Question

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

• A. $\sqrt{2}\left(\frac{h}{2\pi }\right)$
• B. $\sqrt{3}\left(\frac{h}{2\pi }\right)$
• C. $\sqrt{6}\left(\frac{h}{2\pi }\right)$
• D. $\sqrt{12}\left(\frac{h}{2\pi }\right)$

Answer 1

Answer: (C)

The angular momentum is given by
$L=\sqrt{l(l+1)}\left(\frac{h}{2\pi }\right)$
For $3d$ electron, $l=2$
$\therefore L=\sqrt{2(3)}\left(\frac{h}{2\pi }\right)=\sqrt{6}\left(\frac{h}{2\pi }\right)$