A gas absorbs a photon of 355 nm and emits a | vedclass

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(C) 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 emitted photons.$E

Vedclass · Accepted Answer

$743$

Vedclass · Answer

(C) 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 emitted photons.$E_{absorbed} = E_{1} + E_{2}$Since $E = \frac{hc}{\lambda}$,we have:$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$$\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$Given $\lambda = 355 \ nm$ and $\lambda_{1} = 680 \ nm$:$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_{2}}$$\frac{1}{\lambda_{2}} = \frac{1}{355} - \frac{1}{680} = \frac{680 - 355}{355 	imes 680} = \frac{325}{241400}$$\lambda_{2} = \frac{241400}{325} \approx 742.77 \ nm \approx 743 \ nm$

$A$ gas absorbs a photon of $355 \ nm$ and emits at two wavelengths. If one of the emissions is at $680 \ nm,$ the other is at : ................. $nm$

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