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Question
Answer the following question.
Three photodiodes D1, D2, and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?
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Solution
The energy of the incident light
`E = (hC)/lambda`
= `((6.6 xx 10^-34) xx (3 xx 10^8))/((600 xx 10^-9)(1.6 xx 10^-19))`
E = 2.06 eV
The incident radiations can be detected by a photodiode if the energy of incident radiation photon is greater than the band gap. This is true only for D2 (2 eV). Hence, only D2 will detect the light of 600 nm wavelength.
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