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

• A. Zero.
• B. Much less than $0.8 \times 10^{6} eV$
• C. Nearly $0.8 \times 10^{6} eV$
• D. much larger than $0.8 \times 10^{6} eV$

Answer 1

Answer: (C)

Here, $n \rightarrow p +e^{-}+\bar{v}_{e} .$
The $Q$-value of the reaction is
$(K . E .)_{\beta^{-} \max }=0.8 \times 10^{8} eV$
Now, $(K . E .)_{p}+(K . E .)_{\beta^{-}}+(K . E .)_{\bar{v}}= Q$
It is given that $(K . E .)_{p} \approx 0 .$
Therefore, $(K . E .)_{\bar{v})}= Q$ when $(K . E .)_{\beta^{-}}=0 .$
That is, the maximum energy of the anti-neutrino is equal to
approximately the maximum kinetic energy of the electron;
therefore, as given in the paragraph, the maximum $K . E$. of the electron is
$0.8 \times 10^{6} eV$.