Question

Okay I will give you another easy one. Just one last.
Find the angular momentum of the electron in $2^{nd}$ orbit of $H{e}^{+}$ ion.

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

Answer 1

The correct option is B $\frac{h}{\pi }$

Orbital angular momentum for hydrogen like elements are = $\frac{nh}{2\pi }$
Where n is the $n^{th}$ orbital.
Here the electron is in $2^{nd}$ orbital.
Also there is no factor of Z here.
So this is same for all hydrogen like atoms. The angular momentum only depends on the orbital number in which the electron is present in.

Here the electron is in 2nd orbital
So $L=\frac{2h}{2\pi }=\frac{h}{\pi }$