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(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For a $p-n$ junction to detect a signal, the energy of the incident photon must be gre

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$600$

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(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For a $p-n$ junction to detect a signal, the energy of the incident photon must be greater than or equal to the band gap energy $(E_g = 2.8 \ eV)$.If $E Using $h = 6 	imes 10^{-34} \ J \cdot s$ and $c = 3 	imes 10^8 \ m/s$, we have $hc = 18 	imes 10^{-26} \ J \cdot m$.Converting $E_g$ to Joules: $2.8 \ eV = 2.8 	imes 1.6 	imes 10^{-19} \ J = 4.48 	imes 10^{-19} \ J$.The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{E_g} = \frac{18 	imes 10^{-26}}{4.48 	imes 10^{-19}} \approx 4.017 	imes 10^{-7} \ m = 401.7 \ 	ext{nm}$.Any wavelength $\lambda > \lambda_0$ will have energy $E Among the given options, $600 \ nm$ is the only wavelength greater than $401.7 \ nm$.

$A$ $p-n$ junction is fabricated from a semiconductor with a band gap of $2.8 \ eV$. What approximate wavelength can it $NOT$ detect (in $nm$)?
[Use $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]

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$A$ $p-n$ junction is fabricated from a semiconductor with a band gap of $2.8 \ eV$. What approximate wavelength can it $NOT$ detect (in $nm$)?[Use $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]