How many degenerate orbital are present in a subshell if electron associated with that subshell possesses orbital angular momentum $= 2 \sqrt{3} h$ ?

No degenerated orbitals

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Seven degenerated orbitals

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Three degenerated orbitals

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Number of degenerated orbitals are same as in subshell which possesses

μ_{L} = 3.46 B . M

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Solution

The correct option is B Seven degenerated orbitals
Orbital angular momentum $= \sqrt{l (l + 1)} \cdot \frac{h}{2 π}$
where $l = A z i m u t h a l q u a n t u m n o .$
Total no.of degenerate orbitals $= 2 l + 1$
For the given case, $l = 3 \Rightarrow N o . o f d e g e n e r a t e o r b i t a l s = 7$

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= \frac{2 \sqrt{3} h}{2 π}

= \frac{2 \sqrt{3} h}{2 π}

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