Question

# A dye absorbs a photon of wavelength $\lambda$ and re-emits the same energy into two photons of wavelength ${\lambda }_{1}$ and ${\lambda }_{2}$ respectively. The wavelength $\lambda$ is related to ${\lambda }_{1}$ and ${\lambda }_{2}$ as?

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Solution

## Step 1: GivenAssume $E$ to be the energy of photon of wavelength $\lambda$Assume ${E}_{1}$ to be the energy of photon of wavelength ${\lambda }_{1}$Assume ${E}_{2}$ to be the energy of photon of wavelength ${\lambda }_{2}$Step 2: Formula UsedEnergy of a photon is given as $E=\frac{hc}{\lambda }$, where $h$ is Planck's constant and $c$ is speed of light.Step 3: SolutionThe energy of the photon of wavelength $\lambda$ will be equal to the sum of energies of the photons with wavelengths ${\lambda }_{1}$ and ${\lambda }_{2}$, because they are re-emitted from that photon. Thus, $E={E}_{1}+{E}_{2}\phantom{\rule{0ex}{0ex}}⇒\frac{hc}{\lambda }=\frac{hc}{{\lambda }_{1}}+\frac{hc}{{\lambda }_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{hc}{\lambda }=hc\left(\frac{1}{{\lambda }_{1}}+\frac{1}{{\lambda }_{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\lambda }=\frac{1}{{\lambda }_{1}}+\frac{1}{{\lambda }_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\lambda }=\frac{{\lambda }_{1}+{\lambda }_{2}}{{\lambda }_{1}{\lambda }_{2}}\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{{\lambda }_{1}{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}$Hence, the relation is $\lambda =\frac{{\lambda }_{1}{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}$.

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