A photodiode is fabricated from a semiconduct | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $6000 \ \mathring{A}$?

Similar Questions

$A$ photodiode is a device:

The given graph represents $V-I$ characteristic for a semiconductor device. Which of the following statements is correct?

$A$ $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.8 \; eV$. Can it detect a wavelength of $6000 \; nm$?

The semiconductor used for the fabrication of visible LEDs must at least have a band gap of (in $eV$)

Which wavelength of the Sun is primarily used to generate electric energy in solar cells?

Vedclass Test Series

Exam Paper Generator

Online Exam Module