A) \[\frac{h}{\sqrt{2\pi }}\]
B) \[\sqrt{3}\frac{h}{2\pi }\]
C) \[\sqrt{\frac{3}{2}}\frac{h}{\pi }\]
D) \[\sqrt{6}.\frac{h}{2\pi }\]
Correct Answer: A
Solution :
Orbital angular momentum\[=\sqrt{l(l+1)}\times \frac{h}{2\pi }\] \[\because \]For p-electron, \[l=1\] \[\therefore \]Orbital angular momentum, \[=\sqrt{1(1+1)}\times \frac{h}{2\pi }\] \[=\sqrt{2}\times \frac{h}{2\pi }\] \[=\frac{h}{\sqrt{2}\pi }\]
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