Mev. 66. (a) In a-particle emission atomic mass decreases by 4 unit and atomic number decreases by 2 unit. IN -particle emission, atomic mass remains unchanged and atomic number increases by 1 unit. Tje reaction can be shown as nXm ⎯⎯→ n – 2Ym–4 n – 2Ym–4 ⎯⎯⎯→2 nXm–4 Thus, the resulting nucleus is the isotope of parent nucleus and is nXm–4. 67. (b)Since 8 - particles 4 - particles and 2 + - particles are emitted,so new atomic number Z = Z - 8 × 2 + 4 × 1 – 2 × 1 = 92 – 16 + 4 –2 = 92 –14 = 78 68. (d)Order of 1 fermi 1 fermi 69. (c)Nuclear force do not exist when seperation is greater than 1 fermi. 70. (b)Nuclear force is charge independent 71. (c)Neutrino is produced in  + emission. 72. (d)In beta decay, atomic number increases by 1 whereas the mass number remains the same. Therefore, following equation can be possible 64 29Cu ⎯⎯→ 64 30Zn + –1 e 0 73. (a) 5B10+0 n 1 ⎯⎯→ 3 Li7 + 2He4 Total atomic number and mass number should be same on both sides of the equation. 74. (a)In a nuclear reaction conservation of charge +2e and mass 4m. Emission of an - particle reduces the mass of the radionuclide by 4 and its atomic number by2. -particle are negatively charged particles with rest mass as well as charge same as that of electrons. -particles carry no charge and mass. Radioactive reaction will be as follows A A 0 Z Z 1 –1 X Y ⎯⎯→ +  + A A–4 0 4 Z 1 z–1 2 + Y B ⎯⎯→ +  A 4 0 A–4 z 1 z–1 B B − − ⎯⎯→ +  75. (d)Let the radioactive subtance be A Z X Radioactive transition is given by A A–4 A–4 a –2 Z Z–2 Z X X X ⎯⎯⎯→ ⎯⎯⎯→ −  The atoms of element having same atomic number but dirrerent mass numbers are called isotopes so, A Z X and A–4 Z X are isotopes 76. (c)Tavg. = 1   T1/2 = ln2  < Tavg. So more than half the nuclei decay. 77. (b)64 = 26 After 6 half lives activity will become = 1 64 Hence required time = 6 × 2h = 12h . 78. (a)The weight will not change appreciably as the process is  - decay, because no. of nucleons in –decay do not change. 79. (a) 0 0 4 N N N 2 16 = = , % amount remaining = 0 0 0 N 100 100 N N 16 N  =  = 6.25% 80. (d)For stable product , dN dt = –N  0 = –N   = 0 81. (b) dN dt = N  dN N = dt = 10 1/ 2 0.693 0.693 dt 1 t 1.4 10 =   = 4.95 × 10–11 82. (a)R = R0 n 1 2       ....(a) Here R = activity of radioactive substance after n half lives