Answer
The change in photon energy is $~~-(4.9\times 10^{-4})~\%$
Work Step by Step
We can find the Compton shift $\Delta \lambda$:
$\Delta \lambda = \frac{h}{mc}(1-cos~\phi)$
$\Delta \lambda = \frac{h}{mc}(1-cos~90^{\circ})$
$\Delta \lambda = \frac{h}{mc}$
$\Delta \lambda = \frac{6.63\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}$
$\Delta \lambda = 2.43\times 10^{-12}~m$
We can find the fraction of energy loss:
$frac = \frac{\Delta \lambda}{\lambda + \Delta \lambda}$
$frac = \frac{2.43\times 10^{-12}~m}{(500\times 10^{-9}~m) + (2.43\times 10^{-12}~m)}$
$frac = 4.9\times 10^{-6}$
The change in photon energy is $~~-(4.9\times 10^{-4})~\%$
