#### Question

Numerical

**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?

#### 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 D_{2} (2 eV). Hence, only D_{2} will detect the light of 600 nm wavelength.

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