Answer
(B) $2.42eV$
Work Step by Step
We know that
$E=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$E=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{514.5\times 10^{-9}m}$
$E=\frac{3.865889\times 10^{-19}J}{1.6\times 10^{-19}J/eV}$
$E=2.42eV$
Thus, option (B) is correct.