Question

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

Answer 1

Energy band gap of the given photodiode, E_g = 2.8 eV 
Wavelength, λ = 6000 nm =\( 6000 × 10^{−9} m \)
The energy of a signal is given by the relation:
\(E = \frac{hc}{λ}\)
Where, 
h = Planck’s constant = \(6.626 × 10^{−34} Js \)
c = Speed of light = \(3 × 10^{8} \frac{m}{s}\)
\(E =\frac{ 6.626\times10^{-34}\times3\times10^{8}}{6000\times 10^{-9}}\)
\(E = 3.313 × 10^{−20} J\)
But \(1.6 × 10^{−19} J = 1 eV\)
E = \(3.313 × 10^{−20} J\)
E = \(\frac{3.313 × 10^{−20}}{1.6\times10^{-19}} eV\)
E = 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.