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

Question

TestHub · Accepted Answer

To detect a signal, the photon energy ( $E$ ) must be greater than or equal to the band gap ( $E_{g}$ ).

$E = h f = \frac{h c}{λ}$

Given $E_{g} = 2.5$ eV.

The maximum detectable wavelength $λ_{ma x}$ corresponds to $E = E_{g}$ .

$λ_{ma x} = \frac{h c}{E _{g}} = \frac{1240 e Vnm}{2.5 e V} = 496$ nm.

Thus, the photodiode can detect wavelengths $λ \leq 496$ nm.

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