The energy gap of an LED is 2.4 eV. When the | vedclass

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(A) The energy gap of the $LED$ is given as $E_g = 2.4 \ eV$.To convert this energy into Joules,we multiply by $1.6 	imes 10^{-19} \ J/eV$:$E = 2.4 \

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$1.28 	imes 10^{-27} \ kg \ ms^{-1}$

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(A) The energy gap of the $LED$ is given as $E_g = 2.4 \ eV$.To convert this energy into Joules,we multiply by $1.6 	imes 10^{-19} \ J/eV$:$E = 2.4 	imes 1.6 	imes 10^{-19} \ J = 3.84 	imes 10^{-19} \ J$.The momentum $p$ of a photon is related to its energy $E$ by the formula $p = \frac{E}{c}$,where $c$ is the speed of light $(3 	imes 10^8 \ m/s)$.Substituting the values:$p = \frac{3.84 	imes 10^{-19}}{3 	imes 10^8} \ kg \ ms^{-1}$.$p = 1.28 	imes 10^{-27} \ kg \ ms^{-1}$.Therefore,the correct option is $A$.

The energy gap of an $LED$ is $2.4 eV$. When the $LED$ is switched $ON$,the momentum of the emitted photons is

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