Tiêu đề 13.NUCLEI - Explanations.pdf ✅

1 (b) ∆m = 1 − 0.993 = 0.007 gm ∴ E = (∆m)c 2 = (0.007 × 10−3)(3 × 108) 2 = 63 × 1010J 2 (d) Applying principle of energy conservation. Energy of proton=total BE of 2 α −energy of Li 7 = 8 × 7.06 × 7 × 5.6 = 56.48 − 39.2 = 17.28 MeV 3 (c) Energy released from 1 kg of uranium = 200 ×106×1.6×10−19× 6.023 ×1026 235 = 8.2 × 1013 J 4 (c) 1 λ = R ( 1 2 2 − 1 4 2 ) = 3R 16 ⇒ λ = 16 3R = 16 3 × 10−5 cm Frequency n = c λ = 3×1010 16 3 ×10−5 = 9 16 × 1015Hz 5 (d) n = 24 24×138.6 = 1 138.6 ; Now N N0 = ( 1 2 ) n = ( 1 2 ) 1/138.6 ⇒ N = 10,00000 ( 1 2 ) 1/138.6 = 995011 So number of disintegration = 1000000 − 995011 = 4989 = 5000 7 (a) The activity or decay rate R of radioactive substance is the number of decays per second. ∴ R = λN or R = λN0 ( 1 2 ) t/T1/2 or R = R0 ( 1 2 ) t/T1/2 where R0 = λN0 is the activity of radioactive substance at time t = 0. According to question, R R0 = 1 − 75 100 = 25% ∴ 25 100 = ( 1 2 ) t/T1/2 or ( 1 2 ) 2 = ( 1 2 ) t/T1/2 or t T1/2 = 2 ∴ t = 2T1/2 = 2 × 3.20 = 6.40 h or t ≈ 6.38 h 8 (d) In time t = T , N = N0 2 In another half-life,(ie, after 2 half-lives) N = 1 2 N0 2 = N0 4 = N0 ( 1 2 ) 2 After yet another half-life ,(ie, after 3 half-lives) N = 1 2 ( N0 4 ) = N0 8 = N0 ( 1 2 ) 3 and so on. Hence, after n half-lives N = N0 ( 1 2 ) n = N0 ( 1 2 ) t/T where t = n × T = total time of n half-lives. Here, n = t T = 19 3.8 = 5 ∴ The fraction left N N0 = ( 1 2 ) n = ( 1 2 ) 5 = 1 32 = 0.031 9 (b) A = A0 ( 1 2 ) t/T1/2 ⇒ 5 = A0 ( 1 2 ) 2×60 30 = A0 16 ⇒ A0 = 80s −1 10 (b) (Z=92)U (A=238) (8α,6β) → Z ′X A ′ So A ′ = A − 4nα = 238 − 4 × 8 = 206 and Z ′ = nβ − 2nσ + z = 6 − 2 × 8 + 92 = 82 11 (a) Excitation energy ∆E = E2 − E1 = 13.6 Z 2 [ 1 1 2 − 1 2 2 ] ⇒ 40.8 = 13.6 × 3 4 × Z 2 ⇒ Z = 2 Now required energy to remove the electron from ground state = +13.6Z 2 (1)2 = 13.6(Z) 2 = 54.4 eV 12 (d) Number of α −particles emitted = 238−222 4 = 4 This decreases atomic number to 90 − 4 × 2 = 82 Since atomic number of 83Y 222 is 83, this is possible of one β −particle is emitted 14 (a) (4n + 2) series starts from U 238 and it’s stable end product is Pb 206 16 (a) Rest energy of an electron = mec 2 Here me = 9.1 × 10−31kg and c = velocity of light ∴ Rest energy = 9.1 × 10−31 × (3 × 108) 2 joule = 9.1 × 10−31 × (3 × 108) 2 1.6 × 10−19 eV = 510 keV 17 (b) Recoil momentum = momentum of photon = h λ