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

Options

A. zero
B. Much less than `0.8xx10^(6)eV`
C. Nearly `0.8xx10^(6)eV`
D. Much larger than `0.8xx10^(6)eV`

Answer

C

Explanation

`KE_(max)="of"beta^(-)`
`Q=0.8xx10^(6)eV`
`KE_(P)+KE_(beta^(-))+KE_(overset(-)v)=Q`
`KE_(P)` is almost zero
When `KE_(beta^(-))=0`
then `KE_(overset(-)v)=Q-KE_(P)`
`~=Q`