The angular momentum of electron in d -orbital is equal to

The angular momentum of electron in

d

-orbital is equal to

$\sqrt{6} ℏ$
$\sqrt{2} ℏ$
$2 \sqrt{3} ℏ$
$0 ℏ$

Solution

For

d

-electrons,

l = 2

.
Angular orbital momentum

= \sqrt{l (l + 1)} \frac{h}{2 π}

= \sqrt{6} \frac{h}{2 π} = \sqrt{6} ℏ

Asked in: NEET 2015 (Phase 1)

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