Structure of Atom - Result Question 37
####41. The orbital angular momentum of a $p$-electron is given as :
[2012 M]
(a) $\frac{h}{\sqrt{2} \pi}$
(b) $\sqrt{3} \frac{h}{2 \pi}$
(c) $\sqrt{\frac{3}{2}} \frac{h}{\pi}$
(d) $\sqrt{6} \cdot \frac{h}{2 \pi}$
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Solution:
- (a) Orbital angular momentum
$=\frac{h}{2 \pi} \sqrt{\ell(\ell+1)}$
For $p$ orbital $\ell=1$
$ \text{ So, }=\frac{h}{2 \pi} \sqrt{2}=\frac{h}{\sqrt{2} \pi} $
