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The angular momentum of electron in 'd' orbital is equal to:
A
$\sqrt{6}\,\hbar$
B
$\sqrt{2}\,\hbar$
C
$2\sqrt{3}\,\hbar$
D
$0\,\hbar$
Detailed Solution
Orbital angular momentum $= \sqrt{l(l+1)}\,\hbar$
For a d orbital, l = 2: $\sqrt{2(2+1)}\,\hbar = \sqrt{6}\,\hbar$
For a d orbital, l = 2: $\sqrt{2(2+1)}\,\hbar = \sqrt{6}\,\hbar$