Answer
The energy of the photons that are backscattered is $~~58~keV$
Work Step by Step
We can find the original wavelength:
$\lambda_o = \frac{h~c}{E}$
$\lambda_o = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(75,000~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda_o = 1.66\times 10^{-11}~m$
We can find the wavelength shift:
$\Delta \lambda = \frac{h}{mc}~(1 - cos~\theta)$
$\Delta \lambda = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1 - cos~180^{\circ})$
$\Delta \lambda = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~\times (2)$
$\Delta \lambda = 4.85\times 10^{-12}~m$
We can find the final wavelength:
$\lambda_f = \lambda_o+\Delta \lambda$
$\lambda_f = (1.66\times 10^{-11}~m)+(4.85\times 10^{-12}~m)$
$\lambda_f = 2.145\times 10^{-11}~m$
We can find the energy of the photons that are backscattered:
$E = \frac{h~c}{\lambda}$
$E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{2.145\times 10^{-11}~m}$
$E = 9.267\times 10^{-15}~J$
$E = (9.267\times 10^{-15}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 5.8\times 10^4~eV$
$E = 58~keV$
The energy of the photons that are backscattered is $~~58~keV$
