A) \[+\frac{1}{2}.\,\frac{h}{2\pi }\]
B) zero
C) \[\,\frac{h}{2\pi }\]
D) \[\sqrt{2}\,.\,\frac{h}{2\pi }\]
Correct Answer: B
Solution :
For s-electron, \[l=0\]. Therefore, orbital angular momentum = 0You need to login to perform this action.
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