Answer
2.0 eV
Work Step by Step
The minimum-frequency photon that enables conduction has an energy equal to the energy gap.
$$E_g=hf=\frac{hc}{\lambda}$$
$$=\frac{(6.63\times10^{-34}J \cdot s)(3.00\times10^8m/s)}{(1.60\times10^{-19}J/eV)(620\times10^{-9}m)}$$
$$=2.0 eV$$
