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A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelengthof 6000 nm? |
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Answer» Solution :The radiation energy of the WAVELENGTH of 6000 nm. `E=(hc)/(lambda)=(6.625xx10^(-34)xx3xx10^(8))/(6000xx10^(-9)xx1.6xx10^(-19))eV` `=0.00207xx10^(2)eV` `=0.207eV` but `E_(g)=2.8eV` is given `THEREFORE` So, `E gt E_(g)` should be done to test the wavelength of radiation but since there is `E lt E_(g)`the wavelength cannot be tested. |
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