a)
Step 1:Formula required to calculate the energy released during the \alpha -decay of ^{238}_{92} U .
As a nucleus undergoes alpha-decay, it turns it into a separate nucleus by emitting an alpha-particle, For instance, ^{238}_{92} U transforms ^{234}_{90}Th when it undergoes alpha decay.
^{238}_{92} U \rightarrow ^{234}_{90}Th + ^{4}_2He
The decay value for disintegration energy (Q) can be expressed by
Q = (m_X - m_Y - m_{He}) * c^2
or
Q = (m_U - m_{Th} - m_{He}) * c^2
Step 2: Replacing values of atomic masses into disintegration energy
^{238}_{92} U = 238.05079 u
^{234}_{90} Th = 234.04363 u
^{2}_{4} He = 4.00260 u
Q = (m_U - m_{Th} - m_{He}) * c^2
Q = (238.05079 - 234.04363 - 4.00260) * c^2
Q = (0.00456) * (931.5 MeV/u)
Q = 4.25 MeV
b)
Step 1: The decay process, if a proton is spontaneously emitted
^{238}_{92} U \rightarrow ^{237}_{91} Th + ^{1}_1He
Step 2: Replacing values of atomic masses into disintegration energy
^{238}_{92} U = 238.05079 u
^{1}_{1} He = 1.00783 u
^{237}_{91} Pa = 237.05121 u
Q = (m_{U} - m_{Pa} - m_{He}) * c^2
Q = (238.05079 - 237.05121 - 1.00783) * c^2
Q = - (0.00825 u) * c^2
Q = - (0.00825 u)(931.5 MeV/u)
Q = - 7.68 MeV
The Q of the process is thus negative and cannot therefore be spontaneously proceeded. To make it emit one proton, we will have to provide an energy of 7.68 MeV to a ^{238}_{92} U nucleus.