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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Related Book For
Physics for Scientists and Engineers A Strategic Approach with Modern Physics
ISBN: 978-0133942651
4th edition
Authors: Randall D. Knight
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