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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 + + + ν . Ignoring the slight difference in ionization energies, we have 1 3 H atom He atom 2 3 + ν 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 b gb g P44.38 (a) For e + decay, Q M M m c Q X Y e = = = − 2 39 962 591 39 963 999 2 0 000 549 931 5 2 33 2 b g b gb g . . . . . u u u MeV u MeV Since Q < 0, the decay cannot occur spontaneously. (b) For alpha decay, Q M M M c Q X Y = = = − α b g b g 2 91 905 287 4 002 603 93 905 088 931 5 2 24 . . . . . u u u MeV u MeV Since Q < 0, the decay cannot occur spontaneously. (c) For alpha decay, Q M M M c Q X Y = = = α b g
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