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)`

Options

A. The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.75MeV` approximately
B. The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.25MeV` approximately
C. The emission of proton from `._(92)^(238)U` can be spontaneous
D. The emission of proton from `._(92)^(238)U` can not be spontaneous.

Answer

B, D

Explanation

The energy released in `alpha`- decay 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)c^(2)`
`=(0.00456u)(931.5MeV//u)=4.25MeV`
If `._(92)^(238)U` spontaneously emits a proton, the decay process would be `._(92)^(238)Uto_(92)^(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)=-7.68MeV`
Thus, the `Q` of the process is negative and therefore it cannot proceed spiontaneously. We will have to supply an energy of `7.68MeV` to `._(92)^(238)U` nucleus to make it emit a proton.