Question

What is the angular momentum of H-atom orbital if its electron energy is $-3.4 \,eV ?$

• A. $2.1 \times 10^{-34} J s$
• B. $1.05 \times 10^{-34} J s$
• C. $4.2 \times 10^{-34} J s$
• D. $2.6 \times 10^{-35} J s$

Answer 1

Answer: (A)

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

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