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NUMERICALS-ANSWERS QUESTIONS The half life of a radioactive sample is 30 seconds Calculate (i) the decay constant, (AISSCE Delhi 1999) th of its initial value The half life of a radioactive substance is 1.192 × 10–7 s against α decay Calculate the decay rate for 3.18 × 1015 atoms of the substance (ii) time taken for the sample to decay to The half life of a radioactive substance is 30 days What is the time taken for th of the original mass to disintegrate? (AISSCE 1994) A radioactive nucleus undergoes a series of decays according to the sequence given below If the mass number and the atomic number of A3 are 172 and 69 respectively, what are the mass number and the atomic number of A (AISSCE 1994) β α α A→ A1 → A2 → A3 Calculate the binding energy per nucleon for 26Fe56 : Mass of 26Fe56 = 55.934939 amu (AISSCE 1995) A radioactive sample decays to of its initial activity in 25 days Calculate the half life of the 32 substance (AISSCE Delhi 1992) 235 200 MeV of energy is released in the fission of a single 92U occur per second to produce a power of kW? nucleus How many fissions must Obtain approximately the ratio of the nuclear radii of the gold isotope 79Au197 and the silver isotope 47Ag107 (NCERT Book) S Chand & Company Limited In a head on collision between an alpha particle and a gold nucleus, the distance of closest approach is × 10–14 m Calculate the energy of the alpha particle in MeV (Z = 79 for gold) 10 The half life of a radioactive substance is 30 days What is its disintegration constant ? 11 What percentage of a given mass of a radioactive substance will be left undecayed after half lives? (AISSCE 1994) 12 Calculate the binding energy of an alpha particle in MeV, given that mp (mass of proton) = 1.007825 u mn (mass of neutron) = 1.008665 u mass of helium nucleus = 4.002800 u 13 Complete the following nuclear reactions : 17 N14 +  → 8O + 1H (a) (b) 15 30 P30  → 14Si + Rn222  → 2He + 14 Tritium has a half life of 12.5 years against beta decay What fraction of a sample of fine tritium will remain undecayed after 25 years? 15 Calculate the energy released in the following reaction : (c) 86 Li6 + 0n1  → 2He + 1H Given mass of Li6 = 6.015126 u mass of H3 = 3.016049 u mass of He4 = 4.002604 u mass of 0n1 = 1.008665 u S Chand & Company Limited 16 Calculate the half life of a radioactive substance if its activity drops to 116 th of its initial value in 30 years (AISSCE 1998) 17 In some stars, three alpha particles join together in a single fusion to form a 6C12 nucleus Calculate the energy released in this reaction Given that mass of 2He4 = 4.202604 amu 18 An electron-positron pair is produced by a photon of energy 2.32 Mev How much kinetic energy is imparted to each of the charged particles? 234 234 230 226 222 decays successivley to form 234 90Th, 91 Pa, 92 U, 90Th, 88 Ra, and 86 Rn What are the radioactive radiations emitted in each of the decay process (AISSCE 1979) 20 A γ ray photon materializes into a proton-antiproton pair with each particle having an energy of 2.67 MeV Calculate the energy of the γ ray photon Mass of proton or antiproton is 1.007276 amu 21 The fission process of uranium – 235 nucleus by slow neutrons is given by the reaction 19 238 92 U 235 92 U + 01 n  → 141 56 Ba + 92 36 Kr +3 ( n ) Calculate (a) the energy released per fission, (b) the en1 ergy released per kilo-mole of uranium, (c) the energy released per kg of uranium fissioned The atomic masses are : 235 92 U = 235.045733 amu 141 56 Ba = 140.917700 amu 92 36 Kr = 91.885440 amu 1.008665 amu 0n = 22 Calculate the number of disintegrati ons per second of gm of a radioactive substance whose half life is 1.4 × 1016 s The mass number of the substance is 238 23 The half life of radium is 1500 years After how much time will 1/32 part of a piece of radium remain undecayed? S Chand & Company Limited 24 Determine the amount of 210 84 Po necessary to provide a source of alpha particles of mCi strength The half life of polonium is 138 days 25 The half life of 238 92 U against second occur in 1g of alpha decay is 4.5 × 109 years How many disintegrations per 238 92 U ? ANSWERS (i ) 0.693 Decay constant λ = Τ1/2 = 0.693 30 = 0.0235-1 (ii ) Required time N 2.303 log10 λ N 2.303 = log10 0.023 = 10 [0.6021 − 0.4771] t= Decay rate R = λ N Here = 1.25 s T1/2 = 1.192 × 107 s λ= 0.693 0.693 = Τ1/2 1.192 × 107 S Chand & Company Limited N = 3.18 × 1015 R=λN So, = 3.18 × 1015 × 0.693 1.192 × 10 = 1.85 × 108 disintegrations/second Let the original number of radioactive nuclei be N0 Therefore, number of radioactive nuclei left, N0 N= N = N0 (1/2)n, where n represents the number of half lives Now N0 1 = N0    2 or n n=2 ⇒ Time taken = × Half life = × 30 = 60 days The given decay series can be expressed as A ZA → –1 e or ZA A → + −1 e A Z + A1 + → 42 He + A Z − A1 A−4 Z +1− 2A → 42 He + A−4 Z − A2 → 24 He + → 42 He + A−4−4 Z + − − A3 A−8 Z − A3 ⇒ Z – = 69 or Z = 72 A – = 172 or A = 180 S Chand & Company Limited In 28Fe56, number of protons = 26 number of neutrons = 56 – 26 = 30 Mass of 26 protons = 26 × 1.007825 = 26.20345 amu Mass of 30 neutrons = 30 × 1.008665 = 30.25995 amu Total mass of nucleons = 56.46340 amu Mass of Fe nucleus = 55.934939 amu Mass defect = 56.46340 –55.934939 = 0.528461 amu Binding Energy (B.E.) = 0.528461 × 931.5 = 492.26 MeV Binding energy per nucleon = B.E A = N 1 =  N0   ⇒ 492.26 = 8.79 MeV 56 n 1 =  32   n or n = In 25 day, half lives are completed S Chand & Company Limited Therefore, Half life = 25 = days Energy released per fission = 200 × 1.6 × 10–13 J = 3.2 × 10–11 J Power to be produced = kW = 1000 J/s Number of fission required per second 1000 = × 1011 3.2 = 3.125 × 1013 Using the formula R = R0 A1/3 RAu = R0 A1/Au3 RAg = R0 A1/Ag3 RAu  AAu = RAg  AAg = 1.23 or E = = 1/    1/  197  =   107  k ( Ze ) (2 e ) r × 109 × 79 × × (1.6 × 10 −19 ) × 10 −14 × 1.6 × 10 −13 MeV = 2.84 MeV S Chand & Company Limited 10 Disintegration constant λ = 0.693 T1/ = 11 1 N = N0   2 Here n =5 0.693 = 0.0231 per day 30 n N 1 =  N0   = 32 = × 100 % 32 = 3.125 % 12 Number of protons in alpha particle =2 Number of neutrons in alpha particle = mass of protons = × 1.007825 = 2.015650 amu mass of neutrons = × 1.008665 = 2.017330 amu Total mass of nucleons (a) = 4.032980 amu mass of helium nucleus (b) = 4.002800 amu mass defect = (a) – (b) = 0.030180 amu Binding Energy = 0.030180 × 931.5 = 28.11 MeV ⇒ S Chand & Company Limited 13 (a) 14 7N (b) 15 P (c ) 86 Rn + He  → O17 + H1 30  → 14 Si30 + e0 222  → 84 Po 218 + He4 14 Half life of tritium = 12.5 years Number of half lives in 25 years = Fraction remaining undecayed, N 1 =  N0   n 1 =  = 2 15 Li + n1  → He + H3 Total mass of reactants = 6.015126 + 1.008665 = 7.023791 amu Total mass of products = 4.002604 + 3.016049 = 7.018653 amu Mass difference = 7.023791 – 7.018653 = 0.005138 amu Energy released = 0.005138 × 931.5 = 4.786 MeV S Chand & Company Limited n 16 A 1 =  A0   n   1 = or      16    1 1   =    2 n=4 or or Half life T1/2 = n t 30 = = 7.5 years n + He4 + He4  → C12 + Q Mass of reactants = × 4.002604 = 12.007812 amu Mass of product = 12.000000 amu Mass difference = 0.007812 amu Energy released Q = 0.007812 × 931.5 = 7.2276 MeV 18 γ → e– + e+ Total mass of electron and positron = 0.000549 × = 0.001098 amu Total rest mass energy = 0.001098 × 931.5 = 1.0227 = 1.023 MeV Kinetic energy of e– and e+ = 2.32 – 1.023 = 1.297 MeV 17 He S Chand & Company Limited = K.E of each particle 19  → (i ) 238 92 U (ii ) 234 90Th  → 234 91 Pa (iii ) 234 91 Pa  → 234 92 U (iv ) 234 92 U  → 234 90Th 230 90Th 1.297 = 0.648 MeV + 42 He + + –1 e –1 e + 42 He (v ) 230 90Th  → 226 88 Ra + 42 He (vi ) 226 88 Ra  → 222 86 Rn + 24 He 20 γ  → Proton + Antiproton Total mass of products = × 1.007276 = 2.014552 amu Rest mass energy of products = 2.014552 × 931.5 = 1876.6 MeV Energy of γ ray photon = 1876.6 + (2.67) = 1881.9 MeV 21 235 92 U + n1  → 141 56 Ba + Mass of reactants 92 36 Kr +3 ( n) = 235.045733 + 1.008665 = 236.054398 amu S Chand & Company Limited Mass of products = 140.917700 + 91.885440 + (1.008665) = 235.829135 amu Mass difference = 236.054398 – 235.829135 = 0.225263 amu (a) Energy released per fission = 0.225263 × 931.5 = 209.8 MeV (b) kilo mole contains 6.023 × 1026 atoms Therefore energy released in kilo mole of 235U = 6.023 × 1026 × 209.8 = 1.264 × 1029 MeV (c) Mass of kilo mole of 235U = 235 kg Energy Released by kg of 235 1.264 × 1029 235 = 5.38 × 1026 MeV U = 22 Half life T1/2 = 1.4 × 1016 s Number of radioactive nuclei in g S Chand & Company Limited N= 6.023 × 1023 = 25.3 × 1020 238 dN = λN dt 0.693 = × 25.3 × 1020 1.4 × 1016 Number of disintegrations per second = = 1.25 × 105 per second 23 Half life = 1500 years We have Here So, 1 N = N0   2 N0 N= 32 n N0 1 = N0   32 2 n or n=5 Required time = × 1500 = 7500 years 24 Activity A = λN 0.693 = N T1/ S Chand & Company Limited Here A = mCi = × 3.7 × 107 dist./s T1/2 = 138 days = 138 × 8.64 × 104 s × 3.7 × 107 × 138 × 8.64 × 10 0.693 15 = 3.18 × 10 atoms N= Amount of substance required = 210 × 3.18 × 1015 6.02 × 10 23 g = 1.11 × 10–6 g 25 T1/2 = 4.5 × 109 years = 4.5 × 109 × 3.16 × 107 s = 1.42 × 1017 s Number of atoms in gram of 238 92 U 6.023 × 10 23 ×1 238 = 25.3 × 1020 = Number of disintegration per second, A = λN = 0.693 1.42 × 1017 × 25.3 × 10 20 = 1.23 × 104 s −1 S Chand & Company Limited