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?

1. A \(0 \leq K \leq 0.8 \times 10^{6} \mathrm{eV}\)
2. B \(3.0 \mathrm{eV} \leq K \leq 0.8 \times 10^{6} \mathrm{eV}\)
3. C \(3.0 \mathrm{eV} \leq K < 0.8 \times 10^{6} \mathrm{eV}\)
4. D \(0 \leq K < 0.8 \times 10^{6} \mathrm{eV}\)