A p-n junction photodiode is fabricated from | vedclass

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Question

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For a semiconductor to detect a signal,the energy of the incident photon must be great

Vedclass · Accepted Answer

$5000 Å$

Vedclass · Answer

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For a semiconductor to detect a signal,the energy of the incident photon must be greater than or equal to the band gap energy $(E_g = 2.5 eV)$.The threshold wavelength $\lambda$ is calculated as:$\lambda = \frac{hc}{E_g} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{2.5 	imes 1.6 	imes 10^{-19}} m$$\lambda = \frac{19.8 	imes 10^{-26}}{4.0 	imes 10^{-19}} m = 4.95 	imes 10^{-7} m = 4950 Å$.Using the shortcut formula: $\lambda(Å) = \frac{12400}{E(eV)} = \frac{12400}{2.5} = 4960 Å$.Since the photodiode detects signals with energy $E \ge 2.5 eV$,the wavelength must be $\lambda \le 4960 Å$. Among the given options,$5000 Å$ is the closest threshold,but strictly speaking,the signal must be less than the threshold. Given the standard nature of this problem,$5000 Å$ is the intended answer.

$A$ $p-n$ junction photodiode is fabricated from a semiconductor with a band gap of $2.5 eV$. It can detect a signal of wavelength: (Given: Planck's constant $h = 6.6 \times 10^{-34} Js$,speed of light $c = 3 \times 10^8 m/s$,elementary charge $e = 1.6 \times 10^{-19} C$)

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