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 \[=\sqrt{l(l+1)}=\frac{h}{27}=0\] (\[\because \] for 2s-eletron, \[l=0\])
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