Question

The spin angular momentum of an electron has no analog in classical mechanics. However, it turns out that the treatment of spin angular momentum is closely analogous to the treatment of orbital
angular momentum.

Spin angular momentum $=\sqrt{s(s+1)}h$

Orbital angular momentum $=\sqrt{l(l+1)}h$

Total spin of an atom or ion is a multiple of $\frac{1}{2}$. Spin multiplicity isa factor to confirm the electronic configuration of an atom or ion.Spin multiplicity $=(2\sum s+1)$

The orbital angular momentum for a $2p$-electron is:

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

Answer 1

Answer: (D)

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

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

The orbital angular momentum for a 2p-electron $(l=1)$ is

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

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

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

Hence, the correct option is $D$