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Physics Question on Semiconductor electronics: materials, devices and simple circuits

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

Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm =6000×10−9m6000 × 10^{−9} m
The energy of a signal is given by the relation:
E=hcλE = \frac{hc}{λ}
Where,
h = Planck’s constant = 6.626×10−34Js6.626 × 10^{−34} Js
c = Speed of light = 3×108ms3 × 10^{8} \frac{m}{s}
E=6.626×10−34×3×1086000×10−9E =\frac{ 6.626\times10^{-34}\times3\times10^{8}}{6000\times 10^{-9}}
E=3.313×10−20JE = 3.313 × 10^{−20} J
But 1.6×10−19J=1eV1.6 × 10^{−19} J = 1 eV
E = 3.313×10−20J3.313 × 10^{−20} J
E = 3.313×10−201.6×10−19eV\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.