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