Answer
$6.215$ $keV$
Work Step by Step
From the previous section of this problem, we have calculated the momentum of the photon: $p=3.315\times 10^{-24}$ $kg.m/s$
Now, the energy of a photon can be written as
$E=pc$,
where $c$ is the speed of light in free space.
Thus, the energy of the photon is given by
$E=3.315\times 10^{-24}\times 3\times 10^{8}$ $J$
or, $E=\frac{3.315\times 10^{-24}\times 3\times 10^{8}}{1.6\times 10^{-19}}$ $eV$
or, $E\approx 6215$ $eV$
or, $E=6.215$ $keV$
