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2s and 2p orbital energies of hydrogen and lithium - degeneracy, penetration and shielding

A hydrogen atom and a lithium atom in an atom chamber beside an energy ladder. Press start and the bench does the rest: for H 2s, H 2p, Li 2s and Li 2p it solves the radial Schroedinger equation by shooting, draws where the electron is found - orange dots when it is inside the lithium 1s core - puts each energy on the ladder, then tests options (A)-(D) one by one and reveals the true ones. No prediction stage - input, run, watch, result.

Sub-topics covered

• Orbital energy of a one-electron atom: E = -13.6 Z^2/n^2 eV, depends on n only
• Multi-electron atoms: energy depends on n and l
• Shielding by core electrons and effective nuclear charge
• Penetration of s versus p orbitals; radial probability distributions
• Slater's rules
• Radial Schroedinger equation and node counting

What it teaches

• In hydrogen there is one electron and nothing to shield it, so the orbital energy depends only on n: E2s(H) = E2p(H) = -13.6/4 = -3.40 eV (degenerate)
• In lithium the 2s or 2p electron moves in the field of the +3 nucleus and the 1s2 core; outside the core it feels about +1, inside up to +3
• The 2s orbital has a small inner lobe that penetrates the 1s core; the 2p orbital has an angular node at the nucleus and stays mostly outside - so 2s feels a higher effective nuclear charge
• Higher effective nuclear charge means lower energy: E2s(Li) < E2p(Li)
• Zeff(Li, 2s) is greater than 1 (about 1.26 from the measured energy, 1.30 by Slater's rules), so E2s(Li) is below E2s(H): E2s(H) > E2s(Li)
• Since E2p(H) = E2s(H) > E2s(Li), the statement E2p(H) < E2s(Li) is false
• Order: E2s(Li) < E2p(Li) < E2s(H) = E2p(H)

Simulation ID

• ADV-2026-P1-CHE-Q05 · revision 1

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