A free neutron decays according to Eq. 42-26. If the neutron–hydrogen atom mass difference is 840$\mathbf{\mu u}$, what is the maximum kinetic energy${\mathbf{K}}_{\mathbf{max}}$possible for the electron produced in neutron decay?

1. A free neutron decay, $\mathrm{n}\to \mathrm{p}+{\mathrm{e}}^{-}+\mathrm{v}$
2. The neutron-hydrogen atom mass difference,$\left({\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{H}}\right)=840\times {10}^{-6}\mathrm{u}$

The kinetic energy of a particle in a decay process can be given as the maximum binding energy of the disintegrating process.

Formula:

The kinetic energy of a particle is as follows:

${\mathrm{K}}_{\max }=∆{\mathrm{mc}}^2$ …… (i)

The neutron decay according to Eq. 42-26 is given as:

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

From the above equation, it is observed that the kinetic energy is maximum if there is no neutrino emission. Thus, the kinetic energy possible for the electron produced in neutron decay using equation (i) is given as:

${\mathrm{K}}_{\max }=\left({\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{p}}-{\mathrm{m}}_{\mathrm{e}}\right){\mathrm{c}}^2$

As the mass of hydrogen is given by ${\mathrm{m}}_{\mathrm{H}}={\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{e}}$, one can write above equation as:

${\mathrm{K}}_{\max }=\left({\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{H}}\right){\mathrm{c}}^2$

Now, using the given data in the above equation, we can get the maximum kinetic energy of the electron as follows:

$$\begin{gathered} {\mathrm{K}}_{\max }=\left(840\times {10}^{-6}\mathrm{u}\right)\left(931.5\frac{\mathrm{MeV}}{\mathrm{u}}\right) \\ =0.783\mathrm{MeV} \end{gathered}$$

Hence, the value of the maximum energy is 0.783 MeV .