# Ch42Word - Ch. 42 Page 1 CHAPTER 42 Nuclear Physics and...

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Ch. 42 Page 1 CHAPTER 42 – Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e 2 /4p Å 0 = (1.60 × 10 –19 C) 2 /4p(8.85 × 10 –12 C 2 /N · m 2 ) = 2.30 × 10 –28 J · m = 1.44 MeV · fm. 1. To find the rest mass of an α particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m = M He – 2 m e = (4.002603 u)(931.5 MeV/u c 2 ) – 2(0.511 MeV/ c 2 ) = 3727 MeV/ c 2 . 2. We convert the units: m = (139 MeV/ c 2 )/(931.5 MeV/u c 2 ) = 0.149 u . 3. The particle is a helium nucleus: r = (1.2 × 10 –15 m) A 1/3 = (1.2 × 10 –15 m)(4) 1/3 = 1.9 × 10 –15 m = 1.9 fm. 4. The radius of a nucleus is r = (1.2 × 10 –15 m) A 1/3 . If we form the ratio for the two isotopes, we get r 14 / r 12 = (14/12) 1/3 = 1.053. Thus the radius of 14 C is 5.3% greater than that for 12 C. 5. ( a ) The mass of a nucleus with mass number A is A u and its radius is r = (1.2 × 10 –15 m) A 1/3 . Thus the density is ρ = m / V = A (1.66 × 10 –27 kg/u)/ ) p r 3 = A (1.66 × 10 –27 kg/u)/ ) p(1.2 × 10 –15 m) 3 A = 2.3 × 10 17 kg/m 3 , independent of A . ( b ) We find the radius from M = V ; 5.98 × 10 24 kg = (2.3 × 10 17 kg/m 3 ) ) p R 3 , which gives R = 184 m . ( c ) For equal densities, we have = M Earth / ) p R Earth 3 = m U / ) p r U 3 ; ( 5 . 9 8 × 10 24 kg)/(6.38 × 10 6 m) 3 = (238 u)(1.66 × 10 –27 kg/u)/ r U 3 , which gives r U = 2.6 × 10 –10 m . 6. ( a ) The radius of 64 Cu is r = (1.2 × 10 –15 m) A 1/3 = (1.2 × 10 –15 m)(64) 1/3 = 4.8 × 10 –15 m = 4.8 fm. ( b ) We find the value of A from r = (1.2 × 10 –15 m) A 1/3 ; 3 . 9 × 10 –15 m = (1.2 × 10 –15 m) A 1/3 , which gives A = 34 . 7. We find the radii of the two nuclei from r = r 0 A 1/3 ; r = (1.2 fm)(4) 1/3 = 1.9 fm; r U = (1.2 fm)(238) 1/3 = 7.4 fm. If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic energy of the particle: K = U = Z Z U e 2 /4p Å 0 ( r + r U ) = (2)(92)(1.44 MeV · fm)/(1.9 fm + 7.4 fm) = 28 MeV .

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Ch. 42 Page 2 8. We find the radii of the two nuclei from r = r 0 A 1/3 ; r α = (1.2 fm)(4) 1/3 = 1.9 fm; r Am = (1.2 fm)(243) 1/3 = 7.5 fm. We assume that the nucleus is so much heavier than the particle that we can ignore the recoil of the nucleus. We find the kinetic energy of the particle from the conservation of energy: K i + U i = K f + U f ; 0 + Z Z Am e 2 /4p Å 0 ( r + r Am ) = K f + 0; (2)(95)(1.44 MeV · fm)/(1.9 fm + 7.5 fm) = K f , which gives K f = 29 MeV . 9. The radius of a nucleus is r = (1.2 × 10 –15 m) A 1/3 . If we form the ratio for the two nuclei, we get r X / r U = ( A X / A U ) 1/3 ; ! = ( A X /238) 1/3 , which gives A X = 30. From the Appendix, we see that the stable nucleus could be P 15 31 . 10. From Figure 42–1, we see that the average binding energy per nucleon at A = 40 is 8.6 MeV. Thus the total binding energy for 40 Ca is (40)(8.6 MeV) = 340 MeV . 11. ( a ) From Figure 42–1, we see that the average binding energy per nucleon at A = 238 is 7.5 MeV. Thus the total binding energy for 238 U is (238)(7.5 MeV) = 1.8 × 10 3 MeV . ( b ) From Figure 42–1, we see that the average binding energy per nucleon at A = 84 is 8.7 MeV. Thus the total binding energy for 84 Kr is (84)(8.7 MeV) = 7.3 × 10 2 MeV . 12. Deuterium consists of one proton and one neutron. We find the binding energy from the masses: Binding energy = [ M ( 1 H) + m ( 1 n) – M ( 2 H)] c 2 = [(1.007825 u) + (1.008665 u) – (2.014102 u)] c 2 (931.5 MeV/u c 2 ) = 2.22 MeV .
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch42Word - Ch. 42 Page 1 CHAPTER 42 Nuclear Physics and...

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