The orbital angular momentum of a p-electron is given as

The orbital angular momentum of a p-electron is given as
  1. $\frac{h}{\sqrt{2} \pi}$
  2. $\sqrt{3} \frac{h}{2 \pi}$
  3. $\sqrt{\frac{3}{2}} \frac{h}{\pi}$
  4. $\sqrt{6} \cdot \frac{h}{2 \pi}$

Solution

Orbital angular momentum $=\sqrt{l(l+1)} \times \frac{h}{2 \pi}$ $\because$ For $p$-electron, $l=1$ $\therefore$ Orbital angular momentum, $\begin{aligned} & =\sqrt{1(1+1)} \times \frac{h}{2 \pi} \\ & =\sqrt{2} \times \frac{h}{2 \pi} \\ & =\frac{h}{\sqrt{2} \pi} \end{aligned}$

Asked in: NEET 2012 (Mains)

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