= Solution
In <neutron-proton radiative capture>, the initial pair at rest has total <energy> $Mc^2$, where $M=m_p+m_n$. The emitted <photon> carries <momentum>, requiring equal and opposite <nuclear recoil>. Some <binding energy> becomes <deuteron> <kinetic energy>.
Exact two-body kinematics in <natural units> gives
$$
M=E_\gamma+\sqrt{m_d^2+E_\gamma^2},\qquad E_\gamma=\frac{M^2-m_d^2}{2M}.
$$
Since $m_d=M-B$,
$$
E_\gamma=B-\frac{B^2}{2M},\qquad \boxed{B-E_\gamma=\frac{B^2}{2M}.}
$$
With the stated rounded <masses>,
$$
B-E_\gamma=\frac{(2.224)^2}{2(938.272+939.565)}\,\mathrm{MeV}
=0.00131699\,\mathrm{MeV}
=\boxed{1.317\,\mathrm{keV}}.
$$
The <photon> <energy> is $2.222683\,\mathrm{MeV}$. At leading order the difference equals the nonrelativistic recoil estimate $p_\gamma^2/(2m_d)\simeq B^2/(2m_d)$; exact kinematics fixes the small distinction between using $m_d$ and $M$ in that approximation.
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