Question

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

• A. $4.36\times {10}^{-3}\text{m}$
• B. $4.36\times {10}^{-5}\text{m}$
• C. $2.18\times {10}^{-3}\text{m}$
• D. $2.18\times {10}^{-5}\text{m}$

Answer 1

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

Since, the acceptor level is $57\text{meV}$ above the valence band, at least $57\text{meV}$ is needed to create a hole.

Let the $\lambda$ be the wavelength of light, then photon will have energy, $E=\frac{hc}{\lambda }$

So, to create a hole, $\frac{hc}{\lambda }\ge 57\text{meV}$

$\Rightarrow \lambda \le \frac{hc}{57\text{meV}}$

We know,
$h=6.626\times {10}^{-34}\text{Js}$
$c=3\times {10}^8{\text{ms}}^{-1}$
$1\text{eV}=1.6\times {10}^{-19}\text{J}$

$\Rightarrow \lambda \le \frac{6.626\times {10}^{-34}\times 3\times {10}^8}{57\times {10}^{-3}\times 1.6\times {10}^{-19}}$

$\Rightarrow \lambda \le 2.179\times {10}^{-5}\text{m}\approx 2.18\times {10}^{-5}\text{m}$

$\Rightarrow {\lambda }_{max}=2.18\times {10}^{-5}\text{m}$

Hence, option $\left(D\right)$ is correct.