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 :
Orbital angular momentum \[=\sqrt{l(l+1)}.\frac{h}{2\pi }\]for 2s-orbital, \[l=0\] \[\therefore \] Orbital angular momentum \[=\sqrt{0(0+1)}\frac{h}{2\pi }=zero\]
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