MCQ
The angular moment of electron in $d-$ orbital is equal to : $($Where $\not\text{h}=\frac{\text{h}}{2\pi})$
- A$2\sqrt{3}\not{\text{h}}$
- B$0\not{\text{h}}$
- ✓$\sqrt{6\not}\text{h}$
- D$\sqrt{2}\not\text{h}$
✓
Answer
Correct option: C.
$\sqrt{6\not}\text{h}$
$\text{m},\text{vr}=\text{l}\sqrt{(\text{l}+1)}\frac{\text{h}}{2\pi}=\sqrt{2\times3}\frac{\text{h}}{2\pi}$
$=\sqrt{6}\frac{\text{h}}{2\pi}=\sqrt{6}\not\text{h}$
$=\sqrt{6}\frac{\text{h}}{2\pi}=\sqrt{6}\not\text{h}$
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