An LED is constructed from a pn junction base | vedclass

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(C) The energy gap $E_g$ of a semiconductor is related to the wavelength $\lambda$ of the emitted light by the formula: $E_g = \frac{hc}{\lambda}$.Giv

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$6.5 	imes 10^{-7} \ m$

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(C) The energy gap $E_g$ of a semiconductor is related to the wavelength $\lambda$ of the emitted light by the formula: $E_g = \frac{hc}{\lambda}$.Given $E_g = 1.9 \ eV$. Converting this to Joules: $E_g = 1.9 	imes 1.6 	imes 10^{-19} \ J$.Using the values $h = 6.6 	imes 10^{-34} \ J \cdot s$ and $c = 3 	imes 10^8 \ m/s$:$\lambda = \frac{hc}{E_g} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{1.9 	imes 1.6 	imes 10^{-19}} \ m$.$\lambda = \frac{19.8 	imes 10^{-26}}{3.04 	imes 10^{-19}} \ m \approx 6.513 	imes 10^{-7} \ m$.Rounding to the nearest given option,the wavelength is $6.5 	imes 10^{-7} \ m$.

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:

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