In a multielectron atom

1. May 6, 2009

albertsmith

In a multielectron atom, the lowest-l state for each n (2s, 3s, 4s, etc.) is significantly lower in energy than the hydrogen state having the same n. But the highest-l state for each n (2p, 3d, 4f, etc.) is very nearly equal in energy to the hydrogen state with the same n. Explain?

2. May 7, 2009

alxm

Explain why you think this is the case.

3. May 7, 2009

jrlaguna

Never heard about that. Can you specify a little bit more or give numbers?

4. May 7, 2009

Bob S

In multi-electron atoms, the inner electrons hide (shield, screen) the full effect of the nuclear charge from the outer electron(s). This is especially true in atoms that have filled shells +1 electron, like lithium (2 + 1 =3 electrons), sodium (10 + 1 =11 electrons), and potassium (18 + 1 =19 electrons). The outer electron in these atoms effectively "sees" only one nuclear charge.

Last edited: May 7, 2009
5. May 7, 2009

granpa

does the electron in a hydrogen atom ever go into the s subshell except for n=1. I thought (based on the zeeman effect) that the electron in a hydrogen atom only (or at least normally) transitioned between states where l=n-1.

Last edited: May 7, 2009
6. May 7, 2009

Bob S

Electrons are usually captured into high n states, most probably with high l (probability proportional to 2l+1), and as they cascade down, most electrons eventually end up cascading through l=n-1 states. This is why 4f-3d, and 3d-2p transitions are so bright. An electron could find itself stuck in the 2s state, and cannot get to the 1s state by the standard delta-l = +/- 1 rule. It does eventually get there by higher order effects, which allows it to get to the 2p state.