Question

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`.

Answer 1

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`.