The `beta`-decay process, discovered around `1900`, is basically the decay of a neutron `(n)`, In the laboratory, a proton `(p)` and an electron `(e^(-))` are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a tro-body dcay process, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)V_(e ))` to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino has a mass of `3eV//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?

Options

A. `0 le K le 0.8 xx 10^6eV`
B. `3.0eV le K le 0.8 xx 10^6eV`
C. `3.0eV le K lt 0.8 xx 10^6eV`
D. `0 lt K le 0.8 xx 10^6eV`

Answer

D