Question

What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $-3.4eV$?

• A. $L=\frac{h}{\pi }$
• B. $L=\frac{4h}{\pi }$
• C. $L=\frac{3h}{\pi }$
• D. $L=\frac{5h}{\pi }$

Answer 1

The correct option is A $L=\frac{h}{\pi }$

First, we need to identify the quantum number of the energy level.
As $E=-\frac{13.6}{n^2}$, we have
$-3.4=-\frac{13.6}{n^2}$ or $n=2$
The angular momentum quantization gives, $L=mvr=\frac{nh}{2\pi }$
Substituting $n=2$, we get $L=\frac{2h}{2\pi }=\frac{h}{\pi }$