= Solution
The initial total <momentum> is zero, so the outgoing <neutron> recoils with momentum magnitude $E_\gamma/c$. Put $W=(m_p+m_\pi)c^2$. Conservation of <relativistic energy> gives
$$
W=E_\gamma+\sqrt{m_n^2c^4+E_\gamma^2},\qquad
\boxed{E_\gamma=\frac{W^2-m_n^2c^4}{2W}\simeq129.4\,\mathrm{MeV}.}
$$
For example the rest energies $938.2721$, $139.5704$ and $939.5654\,\mathrm{MeV}$ for the proton, charged pion and neutron give this value.<Review of Particle Physics by Particle Data Group (2022)>[Charged-pion, proton and neutron listings, page 1 of each: mass values in MeV]{ref} The result is smaller than the initial-to-final rest-energy difference of about $138.3\,\mathrm{MeV}$, because some released <energy> becomes neutron recoil <kinetic energy>. Atomic binding of a captured pion is neglected in the stated stationary-particle model.
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