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

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$\sqrt{\frac{3}{2}} \frac{h}{π}$

$\sqrt{3} \frac{h}{2 π}$

$\frac{h}{\sqrt{2} π}$

$\sqrt{6} \cdot \frac{h}{2 π}$

answer is A.

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Detailed Solution

Orbital angular momentum $= \sqrt{l (l + 1)} \times \frac{h}{2 π}$
For p-electron $(l = 1) = \sqrt{1 (1 + 1)} \times \frac{h}{2 π} = \sqrt{2} \times \frac{h}{2 π} = \frac{h}{\sqrt{2} π}$

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