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A p-n photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of 6000 run? |
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Answer» Solution :Here energy band GAP `E_(g) = 2.8` eV and WAVELENGTH to be detected `lambda` = 6000 nm = `6 xx 10^(-6)` m ` therefore` Energy of light photonE = `(hc)/(E lambda) `eV ` = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(6 xx 10^(-6) xx 1.6 xx 10^(-19) ) `eV = 0.207 eV As the energy of light photon is much less than the band gap, hence the given PHOTODIODE cannot detect it. |
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