Question

The difference between orbital angular momentum of a $2s$ and a $3d$ electron is:

• A. $\sqrt{6}h$
• B. $\frac{\sqrt{6}h}{2\pi }$
• C. $\sqrt{4}h$
• D. $\sqrt{5}h$

Answer 1

Answer: (B)

$\frac{\sqrt{6}h}{2\pi }$

The expression for the orbital angular momentum is ${\mu }_l=\sqrt{l(l+1)}\frac{h}{2\pi }$.

For $2s$ electron, $l=0$ and for $3d$ electron, $l=2$.

Hence, ${\mu }_{l,3d}-{\mu }_{l,2s}=\sqrt{2(2+1)}\frac{h}{2\pi }-\sqrt{0(0+1)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }$.