Question

What is the total energy (in MeV) released in the beta-minus

decay of ³H?

Short answer

Hence, the energy released is $0.0186298\mathrm{MeV}$

Step-by-step solution

Given information

$$\begin{gathered} m\left({}^3\mathrm{H}\right)=3.016049\mathrm{u} \\ \mathrm{m}\left({}^3\mathrm{He}\right)=3.016029\mathrm{u} \\ \mathrm{\Delta m}=0.000020\mathrm{u} \end{gathered}$$

Explanation

The decay is given as:

${}^3\mathrm{H}\to {}^3\mathrm{He}+e^{-}+\overline{v}$

Formula for energy released is:

$E=\Delta m{c}^2$

Calculations

Substitute the values,

$$\begin{gathered} E=(0.000020\mathrm{u})(931.49\mathrm{MeV}/\mathrm{u}) \\ E=0.0186298\mathrm{MeV} \end{gathered}$$