The orbital angular momentum of an electron i | vedclass

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Question

(D) The orbital angular momentum $(L)$ of an electron is given by the formula: $L = \sqrt{l(l+1)} \hbar$, where $l$ is the azimuthal quantum number an

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$\sqrt{6} \hbar$

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(D) The orbital angular momentum $(L)$ of an electron is given by the formula: $L = \sqrt{l(l+1)} \hbar$, where $l$ is the azimuthal quantum number and $\hbar = \frac{h}{2\pi}$.For a $d$ orbital, the value of the azimuthal quantum number $l$ is $2$.Substituting $l = 2$ into the formula:$L = \sqrt{2(2+1)} \hbar$$L = \sqrt{2(3)} \hbar$$L = \sqrt{6} \hbar$Therefore, the orbital angular momentum of an electron in a $d$ orbital is $\sqrt{6} \hbar$.

The orbital angular momentum of an electron in a $d$ orbital is equal to:

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