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.
What is the maximum energy of the anti-neutrino?

• A. Zero.
• B. Much less than $0.8\times {10}^{6}eV$.
• C. Nearly $0.8\times {10}^{6}eV$.
• D. Much larger than $0.8\times {10}^{6}eV$.

Answer 1

The correct option is C Nearly $0.8\times {10}^{6}eV$.

Anti-neutrino will have maximum energy when electron will have minimuim energy or nearly zero energy.