A) \[\frac{h}{2\pi }\]
B) zero
C) \[\frac{1}{2}.\frac{h}{2\pi }\]
D) \[\frac{1}{3}.\frac{h}{2\pi }\]
Correct Answer: B
Solution :
Orbital angular momentum \[=\frac{h}{2\pi }\,\sqrt{l(l+1)}\] For 3s electron, \[l=0\] \[\therefore \] Orbital angular monentum \[=\frac{h}{2\pi }\sqrt{0(0+1)}\] \[=0\]You need to login to perform this action.
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