Question

The $\beta$-decay process, discovered in 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 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, that is, $n\to p+e^{-}+{\bar{v}}_e$, 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 $0.8\times 10^{6}eV$. The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino had 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?

• A. $0\le K\le 0.8\times 10^{6}eV$
• B. $3.0eV\le K\le 0.8\times 10^{6}eV$
• C. $3.0eV\le K<0.8\times 10^{6}eV$
• D. $0\le K<0.8\times 10^{6}eV$

Answer 1

Answer: (D)

Here, We know that $Q$-value is $(K.E.{)}_p+(K.E.{)}_{{\beta }^{-}}+(K.E.{)}_{\bar{v}}$
Now, $(k\cdot E.{)}_{{\beta }^{-}}{|}_{\min }=0$
Therefore, $0<(K.E.{)}_{{\beta }^{-}}\le Q-(K.E.{)}_{{\beta }^{-}}+(K.E.{)}_{\bar{v}}$
That is, $(K.E.{)}_{{\beta }^{-}}{|}_{\max }=Q$
That is, the range of kinetic energy of the electron lies between zero and less than the kinetic energy of the electron, that is, $0\le K<0.8\times 10^{6}eV$.