14.11 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Step 1: Find the energy of signal.
\(As~E=\frac{hc}{\lambda}=\frac{6.626\times10^{-34}\times3\times10^{8}}{6000\times10^{-9}} =3.313\times10^{-20}~J\)
Step 2: Find the energy of signal in electron-volt.
\(E=\frac{3.313\times10^{-20}}{1.6\times10^{-19}} =0.207~eV\)
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV (the energy band gap of a photodiode.) Hence, the photodiode cannot detect the signal.
© 2024 GoodEd Technologies Pvt. Ltd.
