Question

The orbital angular momentum of an electron is $2s$ orbital is

• A. $+\frac{1}{2}.\frac{h}{2\pi }$
• B. Zero
• C. $\frac{h}{2\pi }$
• D. $\sqrt{2}.\frac{h}{2\pi }$

Answer 1

Answer: (B)

Orbital angular momentum $=\sqrt{l(l+1)}=\frac{h}{27}=0$
( $\because$ for 2s-eletron, $l=0$ )