1247_Physics ProblemsTechnical Physics

1247_Physics ProblemsTechnical Physics - 588 P44.60 Nuclear...

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588 Nuclear Structure P44.60 (a) rrA A == × 0 13 15 120 10 . m . When A = 12, r 275 10 15 m . (b) F kZ e r Z r e = = ×⋅ −× 1 8 9 91 0 11 6 01 0 2 2 9 2 2 af ej .. Nm C C 22 When Z = 6 and r 15 m , F = 152 N . (c) U kqq r r Z r e e = ×− × 12 2 9 2 1 899 10 1 16 10 a f When Z = 6 and r 15 m , U = 419 10 262 13 J M e V . (d) A = 238; Z = 92, r 744 10 15 m F = 379 N and U = 282 10 176 12 M e V . P44.61 (a) Because the reaction p n e →+ + + ν would violate the law of conservation of energy m p = 1007276 u m n = 1008665 u m e u + 549 10 4 . . Note that mm m np +> + e . (b) The required energy can come from the electrostatic repulsion of protons in the nucleus. (c) Add seven electrons to both sides of the reaction for nuclei 7 13 NC e 6 13 →+ + + to obtain the reaction for neutral atoms 7 13 N atom C atom e e 6 13 + + +− Q c m m mmm Q Q = −−− =− × = 21 3 1 3 4 3 931 5 13 005 739 13 003 355 2 5 49 10
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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