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(C) The visible region of the electromagnetic spectrum corresponds to wavelengths approximately between $400 \ nm$ and $700 \ nm$. The energy $E$ of a

Vedclass · Accepted Answer

$1.7 \ eV$ to $3.1 \ eV$

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(C) The visible region of the electromagnetic spectrum corresponds to wavelengths approximately between $400 \ nm$ and $700 \ nm$. The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$.For $\lambda = 700 \ nm = 700 	imes 10^{-9} \ m$:$E_{\min} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{700 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \ eV \approx 1.77 \ eV$.For $\lambda = 400 \ nm = 400 	imes 10^{-9} \ m$:$E_{\max} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{400 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \ eV \approx 3.1 \ eV$.Thus,the energy band gap required for an $LED$ to emit visible light is in the range of $1.7 \ eV$ to $3.1 \ eV$. The correct option is $(c)$.

For an $LED$ to emit light in the visible region of the electromagnetic spectrum,it can have an energy band gap in the range of (Planck's constant,$h = 6.6 \times 10^{-34} \ J s$ and speed of light,$c = 3 \times 10^8 \ m s^{-1}$ in vacuum).

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