A) \[\sqrt{6}\,\frac{h}{2\pi }\]
B) \[\sqrt{2}\,\frac{h}{2\pi }\]
C) \[\frac{\sqrt{h}}{2\pi }\]
D) \[\frac{2h}{2\pi }\]
Correct Answer: A
Solution :
For d-electron, \[l=2\] Orbital angular momentum \[=\sqrt{l(l+1)}\frac{h}{2\pi }\] \[=\sqrt{2(2+1)}\frac{h}{2}=\sqrt{6}\frac{h}{2\pi }\]You need to login to perform this action.
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