Question

What is the total energy (in MeV) released in the beta decay

of a neutron?

Short answer

Therefore, the total energy released is$0.78\mathrm{MeV}$

Step-by-step solution

Given information

Beta decay of a neutron.

Explanation

We need to find the energy released during neutron beta decay.

The decay is given as:

$n\to p^{+}+e^{-}+\overline{v}$

We find the value of $\Delta m$:

$$\begin{gathered} \Delta m=m\left(n\right)-m\left(p^{+}\right)-m\left(e^{-}\right) \\ \Delta m=939.57\mathrm{MeV}/{\mathrm{c}}^2-938.28\mathrm{MeV}/{\mathrm{c}}^2-0.51\mathrm{MeV}/{\mathrm{c}}^2 \\ \mathrm{\Delta m}=0.78\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Calculations

The energy released is calculated as:

$$\begin{gathered} E=\Delta m{c}^2 \\ E=\left(0.78\mathrm{MeV}/{\mathrm{c}}^2\right)\left(c^2\right) \\ E=0.78\mathrm{MeV} \end{gathered}$$