Question

The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron $\left(n\right)$. In the laboratory, a proton $\left(p\right)$ 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 a continuous spectrum. Considering a three-body decay process, i.e. $n\to p+e^{-}+{\overline{v}}_e$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left({\overline{v}}_e\right)$ 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 $0.8\times {10}^{6}eV$. The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino had a mass of 3 $eV/c^2$ (where $c$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?

• A. $0\le K\le 0.8\times {10}^{6}eV$
• B. $3.0eV\le K\le 0.8\times {10}^{6}cV$
• C. $3.0eV\le K<0.8\times {10}^{6}eV$
• D. $0\le K<0.8\times {10}^{6}eV$

Answer 1

The correct option is D $0\le K<0.8\times {10}^{6}eV$

$K$ can't be equal to $0.8\times {10}^{6}eV$ as anti-neutrino must have some energy