The semiconductor used for the fabrication of | vedclass

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

Question

(C) Visible light corresponds to the wavelength range of approximately $380 \, nm$ to $750 \, nm$. Using the relation $E = \frac{hc}{\lambda}$, where

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(C) Visible light corresponds to the wavelength range of approximately $380 \, nm$ to $750 \, nm$. Using the relation $E = \frac{hc}{\lambda}$, where $h = 6.63 	imes 10^{-34} \, J \cdot s$, $c = 3 	imes 10^8 \, m/s$, and $1 \, eV = 1.6 	imes 10^{-19} \, J$. For the longest wavelength of visible light $(\lambda = 750 \, nm)$, the energy is $E = \frac{1240 \, eV \cdot nm}{750 \, nm} \approx 1.65 \, eV$. In practice, to emit visible light efficiently, the semiconductor material used for the fabrication of visible LEDs must have a band gap of at least $1.8 \, eV$.

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

Similar Questions

An $LED$ is constructed from a $pn$ junction based on a certain semiconducting material whose energy gap is $1.9 \ eV$. The wavelength of the emitted light is:

Which one of the following statements is $WRONG$ regarding $LED$?

$A$ light emitting diode $(LED)$ is fabricated using $GaAs$ semiconducting material whose band gap is $1.42 \,eV$. The wavelength of light emitted from the $LED$ is: (in $\,nm$)

Photodiodes are mostly operated in reverse biased condition because

$A$ photodetector is made from a semiconductor $In_{0.53}Ga_{0.47}As$ with $E_g = 0.73 \, eV$. What is the maximum wavelength,which it can detect in $nm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module