Before the neutrino hypothesis, the beta decay process was thought to be the transition, $n \to p + {e^ - }$ If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range. 

Energy of a particle, given by Einstein's special theory of relativity is,

$\mathrm{E}=\sqrt{p^{2} c^{2}+m_{0}^{2} c^{4}}$

$\ldots$ $(1)$

where $p=$ momentum of a particle

$c=$ speed of light in vacuum

$m_{0}=$ rest mass of a particle

For given process, $n \rightarrow p+\bar{e}$, according to law of conservation of momentum,

$\overrightarrow{p_{n}}=\overrightarrow{p_{p}}+\overrightarrow{p_{e}}$ $\therefore \overrightarrow{0}=\overrightarrow{p_{p}}+\overrightarrow{p_{e}} \quad(\because \text { neutron is at rest) }$ $\therefore \overrightarrow{p_{p}}=-\overrightarrow{p_{e}}$

$\therefore \overrightarrow{p_{p}}=-\overrightarrow{p_{e}}$

Taking magnitude on both the sides, $p_{p}=p_{e}=p$ (Suppose)

... $(2)$

Energy of neutron from equation $(1)$,

$\mathrm{E}_{n}=\sqrt{p_{n}^{2} c^{2}+m_{n}^{2} c^{4}}$ $\therefore \mathrm{E}_{n}=m_{n} c^{2}\left(\because p_{n}=0\right)$ $...(3)$

Energy of proton from equation $(1) $,

$\mathrm{E}_{p}=\sqrt{p_{p}^{2} c^{2}+m_{p}^{2} c^{4}}$

$\therefore \mathrm{E}_{p}=\left(p^{2} c^{2}+m_{p}^{2} c^{4}\right)^{1 / 2}$ $\quad\left(\because p_{p}=p\right)$$...(4)$

Energy of electron from equation $(1)$,

$\mathrm{E}_{e}=\sqrt{p_{e}^{2} c^{2}+m_{e}^{2} c^{4}}$

$\therefore \mathrm{E}_{e}=\left(p^{2} c^{2}+m_{e}^{2} c^{4}\right)^{1 / 2} \quad\left(\because p_{e}=p\right)$$...(5)$