 Solution

Work first in units $c=1$. In the neutron rest frame, let $s_{p\nu }=(p_p+p_{\nu }{)}^2$. From $p_n-p_e=p_p+p_{\nu }$,The electron's minimum possible total energy is its rest energy. This configuration has the electron at rest while the proton and massless antineutrino move back to back with equal momentum magnitudes. Their common momentum satisfies a two-body energy balance for the remaining mass $m_n-m_e$:For the maximum electron energy, minimize $s_{p\nu }$. Since the massless antineutrino has nonnegative energy, $s_{p\nu }\ge m_p^2$, with the boundary reached as its energy tends to zero. ThenAt this limiting endpoint the electron and proton have equal and opposite momenta. The proton recoils; it is not at rest. A massless neutrino has no rest frame, so “neutrino at rest” must be replaced by the precise soft-energy limit $E_{\nu }\to 0$. The electron's maximum kinetic energy is the displayed total energy minus $m_e{c}^2$, rather than the same number. These endpoint configurations lie on phase-space boundaries; their kinematic existence does not imply a nonzero event density exactly at the boundary.