A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Energy band gap E=2.8 eV E=λhc ⇒6000×10−9×1.6×10−196.6×10−34×3×108=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.
