Answer
$E = 2.05~eV$
Work Step by Step
$\lambda = \frac{1}{1,650,763.73}~m = 6.0578\times 10^{-7}~m$
We can find the photon energy:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{6.0578\times 10^{-7}~m}$
$E = 3.28\times 10^{-19}~J$
$E = (3.28\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 2.05~eV$
