1240_Physics ProblemsTechnical Physics

# 1240_Physics ProblemsTechnical Physics - Chapter 44 P44.37...

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Chapter 44 581 P44.37 1 3 H nucleus He nucleus e 2 3 →+ + ν becomes 1 3 2 3 2 H nucleus e He nucleus e +→ + + −− . Ignoring the slight difference in ionization energies, we have 1 3 H atom He atom 2 3 3 016 049 3 016 029 0 3 016 049 3 016 029 931 5 0 018 6 18 6 2 .. . . . u u u u MeV u MeV keV =+ + =− = = Q c Q bg b g P44.38 (a) For e + decay, QM M m c Q XY e = 2 39 962 591 39 963 999 2 0 000 549 931 5 233 2 b g b g . . . u u M e V u MeV Since Q < 0, the decay cannot occur spontaneously. (b) For alpha decay, QM M Mc Q = α b g 2 91 905 287 4 002 603 93 905 088 931 5 224 . . . u u M e V u MeV Since Q < 0, the decay (c) For alpha decay, Q = = b g 2 143 910 083 4 002 603 139 905 434 931 5 191 . . . u u M e V u MeV
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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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