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Page 1 In tr o duc tor y Ex er cis e 30.2 1. D D E Mc = 2 Here, P = = ´ ´ 1 0 1 0 2 4 6 0 9 9 J / s J / d a y Þ D M = ´ ´ ´ = ´  1 0 2 4 6 0 3 1 0 1 0 9 8 2 4 ( ) 9 . 6 k g 2. Number of fis sion = ´ ´ ´  10 200 10 10 9 6 19 J/s 1.6 = ´ 3.125 10 19 3. Giv en re ac tion is 92 238 90 234 2 4 U T h H e ¾® + D M =  ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3  4 . 0 0 2 6 0 2 ) u D M = 0 . 0 0 0 4 5 8 9 u D D E M = ´ 9 3 1 . 5 M eV = ´ 0.000 931.5 4589 = 4.27 MeV 4. Com plete reactions are (a) 3 6 1 2 4 7 0 1 Li H Be + ® + n (b) 17 35 1 1 16 32 2 4 Cl H S He + ® + (c) 4 9 2 4 2 0 1 3 Be He He 4 + ® + ( ) n (d) 35 79 1 2 36 79 0 1 2 Br H Kr + ® + ( ) n AIEEE Corner ¢ Subjective Questions (Level 1) Radioactivity 1. (a) Initially the rate of distingration is  æ è ç ö ø ÷ = dN dt N 0 0 l After 5 min  = dN dt N l Þ N N dN dt dN dt 0 0 4750 2700 = æ è ç ö ø ÷ = =1.76 Now N N e t =  0 l or l =  æ è ç ç ö ø ÷ ÷ log N N t 0 Þ l = 2.3026 10 0 t N N log = 2.3026 1.76 5 10 log ( ) Þ l = 0.113 /min (b) Halflife = = = 0.693 0.693 0.113 6.132 l min 2. We have A N = l 6 10 11 ´ = ´ ´ l 1 10 15 Þ l = ´   6 10 4 1 s T 1 2 4 6 10 1155 / = = ´ =  0.693 0.693 l s = 19.25 min 3. A N = l A = = ´ ´ 8 10 10 Ci 8 3.7 decay/s l = = 0.693 0.693 5.3 yr T 1 2 / = ´ ´ ´ ´ 0.693 5.3 s 365 24 60 60 Þ N A = l = ´ ´ ´ ´ ´ ´ 8 10 365 24 3600 10 3.7 5.3 0.693 Þ N = ´ 7.2 10 19 83 Page 2 In tr o duc tor y Ex er cis e 30.2 1. D D E Mc = 2 Here, P = = ´ ´ 1 0 1 0 2 4 6 0 9 9 J / s J / d a y Þ D M = ´ ´ ´ = ´  1 0 2 4 6 0 3 1 0 1 0 9 8 2 4 ( ) 9 . 6 k g 2. Number of fis sion = ´ ´ ´  10 200 10 10 9 6 19 J/s 1.6 = ´ 3.125 10 19 3. Giv en re ac tion is 92 238 90 234 2 4 U T h H e ¾® + D M =  ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3  4 . 0 0 2 6 0 2 ) u D M = 0 . 0 0 0 4 5 8 9 u D D E M = ´ 9 3 1 . 5 M eV = ´ 0.000 931.5 4589 = 4.27 MeV 4. Com plete reactions are (a) 3 6 1 2 4 7 0 1 Li H Be + ® + n (b) 17 35 1 1 16 32 2 4 Cl H S He + ® + (c) 4 9 2 4 2 0 1 3 Be He He 4 + ® + ( ) n (d) 35 79 1 2 36 79 0 1 2 Br H Kr + ® + ( ) n AIEEE Corner ¢ Subjective Questions (Level 1) Radioactivity 1. (a) Initially the rate of distingration is  æ è ç ö ø ÷ = dN dt N 0 0 l After 5 min  = dN dt N l Þ N N dN dt dN dt 0 0 4750 2700 = æ è ç ö ø ÷ = =1.76 Now N N e t =  0 l or l =  æ è ç ç ö ø ÷ ÷ log N N t 0 Þ l = 2.3026 10 0 t N N log = 2.3026 1.76 5 10 log ( ) Þ l = 0.113 /min (b) Halflife = = = 0.693 0.693 0.113 6.132 l min 2. We have A N = l 6 10 11 ´ = ´ ´ l 1 10 15 Þ l = ´   6 10 4 1 s T 1 2 4 6 10 1155 / = = ´ =  0.693 0.693 l s = 19.25 min 3. A N = l A = = ´ ´ 8 10 10 Ci 8 3.7 decay/s l = = 0.693 0.693 5.3 yr T 1 2 / = ´ ´ ´ ´ 0.693 5.3 s 365 24 60 60 Þ N A = l = ´ ´ ´ ´ ´ ´ 8 10 365 24 3600 10 3.7 5.3 0.693 Þ N = ´ 7.2 10 19 83 6.023 ´ 10 23 nuclei = 60 g 1 60 10 23 = ´ g 6.023 Hence 7.2 ´ 10 19 nuclei = ´ ´ ´ 60 10 10 19 23 7.2 6.023 = ´  7.11 10 3 g 4. Number of decay per second m M N A ´ ´ = ´ ´ l 1 238 6 10 23 ´ ´ 0.693 4.5 yr 10 9 = ´ 6 10 238 23 ´ ´ ´ ´ ´ ´ 0.693 4.5 s 10 365 24 60 60 9 = ´ 1.23 10 4 decay/s 5. Probability of decay P e e t t T =  =    ( ) ( ) / 1 1 l mean Þ P e e =  =  =   ( ) / 1 1 5 10 2 0.39 6. Since initially no Pb 206 nuclei is present and after time t the ratio of N N v Pb = 3 It means 1 4 of original U 238 nuclei decays. Hence N N e t =  0 l N N N e 0 0 0 4  æ è ç ö ø ÷ =  l 3 4 =  e t l Þ t =  log log 4 3 l Þ t =  ´ ´ (log log ) 4 3 10 9 0.693 4.5 yr Þ t = ´ 1.88 yr 10 9 7. ( ) R N 1 0 1 0 32 4 P = l ( ) R N 2 0 2 0 33 P = l R R R 0 1 0 2 0 32 33 = + ( ) ( ) P P = + = N 0 1 2 4 3 ( ) l l mCi ( ) R N N e t t 1 1 1 0 32 1 4 P = =  l l l ( ) log R N e t 1 1 0 2 14 60 365 32 4 P = ´  ´ ´ l ( ) log R N e t 2 2 0 2 25 6 0 3 6 5 33 P = ´  ´ ´ l R R R t t = + ( ) ( ) 1 2 32 33 P P = + é ë ê ê ù û ú ú  ´ ´  ´ ´ N e e 0 1 2 14 60 365 2 2 25 60 365 4 l l log log = + 3 4 1 2 mCi ( ) l l 4 1 2 60 365 14 2 2 25 60 365 l l e e  ´ ´  ´ ´ + é ë ê ê ù û ú ú log log = ´ + é ë ê ù û ú 3 4 2 14 2 25 mCi log log 4 2 14 2 25 2 60 365 14 2 25 60 365 log log log log e e  ´ ´  ´ ´ + ´ é ë ê ê ù û ú ú » 0.205mCi 8. Complete reactions are (a) 88 226 86 222 Ra RN ¾¾® + a (b) 8 19 9 19 O F ¾¾® + + e n (c) 13 25 12 25 Al Mg ¾¾® + + + e n 9. Only reaction (b) is possible. 10. DE = ´ + ´  ( ) 7 7 1.000783 1.00867 14.00307 ´ 931.5 MeV Þ DE = 104.72 MeV 11. DE m m m O p n = +  ´ [ ( )] 8 8 8 16 931.5 = ´ + ´  ( . 8 825 8 1 008665 1.007 15.994915) ´ 931.5 = 127.6 MeV 12. (a) Number of nuclei in kg = ´ ´ 6.023 10 235 1 23 Energy = ´ ´ ´ 6.023 10 235 200 10 23 6 ´ ´  1.6 10 19 = ´ 8.09 J 10 13 84 Page 3 In tr o duc tor y Ex er cis e 30.2 1. D D E Mc = 2 Here, P = = ´ ´ 1 0 1 0 2 4 6 0 9 9 J / s J / d a y Þ D M = ´ ´ ´ = ´  1 0 2 4 6 0 3 1 0 1 0 9 8 2 4 ( ) 9 . 6 k g 2. Number of fis sion = ´ ´ ´  10 200 10 10 9 6 19 J/s 1.6 = ´ 3.125 10 19 3. Giv en re ac tion is 92 238 90 234 2 4 U T h H e ¾® + D M =  ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3  4 . 0 0 2 6 0 2 ) u D M = 0 . 0 0 0 4 5 8 9 u D D E M = ´ 9 3 1 . 5 M eV = ´ 0.000 931.5 4589 = 4.27 MeV 4. Com plete reactions are (a) 3 6 1 2 4 7 0 1 Li H Be + ® + n (b) 17 35 1 1 16 32 2 4 Cl H S He + ® + (c) 4 9 2 4 2 0 1 3 Be He He 4 + ® + ( ) n (d) 35 79 1 2 36 79 0 1 2 Br H Kr + ® + ( ) n AIEEE Corner ¢ Subjective Questions (Level 1) Radioactivity 1. (a) Initially the rate of distingration is  æ è ç ö ø ÷ = dN dt N 0 0 l After 5 min  = dN dt N l Þ N N dN dt dN dt 0 0 4750 2700 = æ è ç ö ø ÷ = =1.76 Now N N e t =  0 l or l =  æ è ç ç ö ø ÷ ÷ log N N t 0 Þ l = 2.3026 10 0 t N N log = 2.3026 1.76 5 10 log ( ) Þ l = 0.113 /min (b) Halflife = = = 0.693 0.693 0.113 6.132 l min 2. We have A N = l 6 10 11 ´ = ´ ´ l 1 10 15 Þ l = ´   6 10 4 1 s T 1 2 4 6 10 1155 / = = ´ =  0.693 0.693 l s = 19.25 min 3. A N = l A = = ´ ´ 8 10 10 Ci 8 3.7 decay/s l = = 0.693 0.693 5.3 yr T 1 2 / = ´ ´ ´ ´ 0.693 5.3 s 365 24 60 60 Þ N A = l = ´ ´ ´ ´ ´ ´ 8 10 365 24 3600 10 3.7 5.3 0.693 Þ N = ´ 7.2 10 19 83 6.023 ´ 10 23 nuclei = 60 g 1 60 10 23 = ´ g 6.023 Hence 7.2 ´ 10 19 nuclei = ´ ´ ´ 60 10 10 19 23 7.2 6.023 = ´  7.11 10 3 g 4. Number of decay per second m M N A ´ ´ = ´ ´ l 1 238 6 10 23 ´ ´ 0.693 4.5 yr 10 9 = ´ 6 10 238 23 ´ ´ ´ ´ ´ ´ 0.693 4.5 s 10 365 24 60 60 9 = ´ 1.23 10 4 decay/s 5. Probability of decay P e e t t T =  =    ( ) ( ) / 1 1 l mean Þ P e e =  =  =   ( ) / 1 1 5 10 2 0.39 6. Since initially no Pb 206 nuclei is present and after time t the ratio of N N v Pb = 3 It means 1 4 of original U 238 nuclei decays. Hence N N e t =  0 l N N N e 0 0 0 4  æ è ç ö ø ÷ =  l 3 4 =  e t l Þ t =  log log 4 3 l Þ t =  ´ ´ (log log ) 4 3 10 9 0.693 4.5 yr Þ t = ´ 1.88 yr 10 9 7. ( ) R N 1 0 1 0 32 4 P = l ( ) R N 2 0 2 0 33 P = l R R R 0 1 0 2 0 32 33 = + ( ) ( ) P P = + = N 0 1 2 4 3 ( ) l l mCi ( ) R N N e t t 1 1 1 0 32 1 4 P = =  l l l ( ) log R N e t 1 1 0 2 14 60 365 32 4 P = ´  ´ ´ l ( ) log R N e t 2 2 0 2 25 6 0 3 6 5 33 P = ´  ´ ´ l R R R t t = + ( ) ( ) 1 2 32 33 P P = + é ë ê ê ù û ú ú  ´ ´  ´ ´ N e e 0 1 2 14 60 365 2 2 25 60 365 4 l l log log = + 3 4 1 2 mCi ( ) l l 4 1 2 60 365 14 2 2 25 60 365 l l e e  ´ ´  ´ ´ + é ë ê ê ù û ú ú log log = ´ + é ë ê ù û ú 3 4 2 14 2 25 mCi log log 4 2 14 2 25 2 60 365 14 2 25 60 365 log log log log e e  ´ ´  ´ ´ + ´ é ë ê ê ù û ú ú » 0.205mCi 8. Complete reactions are (a) 88 226 86 222 Ra RN ¾¾® + a (b) 8 19 9 19 O F ¾¾® + + e n (c) 13 25 12 25 Al Mg ¾¾® + + + e n 9. Only reaction (b) is possible. 10. DE = ´ + ´  ( ) 7 7 1.000783 1.00867 14.00307 ´ 931.5 MeV Þ DE = 104.72 MeV 11. DE m m m O p n = +  ´ [ ( )] 8 8 8 16 931.5 = ´ + ´  ( . 8 825 8 1 008665 1.007 15.994915) ´ 931.5 = 127.6 MeV 12. (a) Number of nuclei in kg = ´ ´ 6.023 10 235 1 23 Energy = ´ ´ ´ 6.023 10 235 200 10 23 6 ´ ´  1.6 10 19 = ´ 8.09 J 10 13 84 (b) Mass = ´ ´ 8.09 J J/g 10 30 10 13 3 = ´ ´ 8.09 g 10 30 10 13 3 = ´ ´ 8.09 kg 3 10 1 10 9 3 = ´ 2.7 kg 10 6 13. Applying conservation of momentum M v M v a a = Ti Ti Þ v M v M Ti Ti = a a K M v a a a = = 1 2 2 6.802 MeV K M v M M v M L Ti Ti Ti Ti Ti = = 1 2 1 2 2 2 2 a a Þ K M M M v Ti Ti = æ è ç ö ø ÷ a a a 1 2 2 = ´ 4 208 6.802 MeV = ´ = 1 52 6.802 0.1308 MeV 14. Power = = = 100 10 10 8 8 MW W J/s = ´ =  10 10 10 8 13 21 1.6 J/MeV 1.6 MeV s Energy per fission = 185 MeV Hence number of fissions = ´ 10 185 21 1.6 / s. Number of nuclei in 1 kg U 6.023 235 26 10 235 = ´ Hence t = ´ ´ ´ ´ 6.023 1.6 10 185 235 10 26 21 = 8.78 days 15. (a) The given reaction is 1 2 1 2 1 1 1 H H H H 3 + ¾® + DM m m m =   [ ( ) ( ) ( )] 2 1 2 1 3 1 1 H H H =   [ ( ] 2 7825 2.014102) 3.016049 1.00 u DM = 0.000433 u Q M = ´ = ´ D 931.5 0.000433 931.5 » 4.05 MeV (b) DM = ´  (2 2.014102 3.016049  ´ 1.008665) 931.5 DM = 3.25 MeV (c) DM = (2.014102 3.016049 4.002603 +   ´ 1.008665 931.5 ) D M = 1 7 . 5 7 M eV 16. The given reaction is He He Be 4 4 8 + ¾® DM = ´  ( ) 2 4.0026 8.0053 u =  (8.0052 8.0053) u DM =  (0.0001) u Q DM is negative this reaction is not energetically favourable D D E M = ´ 931.5 =  ´ ´  1 10 4 931.5 MeV =  93.15 keV 17. Number of nuclei in 1 kg water = ´ 6.023 10 18 26 Heavy water = ´ ´ ´  6.023 1.5 10 18 10 100 26 2 = ´ ´ 6.023 1.5 18 10 22 Energy realesed per fission = ´   ´ ( ) 2 2.014102 3.016049 1.007825 931.5 = ´ ´ ´ ´ ´   4.33 931.5 1.6 J 10 10 10 3 6 19 Hence total energy = ´ ´ 6.023 1.5 10 18 22 ´ ´ ´ ´  4.33 931.5 1.6 10 16 = 3200 MJ 85 Page 4 In tr o duc tor y Ex er cis e 30.2 1. D D E Mc = 2 Here, P = = ´ ´ 1 0 1 0 2 4 6 0 9 9 J / s J / d a y Þ D M = ´ ´ ´ = ´  1 0 2 4 6 0 3 1 0 1 0 9 8 2 4 ( ) 9 . 6 k g 2. Number of fis sion = ´ ´ ´  10 200 10 10 9 6 19 J/s 1.6 = ´ 3.125 10 19 3. Giv en re ac tion is 92 238 90 234 2 4 U T h H e ¾® + D M =  ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3  4 . 0 0 2 6 0 2 ) u D M = 0 . 0 0 0 4 5 8 9 u D D E M = ´ 9 3 1 . 5 M eV = ´ 0.000 931.5 4589 = 4.27 MeV 4. Com plete reactions are (a) 3 6 1 2 4 7 0 1 Li H Be + ® + n (b) 17 35 1 1 16 32 2 4 Cl H S He + ® + (c) 4 9 2 4 2 0 1 3 Be He He 4 + ® + ( ) n (d) 35 79 1 2 36 79 0 1 2 Br H Kr + ® + ( ) n AIEEE Corner ¢ Subjective Questions (Level 1) Radioactivity 1. (a) Initially the rate of distingration is  æ è ç ö ø ÷ = dN dt N 0 0 l After 5 min  = dN dt N l Þ N N dN dt dN dt 0 0 4750 2700 = æ è ç ö ø ÷ = =1.76 Now N N e t =  0 l or l =  æ è ç ç ö ø ÷ ÷ log N N t 0 Þ l = 2.3026 10 0 t N N log = 2.3026 1.76 5 10 log ( ) Þ l = 0.113 /min (b) Halflife = = = 0.693 0.693 0.113 6.132 l min 2. We have A N = l 6 10 11 ´ = ´ ´ l 1 10 15 Þ l = ´   6 10 4 1 s T 1 2 4 6 10 1155 / = = ´ =  0.693 0.693 l s = 19.25 min 3. A N = l A = = ´ ´ 8 10 10 Ci 8 3.7 decay/s l = = 0.693 0.693 5.3 yr T 1 2 / = ´ ´ ´ ´ 0.693 5.3 s 365 24 60 60 Þ N A = l = ´ ´ ´ ´ ´ ´ 8 10 365 24 3600 10 3.7 5.3 0.693 Þ N = ´ 7.2 10 19 83 6.023 ´ 10 23 nuclei = 60 g 1 60 10 23 = ´ g 6.023 Hence 7.2 ´ 10 19 nuclei = ´ ´ ´ 60 10 10 19 23 7.2 6.023 = ´  7.11 10 3 g 4. Number of decay per second m M N A ´ ´ = ´ ´ l 1 238 6 10 23 ´ ´ 0.693 4.5 yr 10 9 = ´ 6 10 238 23 ´ ´ ´ ´ ´ ´ 0.693 4.5 s 10 365 24 60 60 9 = ´ 1.23 10 4 decay/s 5. Probability of decay P e e t t T =  =    ( ) ( ) / 1 1 l mean Þ P e e =  =  =   ( ) / 1 1 5 10 2 0.39 6. Since initially no Pb 206 nuclei is present and after time t the ratio of N N v Pb = 3 It means 1 4 of original U 238 nuclei decays. Hence N N e t =  0 l N N N e 0 0 0 4  æ è ç ö ø ÷ =  l 3 4 =  e t l Þ t =  log log 4 3 l Þ t =  ´ ´ (log log ) 4 3 10 9 0.693 4.5 yr Þ t = ´ 1.88 yr 10 9 7. ( ) R N 1 0 1 0 32 4 P = l ( ) R N 2 0 2 0 33 P = l R R R 0 1 0 2 0 32 33 = + ( ) ( ) P P = + = N 0 1 2 4 3 ( ) l l mCi ( ) R N N e t t 1 1 1 0 32 1 4 P = =  l l l ( ) log R N e t 1 1 0 2 14 60 365 32 4 P = ´  ´ ´ l ( ) log R N e t 2 2 0 2 25 6 0 3 6 5 33 P = ´  ´ ´ l R R R t t = + ( ) ( ) 1 2 32 33 P P = + é ë ê ê ù û ú ú  ´ ´  ´ ´ N e e 0 1 2 14 60 365 2 2 25 60 365 4 l l log log = + 3 4 1 2 mCi ( ) l l 4 1 2 60 365 14 2 2 25 60 365 l l e e  ´ ´  ´ ´ + é ë ê ê ù û ú ú log log = ´ + é ë ê ù û ú 3 4 2 14 2 25 mCi log log 4 2 14 2 25 2 60 365 14 2 25 60 365 log log log log e e  ´ ´  ´ ´ + ´ é ë ê ê ù û ú ú » 0.205mCi 8. Complete reactions are (a) 88 226 86 222 Ra RN ¾¾® + a (b) 8 19 9 19 O F ¾¾® + + e n (c) 13 25 12 25 Al Mg ¾¾® + + + e n 9. Only reaction (b) is possible. 10. DE = ´ + ´  ( ) 7 7 1.000783 1.00867 14.00307 ´ 931.5 MeV Þ DE = 104.72 MeV 11. DE m m m O p n = +  ´ [ ( )] 8 8 8 16 931.5 = ´ + ´  ( . 8 825 8 1 008665 1.007 15.994915) ´ 931.5 = 127.6 MeV 12. (a) Number of nuclei in kg = ´ ´ 6.023 10 235 1 23 Energy = ´ ´ ´ 6.023 10 235 200 10 23 6 ´ ´  1.6 10 19 = ´ 8.09 J 10 13 84 (b) Mass = ´ ´ 8.09 J J/g 10 30 10 13 3 = ´ ´ 8.09 g 10 30 10 13 3 = ´ ´ 8.09 kg 3 10 1 10 9 3 = ´ 2.7 kg 10 6 13. Applying conservation of momentum M v M v a a = Ti Ti Þ v M v M Ti Ti = a a K M v a a a = = 1 2 2 6.802 MeV K M v M M v M L Ti Ti Ti Ti Ti = = 1 2 1 2 2 2 2 a a Þ K M M M v Ti Ti = æ è ç ö ø ÷ a a a 1 2 2 = ´ 4 208 6.802 MeV = ´ = 1 52 6.802 0.1308 MeV 14. Power = = = 100 10 10 8 8 MW W J/s = ´ =  10 10 10 8 13 21 1.6 J/MeV 1.6 MeV s Energy per fission = 185 MeV Hence number of fissions = ´ 10 185 21 1.6 / s. Number of nuclei in 1 kg U 6.023 235 26 10 235 = ´ Hence t = ´ ´ ´ ´ 6.023 1.6 10 185 235 10 26 21 = 8.78 days 15. (a) The given reaction is 1 2 1 2 1 1 1 H H H H 3 + ¾® + DM m m m =   [ ( ) ( ) ( )] 2 1 2 1 3 1 1 H H H =   [ ( ] 2 7825 2.014102) 3.016049 1.00 u DM = 0.000433 u Q M = ´ = ´ D 931.5 0.000433 931.5 » 4.05 MeV (b) DM = ´  (2 2.014102 3.016049  ´ 1.008665) 931.5 DM = 3.25 MeV (c) DM = (2.014102 3.016049 4.002603 +   ´ 1.008665 931.5 ) D M = 1 7 . 5 7 M eV 16. The given reaction is He He Be 4 4 8 + ¾® DM = ´  ( ) 2 4.0026 8.0053 u =  (8.0052 8.0053) u DM =  (0.0001) u Q DM is negative this reaction is not energetically favourable D D E M = ´ 931.5 =  ´ ´  1 10 4 931.5 MeV =  93.15 keV 17. Number of nuclei in 1 kg water = ´ 6.023 10 18 26 Heavy water = ´ ´ ´  6.023 1.5 10 18 10 100 26 2 = ´ ´ 6.023 1.5 18 10 22 Energy realesed per fission = ´   ´ ( ) 2 2.014102 3.016049 1.007825 931.5 = ´ ´ ´ ´ ´   4.33 931.5 1.6 J 10 10 10 3 6 19 Hence total energy = ´ ´ 6.023 1.5 10 18 22 ´ ´ ´ ´  4.33 931.5 1.6 10 16 = 3200 MJ 85 ¢ Objective Questions (Level1) 1. Since during b  decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino as n P e ® + +  g. Hence Correct option is (c). 2. Since nuclear force is same for all nucleons. Hence F F F 1 2 3 = = Correct option is (a).. 3. Given reaction is 90 200 80 168 X Y ¾® Difference in mass number =  = 200 168 32 Hence Number of aparticles = = 32 4 8 Difference in atomic number = 10 hence number of b particles = 6 Hence correct option is (d). 4. The reaction is 92 235 0 1 54 138 38 94 0 1 3 U Xe Sr + ¾® + + n n ( ) The correct option is (b) three neutrons. 5. The reactions are A ® + b a and b b ® + C 2 After one a atomic number reduced by 2 and after 2b atomic number increased by. Hence A and C are isotopes Correct option is (d). 6. Here m m p n = = 1.00785 u 1.00866 u , and m a = 4.00274 u Dm m m m p n = +  2( ) a Þ Dm = +  [ ( ) ] 2 1.00785 1.00866 4.00274 u Þ Dm = 00 . 3028 u D D E m = ´ 931.5 MeV = ´ = 0.3028 931.5 28.21 MeV Hence correct option is (c). 7. N N N N =  = 0 0 0 7 8 1 8 But N N n = æ è ç ö ø ÷ 0 1 2 Þ 1 8 1 2 0 0 N N n = æ è ç ö ø ÷ Þ n = 3 Þ 3 8 1 2 ´ = T / s Þ T 1 2 8 3 / = s Hence correct option is (d). 8. N N e t =  0 l for mean life t = 1 l Þ N N e N e = =  ´ 0 1 0 l l Hence the fraction disintegrated =  =  æ è ç ö ø ÷ N N N e 0 0 1 1 Correct option is (b). 9. N N N N =  = 0 0 0 7 8 1 8 But N N n = æ è ç ö ø ÷ 0 1 2 Þ 1 8 1 2 = æ è ç ö ø ÷ n Þn = 3 Hence half life T 1 2 15 3 5 / = = min min Correct option is (a). 10. Since radioactive substance loses half of its activity in 4 days it means its half life T 12 4 / = days Now A = 5% of A 0 Þ A A = 1 20 0 Þ A A 0 1 20 = But A A e t =  0 l Þ 1 20 =  e t l Þ log 20 = lt Þ t = log 20 l 86 Page 5 In tr o duc tor y Ex er cis e 30.2 1. D D E Mc = 2 Here, P = = ´ ´ 1 0 1 0 2 4 6 0 9 9 J / s J / d a y Þ D M = ´ ´ ´ = ´  1 0 2 4 6 0 3 1 0 1 0 9 8 2 4 ( ) 9 . 6 k g 2. Number of fis sion = ´ ´ ´  10 200 10 10 9 6 19 J/s 1.6 = ´ 3.125 10 19 3. Giv en re ac tion is 92 238 90 234 2 4 U T h H e ¾® + D M =  ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3  4 . 0 0 2 6 0 2 ) u D M = 0 . 0 0 0 4 5 8 9 u D D E M = ´ 9 3 1 . 5 M eV = ´ 0.000 931.5 4589 = 4.27 MeV 4. Com plete reactions are (a) 3 6 1 2 4 7 0 1 Li H Be + ® + n (b) 17 35 1 1 16 32 2 4 Cl H S He + ® + (c) 4 9 2 4 2 0 1 3 Be He He 4 + ® + ( ) n (d) 35 79 1 2 36 79 0 1 2 Br H Kr + ® + ( ) n AIEEE Corner ¢ Subjective Questions (Level 1) Radioactivity 1. (a) Initially the rate of distingration is  æ è ç ö ø ÷ = dN dt N 0 0 l After 5 min  = dN dt N l Þ N N dN dt dN dt 0 0 4750 2700 = æ è ç ö ø ÷ = =1.76 Now N N e t =  0 l or l =  æ è ç ç ö ø ÷ ÷ log N N t 0 Þ l = 2.3026 10 0 t N N log = 2.3026 1.76 5 10 log ( ) Þ l = 0.113 /min (b) Halflife = = = 0.693 0.693 0.113 6.132 l min 2. We have A N = l 6 10 11 ´ = ´ ´ l 1 10 15 Þ l = ´   6 10 4 1 s T 1 2 4 6 10 1155 / = = ´ =  0.693 0.693 l s = 19.25 min 3. A N = l A = = ´ ´ 8 10 10 Ci 8 3.7 decay/s l = = 0.693 0.693 5.3 yr T 1 2 / = ´ ´ ´ ´ 0.693 5.3 s 365 24 60 60 Þ N A = l = ´ ´ ´ ´ ´ ´ 8 10 365 24 3600 10 3.7 5.3 0.693 Þ N = ´ 7.2 10 19 83 6.023 ´ 10 23 nuclei = 60 g 1 60 10 23 = ´ g 6.023 Hence 7.2 ´ 10 19 nuclei = ´ ´ ´ 60 10 10 19 23 7.2 6.023 = ´  7.11 10 3 g 4. Number of decay per second m M N A ´ ´ = ´ ´ l 1 238 6 10 23 ´ ´ 0.693 4.5 yr 10 9 = ´ 6 10 238 23 ´ ´ ´ ´ ´ ´ 0.693 4.5 s 10 365 24 60 60 9 = ´ 1.23 10 4 decay/s 5. Probability of decay P e e t t T =  =    ( ) ( ) / 1 1 l mean Þ P e e =  =  =   ( ) / 1 1 5 10 2 0.39 6. Since initially no Pb 206 nuclei is present and after time t the ratio of N N v Pb = 3 It means 1 4 of original U 238 nuclei decays. Hence N N e t =  0 l N N N e 0 0 0 4  æ è ç ö ø ÷ =  l 3 4 =  e t l Þ t =  log log 4 3 l Þ t =  ´ ´ (log log ) 4 3 10 9 0.693 4.5 yr Þ t = ´ 1.88 yr 10 9 7. ( ) R N 1 0 1 0 32 4 P = l ( ) R N 2 0 2 0 33 P = l R R R 0 1 0 2 0 32 33 = + ( ) ( ) P P = + = N 0 1 2 4 3 ( ) l l mCi ( ) R N N e t t 1 1 1 0 32 1 4 P = =  l l l ( ) log R N e t 1 1 0 2 14 60 365 32 4 P = ´  ´ ´ l ( ) log R N e t 2 2 0 2 25 6 0 3 6 5 33 P = ´  ´ ´ l R R R t t = + ( ) ( ) 1 2 32 33 P P = + é ë ê ê ù û ú ú  ´ ´  ´ ´ N e e 0 1 2 14 60 365 2 2 25 60 365 4 l l log log = + 3 4 1 2 mCi ( ) l l 4 1 2 60 365 14 2 2 25 60 365 l l e e  ´ ´  ´ ´ + é ë ê ê ù û ú ú log log = ´ + é ë ê ù û ú 3 4 2 14 2 25 mCi log log 4 2 14 2 25 2 60 365 14 2 25 60 365 log log log log e e  ´ ´  ´ ´ + ´ é ë ê ê ù û ú ú » 0.205mCi 8. Complete reactions are (a) 88 226 86 222 Ra RN ¾¾® + a (b) 8 19 9 19 O F ¾¾® + + e n (c) 13 25 12 25 Al Mg ¾¾® + + + e n 9. Only reaction (b) is possible. 10. DE = ´ + ´  ( ) 7 7 1.000783 1.00867 14.00307 ´ 931.5 MeV Þ DE = 104.72 MeV 11. DE m m m O p n = +  ´ [ ( )] 8 8 8 16 931.5 = ´ + ´  ( . 8 825 8 1 008665 1.007 15.994915) ´ 931.5 = 127.6 MeV 12. (a) Number of nuclei in kg = ´ ´ 6.023 10 235 1 23 Energy = ´ ´ ´ 6.023 10 235 200 10 23 6 ´ ´  1.6 10 19 = ´ 8.09 J 10 13 84 (b) Mass = ´ ´ 8.09 J J/g 10 30 10 13 3 = ´ ´ 8.09 g 10 30 10 13 3 = ´ ´ 8.09 kg 3 10 1 10 9 3 = ´ 2.7 kg 10 6 13. Applying conservation of momentum M v M v a a = Ti Ti Þ v M v M Ti Ti = a a K M v a a a = = 1 2 2 6.802 MeV K M v M M v M L Ti Ti Ti Ti Ti = = 1 2 1 2 2 2 2 a a Þ K M M M v Ti Ti = æ è ç ö ø ÷ a a a 1 2 2 = ´ 4 208 6.802 MeV = ´ = 1 52 6.802 0.1308 MeV 14. Power = = = 100 10 10 8 8 MW W J/s = ´ =  10 10 10 8 13 21 1.6 J/MeV 1.6 MeV s Energy per fission = 185 MeV Hence number of fissions = ´ 10 185 21 1.6 / s. Number of nuclei in 1 kg U 6.023 235 26 10 235 = ´ Hence t = ´ ´ ´ ´ 6.023 1.6 10 185 235 10 26 21 = 8.78 days 15. (a) The given reaction is 1 2 1 2 1 1 1 H H H H 3 + ¾® + DM m m m =   [ ( ) ( ) ( )] 2 1 2 1 3 1 1 H H H =   [ ( ] 2 7825 2.014102) 3.016049 1.00 u DM = 0.000433 u Q M = ´ = ´ D 931.5 0.000433 931.5 » 4.05 MeV (b) DM = ´  (2 2.014102 3.016049  ´ 1.008665) 931.5 DM = 3.25 MeV (c) DM = (2.014102 3.016049 4.002603 +   ´ 1.008665 931.5 ) D M = 1 7 . 5 7 M eV 16. The given reaction is He He Be 4 4 8 + ¾® DM = ´  ( ) 2 4.0026 8.0053 u =  (8.0052 8.0053) u DM =  (0.0001) u Q DM is negative this reaction is not energetically favourable D D E M = ´ 931.5 =  ´ ´  1 10 4 931.5 MeV =  93.15 keV 17. Number of nuclei in 1 kg water = ´ 6.023 10 18 26 Heavy water = ´ ´ ´  6.023 1.5 10 18 10 100 26 2 = ´ ´ 6.023 1.5 18 10 22 Energy realesed per fission = ´   ´ ( ) 2 2.014102 3.016049 1.007825 931.5 = ´ ´ ´ ´ ´   4.33 931.5 1.6 J 10 10 10 3 6 19 Hence total energy = ´ ´ 6.023 1.5 10 18 22 ´ ´ ´ ´  4.33 931.5 1.6 10 16 = 3200 MJ 85 ¢ Objective Questions (Level1) 1. Since during b  decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino as n P e ® + +  g. Hence Correct option is (c). 2. Since nuclear force is same for all nucleons. Hence F F F 1 2 3 = = Correct option is (a).. 3. Given reaction is 90 200 80 168 X Y ¾® Difference in mass number =  = 200 168 32 Hence Number of aparticles = = 32 4 8 Difference in atomic number = 10 hence number of b particles = 6 Hence correct option is (d). 4. The reaction is 92 235 0 1 54 138 38 94 0 1 3 U Xe Sr + ¾® + + n n ( ) The correct option is (b) three neutrons. 5. The reactions are A ® + b a and b b ® + C 2 After one a atomic number reduced by 2 and after 2b atomic number increased by. Hence A and C are isotopes Correct option is (d). 6. Here m m p n = = 1.00785 u 1.00866 u , and m a = 4.00274 u Dm m m m p n = +  2( ) a Þ Dm = +  [ ( ) ] 2 1.00785 1.00866 4.00274 u Þ Dm = 00 . 3028 u D D E m = ´ 931.5 MeV = ´ = 0.3028 931.5 28.21 MeV Hence correct option is (c). 7. N N N N =  = 0 0 0 7 8 1 8 But N N n = æ è ç ö ø ÷ 0 1 2 Þ 1 8 1 2 0 0 N N n = æ è ç ö ø ÷ Þ n = 3 Þ 3 8 1 2 ´ = T / s Þ T 1 2 8 3 / = s Hence correct option is (d). 8. N N e t =  0 l for mean life t = 1 l Þ N N e N e = =  ´ 0 1 0 l l Hence the fraction disintegrated =  =  æ è ç ö ø ÷ N N N e 0 0 1 1 Correct option is (b). 9. N N N N =  = 0 0 0 7 8 1 8 But N N n = æ è ç ö ø ÷ 0 1 2 Þ 1 8 1 2 = æ è ç ö ø ÷ n Þn = 3 Hence half life T 1 2 15 3 5 / = = min min Correct option is (a). 10. Since radioactive substance loses half of its activity in 4 days it means its half life T 12 4 / = days Now A = 5% of A 0 Þ A A = 1 20 0 Þ A A 0 1 20 = But A A e t =  0 l Þ 1 20 =  e t l Þ log 20 = lt Þ t = log 20 l 86 But l = log / e T 2 1 2 Þ = = ´ t T T e e 1 2 1 2 10 10 20 2 20 2 4 / / log log log log = 4.32 Þ t = 17.3 days Hence correct option is (c). 11. Total energy released per sec = = ´ 1.6 MW 1.6 J/s 10 6 Energy released per fission= 200 MeV = ´ ´ ´  200 10 10 6 19 1.6 J = ´ ´  2 10 11 1.6 J Number of fission per second = ´ ´ ´ = ´  1.6 1.6 10 2 10 5 10 6 11 16 /s Hence correct option is (a). 12. Q R R A = 0 1 3 / Volume = = 4 3 4 3 3 0 3 p p R R A Mass of nucleus = ´ ´  A 1.67 kg 10 27 Density r p = = ´ ´  mass volume 1.67 kg A R A 10 4 3 27 0 3 Þ r p = ´  1.67 kg 10 4 3 27 0 3 R Q r is independent of A, hence ratio of densities r r 1 2 1 = . Correct option is (d). 13. N N e t = 0 – l Þ N N e t 0 = – l Þ N N e e e 0 0 693 6 93 10 1 1 = = = ´ – . . – Fractional change =  =  æ è ç ö ø ÷  N N N e 0 0 1 1 063 ~ . Hence correct option is (b). 14. Since radioactive substance reduce to about 6% it means N N = 0 16 We have N N n = æ è ç ö ø ÷ ( ) 0 1 2 Þ 1 16 1 2 = æ è ç ö ø ÷ n Þ n = 4 4 2 1 2 ´ = T h / Þ T 1 2 30 / = min Hence correct option is (a). 15. Probability of a nucleus for survival of time t. p N N N e N e t t (survival) = = =   0 0 0 l l For one mean life t = 1 l Þ P survival = = =  ´  e e e l l 1 1 1 Hence Correct option is (a). ¢ Assertion and Reason 1. Here both assertion and reason are true but reason does not explain assertion. Hence correct option is (b). 2. Here assertion is false but reason is true since for heavier nucleus binding energy per nucleon is least. Correct option is (d). 3. Here both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 4. Here a ssertion is true but reason is false since electromagnetic waves are produced by accelerating charge particles. Correct option is (c). 87Read More

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