A) \[\frac{2h}{2\pi }\]
B) \[\frac{\sqrt{6}h}{2\pi }\]
C) \[\frac{\sqrt{2}h}{2\pi }\]
D) \[\frac{6h}{2\pi }\]
Correct Answer: B
Solution :
Orbital angular momentum \[=\sqrt{l(l+1)}\frac{h}{2\pi }\]where\[l=\]azimuthal quantum number Here, \[l=2\] \[\therefore \]Orbital angular momentum\[=\sqrt{2(2+1)}\frac{h}{2\pi }\] \[=\frac{\sqrt{6}h}{2\pi }\]
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