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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 6000 nm?

Q. 14.7     A p-n photodiode is fabricated from a semiconductor with bandgap of  2.8\; eV.  Can it detect a wavelength of 6000 \; nm\; ?

Answers (1)

best_answer

Given 

energy band gap of photodiode is 2.8eV.

wavelength = \lambda = 6000nm = 6000*10^{-9} 

Energy of signal will be          \frac{hc}{\lambda }

where c is speed of light(300000000m/s) , h is planks constant ( = 6.626 * 10^{-34}Js )

putting the corresponding value 

Energy of signal =    \frac{(6.626 * 10^{-34} * 3*10^8)}{6000*10^{-9}}

                           =   3.313*10^{-20}J

                           =   0.207eV (since 1.6*10^{-20}= 1eV)

The energy of the signal is 0.207eV which is less than 2.8eV ( the energy and gap of photodiode). Hence signal can not be detected by the photodiode.

Posted by

Pankaj Sanodiya

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