A p-n photodiode is fabricated from a semicon | vedclass

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(B) The energy of the incident photon must be greater than or equal to the band gap energy of the semiconductor for detection to occur.$E_{ph} \geq E_

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$8000 \ \mathring{A}$

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(B) The energy of the incident photon must be greater than or equal to the band gap energy of the semiconductor for detection to occur.$E_{ph} \geq E_g$Since $E_{ph} = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} \geq E_g$,which implies $\lambda \leq \frac{hc}{E_g}$.Using the relation $\lambda (	ext{in } \mathring{A}) \approx \frac{12400}{E_g (	ext{in } eV)}$,we substitute $E_g = 1.4 \ eV$:$\lambda \leq \frac{12400}{1.4} \ \mathring{A} \approx 8857 \ \mathring{A}$.Among the given options,the signal with a wavelength of $8000 \ \mathring{A}$ is less than $8857 \ \mathring{A}$,so it can be detected.

$A$ $p-n$ photodiode is fabricated from a semiconductor with a band gap of $1.4 \ eV$. It can detect a signal of wavelength:

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