Answer
See answers.
Work Step by Step
$$ ^{238}_{92}U \rightarrow ^{234}_{90}Th+^{4}_{2}He $$
The kinetic energy is low enough so that classical physics is applicable, i.e., $KE=(p^2)/2m$.
Calculate the thorium’s recoil KE.
$$KE_{Th}=\frac{p_{Th}^2}{2m_{Th}}$$
Assuming the uranium nucleus is initially at rest, the two daughter nuclei have equal and opposite momentum.
$$KE_{Th}=\frac{p_{\alpha}^2}{2m_{Th}}=\frac{2m_{\alpha}KE_{\alpha}}{2m_{Th}}$$
$$KE_{Th}=\frac{m_{\alpha} }{m_{Th}} KE_{\alpha}=\frac{4u}{234u}(4.20MeV)=0.0718MeV$$
The Q-value is the total kinetic energy of the thorium and of the alpha particle.
Q=4.20 MeV+0.0718 MeV=4.27 MeV