Atoms And Nuclei Question 444

Question: The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is $ 20.397,cm $ . The wavelength of energy for the same transition in $ H{{e}^{+}} $ is

Options:

A) $ 5.099\ c{{m}^{-1}} $

B) $ 20.497\ c{{m}^{-1}} $

C) $ 40.994\ c{{m}^{-1}} $

D) $ 81.988\ c{{m}^{-1}} $

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Answer:

Correct Answer: A

Solution:

  • $ E( =\frac{hc}{\lambda } )\propto \frac{Z^{2}}{n^{2}} $
    Þ $ \lambda \propto \frac{1}{Z^{2}} $ Hence $ {\lambda_{H{{e}^{+}}}}=\frac{20.397}{4}=5.099,cm $


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