Atoms And Nuclei Question 438

Question: The following diagram indicates the energy levels of a certain atom when the system moves from 2E level to E, a photon of wavelength $ \lambda $ is emitted. The wavelength of photon produced during its transition from $ \frac{4E}{3} $ level to E is

Options:

A) $ \lambda /3 $

B) $ 3\lambda /4 $

C) $ 4\lambda /3 $

D) $ 3\lambda $

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Answer:

Correct Answer: D

Solution:

  • $ 2E-E=\frac{hc}{\lambda }\Rightarrow E=\frac{hc}{\lambda } $ $ \frac{4E}{3}-E=\frac{hc}{\lambda ‘}\Rightarrow \frac{E}{3}=\frac{hc}{\lambda ‘}$

$\therefore \frac{\lambda ‘}{\lambda }=3\Rightarrow \lambda ‘=3\lambda $



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