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Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5eV, 2eV and 3eV respectively. Which of them will not be able to detect light of wavelength 600nm?

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1 Answer

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E = \(\frac{hc}{\lambda}\) 

E = \(\frac{6.64\times10^{-34}\times3\times10^8}{600\times10^{-9}}\) 

E = \(\frac{19.92}{600}\times10^{-19}\)

E = \(\frac{0.0332\times10^{-19}}{1.6\times10^{-19}}\)

E = 0.020 ev

For the incident radiation to be detected by photo diode, energy of incident radiation photon should be greater then the band gap. In this case all band gap are greater then photodiode. So any diode will not able to detect light of wavelength.

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