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Q. If an electron has orbital angular momentum quantum number $ l=7, $ then it will have an orbital angular momentum equal to

BHUBHU 2011

Solution:

$ L=\sqrt{l(l+1)}\frac{n}{2\pi } $
$ L=\sqrt{7(7+1)}\frac{h}{2\pi } $
$ L=\sqrt{7\times 8}\frac{h}{2\pi }=\sqrt{56}\left( \frac{h}{2\pi } \right) $