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- Thread starter Chemist20
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Bill_K

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_{r}+ 1 where n_{r}is the radial quantum number, i.e. the number of radial nodes. And so more nodes in the angular direction tends to increase the energy, but it is accompanied by fewer nodes in the radial direction which tends to decrease it.

Uhm okay, but I still don't get why for the 2s being more core-like than 2p, for the hydrogenic atom this isn't taken into account and hence both orbitals have the same energy.

why does it only take it into account when we talk about multi electron atoms??

- #4

jtbell

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It's an "accidental" degeneracy of course, but qualitatively here's why: the energy depends only on the principal quantum number n = ℓ + n_{r}+ 1 where n_{r}is the radial quantum number

This holds only for the Coulomb potential, V ≈ -1/r, IIRC.

why does it only take it into account when we talk about multi electron atoms??

In a multi-electron atom, an individual electron "feels" not only the attraction of the nucleus,but also the repulsion of the other electrons. The "effective" potential is not -1/r as with a one-electron atom.

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phyzguy

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Your intuition is correct. In reality they

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Your intuition is correct. In reality theydon'thave the same energy. The relativistic corrections and the spin-orbit coupling breaks the degeneracy.

Right... so then why in the H atom the 2s has the same energy as the 2p ?

also... how does spin coupling affect the energies. as far as I knew, spin coupling arises due to the interaction of the orbital angular momentum and spin angular momentum. So....?

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Vanadium 50

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Potentials of the form r

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