Question
For a $$d$$-electron, the orbital angular momentum is
A.
$$\sqrt 6 \left( {\frac{h}{{2\pi }}} \right)$$
B.
$$\sqrt 2 \left( {\frac{h}{{2\pi }}} \right)$$
C.
$$\left( {\frac{h}{{2\pi }}} \right)$$
D.
$$2\left( {\frac{h}{{2\pi }}} \right)$$
Answer :
$$\sqrt 6 \left( {\frac{h}{{2\pi }}} \right)$$
Solution :
The expression for orbital angular momentum is
Angular momentum $$ = \sqrt {l\left( {l + 1} \right)} \left( {\frac{h}{{2\pi }}} \right)$$
$$\eqalign{
& {\text{For }}d{\text{ orbital,}}\,l = 2. \cr
& {\text{Hence,}}\,L = \sqrt {2\left( {2 + 1} \right)} \left( {\frac{h}{{2\pi }}} \right) = \sqrt 6 \left( {\frac{h}{{2\pi }}} \right) \cr} $$