The $\beta$-decay process, discovered around 1900 , is basically the decay of a neutron $(n)$. In the…
The $\beta$-decay process, discovered around 1900 , is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $\left(e^{-}\right)$are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.
$n \rightarrow p+e^{-}+\bar{v}_{e}$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{e}\right)$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10^{6} \mathrm{eV}$. The kinetic energy carried by the proton is only the recoil energy.
Question: If the anti-neutrino had a mass of $3 \mathrm{eV} / \mathrm{c}^{2}$ (where $\mathrm{c}$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?
$0 \leq K \leq 0.8 \times 10^{6} \mathrm{eV}$
$3.0 \mathrm{eV} \leq K \leq 0.8 \times 10^{6} \mathrm{eV}$
$3.0 \mathrm{eV} \leq K < 0.8 \times 10^{6} \mathrm{eV}$
$0 \leq K < 0.8 \times 10^{6} \mathrm{eV}$
Solution
Kinetic energy $(K)$ of electron will be minimum or zero when total energy is shared by proton and anti-neutrino $\therefore 0 \leq \mathrm{K} < 0.8 \times 10^{6} \mathrm{eV}$
.