Question

The angular momentum of electron in $f-$orbital is equal to:

• A. 0
• B. $\sqrt{6\pi }$
• C. $\sqrt{2\pi }$
• D. $2\sqrt{3\pi }$

Answer 1

The correct option is D $2\sqrt{3\pi }$

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

For electron in $f$ orbital, $l=3$

$\therefore$ Angular momentum $=\sqrt{3\times 4}\frac{h}{2\pi }=2\sqrt{3}\frac{h}{2\pi }$