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The angular momentum of an electron in p orbitals is equal to

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zero

$(h / 2 π)$

$\sqrt{2} (h / 2 π)$

$\sqrt{6} (h / 2 π)$

answer is C.

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Detailed Solution

A p-orbiting electron's orbital angular momentum is $\frac{h}{\sqrt{2} π}$

$μ_{L} = \sqrt{1 (1 + 1)} \frac{h}{2 π}$

A p orbital's l=1

Hence,

$\begin{array}{l} μ_{L} = \sqrt{1 (1 + 1)} \frac{h}{2 π} \\ μ_{L} = \sqrt{2} \frac{h}{2 π} \end{array}$

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