In a multielectron atom, the lowest-l state for each n (2s, 3s, 4s, etc.) is significantly lower

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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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