Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The s orbital have higher probability to be closer to the core and feels larger attraction than the p orbital and on average is further away and in addition p has repulsive potentilal l(l+1)h^2/2mr^2. But is there name for this effect and is it this the only way to explain ?

What is the usual energy difference between these levels and how is measured, the transitions are highly forbidden ?

Does quantum electrodynamics separate s from p ? In Lamb shift I think I does not but how about the multielectron atoms ?

In real life if hydrogen was not a molecule but atom is s lower that p, and is it opposite in all multielectron atoms ? There is no hund rule for this ?

share|improve this question
This course might help you docbrown.info/page07/ASA2ptable2.htm#2.2 –  anna v Sep 26 '12 at 5:26

2 Answers 2

The confusion in your question comes from the fact that there are many S states, not just one. The lowest S-state is the ground state, n=1, then there is an S and P state at n=2, then an S,P,D state at n=3, and S,P,D,F state at n=4, corresponding to l=0,1,2,3 (the letter names are not particularly informative).

The l states do have a stronger repulsion from the core, this is called the "centrifugal barrier", but this just makes the l states degenerate with higher S-states than the ground state, and there are infinitely many S-states.

There is a simple theorem that guarantees the lowest energy state is everywhere positive, this is because the ground state minimizes the energy function, but a wavefunction and it's absolute value have the same integrated average energy, and if there is a sign change, the absolute value has a kink.

share|improve this answer

First, your explanation at the beginning isn't right. The centrifugal potential is counted in the hydrogen atom calculation but the energy of the state only turns out to be a function of $n$, not $l$ or $m$.

The preference for $l=0$ states only applies to multi-electron atoms and it boils down to electron-electron interactions. Electrons with lower values $l$ are closer to the nucleus so they feel the full electric charge of it which gives these bound states a higher, more negative potential energy. On the other hand, electrons with higher $l$ are further and the nuclei charges are partially "screened" or "shielded" by other electrons, so the "gain" for the binding energy is smaller.

More detailed questions about the ordering of the electron states and their spin in multi-electron atoms are answered by Hund's rules.

The energy difference between $s$ and $p$ electron states is substantial, comparable to other energy differences between major levels, i.e. nearly an electronvolt (values of course depend on the exact atoms and states), and may only be suppressed for higher $Z$ (the relative difference i.e. ratio is suppressed, the absolute energy difference is large, too).

The energy of all levels of an atom may always be measured from the spectra. Each atom only has one truly stable state, the ground state with the lowest energy, and there exist transitions from all higher states that emit photons of measurable frequencies.

The Lamb shift is a very subtle effect in the hydrogen atom implied by QED. It only makes qualitative sense to calculate it for the hydrogen atom. For multi-electron atoms, the degeneracy (originally equal energies of levels) is lifted (removed) by much stronger effects than the Lamb shift. So although the Lamb shift is present even for multi-electron atoms, it almost never influences the qualitative ordering of levels.

For non-relativistic hydrogen atom, as I said, the energy depends on the quantum number $n$ via $-1/n^2$ only. It doesn't depend on $l,m,s$. The Dirac equation adds some small, relativistic correction dependence on $j$ which comes from $\vec L+\vec S$ but keeps some accidental degeneracy for different values (pairs) of $l$. This degeneracy is removed in QED by the Lamb shift. Those things are calculable, fine, and predictable for the hydrogen atom only. More complicated atoms already lift the degeneracy at the non-relativistic level, before we even consider the Dirac equation or QED, and Dirac equation or QED only make small corrections to a "maximally split" spectrum.

share|improve this answer
Lubos some clarification necessary, you say: "the energy of the state only turns out to be a function of n, not l or m." and in the next paragraph "The energy difference between s and p electron states is substantial, comparable to other energy differences between major levels, i.e. nearly an electronvolt" . In the link I gave above the stateent is made :" The principal quantum electronic energy levels (n) can be split into sub-levels denoted by s, b, d and f depending on the number of electrons in the 'system'." –  anna v Sep 26 '12 at 5:27
maybe the contrast Hydrogen_atom/multielectron_atoms should have "n level energies are split depending on the l,m,spin content" or some such. –  anna v Sep 26 '12 at 6:27
Dear anna, I think that my wording made it clear that one of the sentences applies to hydrogen and the other sentence applies to multi-electron atoms. –  Luboš Motl Sep 26 '12 at 11:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.