We are given the following atomic masses:
$^{238}_{92}U = 238.05079 \; u$
$^{4}_{2}He = 4.00260 \; u$
$^{234}_{90}Th = 234.04363 \; u$
$^{1}_{1}H = 1.00783 \; u$
$^{237}_{91}Pa = 237.05121 \; u$
Here,the symbol $Pa$ represents the element protactinium $(Z=91)$.
$(a)$ Calculate the energy released during the alpha decay of $^{238}_{92}U$.
$(b)$ Show that $^{238}_{92}U$ cannot spontaneously emit a proton.

(N/A) The alpha decay of $^{238}_{92}U$ is represented by the equation: $^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He$. The energy released ($Q$-value) in this process is given by $Q = (M_{U} - M_{Th} - M_{He})c^{2}$.
Substituting the given atomic masses:
$Q = (238.05079 - 234.04363 - 4.00260) \; u \times c^{2}$
$Q = 0.00456 \; u \times c^{2}$
Using the conversion factor $1 \; u = 931.5 \; MeV/c^{2}$:
$Q = 0.00456 \times 931.5 \; MeV = 4.25 \; MeV$.
$(b)$ If $^{238}_{92}U$ were to spontaneously emit a proton,the decay process would be: $^{238}_{92}U \rightarrow ^{237}_{91}Pa + ^{1}_{1}H$.
The $Q$-value for this process is:
$Q = (M_{U} - M_{Pa} - M_{H})c^{2}$
$Q = (238.05079 - 237.05121 - 1.00783) \; u \times c^{2}$
$Q = -0.00825 \; u \times c^{2}$
$Q = -0.00825 \times 931.5 \; MeV = -7.68 \; MeV$.
Since the $Q$-value is negative,the process is endothermic and cannot proceed spontaneously. An external energy of $7.68 \; MeV$ must be supplied to the $^{238}_{92}U$ nucleus to trigger proton emission.