p43_011 - 11. (a) For 239 Pu, Q = 94e and R = 6.64 fm....

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11. (a) For 239 Pu, Q =94 e and R =6 . 64 fm. Including a conversion factor for J eV, we obtain U = 3 Q 2 20 πε 0 r = 3[94(1 . 60 × 10 19 C)] 2 (8 . 99 × 10 9 N · m 2 / C 2 ) 5(6 . 64 × 10 15 m) µ 1eV 1 . 60 × 10 19 J =1 . 15 × 10 9 eV = 1 . 15 GeV . (b) Since Z =94and A = 239, the electrostatic potential per nucleon is 1 . 15 GeV / 239 = 4 . 81 MeV / nucleon, and per proton is 1
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