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 λ is related to $ {{\lambda }_{1}} $ and $ {{\lambda }_{2}} $ as:
(A) $ \lambda =\dfrac{{{\lambda }_{1}}{{\lambda }_{2}}}{{{\left( {{\lambda }_{1}}{{\lambda }_{2}} \right)}^{2}}} $
(B) $ \lambda =\dfrac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}} $
(C) $ \lambda =\dfrac{{{\lambda }_{1}}{{\lambda }_{2}}}{{{\lambda }_{1}}+{{\lambda }_{2}}} $
(D) $ \dfrac{\lambda _{1}^{2}\lambda _{2}^{2}}{{{\lambda }_{1}}+{{\lambda }_{2}}} $

Answer 1

Hint :We know that Photon vitality is the vitality conveyed by a solitary photon. The measure of vitality is straightforwardly relative to the photon's electromagnetic recurrence and accordingly, comparably, is conversely corresponding to the frequency.

Complete Step By Step Answer:
The higher the photon's recurrence, the higher it is energy. Photon vitality is more valuable than heat vitality, as the vast majority of the sunlight based pillar will in general change into warmth of ambient. The most effective work of sun based bars will be finished by using the quantum parts of it as in the photosynthetic elements of plants. It is imperative to examine the utilization of sunlight based photon energies to part water accepting plants as a model.
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}} $
 $ \therefore E={{E}_{1}}+{{E}_{2}} $
Since we know that $ E=\dfrac{hc}{\lambda }; $ $ \therefore \dfrac{hc}{\lambda }=\dfrac{hc}{{{\lambda }_{1}}}+\dfrac{hc}{{{\lambda }_{2}}} $
 $ \Rightarrow \dfrac{1}{\lambda }=\dfrac{1}{{{\lambda }_{1}}}+\dfrac{1}{{{\lambda }_{2}}} $
On further solving we get; $ \Rightarrow \dfrac{1}{\lambda }=\dfrac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}} $
By taking reciprocal the above equation we get is;
 $ \lambda =\dfrac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}}. $
Therefore, 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 λ is related to $ {{\lambda }_{1}} $ and $ {{\lambda }_{2}} $ as $ \lambda =\dfrac{{{\lambda }_{1}}{{\lambda }_{2}}}{{{\lambda }_{1}}+{{\lambda }_{2}}} $
So the correct answer is option C.

Note :
Remember the formula and as in the question we are given a percentage of absorbed energy and re-emitted energy so use it to equate the two energies and calculate the ratio that is needed. This was an improvement to the Rutherford model and can be considered as a quantum physical explanation for it.