Answer
${\bf 0.0186 } \, \text{MeV}$
Work Step by Step
We need to determine the energy released in the beta-minus decay of tritium ($^3\text{H}$).
In beta-minus decay,
$$
^3\text{H} \to \; ^3\text{He} + e^- + \bar{\nu}\tag 1
$$
Tritium ($^3\text{H}$) decays into helium-3 ($^3\text{He}$), releasing an electron and an antineutrino.
And we need to calculate the total energy released in this decay process in MeV. The energy released corresponds to the mass difference between the parent and daughter nuclei (or atoms).
So, the total energy released is given by
$$E=\Delta m c^2\tag 2$$
From (1), we can see the reaction for beta-minus decay; where $ e^- $ is the emitted electron and $ \bar{\nu} $ is the antineutrino.
The known masses are:
- Mass of $^3\text{H}= 3.016049\;\rm u$
- Mass of $^3\text{He}= 3.016029 \;\rm u$
The mass difference $\Delta m$ is therefore:
$$
\Delta m = m_{(^3\text{H})} - m_{(^3\text{He}) }= 3.016049 \, \text{u} - 3.016029 \, \text{u} $$
$$
\Delta m = \bf 0.000020 \rm \, {u}
$$
Plug into (2);
$$ E=(0.000020 ) (931.49 )$$
where $ 1 \, \text{u} = 931.49 \, \text{MeV}/c^2 $,
$$
E = \color{red}{\bf 0.0186 } \, \text{MeV}
$$
:
