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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

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 $(m v r)=\frac{n h}{2 \pi} \pi$

$=\frac{2 \times 6.6 \times 10^{-34}}{2 \times 3.14}=2.1 \times 10^{-34} Js$