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A p-n photodiode is fabricated from a semiconductor with band - gap of 2.8 eV . Can it detect a wavelength of 6000nm? |
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Answer» Energy band gap of the given photodiode, `E_(g) = 2.8 eV` Wavelength, `lamda = 6000 nm = 6000 × 10^(−9) m` The energy of a signal is given by the relation: `E=overset((hc)/(lamda)` Where, h = Planck’s constant `= 6.626 xx 10−^(34) "Js"` c = Speed of light `= 3 × 10^(8) " m/s"` `=(6.626xx10^(-34)xx3xx10^(8))/(600xx10^(-9))` E `= 3.313 xx 10^(−20) J` But `1.6 xx 10^(−19) J = 1 eV` `thereforeE = 3.313 xx 10^(−20) J` `=(3.313xx10^(-20))/(1.6xx10^(-19))=0.207 " eV"` The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal. |
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