Question

If a dye absorbs a photon of wavelength $\lambda$ and re-emits the absorbed energy as two photons of wavelength ${\lambda }_1$ and ${\lambda }_2$. Then:

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

Answer 1

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

According to the question,
$E=E_1+E_2$
$\Rightarrow \frac{hc}{\lambda }=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}[\because E=hv=\frac{hc}{\lambda }]$
$\Rightarrow \frac{hc}{\lambda }=hc\left(\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}\right)$
$\Rightarrow \lambda =\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$