Atomic Structure Question 381
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 with $ {\lambda_1} $ and $ {\lambda_2} $ as:
Options:
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}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ E=E_1+E_2; $
$ \frac{hc}{\lambda }=\frac{hc}{{\lambda_1}}+\frac{hc}{{\lambda_2}} $
$ \Rightarrow \frac{hc}{\lambda }=hc( \frac{{\lambda_2}+{\lambda_1}}{{\lambda_1}{\lambda_2}} );\lambda =\frac{{\lambda_1}{\lambda_2}}{{\lambda_1}+{\lambda_2}} $
