Answer
${\bf 0.774} \, \text{MeV}$
Work Step by Step
In beta decay, a neutron decays into a proton, releasing an electron and an antineutrino. The goal is to calculate the energy released in this decay (in MeV).
The reaction for neutron beta decay is:
$$
n \rightarrow p^+ + e^- + \bar{\nu}\tag 1
$$
where $ p^+ $ is the proton, $ e^- $ is the electron, and $ \bar{\nu} $ is the antineutrino.
The total energy released is given by
$$E=\Delta m c^2\tag 2$$
The known masses are (see Appendix C):
- Mass of neutron $ n = 1.008665\;\rm u$
- Mass of proton $ p^+ = 1.007285 \;\rm u$ (it is written in your textbook as 1.007825 which is wrong and the right value is here, they mistakenly replaced 2 with 8).
- Mass of electron $ e^-= 0.00054858 \;\rm u$
The mass difference $\Delta m$ is:
$$
\Delta m = m_{n} - m_{p^+} - m_{e^-} = 1.008665- 1.007285 -0.00054858
$$
$$
\Delta m = \bf 0.00083142\, \rm u
$$
Plug into (2);
$$ E=(0.00083142) (931.49 )$$
where $ 1 \, \text{u} = 931.49 \, \text{MeV}/c^2 $,
$$
E = \color{red}{\bf 0.774} \, \text{MeV}
$$
