Question

The angular momentum of an electron in the $2^{nd}$ excited state ofa Helium ion $\left(H{e}^{+}\right)$ is:

• A. $\frac{h}{2\pi }$
• B. $\frac{2h}{2\pi }$
• C. $\frac{3h}{2\pi }$
• D. $\frac{4h}{2\pi }$

Answer 1

The correct option is C $\frac{3h}{2\pi }$

Angular momentum of an electron is given by $\frac{nh}{2\pi }$, where $n$ is the energy orbit.

For second excited state $n=3$

Therefore, $L=\frac{3h}{2\pi }$