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Question
Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which 0 ones will be able to detect light of wavelength 6000 Å?
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Solution
In Photodiodes, electron and hole pairs are created by the junction photoelectric effect. That is the covalent bonds are broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals.
According to the problem,
The wavelength of light λ = 6000 Å = 6000 × 10-10 m
The energy of the light photon `E = (hc)/λ = (6.6 xx 10^-34 xx 3 xx 10^8)/(6000 xx 10^-10 xx 1.6 xx 10^-19)` eV = 2.06 eV
The incident radiation is detected by the photodiode D2 because the energy of incident radiation is greater than the band gap.
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