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(A) The energy bandgap is given as $E_g = 2.5 \ eV$. To detect a photon,the energy of the incident photon $E = \frac{hc}{\lambda}$ must be greater tha

Vedclass · Accepted Answer

$4000 \ \mathring{A}$

Vedclass · Answer

(A) The energy bandgap is given as $E_g = 2.5 \ eV$. To detect a photon,the energy of the incident photon $E = \frac{hc}{\lambda}$ must be greater than or equal to the bandgap $E_g$. First,calculate the threshold wavelength $\lambda_{max}$ corresponding to $E_g$: $\lambda_{max} = \frac{hc}{E_g} = \frac{12400 \ eV \cdot \mathring{A}}{2.5 \ eV} = 4960 \ \mathring{A}$. Since the photodiode can only detect wavelengths shorter than or equal to the threshold wavelength $(\lambda \leq \lambda_{max})$,it can detect wavelengths up to $4960 \ \mathring{A}$. Therefore,it can detect $4000 \ \mathring{A}$.

$A$ photodiode is fabricated from a semiconductor with a bandgap of $2.5 \ eV$. Can it detect a wavelength of $6000 \ \mathring{A}$?

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