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In a multielectron atom

  • #1
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. Can someone please explain this?
 

Answers and Replies

  • #2
674
2
Think of it this way, the hydrogen atom has Z=1. So in a multielectron system you would assume the electron furthest out would only see Z=1, because all the inner electrons are shielding the rest of the Z-1 nuclear charge. This assumption states that the outer electron's energy would behave like a hydrogen electron at each 'n' and not caring about the 'l'.

But, if you look at the wavefunctions for the different l's you notice that the 2s, 3s, 4s and so on spend a lot of time near the core. So they feel a stronger attraction than just Z=1 when they are that close to the nucleus. Whereas the larger l's spend more time on the edge of the atom where Z=1 is dominant. That is why the lower angular momentums tend to be more tightly bound and the higher angular momentums tend to behave like hydrogen atoms.
 

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