Answer
$\Delta E = -4.45~keV$
Work Step by Step
In part (e) we found that $\frac{\Delta \lambda}{\lambda} = 9.76\times 10^{-2}$
We can find an expression for the new wavelength:
$\lambda_f = (1+9.76\times 10^{-2})~\lambda$
$\lambda_f = (1.0976)~\lambda$
We can find an expression for the new energy:
$E_f = \frac{hc}{\lambda_f}$
$E_f = \frac{hc}{(1.0976)~\lambda}$
$E_f = 0.911~\frac{hc}{\lambda}$
$E_f = 0.911~E$
We can find $\Delta E$:
$\Delta E = E_f - E$
$\Delta E = 0.911~E - E$
$\Delta E = -0.089~E$
$\Delta E = (-0.089)~(50.0~keV)$
$\Delta E = -4.45~keV$
