A) \[\frac{h}{2\pi }\]
B) \[\frac{h}{\pi }\]
C) \[\sqrt{2(2+l)}\frac{h}{2\pi }\]
D) none of these
Correct Answer: B
Solution :
The angular momentum of an electron \[=mvr=\frac{nh}{2\pi }\] For the first excited state\[n=2\] then, \[mvr=\frac{2h}{2\pi }\] \[=\frac{h}{\pi }\]
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