Sol-Ch43 - CHAPTER 43 E xercises 1. R = 1 .2A 1 / 3 : (a)...

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Exercises 1. R = 1.2A 1/3 : (a) 3.0 fm; (b) 4.6 fm; (c)7.4 fm 2. M = 6.0 ×  10 24 = 4π ρ R 3 /3, R = 184 m 3. ρ = 3M/4πr\R 3 = 4.8 ×  10 17 kg/m 3 4. R 3 = 3M/4π ρ = 103.7 ×  10 30 , so R = 4.7 ×  10 10 m 5. (A 2 /A 1 ) 1/3 = 2, so A 2 /A 1 = 8 6. m = 235 u = 4π ρ R 3 /3, find R = 2.6 ×  10 –10 m 7. 63.55 = 62.95x + 64.95(1 – x), thus x = 0.7, so 70% is 63 Cu 8. (0.91)20 + (0.09)22 = 20.18 u 9. R Au = 7 fm, then K = U = k(79e)(2e)/(R Au + R α ) = 26 MeV 10. ρ = 3(26e)/4πR 3 , where R 3 = (1.2 ×  10 –15 ) 3 (56) ρ = 1.03 ×  10 25 C/m 3 11. (a) 8.55 MeV; (b) 7391 MeV 12. (a) BE = 31.99 MeV; BE/A = 5.33 MeV (b) BE = 1118 MeV, BE/A = 8.41 MeV 13. BE( 13 C) = 97.1 MeV, BE( 13 N) = 94.1 MeV 14. (a) 115.5 MeV; (b) 112 MeV 15. (a) BE( 7 Li) = 39.24 MeV; EB( 6 Li) = 31.99 MeV; so ∆BE = 7.25 MeV; (b) EB/A = 5.61 MeV 16. (a) BE( 12 C) = 92.16 Mev; Be( 11 B) = 76.21 MeV, CHAPTER 43

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so ∆BE = 15.95 MeV; (b) EB/A = 7.68 MeV 17. N o = mN A /M = 1.0037 ×  10 20 ; R o = λ N o = 0.693N o /T 1/2 = 4.2 ×  10 11 Bq 18. N o = mN A /M = 1.88 ×  10 19 ; R o = λ N o = 0.693N o /T 1/2 = 4.2 ×  10 13 Bq 19. N o = R o T 1/2 /0.693 = 1.52 ×  10 8 N = N o exp(– λ t) = N o exp(–0.693/3.82) = 1.27 ×  10 8 20. (a) (m Be – 2m He ) = 9.9 ×  10 –5 u; Q = ∆mc 2 = 92.2 keV (b) (m C – 3m He ) = –7.8 ×  10 –3 u. Q <0, so not possible. 21. 11 B, yes 22. In 1 m 2 , N o = mN A /M = 6.69 ×  10 15 . R o = 0.693N o /T 1/2 = 5.07 ×  10 6 Bq, and R = 3.7 ×  10 4 Bq Thus, 3.7 ×  10 4 = 5.07 ×  10 6 exp(– λ t), find t = 206 y. 23. (a) – λ t = ln(9/15), so T 1/2 = 0.693/ λ = 3.4 h. (b) N o = R o T 1/2 /0.693 = 9.8 ×  10 9 24. N o = mN A /M = 2.52 ×  10 21 . R
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch43 - CHAPTER 43 E xercises 1. R = 1 .2A 1 / 3 : (a)...

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