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(D) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ for the material to absorb it.Given $E_g = 2.0\, eV$.The r

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

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(D) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ for the material to absorb it.Given $E_g = 2.0\, eV$.The relationship between energy and frequency is given by $E = h
u$,where $h$ is Planck's constant $(6.63 	imes 10^{-34}\, J\cdot s)$ and $
u$ is the frequency.First,convert the energy from $eV$ to Joules:$E = 2.0 	imes 1.6 	imes 10^{-19}\, J = 3.2 	imes 10^{-19}\, J$.Now,calculate the minimum frequency $
u$:$
u = \frac{E}{h} = \frac{3.2 	imes 10^{-19}}{6.63 	imes 10^{-34}} \approx 0.4826 	imes 10^{15}\, Hz$.$
u \approx 4.83 	imes 10^{14}\, Hz$.Rounding to the nearest given option,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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