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

Options

A. 0.9
B. 0.7
C. 0.5
D. 1.1

Answer

C

Explanation

The band gap is given by `E_g=hv=(hc)/lamda`
Here, `lamda = 2480 nm = 2480 xx10^(-9) m = 248 xx10^(-8)m`
`:.E_g=(6.6xx10^(-34)xx3xx10^(8))/(248xx10^(-8))=7.984xx10^(-20)J`
`=(7.984xx10^(-20))/(1.6xx10^(-19))eV=0.499eV = 0.5 eV`