A) \[\frac{\sqrt{6}\,h}{2\pi }\]
B) \[\frac{\sqrt{2}\,h}{2\pi }\]
C) \[\frac{\,h}{2\pi }\]
D) \[\frac{2\,h}{\pi }\]
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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