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Chemistry
What is the angular momentum of H-atom orbital if its electron energy is -3.4 eV ?
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Q. What is the angular momentum of H-atom orbital if its electron energy is $-3.4 \,eV ?$
Structure of Atom
A
$2.1 \times 10^{-34} J s$
22%
B
$1.05 \times 10^{-34} J s$
37%
C
$4.2 \times 10^{-34} J s$
11%
D
$2.6 \times 10^{-35} J s$
30%
Solution:
We know, $E_{n}=-\frac{13.6}{n^{2}} eV$
For H-atom.
$-3.4=-\frac{13.6}{n^{2}}$
$\Rightarrow n^{2}=\frac{13.6}{3.4}=4 \Rightarrow n=2$
Now, angular momentum $(m v r)=\frac{n h}{2 \pi} \pi$
$=\frac{2 \times 6.6 \times 10^{-34}}{2 \times 3.14}=2.1 \times 10^{-34} Js$