Answer
${\bf 5.513} \, \text{MeV}$
Work Step by Step
In beta-minus decay, $^{24}\text{Na}$ decays into $^{24}\text{Mg} $, releasing an electron and an antineutrino. We need to find the total energy released in this decay process (in MeV).
$$
^{24}\text{Na} \rightarrow ^{24}\text{Mg} + e^- + \bar{\nu}\tag 1
$$
where $ e^- $ is the emitted electron and $ \bar{\nu} $ is the antineutrino.
So, the total energy released is given by
$$E=\Delta m c^2\tag 2$$
The energy released corresponds to the difference in mass between the parent and daughter atoms.
The known masses are (see Appendix C):
- Mass of $^{24}\text{Na}= 23.990961\;\rm u$
- Mass of $^{24}\text{Mg}=23.985042 \;\rm u$
The mass difference $\Delta m$ is:
$$
\Delta m = m_{(^{24}\text{Na}) }- m_{(^{24}\text{Mg})} = 23.990961 \, \text{u} - 23.985042 \, \text{u}$$
$$
\Delta m = \bf 0.005919 \, \rm {u}
$$
Plug into (2);
$$ E=(0.005919 ) (931.49 )$$
where $ 1 \, \text{u} = 931.49 \, \text{MeV}/c^2 $,
$$
E = \color{red}{\bf 5.513} \, \text{MeV}
$$
