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Degeneracy

This problem was asked in JEE 2015. It states:
If the degeneracy of hydrogen atom in second excited state (\(n=3\)) is 9 then find the degeneracy of hydride anion i.e. \(\ce{H^-}\) ion in second excited state.

Note: electronic spin is not to be considered here.

Note by Dipanjan Chowdhury
8 months, 2 weeks ago

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Before I answer this, let me ask you a question that will be key to understanding what's happening here:
Why is the degeneracy of hydrogen atom 9 when n = 3? Ameya Daigavane · 8 months, 2 weeks ago

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@Ameya Daigavane I think because of the one 3s three 3p and five 3d orbitals which are not excited as have no electron and as a whole exhibit degeneracy Dipanjan Chowdhury · 8 months, 2 weeks ago

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@Dipanjan Chowdhury Okay, good! The fact that hydrogen has only one electron means that orbitals with the same value of n have the same energies. n = 3 has 9 orbitals, as you said, so 9 is the degeneracy.
But for H-, there are two electrons. The orbitals' energies now depend on l (the azimuthal quantum number that differentiates between s,p, d, f and so on.) like, \( 1s < 2s < 2p < 3s < 3p ...\)
Now the ground state of H- is \(1s^2 \).
The 1st excited state is \( 1s^1 2s^1 \).
The second excited state is \( 1s^1 2s^0 2p^1 \) (Compare with H atom for which the 2nd excited state is \( 3s^1 \).)
But there are 3 p orbitals for the electron to 'choose' between when making the transition from first excited state to second excited state.
So the degeneracy is 3. Hope this helps. Ameya Daigavane · 8 months, 2 weeks ago

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@Ameya Daigavane Absolutely clear, Sir. Thank You . :D I was unaware of this sort of splitting. Dipanjan Chowdhury · 8 months, 2 weeks ago

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