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`.

Explanation

The energy released in the decay is the disintegration energy Q, which we can calculate from the change in mass `DeltaM` due to the `""^(238)U` decay.
Calculation: To this we use Eq 36-50
`Q=M_(i)c^(2)-M_(f)c^(2)` 40-24
where the initial mas `M_(i)` is that of `""^(238)U` and the final mass `M_(f)` is the sum of the `""^(238)Th` and `""^(4)He` masses. Using the atomic masses given in the problem statement. Eq. 40-24 becomes
`Q=(238.05079u)c^(2)-(234.043 63u+4.002 60u)c^(2)`
`=(0.004 56u)c^(2)=(0.00456u)(931.494 0.13 MeV//u)`
`=4.25MeV`
Note that using atomic masses instead of nuclear masses does to affect the result because the total mass of the electrons in the products subtracts out from themass of the nucleons+electrons in the original `""^(238)U`.
b. Show that `""^(238)U` cannot spontaneously emit a proton, that is protons donot leak out of the nucleus in spite of the proton proton repulsion within the nucleus.
Calculation: If this happened , the decay process would be
`""^(238)Uto""^(237)Pa+""^(.)H`
(You should verify that both nuclear charge and the number of necleons and conserved in this process). Using the same Key idea as in part (a) and proceeding as we did there, we would find that the mass of the two decay products
`237051u+1.00783u`
would exceed the mass of `""^(238)U` by `Deltam=0.00825u`, with disintegration energy
`Q=-7.68 MeV`
The minus sign indicates that we must add 7.68 MeV to a `""^(238)U` nucleus before it will emit a proton, it will certainly not do so spontaneously.