InterviewSolution
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InterviewSolution
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We are given the following atomic masses : `._(92)^(238)U = 238.05079 u " " ._(2)^(4)He=4.00260 u` `._(90)^(234)Th=234.04363u " " ._(1)^(1)H=1.00783 u` `._(91)^(237)Pa=237.05121 u` Here the symbol Pa is for the element protactinium (Z = 91) . Calculate the energy released during the alpha decay of `._(92)^(238)U`. |
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Answer» The alpha decay of `._(92)^(238)U` is given by `._(Z)^(Delta)X to ._(Z-2)^(A-4)Y_(2)^(4)+ ._(2)^(4)He`. The energy released in this process is given by `Q=(M_(U)-M_(Th)-M_(He))c^(2)` Substituting the atomic masses as given in the data, we find `Q=(238.05079-234.04363-4.00260)u xx c^(2)` `= (0.00456 u)c^(2)` `= (0.00456 u) (931.5 MeV//u)` `= 4.25 MeV`. |
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