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For a d electron,the orbital angular momentum is :

$(a)\;\sqrt 6h\qquad(b)\;\sqrt 2h\qquad(c)\;h\qquad(d)\;2h$

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Angular momentum in 3d-orbital=$\large\frac{h}{2\pi}$$\sqrt{l(l+1)}$.
$\Rightarrow \large\frac{h}{2\pi}$$\sqrt{2(2+1)}$
$\Rightarrow \sqrt{6}\large\frac{h}{2\pi}$
$h=\large\frac{h}{2\pi}$
$\Rightarrow \sqrt{6}h$
Hence (a) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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