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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