The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap for the semiconductor is:

0.5 eV

The correct answer is Option (1) → 0.5 eV

The band gap (Eg) of the semiconductor,

$Eg=\frac{hc}{λ}=\frac{(6.63×10^{-34})(3×10^8)}{2480×10^{-9}}$

$=\frac{6.63×3×10^{-26}}{2480×10^{-9}×1.6×10^{-19}}$

$=0.5eV$