We are given the following atomic masses:
`._(92)^(238)U=238.05079u ._(2)^(4)He=4.00260u`
`._(90)^(234)Th=234.04363u ._(1)^(1)H=1.00783u`
`._(91)^(237)Pa=237.05121u`
Here the symbol `Pa` is for the element protactinium `(Z=91)`

Explanation

(a) The alpha decay of `._(92)^(238)U` is given by Eq. (13.20). 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)uxxc^(2)`
`=(0.00456u)c^(2)`
`=(0.00456u)(931.5MeV//u)`
`4.25MeV`.
(b) If `._(92)^(238)U` spontaneously emits a proton, the decay process would be
`._(92)^(238)Uto._(91)^(237)Pa+._(1)^(1)H`
The Q for this process to happen is
`=(M_(U)-M_(Pa)-M_(H))c^(2)`
`=(238.05079-237.05121-1.00783)uxxc^(2)`
`=(-0.00825u)c^(2)`
`= -(0.00825u)(931.5MeV//u)`
`= -7.86 MeV`
`Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a `._(92)^(238)U` nucleus to make it emit a proton.