Question

A p-type semiconductor has acceptor level of $57\text{meV}$ above the valence band. Find the maximum wavelength of light which can create a hole

• A. $1.19\times {10}^{-6}m$
• B. $1.19\times {10}^{-5}m$
• C. $2.18\times {10}^{-6}m$
• D. $2.18\times {10}^{-5}m$

Answer 1

The correct option is D $2.18\times {10}^{-5}m$

Given,
minimum energy of the hole= $57meV$
Now the energy associated by the photon with wavelength $\lambda$ is given by
$E=\frac{hc}{\lambda }>57\times {10}^{-3}eV$
So, $\frac{hc}{\lambda }=57\times {10}^{-3}eV\lambda =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{57\times {10}^{-3}\times 1.6\times {10}^{-19}}=2.18\times {10}^{-5}m$