Question

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480nm is incident on it. The band gap in (eV) for the semiconductor is
(a) 0.5eV 

(b) 0.7eV

(c) 1.1eV 

(d) 2.5eV

Answer 1

(a) : Band gap = Energy of photon of λ = 2480 nm
Therefore,  Energy =(hc/λ)J = (hc/λe)(eV)