Question

A dye absorbs a photon of wave length lamda and re - emits the same energy into two photons of wavelength lamda 1 and lamda 2 respectively. The wave length lamha is related with lamda 1 and lamda 2 as-

Answer 1

Lamda=lamda1 ×lamda2/lamda1 +lamda 2

Answer 2

Answer: The wave length $\lambda '$ is related with $\lambda _1$ and $\lambda _2$ as:

$\frac{1}{\lambda '}=\frac{1}{\lambda _1}+\frac{1}{\lambda _2}$

Explanation:

A dye absorbs a photon of wave length =$\lambda '$

$E'=\frac{hc}{\lambda '}$...(1) ( from Planck's Equation)

Dye emits the same energy into two photons of wavelength$\lambda _1$ and $\lambda _2$, the energies of the emitted photons are:

$E_1=\frac{hc}{\lambda _1}$...(2)

$E_2=\frac{hc}{\lambda _2}$...(3)

Since, energy remained conserved in above process of absorption and emission of the photons.

So,(1)=(2)+(3)

$E'=E_1+E_2$

$\frac{hc}{\lambda '}=\frac{hc}{\lambda _1}+\frac{hc}{\lambda _2}$

The above equation reduces to:

$\frac{1}{\lambda '}=\frac{1}{\lambda _1}+\frac{1}{\lambda _2}$

The wave length $\lambda '$ is related with $\lambda _1$ and $\lambda _2$ as:

$\frac{1}{\lambda '}=\frac{1}{\lambda _1}+\frac{1}{\lambda _2}$