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.
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, $n\to p+e^{-}+{\bar{v}}_e.$
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
$0.8\times 10^{6}eV$.