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The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations show whether it can detect light of wavelength 400 nm. |
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Answer» Solution :Here energy gap of photodiode `E_(g)` = 1.2 eV, and WAVELENGTH of incident light `lambda` = 400 nm = ` 400 XX 10^(-9)` m. `THEREFORE ` Energy of incident light photon E = `(hc)/(e lambda) eV = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx 400 xx 10^(-9)) `eV = 3.1 eV As energy of incident light photon is greater than the energy gap `(E gt E_(g))` of given photodiode, the photodiode will definitely DETECT the light. |
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