Question

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

decay of ²⁴Na?

Short answer

Hence, the energy is$5.513489\mathrm{MeV}$

Step-by-step solution

Given information

$$\begin{gathered} \Delta m=0.005919\mathrm{u} \\ \mathrm{m}\left({}^{24}\mathrm{N}_{\mathrm{a}}\right)=23.990961\mathrm{u} \\ \mathrm{m}\left({}^{24}\mathrm{M}_{\mathrm{g}}\right)=23.985042\mathrm{u} \end{gathered}$$

Explanation

The decay is given as:

${}^{24}N_a\to {}^{24}M_g+e^{-}+\overline{v}$

Formula for finding energy is:

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

Calculations

Substituting the values:

$$\begin{gathered} E=(0.005919\mathrm{u})c^2\times \frac{931.49\frac{\mathrm{MeV}}{{\mathrm{c}}^2}}{1\mathrm{u}} \\ E=5.513489\mathrm{MeV} \end{gathered}$$