MCQ
For a $d$ electron, the orbital angular momentum is
- ✓$\sqrt {6\hbar } $
- B$\sqrt {2\hbar } $
- C$\hbar $
- D$2\hbar $
✓
Answer
Correct option: A.
$\sqrt {6\hbar } $
a
We know that the orbital angular momentum $(L)=\sqrt{1(1+1)} \frac{h}{2 \pi}$
We know that the orbital angular momentum $(L)=\sqrt{1(1+1)} \frac{h}{2 \pi}$
For $d -$ orbital, $1=2$
Thus, the orbital angular momentum is $=\sqrt{6} \frac{ h }{2 \pi}$
$\sqrt {6\hbar } $
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