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(D) The energy of the photon required to excite an electron across the band gap is given by $E = h
u$.For the minimum frequency $
u$,the energy of t

Vedclass · Accepted Answer

$5 	imes 10^{14}\, Hz$

Vedclass · Answer

(D) The energy of the photon required to excite an electron across the band gap is given by $E = h
u$.For the minimum frequency $
u$,the energy of the photon must be equal to the band gap energy $E_g$.Given $E_g = 2.0\, eV$.We know that $1\, eV = 1.6 	imes 10^{-19}\, J$,so $E_g = 2.0 	imes 1.6 	imes 10^{-19} = 3.2 	imes 10^{-19}\, J$.Using the relation $E = h
u$,we have $
u = \frac{E_g}{h}$.Substituting the values $h = 6.63 	imes 10^{-34}\, J\cdot s$:$
u = \frac{3.2 	imes 10^{-19}}{6.63 	imes 10^{-34}} \approx 0.482 	imes 10^{15}\, Hz$.$
u \approx 4.82 	imes 10^{14}\, Hz$.Rounding this value,we get $
u \approx 5 	imes 10^{14}\, Hz$.

$A$ $p-n$ photodiode is made of a material with a band gap of $2.0\, eV$. The minimum frequency of the radiation that can be absorbed by the material is nearly:

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