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Three photo diodes D_(1) D_(2) and D_(3) 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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Answer» Solution : Energy of light photon of wavelength `lambda = 600 NM = 600 xx 10^(-19)` m in eV is `E = (hc)/(e lambda) eV = ((6.63 xx 10^(-34))xx (3 xx 10^(8)))/((1.6 xx 10^(-19) xx (600 xx 10^(-9))) eV = 2.07` eV As band gaps of photodiodes `D_(1) D_(2) and D_(3)` are 2.5 eV, 2 eV and 3 eV respectively, it is clear that given light photon can be detected only by photodiode, `D_(2)` WHOSE band gap is less than E. THEREFORE `D_(1) and D_(3)` will not detect the given light. |
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