Consider the decay of a free neutron at rest: $n \rightarrow p + e^-$. 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.

(N/A) For a two-body decay of a free neutron at rest $(n \rightarrow p + e^-)$,the conservation of energy and momentum must hold.
Let $M_n$,$M_p$,and $m_e$ be the masses of the neutron,proton,and electron,respectively.
From the conservation of energy: $M_n c^2 = E_p + E_e$,where $E_p$ and $E_e$ are the total energies of the proton and electron.
From the conservation of momentum: $\vec{p}_p + \vec{p}_e = 0$,which implies $|\vec{p}_p| = |\vec{p}_e| = p$.
Using the relativistic energy-momentum relation $E^2 = p^2 c^2 + m^2 c^4$,we have $E_p^2 = p^2 c^2 + M_p^2 c^4$ and $E_e^2 = p^2 c^2 + m_e^2 c^4$.
Substituting $p^2 c^2 = E_e^2 - m_e^2 c^4$ into the proton energy equation: $E_p = \sqrt{E_e^2 - m_e^2 c^4 + M_p^2 c^4}$.
Substituting this into the energy conservation equation: $M_n c^2 = E_e + \sqrt{E_e^2 - m_e^2 c^4 + M_p^2 c^4}$.
Rearranging and solving for $E_e$ shows that $E_e$ is uniquely determined by the masses of the particles involved,meaning the electron must have a fixed (discrete) energy.
Since the observed $\beta$-decay spectrum is continuous,this two-body model is insufficient. The actual decay is a three-body process: $n \rightarrow p + e^- + \bar{\nu}_e$,where the antineutrino carries away the remaining energy,allowing the electron to have a continuous range of energies.