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Three photodiodes 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 ones will be able to detect light of wavelength 6000 Å? |
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Answer» Solution :When energy of photon (E ), made INCIDENT on photodiode, is greater than its bnd gap energy `(E_(g))`, no. of minority charge carriers increases due to increase in electron-hole pairs in the region near the junction and so reverse saturation current passing through photodiodeincreases, which can be NOTED by microammeter, connected with photodiode in the external circuit. In the present case, `E=(hc)/(lambda)=(1242.19(eV)(nm))/(6000xx10^(-10)xx10^(9)(nm))` `therefore E = 2.07eV` Among given PHOTODIODES, for `D_(2)`, we have `E_(g)=2eV rArr ` For `D_(2), E gt E_(g)` and for other diodes `E LT E_(g)`. Hence only `D_(2)` can detect given light. |
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