A) \[\frac{1}{2}\frac{h}{2\pi }\]
B) \[\frac{h}{2\pi }\]
C) \[\sqrt{2}\frac{h}{2\pi }\]
D) Zero
Correct Answer: D
Solution :
Since s-orbital have \[l=0\] Angular momentum = \[\sqrt{l\,(l+1)}\times \frac{h}{2\pi }\] = \[0\times \frac{h}{2\pi }=0\] .You need to login to perform this action.
You will be redirected in
3 sec