Question

A p-n photo diode is fabricated from a semiconductor with band gap of 2.8e V. Can it detect a wavelength of 6000 nm?

Answer 1

Band gap E_g = 2.8 eV

Energy band gap corresponding to wavelength 6000 nm is given by

E_g = \(\frac{hc}{λ} = \frac{6.62\times10^{-34}\times3\times10^8}{6000\times10^{-9}}J\)

= \(\frac{19.86\times10^{-26}}{6\times10^{-6}\times11.6\times10^{-19}}eV\)

= \(\frac{19.86}{9.6}\times10^{-1}\) = 0.207eV

Since E_g , < E_g , so it can not detect a wave length of 6000 nm