Answer
The energy of the backscattered photon is $~~41.8~keV$
Work Step by Step
We can find the original wavelength:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(50.0\times 10^3~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 2.486\times 10^{-11}~m$
We can find the wavelength shift $\Delta \lambda$:
$\Delta \lambda = \frac{h}{mc}(1-cos~\phi)$
$\Delta \lambda = \frac{h}{mc}(1-cos~180^{\circ})$
$\Delta \lambda = \frac{2h}{mc}$
$\Delta \lambda = \frac{(2)(6.63\times 10^{-34}~J~s)}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}$
$\Delta \lambda = 4.85\times 10^{-12}~m$
We can find the fractional loss in energy:
$frac = \frac{\Delta \lambda}{\lambda + \Delta \lambda}$
$frac = \frac{4.85\times 10^{-12}~m}{2.486\times 10^{-11}~m + 4.85\times 10^{-12}~m}$
$frac = 0.163$
We can find the energy of the backscattered photon:
$E_f = E - 0.163~E$
$E_f = 0.837~E$
$E_f = (0.837)~(50.0~keV)$
$E_f = 41.8~keV$
The energy of the backscattered photon is $~~41.8~keV$
