Answer
$\Delta E = -40.6~keV$
Work Step by Step
In part (a) we found that $~~\Delta \lambda = 4.85\times 10^{-12}~m$
We can find the fraction of energy loss:
$frac = \frac{\Delta \lambda}{\lambda + \Delta \lambda}$
$frac = \frac{4.85\times 10^{-12}~m}{(1.00\times 10^{-11}~m) + (4.85\times 10^{-12}~m)}$
$frac = 0.3266$
We can find the change in energy of the photon:
$\Delta E = (-0.3266)~E$
$\Delta E = (-0.3266)~\frac{hc}{\lambda}$
$\Delta E = (-0.3266)~\frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{1.00\times 10^{-11}~m}$
$\Delta E = -6.5\times 10^{-15}~J$
$\Delta E = (-6.50\times 10^{-15}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$\Delta E = -40.6~keV$
