Answer
${\bf 5.29} \;\rm MeV$
Work Step by Step
The alpha decay process for $^{239}\text{Pu}$ is
$$
^{239}\text{Pu} \rightarrow ^{235}\text{U} + ^{4}\text{He}
$$
where $^{4}\text{He}$ is the alpha particle emitted during the decay.
The energy released ($E$) during this decay is due to the difference in mass (mass defect) between the original $^{239}\text{Pu}$ nucleus and the combined mass of the products, $^{235}\text{U}$ and $^{4}\text{He}$.
The formula for the energy released is:
$$
E = \Delta mc^2\tag 1
$$
where $\Delta m$ is the mass difference.
From appendix c;
- Mass of $^{239}\text{Pu}=\rm 239.052157 \;u$
- Mass of $^{235}\text{U}=\rm 235.043924 \;u$
- Mass of $^{4}\text{He}=\rm4.002602 \;u$
Calculate the mass defect $\Delta m$:
$$
\Delta m = \left( \text{Mass of }^{239}\text{Pu} \right) - \left( \text{Mass of }^{235}\text{U} + \text{Mass of }^{4}\text{He} \right)
$$
\begin{align}
\Delta m &= 239.052157 \, \text{u} - (235.043924 \, \text{u} + 4.002602 \, \text{u}) \\\\
\Delta m &=\bf 0.005631 \, \rm {u} \tag 2
\end{align}
Plug into (1);
$$
E =( 0.005631 )(1.67\times 10^{-27}) (3\times 10^8)^2=\bf 8.46339\times 10^{-13}\;\rm J
$$
$$
E =\color{red}{\bf 5.29} \;\rm MeV
$$
