Understanding the Energy Differences in Multielectron Atom States

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Discussion Overview

The discussion centers on the energy differences observed in multielectron atom states compared to hydrogen atom states, particularly focusing on the lowest-l and highest-l states for each principal quantum number n. The scope includes theoretical explanations and conceptual clarifications regarding electron shielding and energy levels in multielectron systems.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant notes that the lowest-l state for each n in multielectron atoms is significantly lower in energy than the corresponding hydrogen state, while the highest-l state is nearly equal in energy to the hydrogen state with the same n.
  • Another participant requests clarification and numerical examples regarding the energy differences mentioned.
  • A participant explains that in multielectron atoms, inner electrons shield the nuclear charge from outer electrons, which affects the energy levels, particularly in atoms with filled shells plus one additional electron.
  • There is a question about whether an electron in a hydrogen atom can occupy the s subshell for n greater than 1, referencing the Zeeman effect and the typical transitions between states.
  • One participant discusses the behavior of electrons being captured into high n states, often with high l, and how they transition through lower l states, explaining the brightness of certain transitions.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of electrons in hydrogen and multielectron atoms, particularly regarding energy levels and transition rules. The discussion does not reach a consensus on the specifics of these behaviors.

Contextual Notes

Some claims depend on assumptions about electron shielding and the definitions of energy states, which are not fully resolved in the discussion. The relationship between quantum numbers and energy levels is also a point of contention.

albertsmith
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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?
 
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Explain why you think this is the case.
 
Never heard about that. Can you specify a little bit more or give numbers?
 
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.
 
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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.
 
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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.
 

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