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

• A

  Zero
• B

  Much less than $0.8 \times 10^{6} \mathrm{eV}$.
• Correct answer
  C

  Nearly $0.8 \times 10^{6} \mathrm{eV}$
• D

  Much larger than $0.8 \times 10^{6} \mathrm{eV}$

Hence, the correct answer is option: 'C'.