LED is constructed from GaAsP semiconducting | vedclass

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(D) The energy gap $E_g$ is given as $1.9\, eV$.To convert this into Joules,we multiply by $1.6 	imes 10^{-19} \, J/eV$:$E = 1.9 	imes 1.6 	imes 10

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$654\, nm$ and red colour

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(D) The energy gap $E_g$ is given as $1.9\, eV$.To convert this into Joules,we multiply by $1.6 	imes 10^{-19} \, J/eV$:$E = 1.9 	imes 1.6 	imes 10^{-19} \, J = 3.04 	imes 10^{-19} \, J$.The wavelength $\lambda$ is given by the formula $\lambda = \frac{hc}{E}$.Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{3.04 	imes 10^{-19}}$.$\lambda = \frac{19.89 	imes 10^{-26}}{3.04 	imes 10^{-19}} \approx 6.54 	imes 10^{-7} \, m$.$\lambda = 654 \, nm$.$A$ wavelength of $654 \, nm$ corresponds to the red region of the visible spectrum.

$LED$ is constructed from $GaAsP$ semiconducting material. The energy gap of this $LED$ is $1.9\, eV$. Calculate the wavelength of light emitted and its colour.
$[h = 6.63 \times 10^{-34} \, Js$ and $c = 3 \times 10^{8} \, ms^{-1}]$

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$LED$ is constructed from $GaAsP$ semiconducting material. The energy gap of this $LED$ is $1.9\, eV$. Calculate the wavelength of light emitted and its colour.$[h = 6.63 \times 10^{-34} \, Js$ and $c = 3 \times 10^{8} \, ms^{-1}]$