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, $n \rightarrow p +e^{-}+\bar{v}_{e}$, 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 $0.8 \times 10^{6} eV$. The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino had a mass of $3 eV / c ^{2}$ (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. $0 \leq K \leq 0.8 \times 10^{6} eV$
• B. $3.0 eV \leq K \leq 0.8 \times 10^{6} eV$
• C. $3.0 eV \leq K < 0.8 \times 10^{6} eV$
• D. $0 \leq K < 0.8 \times 10^{6} eV$

Answer 1

Answer: (D)

Here, We know that $Q$-value is $(K . E .)_{p}+(K . E .)_{\beta^{-}}+(K . E .)_{\bar{v}}$
Now, $(k \cdot E .)_{\beta ^-}\left.\right|_{\min }=0$
Therefore, $0<(K . E .)_{\beta^{-}} \leq Q -(K . E .)_{\beta^{-}}+(K . E .)_{\bar{v}}$
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, $0 \leq K <0.8 \times 10^{6} eV$.