Question

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 the semiconductor is

• A. 1.1 $eV$
• B. $2.5eV$
• C. $0.5$ $eV$
• D. $0.7$ $eV$

Answer 1

The correct option is C $0.5$ $eV$

Threshold wavelength is 2480 $nm$.
So energy associated with this wavelength is threshold energy which will be same as energy of bandgap for the electron to jump in that bandgap.
$E_{bandgap}=\frac{hc}{\lambda }=\frac{\left(6.626\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{\left(}J$
$=\frac{\left(6.626\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{\left(2480\times {10}^{-9}\right)\times \left(1.6\times {10}^{-19}\right)}eV$
$=0.5eV$