Question

The angular momentum of an electron in $3d$ orbital is:

• A. $\frac{2h}{\pi }$
• B. $\frac{\sqrt{5}h}{\pi }$
• C. $\frac{\sqrt{3}h}{2\pi }$
• D. $\frac{\sqrt{6}h}{2\pi }$

Answer 1

The correct option is D $\frac{\sqrt{6}h}{2\pi }$

$\text{Angular momentum of an electron in an orbital}=\sqrt{(l+1)}\frac{h}{2\pi }$

$\text{For d orbital,}l=2$

$\therefore \text{Angular momentum}=\sqrt{2(2+1)}\frac{h}{2\pi }=\frac{\sqrt{6}h}{2\pi }$