The bandgap of the material of a photodiode i | vedclass

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(D) The energy of the incident photon must be at least equal to the bandgap energy $(E_g)$ of the material for absorption to occur.Given: $E_g = 2.0 \

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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 bandgap energy $(E_g)$ of the material for absorption to occur.Given: $E_g = 2.0 \ eV$.Convert the energy into Joules: $E_g = 2.0 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-19} \ J$.The relationship between energy and frequency $(f)$ is given by $E_g = hf$,where $h$ is Planck's constant $(h \approx 6.626 	imes 10^{-34} \ J \cdot s)$.Therefore,the minimum frequency is $f = \frac{E_g}{h}$.$f = \frac{3.2 	imes 10^{-19}}{6.626 	imes 10^{-34}} \approx 0.4829 	imes 10^{15} \ Hz$.$f \approx 4.83 	imes 10^{14} \ Hz$.Rounding to the nearest given option,we get $f \approx 5 	imes 10^{14} \ Hz$.

The bandgap of the material of a photodiode is $2.0 \ eV$. What is the minimum frequency of radiation that this material can absorb?

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