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

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(C) The energy of a photon required to excite an electron across the band gap is given by $E_g = 2.5 \, eV$.For a photodiode to detect a signal,the en

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$4000 \, \mathring{A}$

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(C) The energy of a photon required to excite an electron across the band gap is given by $E_g = 2.5 \, eV$.For a photodiode to detect a signal,the energy of the incident photon $(E = \frac{hc}{\lambda})$ must be greater than or equal to the band gap energy $(E_g)$.Thus,$\frac{12400 \, eV \cdot \mathring{A}}{\lambda} \geq 2.5 \, eV$.Calculating the threshold wavelength $\lambda_{max} = \frac{12400}{2.5} \, \mathring{A} = 4960 \, \mathring{A}$.Any signal with a wavelength less than or equal to $4960 \, \mathring{A}$ can be detected.Among the given options,$4000 \, \mathring{A}$ is the only wavelength smaller than $4960 \, \mathring{A}$.

$A$ $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.5 \, eV$. It can detect a signal of wavelength:

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