Question

The orbital angular momentum of $3p$ electron is :

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

Answer 1

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

$\text{Orbital angular momentum}=\sqrt{l(l+1)}\frac{h}{2\pi }l=1\text{for p-orbital.}\therefore \sqrt{l(l+1)}\frac{h}{2\pi }=\sqrt{1(1+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi }$