A) \[\frac{6h}{2\pi }\]
B) \[\frac{\sqrt{6}\,h}{2\pi }\]
C) \[\frac{12h}{2\pi }\]
D) \[\frac{\sqrt{12}\,h}{2\pi }\]
Correct Answer: B
Solution :
| [b] We know that for d-electron \[l=2.\] |
| \[\mu =\sqrt{l(l+1)}\,\frac{h}{2\pi }\]; \[\mu =\sqrt{2\,(2+1)}\,\frac{h}{2\pi }\] |
| \[\mu =\sqrt{2\,(2+1)}\frac{h}{2\pi }\]; \[\mu =\,\sqrt{6}\frac{h}{2\pi }.\] |
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