Dual Nature Of Light Question 392
Question: An atom emits a photon of wavelength X = 600 m by transition from an excited state of life time $ 8\times {{10}^{-9}}s $ . If $ \Delta ,v $ represents the minimum uncertainty in the frequency of the photon, the fractional width $ \frac{\Delta v}{v} $ of the spectral line is of the order of
Options:
A) $ {{10}^{-,4}} $
B) $ {{10}^{-6}} $
C) $ {{10}^{-,8}} $
D) $ {{10}^{-10}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \Delta E.\Delta t\sim h $ $ E=\frac{hc}{\lambda }\text{ }\Delta \Epsilon =\frac{hc}{\lambda }\frac{\Delta \lambda }{\lambda } $
$ \therefore \frac{hc}{\lambda }\frac{\Delta \lambda }{\lambda }.\Delta t\sim h $ Now, $ c=v\lambda $ $ v\Delta \lambda +\lambda \Delta v=0 $ $ vD\lambda =-\lambda \Delta v $
$ \therefore \frac{\Delta \lambda }{\lambda }=-\frac{\Delta v}{v}\text{ }\therefore \frac{ch}{\lambda }\frac{\Delta v}{v}\Delta t\sim h $
$ \therefore \frac{\Delta v}{v}\sim \frac{h}{\Delta t}.\frac{\lambda }{hc}\sim \frac{\lambda }{\Delta tc}=\frac{600\times {{10}^{-9}}}{8\times {{10}^{-9}}\times 3\times 10^{8}} $ $ \frac{\Delta v}{v}\sim {{10}^{-6}} $
BETA
