JEE Main On 16 April 2018 Question 16
Question: Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths $ {\lambda_N},{\lambda_A} $ respectively. The ratio $ \frac{\lambda N}{\lambda A} $ is closest to [JEE Main 16-4-2018]
Options:
A) $ {10^{-6}} $
B) 10
C) $ {10^{-1}} $
D) $ {10^{-10}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{E _{A}}{E _{B}}=\frac{\frac{hc}{{\lambda_A}}}{\frac{hc}{{\lambda_N}}} $
So, $ \frac{E _{A}}{E _{N}}=\frac{{\lambda_N}}{{\lambda_A}} $
As order of $ E _{A}=ev $ and order of $ E _{N}=Mev $
$ \therefore $ $ \frac{{\lambda_N}}{{\lambda_A}}=\frac{1ev}{1Mev}=\frac{1}{10^{6}}={10^{-6}} $
So option A is correct Answer
