Question

Consider the decay of a free neutron at rest : $n\to p+e^{-}$
Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the $\beta$−decay of a neutron or a nucleus (Fig.).

Note: The simple result of this exercise was one among the several arguments advanced by $W$. Pauli to predict the existence of a third particle in the decay products of $\beta$−decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin $1/2$ (like $e^{-}$, $p$ or $n$), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is : $n\to p+e^{-}+v$]

Answer 1

Step 1: As per the neutron decay equation,
$n\to p+e^{-}$

Energy of electron is equal to $E=\Delta m{c}^2$

Where $\Delta m$ = mass defect

i.e. $\Delta m$ = mass of neutron – mass of proton and electron
This is fixed. Therefore, two body decay of this type cannot explain the observed continuous energy distribution in the proton decay of a neutron or a nucleus.

Step 2: Actually, Pauli explained the existence of a third particle in the −decay , which is known as neutrino. It has an intrinsic spin $1/2$, has a small mass and it is neutral. So, actual neutron decay is as shown below,

$n\to p+e^{-}+v^{\prime }$