A photon of 300 , nm is absorbed by a gas and | vedclass

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(A) According to the law of conservation of energy,the energy of the absorbed photon is equal to the sum of the energies of the two re-emitted photons

Vedclass · Accepted Answer

$759$

Vedclass · Answer

(A) According to the law of conservation of energy,the energy of the absorbed photon is equal to the sum of the energies of the two re-emitted photons.$E = E_{1} + E_{2}$Since $E = \frac{hc}{\lambda}$,we have:$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$Dividing by $hc$:$\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$Given $\lambda = 300 \, nm$ and $\lambda_{1} = 496 \, nm$:$\frac{1}{300} = \frac{1}{496} + \frac{1}{\lambda_{2}}$$\frac{1}{\lambda_{2}} = \frac{1}{300} - \frac{1}{496} = \frac{496 - 300}{300 	imes 496} = \frac{196}{148800}$$\lambda_{2} = \frac{148800}{196} \approx 759.18 \, nm$Rounding to the nearest integer,we get $759 \, nm$.

$A$ photon of $300 \, nm$ is absorbed by a gas and then re-emits two photons. One re-emitted photon has a wavelength of $496 \, nm$. The wavelength of the second re-emitted photon is: (in $, nm$)

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