Question

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

Answer 1

The energy bandgap of the given photodiode, $E_g=2.8eV$
The wavelength is given by $\lambda =6000nm=6000\times {10}^{-9}m$
We can find the energy of the signal from the following relation:
$E=\frac{hc}{\lambda }...\left(i\right)$
Where,
$h=6.63\times {10}^{-34}Js$
$c=3\times {10}^{8}m/s$
Putting values in (i),
$E=\frac{6.63\times {10}^{-34}\times 3\times {10}^9}{6000\times {10}^{-9}}$
$E=3.315\times {10}^{-20}J$
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 6000 nm 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 ($hv$ has to be greater than $E_g)$.