A common example of alpha decay is

${{}^{238}}_{92}\mathrm{U}{{\to }^{234}}_{90}\mathrm{Th}{+}_2{\mathrm{He}}^4+Q$

Given:

${{}^{238}}_{92}\mathrm{U}=238.05060 \mathrm{u}$

${{}^{234}}_{90}\mathrm{Th}=234.04360 \mathrm{u}$

${{}^4}_2\mathrm{He}=4.00260 \mathrm{u}$ and $1\mathrm{u}=931.5\frac{\mathrm{MeV}}{c^2}$

The energy released $\left(Q\right)$ during the alpha decay of ${{}^{238}}_{92}\mathrm{U}$ is _____ $\mathrm{MeV}$.

The formula to calculate the Q-value of the given nuclear reaction can be written as

$Q=\left(m_{\mathrm{U}}-m_{\mathrm{Th}}-m_{\mathrm{He}}\right)c^2 ...\left(1\right)$

Substitute the values of the known parameters into equation (1) to calculate the required Q-value for the reaction.

$\begin{array}{rcl}Q& =& \left(238.05060 \mathrm{u}-234.04360 \mathrm{u}-4.00260 \mathrm{u}\right)c^2\\ & =& 0.0044c^2\times 931.5 \mathrm{MeV}/c^2\\ & \approx & 4 \mathrm{MeV}\end{array}$