Question

The orbital angular momentum of an electron in a 3d-subshell is:

• A. Zero
• B. $\frac{\mathrm{h}}{2\mathrm{\pi }}$
• C. $\frac{\mathrm{h}}{4\mathrm{\pi }}$
• D. $\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

Answer 1

The correct option is D

$\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

The explanation for the correct option

(D)$\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

Step1. The formula for orbital angular momentum

$\mathrm{L}=\sqrt{\mathrm{l}(\mathrm{l}+1)}\frac{\mathrm{h}}{2\mathrm{\pi }}$

Step2. The calculation for orbital angular momentum

For the 3d subshell, $\mathrm{l}=2$

$$\begin{gathered} \mathrm{L}=\sqrt{2(2+1)}\frac{\mathrm{h}}{2\mathrm{\pi }} \\ =\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }} \end{gathered}$$

Explanation for incorrect option

The orbital angular momentum of an electron in a 3d-subshell is $\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$. Hence, options (A), (B), and (C) are incorrect.

Hence, option (D) is correct. The orbital angular momentum of an electron in a 3d-subshell is $\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$.