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:

• A. $\lambda =\frac{{\lambda }_1{\lambda }_2}{{\left({\lambda }_1{\lambda }_2\right)}^2}$
• B. $\lambda =\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}$
• C. $\lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
• D. $\frac{{\lambda }_1^2{\lambda }_2^2}{{\lambda }_1+{\lambda }_2}$

Answer 1

The correct option is C $\lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

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
$\lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
Let E be the energy of photon with wavelength $\lambda$,
$E_1$ be the energy of photon with wavelength ${\lambda }_1$
$E_2$ be the energy of photon with wavelength ${\lambda }_2$
$E=E_1+E_2$
$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$
$\frac{1}{\lambda }=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$
$\frac{1}{\lambda }=\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}$
$\lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$