Question

# A gas absorbs a photon of 355 nm and emits it at two wavelengths. If one of the emissions is at 680 nm, what will be the next wavelength emitted?

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Solution

## Conservation of EnergyAccording to the Law of conservation energy, the absorbed photon must be equal to the combined energy of the emitted photons${\mathrm{E}}_{\mathrm{A}}={\mathrm{E}}_{1}+{\mathrm{E}}_{2}+\dots$Here, ${\mathrm{E}}_{\mathrm{A}}$ is the energy of the absorbed photon. ${\mathrm{E}}_{1}\mathrm{and}{\mathrm{E}}_{2}$are the energy of the emitted photons. Calculation of the second wavelengthStep 1: Relationship between energy and the wavelength$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$Here, E represents energyh represents Planck's constantc represents velocity $\mathrm{\lambda }$ represents wavelengthStep 2: Determination of the second wavelengthThe wavelength of the absorbed photon (${\mathrm{\lambda }}_{\mathrm{A}}$) = 355 nmThe wavelength of the emitted photon (${\mathrm{\lambda }}_{1}$) = 680 nm $\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{1}}+\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{1}{{\mathrm{\lambda }}_{1}}+\frac{1}{{\mathrm{\lambda }}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{355}=\frac{1}{680}+\frac{1}{{\mathrm{\lambda }}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{{\mathrm{\lambda }}_{2}}=\frac{1}{355}-\frac{1}{680}\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda }}_{2}=743\mathrm{nm}$Therefore, the wavelength of the second emitted photon is 743 nm.

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