Answer
$E = 0.78~MeV$
Work Step by Step
We can find the energy of the photon:
$E = m_e~c^2+m_p~c^2+2K$
$E = (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(2)(0.22~MeV)$
$E = 5.46\times 10^{-14}~J+0.44~MeV$
$E = (5.46\times 10^{-14}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})+0.44~MeV$
$E = 3.4\times 10^5~eV+0.44~MeV$
$E = 0.34\times 10^6~eV+0.44~MeV$
$E = 0.34~MeV+0.44~MeV$
$E = 0.78~MeV$
