A) \[\frac{\sqrt{6}h}{2\pi }\]
B) \[\frac{\sqrt{2}h}{2\pi }\]
C) \[\frac{h}{2\pi }\]
D) \[\frac{2h}{\pi }\]
Correct Answer: A
Solution :
Angular momentum \[=\sqrt{l(l+1)}\frac{h}{2\pi }\] For d-orbital;\[l=2\] Angular momentum\[=\sqrt{2(2+1)}\frac{h}{2\pi }\] \[=\sqrt{6}\frac{h}{2\pi }\]
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