Atoms And Nuclei Question 79
The wavelength of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is $ 20.397,\text{nm} $ . The wavelength of energy for the same transition in $ H{{e}^{+}} $ is [AIIMS 1997; JIPMER 2000]
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,\text{nm} $
BETA
