Answer
The scattering angle is $~~\theta = 86^{\circ}$
Work Step by Step
We can find the wavelength of photons with energy $55~keV$:
$\lambda = \frac{h~c}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(55,000~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 2.25886\times 10^{-11}~m$
We can find the wavelength of photons with energy $50~keV$:
$\lambda = \frac{h~c}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(50,000~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 2.48475\times 10^{-11}~m$
We can find $\Delta \lambda$:
$\Delta \lambda = \lambda_f - \lambda_o$
$\Delta \lambda = (2.48475\times 10^{-11}~m) - (2.25886\times 10^{-11}~m)$
$\Delta \lambda = 2.2589\times 10^{-12}~m$
We can find the scattering angle:
$\Delta \lambda = \frac{h}{mc}~(1 - cos~\theta)$
$1 - cos~\theta = \frac{\Delta \lambda~m~c}{h}$
$cos~\theta = 1 - \frac{\Delta \lambda~m~c}{h}$
$cos~\theta = 1 - \frac{(2.2589\times 10^{-12}~m)(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}{6.626\times 10^{-34}~J~s}$
$cos~\theta = 1 - 0.931617$
$cos~\theta = 0.06838$
$\theta = cos^{-1}~(0.06838)$
$\theta = 86^{\circ}$
The scattering angle is $~~\theta = 86^{\circ}$
