Atoms And Nuclei Question 291
Question: A hydrogen atom in its ground state absorbs 10.2 eV of energy The orbital angular momentum is increased by
Options:
A) $ 1.05\times {{10}^{-34}}J-s $
B) $ 3.16\times {{10}^{-34}}J-s $ [b] $ E_{p}=\frac{2K_{ex}}{{{( x^{2}+\frac{b^{2}}{4} )}^{3/2}}} $ For maximum $ E_{p} $ $ \frac{dE_{p}}{dx}=0 $
$ \Rightarrow x=\frac{b}{2\sqrt{2}} $
C) $ 2.11\times {{10}^{-34}}J-s $
D) $ 4.22\times {{10}^{-34}}J-s $
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Answer:
Correct Answer: A
Solution:
- Electron after absorbing 10.2 eV energy goes to its first excited state (n = 2) from ground state (n = 1).
$ \therefore $ Increase in momentum $ =\frac{h}{2\pi } $ $ =\frac{6.6\times {{10}^{-34}}}{6.28}=1.05\times {{10}^{-34}}J-s $
