Answer
$\Delta E = -8.66\times 10^{-6}~eV$
Work Step by Step
In part (b) we found that $~~\frac{\Delta \lambda}{\lambda} = 4.11\times 10^{-6}$
We can find an expression for the new wavelength:
$\lambda_f = \lambda+(4.11\times 10^{-6})~\lambda$
$\lambda_f = 1.00000411~\lambda$
We can find an expression for the new energy:
$E_f = \frac{hc}{\lambda_f}$
$E_f = \frac{hc}{1.00000411~\lambda}$
$E_f = 0.99999589~\frac{hc}{\lambda}$
$E_f = 0.99999589~E$
We can find $\Delta E$:
$\Delta E = E_f - E$
$\Delta E = 0.99999589~E - E$
$\Delta E = -0.00000411~E$
$\Delta E = (-0.00000411) \frac{hc}{\lambda}$
$\Delta E = (-0.00000411) ~\frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{590\times 10^{-9}~m}$
$\Delta E = -1.38556\times 10^{-24}~J$
$\Delta E = (-1.38556\times 10^{-24}~J) (\frac{1~eV}{1.6\times 10^{-19}~J})$
$\Delta E = -8.66\times 10^{-6}~eV$
