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  • . (a) 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 of them will not be able to detect light of wavelength 600 nm? <

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 of them will not be able to detect light of wavelength 600 nm?

 

Answers (1)

best_answer

According to the question,

The band gap of D1 = 2.5eV, D2 = 2 eV, and for D3 it is 3eV

The energy of the incident light can be calculated by using the below-mentioned formula,

E\frac{hc}{\lambda }

=\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(600 \times 10^{-9}\right)\left(1.6 \times 10^{-19}\right)}

E = 2.06 eV

The incident radiation can only be detected by a photodiode if the energy of the photon with incident radiation is greater than the band gap. Here among D1, D2, and D3, it can only be achieved by D2 with the value of the band gap being 2eV.

Posted by

Divya Sharma

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