The angular momentum of an electron in a 3d o | vedclass

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Question

(B) The angular momentum $(L)$ of an electron in an orbital is calculated using the formula: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$.For a $d$ orbital,the

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$\sqrt{6}h/2\pi$

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(B) The angular momentum $(L)$ of an electron in an orbital is calculated using the formula: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$.For a $d$ orbital,the azimuthal quantum number $(l)$ is $2$.Substituting $l = 2$ into the formula:$L = \sqrt{2(2+1)} \frac{h}{2\pi} = \sqrt{2 	imes 3} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$.Since $\sqrt{6} = \sqrt{2 	imes 3} = \sqrt{2} 	imes \sqrt{3}$,we can write this as $\sqrt{2} 	imes \sqrt{3} \frac{h}{2\pi} = \frac{\sqrt{3}h}{\sqrt{2}\pi}$.However,looking at the standard representation $\sqrt{6} \frac{h}{2\pi}$,if we simplify $\frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}} = \sqrt{1.5}$,the value is $\sqrt{1.5} \frac{h}{\pi}$.Given the options provided in the prompt,there seems to be a discrepancy in the standard notation. Re-evaluating the provided options,option $B$ is the closest mathematical equivalent to $\sqrt{6} \frac{h}{2\pi}$ if written as $\frac{\sqrt{3}h}{\sqrt{2}\pi}$.

The angular momentum of an electron in a $3d$ orbital is given by:

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