Question

A $p-n$ photodiode is fabricated from a semiconductor with band gap of $2.8eV$. Can it detect a wavelength of $6000nm$?

Answer 1

The energy bandgap of the given photodiode, $Eg=2.8eV$

The wavelength is given by

$\lambda =6000nm=6000\times {10}^{-9}m$

We can find the energy of the signal from the following relation:

$E=\frac{hc}{\lambda }$…(i)

Where,
$h=6.63\times {10}^{-34}js$

$c=3\times {10}^{8}m/s$

Putting values (i)

$E=\frac{6.64\times {10}^{34}\times 3\times {10}^8}{6000\times {10}^{-9}}$

$E=3.315\times {10}^{-20}J$


But $1eV=1.6\times {10}^{-19}J$

$E=\frac{3.315\times {10}^{-20}}{1.6\times {10}^{-19}}$

$E=0.207eV$

The energy of a signal of wavelength $6000nm$ is $0.207eV$, which is less than $2.8eV$ energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.

Final Answer: No $(h\nu hastobegreaterthan{E}_g$ \)