A) \[\sqrt{6}(h/2\pi )\]
B) \[\sqrt{2}(h/2\pi )\]
C) \[(h/2\pi )\]
D) \[2(h/2\pi )\]
Correct Answer: C
Solution :
Orbital angular momentum \[=\frac{h}{2\pi }\sqrt{\ell (\ell +1)}\]\[\ell \]-orbital has 2 nodal plane
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