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

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#### Energy band gap E=2.8 eVE=hcλ⇒6.6×10−34×3×1086000×10−9×1.6×10−19=0.207 eVThe 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.

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