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?

Answer 1

Step 1: Given

Assume $E$ 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 Used

Energy of a photon is given as $E=\frac{hc}{\lambda }$, where $h$ is Planck's constant and $c$ is speed of light.

Step 3: Solution

The 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,

$$\begin{gathered} E=E_1+E_2 \\ \Rightarrow \frac{hc}{\lambda }=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2} \\ \Rightarrow \frac{hc}{\lambda }=hc\left(\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}\right) \\ \Rightarrow \frac{1}{\lambda }=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2} \\ \Rightarrow \frac{1}{\lambda }=\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2} \\ \Rightarrow \lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2} \end{gathered}$$

Hence, the relation is $\lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$.