Atoms And Nuclei Question 86
Question: When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck’s constant: $ h=6.6\times {{10}^{-34}}J\text{-s} $ ) [AFMC 2000; Pb. PET 2001]
Options:
A) $ 4.16\times {{10}^{-34}},J\text{-}s $
B) $ 3.32\times {{10}^{-34}},J\text{-}s $
C) $ 1.05\times {{10}^{-34}},J\text{-}s $
D) $ 2.08\times {{10}^{-34}},J\text{-}s $
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Answer:
Correct Answer: C
Solution:
Change in the angular momentum $ \Delta L=L_{2}-L_{1}=\frac{n_{2}h}{2\pi }-\frac{n_{1}h}{2\pi } $
$ \Rightarrow \Delta L=\frac{h}{2\pi }(n_{2}-n_{1}) $ $ =\frac{6.6\times {{10}^{-34}}}{2\times 3.14}(5-4) $ $ =1.05\times {{10}^{-34}}J\text{-}S $
