A) \[+\frac{1}{2}.\frac{h}{2\pi }\]
B) \[zero\]
C) \[\frac{h}{2\pi }\]
D) \[\sqrt{2}\cdot \frac{h}{2\pi }\]
Correct Answer: B
Solution :
Orbital angular momentum\[\sqrt{l(l+1)}\frac{h}{27}=0\]\[(\because \,\,for\,\,2s-electron,\,\,l=0)\]
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