For LEDs to emit light in the visible region | vedclass

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(D) The energy of a photon emitted by an $LED$ is approximately equal to the energy band gap $E_g$ of the semiconductor material.The relationship betw

Vedclass · Accepted Answer

$1.7 \, eV$ to $3.0 \, eV$

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(D) The energy of a photon emitted by an $LED$ is approximately equal to the energy band gap $E_g$ of the semiconductor material.The relationship between energy and wavelength is given by $E_g = \frac{hc}{\lambda}$.The visible region of the electromagnetic spectrum corresponds to wavelengths ranging from approximately $400 \, nm$ to $700 \, nm$ (i.e.,$4 	imes 10^{-7} \, m$ to $7 	imes 10^{-7} \, m$).For $\lambda = 700 \, nm$ $(7 	imes 10^{-7} \, m)$:$E_g = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{7 	imes 10^{-7}} \approx 2.84 	imes 10^{-19} \, J \approx 1.77 \, eV$.For $\lambda = 400 \, nm$ $(4 	imes 10^{-7} \, m)$:$E_g = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{4 	imes 10^{-7}} \approx 4.97 	imes 10^{-19} \, J \approx 3.1 \, eV$.Thus,for an $LED$ to emit light in the visible region,the energy band gap should be in the range of approximately $1.7 \, eV$ to $3.0 \, eV$.

For $LED$s to emit light in the visible region of the electromagnetic spectrum,they should have an energy band gap in the range of:

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