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Q. The angular momentum of electron in $3d$ orbital of an atom is

JIPMERJIPMER 2012Atoms

Solution:

The angular momentum is given by
$L = \sqrt{l(l+1)} \left(\frac{h}{2\pi} \right)$
For $3d$ electron, $l = 2$
$\therefore \:\:\: L = \sqrt{2(3)} \left(\frac{h}{2\pi}\right) = \sqrt{6} \left(\frac{h}{2\pi}\right) $