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(B) $p-n$ photodiode is a semiconductor device that generates current when illuminated by light. For a photon to be absorbed by the semiconductor mate

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$5 	imes 10^{14} 	ext{ Hz}$

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(B) $p-n$ photodiode is a semiconductor device that generates current when illuminated by light. For a photon to be absorbed by the semiconductor material, its energy $E$ must be at least equal to the band gap energy $E_g$.Given, $E_g = 2.0 	ext{ eV}$.The relationship between energy and frequency is $E = h
u$, where $h$ is Planck's constant $(6.63 	imes 10^{-34} 	ext{ J} \cdot 	ext{s})$ and $
u$ is the frequency.Converting the band gap energy into Joules: $E_g = 2.0 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-19} 	ext{ J}$.Setting $h
u = E_g$, we get $
u = \frac{E_g}{h} = \frac{3.2 	imes 10^{-19} 	ext{ J}}{6.63 	imes 10^{-34} 	ext{ J} \cdot 	ext{s}}$.Calculating the value: $
u \approx 0.482 	imes 10^{15} 	ext{ Hz} = 4.82 	imes 10^{14} 	ext{ Hz}$.Rounding to the nearest option, we get $
u \approx 5 	imes 10^{14} 	ext{ Hz}$.

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

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