University Physics V · Multi-Electron Atoms · 12.6
Screening & Shell Structure
Hydrogen's levels depend on n alone. Put ten more electrons in the way and that stops being true: the inner shells hide part of the nucleus, low-l states slip inside them, and each l acquires its own binding energy. This lesson turns that one fact into the shape of the periodic table.
Build the model
Connect the measurement to the mechanism.
A many-electron atom has no exact one-electron solution, so the valence electron is modelled as moving in a spherically averaged field: a central potential −e²Zeff(r)/4πε₀r whose screening charge runs from the bare Z at the nucleus down to the net core charge far out. Because that field is still central, [H, L²] = [H, Lz] = 0 and the eigenstates keep their spherical-harmonic labels, so the 2l+1 degeneracy in mₗ survives untouched. What does not survive is the Runge–Lenz vector, conserved only for an exact 1/r law, whose closure into so(4) is what forced hydrogen's energies to depend on n alone.
Deform the potential and the algebra drops back to so(3): E = Eₙₗ, and in sodium the 3s and 3d terms separate by 3.62 eV. Empirically the whole deformation is carried by one number per l, the quantum defect δₗ in E = −Rhc/(n − δₗ)², nearly independent of n because it is the phase a threshold wave picks up crossing a fixed core. Its ordering δₛ > δₚ > δd comes from the centrifugal barrier, which only low-l states can ignore.
The cost is honesty about what has been assumed: Zeff(r) is fitted or computed self-consistently rather than derived, one electron is treated at a time so correlation is invisible, and the filling order the picture predicts is an empirical fit that chromium, copper and palladium break.
- Simple definition
- Screening is the reduction of the nuclear charge felt by an outer electron by the electrons lying between it and the nucleus, leaving an effective charge Zeff(r) that falls from Z at the nucleus towards the net core charge far out.
- Example
- Sodium's valence electron is bound by 5.139 eV, not the hydrogenic 13.606/3² = 1.512 eV; matching E = −13.606/(3 − δ)² eV gives δₛ = 1.373, an energy-equivalent Zeff of 3/1.627 = 1.84 rather than 1.
One fitted number per l replaces the whole radial eigenvalue problem, and it returns sodium's series to four figures.
R∞hc = 13.606 eV; Zc is the core's net charge, 1 for a neutral atom; δₗ is dimensionless and nearly n-independent
Screening is a function of r, so no single constant Zeff can fit both ends of the well at once.
Zeff(r) → Z as r → 0 and → 1 as r → ∞ for a neutral core; r in a₀, energies in hartree, 1 Ha = 27.211 eV
Two extra powers of rc per unit of l, so with rc ≈ 0.5 a₀ each step up in l cuts the core weight by at least four — hence δₛ > δₚ > δd.
u = rR is the reduced radial function; rc is the core radius, near 0.5 a₀ in sodium
Period p is filled by the group k = p + 1, so the period lengths are group capacities: 2, 8, 8, 18, 18, 32, 32.
m = ⌈(n+l)/2⌉ counts the l values in one n + l group, so 2m² gives 2, 8, 18, 32, 50
It draws the whole table, then fails at Cr, Cu, Nb, Mo, Ru, Rh, Pd, Ag, Pt, Au and about ten more.
Empirical, and only for ground configurations of neutral atoms: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p
Sodium 3s gives 3/1.627 = 1.84, where Slater's rules predict 2.2 — an energy 42% too deep.
An energy-equivalent constant charge in units of e, not the r-dependent Zeff(r) of the real potential
The symmetry that survives and the one that dies
In any central field the Hamiltonian commutes with L² and Lz, so eigenstates carry the labels l and mₗ and every level keeps its 2l+1-fold m-degeneracy. That follows from rotational invariance alone, and screening cannot touch it. Hydrogen has more than that. For an exact 1/r potential the Runge–Lenz vector M = (p × L − L × p)/2mₑ − e²r̂/4πε₀ is also conserved, and on the E < 0 subspace the rescaled M and L close into so(4); its multiplets are labelled by n = nᵣ + l + 1 alone, which is why 3s, 3p and 3d coincide in hydrogen. Screening makes Zeff depend on r, the potential is no longer 1/r, [H, M] ≠ 0, and the extra symmetry is gone. What remains is so(3), so E = Eₙₗ. Nothing about the split is small: in sodium the 3s and 3d terms lie 5.139 eV and 1.522 eV below the ionisation limit, a gap of 3.62 eV — more than five times hydrogen's whole n = 3 to n = 4 spacing of 0.661 eV.
Penetration, and why the order of the defects is fixed
Near the origin the centrifugal term dominates the radial equation, and the regular solution behaves as uₙₗ(r) ∝ r(l+1); the second Frobenius root r(−l) is rejected by the boundary condition u(0) = 0. The weight a state carries inside a core of radius rc is therefore ∫₀(r_c) r(2l+2) dr = rc(2l+3)/(2l+3) — two extra powers of rc for every unit of l. With rc ≈ 0.5 a₀ in sodium that is a factor of four lost per unit of l, and it is precisely the amplitude in that region which samples an unscreened charge climbing towards +11 instead of +1. Read the same fact through the barrier: ħ²l(l+1)/2mₑᵣ² at r = 0.5 a₀ is 12 hartree, 327 eV, for l = 2 and 4 hartree for l = 1, against nothing at all for l = 0. Be careful with the argument, though. Hydrogen carries the identical barrier and shows no l-splitting whatever. What creates a defect is the barrier combined with a Zeff(r) that varies across the very region the barrier excludes.
Reading the defect off a spectrum
The quantum defect is measured, not postulated. Fit sodium's series to Eₙₗ = −13.606 eV/(n − δₗ)²: the sharp series (ns → 3p) returns δₛ ≈ 1.37, the principal series (np → 3s) gives δₚ = 0.883 at n = 3, 0.866 at n = 4 and 0.861 at n = 5, and the diffuse series (nd → 3p) gives δd = 0.010. Two things deserve attention. First, δₗ barely moves with n — it falls 2.5% from 3p to 5p — because it describes the core the electron crosses, not the orbit; quantum-defect theory sharpens this by identifying πδₗ with the threshold scattering phase shift of the core, and what residual drift is left is core polarisation, the valence electron's own field distorting the core it flies past, which a Rydberg–Ritz expansion δ(n) = δ₀ + δ₁/(n − δ₀)² absorbs. Second, δd ≈ 0.01 says a sodium 3d electron is hydrogenic to better than 1%, a claim about geometry rather than about nuclear size. For δₗ from a model rather than a fit, put Zeff(r) on a radial grid of a few thousand points, assemble the tridiagonal radial Hamiltonian in numpy, and hand it to scipy.linalg.eigh_tridiagonal; the eigenvalues invert straight back to δₗ.
From subshell capacities to period lengths
A subshell (n, l) holds 2(2l+1) electrons: 2l+1 values of mₗ times two spin projections, with antisymmetry forbidding any repeat. Group the subshells by n + l, breaking ties by increasing n, and the Madelung order appears: 1s | 2s | 2p 3s | 3p 4s | 3d 4p 5s | 4d 5p 6s | 4f 5d 6p 7s. The group with n + l = k contains l = 0 up to m − 1, where m = ⌈k/2⌉, because l ≤ n − 1 forces n ≥ l + 1, and its capacity is Σ 2(2l+1) = 2m². A period opens at an ns and closes at the np of the group above, and since an ns and an (n+1)s both carry 2, the length of period p is the capacity of group k = p + 1: 2, 8, 8, 18, 18, 32, 32 for the seven periods that exist. The repeats are the pairing k = 2j − 1 and k = 2j sharing m = j; only the leading 2 stands alone, because the group that would pair with it is the 1s that opens the table rather than closing a period. Notice what 2m² is not: it is a group capacity, not the 2n² of a hydrogenic shell. Both are the same sum Σ2(2l+1), but hydrogen's runs over l = 0…n−1 at fixed n while this one runs over l = 0…m−1 at fixed n + l — which is why period lengths repeat and hydrogenic shell capacities never do.
Where the filling order breaks
About twenty of the first 103 elements disobey the n + l rule, and every one of them sits in the d or f block: chromium is [Ar]3d⁵4s¹ rather than 3d⁴4s², copper is [Ar]3d¹⁰4s¹, niobium is [Kr]4d⁴5s¹, and palladium drops the 5s subshell altogether at [Kr]4d¹⁰. No s- or p-block element breaks it, and that pattern is the clue: the exceptions appear only where two configurations are already nearly degenerate in total energy — in nickel the lowest 3d⁹4s¹ level lies just 205 cm⁻¹, 0.025 eV, above the 3d⁸4s² ground level. Two effects the rule never mentions then decide the winner. One is exchange: each pair of parallel-spin d electrons gains an exchange integral K, and 3d⁵4s¹ has ten such pairs against six for 3d⁴4s². The other is the compactness of 3d, where ⟨1/r₁₂⟩ between two d electrons is far larger than between two diffuse 4s electrons, so double occupancy of 3d is expensive. Both are two-electron quantities, invisible to any ordering of one-electron levels — which is the honest description of the Madelung rule: an empirical regularity of measured ground states with no derivation from the Schrödinger equation behind it.
Valence configuration, group chemistry, and where Dirac enters
What the shell model buys is a reason for the columns. Lithium, sodium, potassium, rubidium and caesium share an ns¹ valence configuration outside a closed core, and their first ionisation energies — 5.39, 5.14, 4.34, 4.18 and 3.89 eV — span just 1.5 eV while Z runs from 3 to 55. The nuclear charge changes by a factor of eighteen and the chemistry does not, because the valence electron still sees a screened charge near +1 and a fixed l. Halogens repeat ns²np⁵, noble gases close at np⁶, and the table becomes a statement about Zeff and l rather than about Z. Two limits bite. Screening says nothing about correlation, and a single determinant misses real energy — 1.14 eV in helium alone, where Hartree–Fock gives −2.8617 Ha against −2.9037 Ha exact — though cancellation between atom and ion leaves Koopmans ionisation energies only a few tenths of an electron-volt out. And in heavy atoms the 1s speed is of order Zαc — 0.58c in gold — so relativistic contraction pulls 6s down; gold's 5d → 6s gap falls near 2.4 eV, absorbing blue at 517 nm and leaving the metal yellow, while mercury's contracted, closed 6s² leaves it liquid at 234 K. Neither follows from Schrödinger; both need Dirac.
Change one variable at a time
Make the relationship visible.
Keep l = 0 and slide the core radius out: inside r = d the solid curve dives far below the dashed hydrogenic one, and that extra depth is the whole origin of the quantum defect. Now set l = 2 with d back at 0.5 a₀ — the barrier throws both curves off the top of the frame before r = d, and past about 2.5 a₀ they lie on one line, because the screening term has died out where an l = 2 state actually lives.
Zeff AT r = 1 a₀2.35 e
CORE PULL AT r = d7.36 Ha
BARRIER AT r = d0.00 Ha
BARRIER / CORE PULL0.00
Live interpretationZeff AT r = 1 a₀: 2.35 e. CORE PULL AT r = d: 7.36 Ha. BARRIER AT r = d: 0.00 Ha. BARRIER / CORE PULL: 0.00
Catch the common trap
Explain before calculating.
In sodium the 3s term lies 5.139 eV below the ionisation limit while the 3d term lies 1.522 eV below it, against a hydrogenic 13.606/3² = 1.512 eV. Which statement accounts for both numbers?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium's first ionisation energy is 5.139 eV. Using Eₙₗ = −13.606 eV/(n − δₗ)² for the 3s valence term, find δₛ, the effective quantum number n − δₛ, and the energy-equivalent effective charge, then compare with the hydrogenic n = 3 value.
- The 3s term value is its binding energy below the ionisation limit, 5.139 eV, so 13.606/(3 − δₛ)² = 5.139.
- Rearranged: (3 − δₛ)² = 13.606/5.139 = 2.6476, so 3 − δₛ = 1.6271 and δₛ = 1.373.
- The effective quantum number is n* = 1.627. Sodium's valence electron is bound like a hydrogen state between n = 1 and n = 2, not like n = 3.
- Energy-equivalent charge: setting E = −13.606 Zeff²/n² gives Zeff = n/(n − δₛ) = 3/1.6271 = 1.844.
- Hydrogenic n = 3 would give 13.606/9 = 1.512 eV, so penetration binds the 3s electron 5.139/1.512 = 3.40 times more deeply.
Answerδₛ = 1.373, n* = 1.627 and Zeff = 1.84; the 3s term is 3.40 times deeper than the hydrogenic n = 3 value.
MediumSodium's principal series runs np → 3s, with its first two lines at 589.3 nm and 330.3 nm. The 3s term lies 5.139 eV below the ionisation limit. Find δₚ from each line, say what the comparison shows, and predict the wavelength of the third line. Use hc = 1239.84 eV nm and R = 13.606 eV.
- Photon energies: 1239.84/589.3 = 2.1039 eV and 1239.84/330.3 = 3.7537 eV.
- Both lines end on 3s, which is 5.139 eV below the limit, so the np binding energies are 5.139 − 2.1039 = 3.035 eV and 5.139 − 3.7537 = 1.385 eV.
- For n = 3: (3 − δₚ)² = 13.606/3.035 = 4.4830, so 3 − δₚ = 2.1173 and δₚ = 0.883.
- For n = 4: (4 − δₚ)² = 13.606/1.385 = 9.8238, so 4 − δₚ = 3.1343 and δₚ = 0.866. The defect moved 2% while n changed by one, so it is a property of l and of the core, not of n.
- Predict with δₚ = 0.87 at n = 5: binding = 13.606/(5 − 0.87)² = 13.606/17.057 = 0.798 eV, so the photon carries 5.139 − 0.798 = 4.341 eV and λ = 1239.84/4.341 = 285.6 nm.
Answerδₚ = 0.883 from 3p and 0.866 from 4p; the 5p → 3s line is predicted at 285.6 nm, against 285.3 nm measured.
HardUse the n + l filling rule to work out the length of the eighth period and the predicted ground configuration of element 119, then give the reason the prediction should not be trusted.
- Sort the subshells by n + l, ties broken by increasing n. Because l ≤ n − 1 forces n ≥ l + 1, the group with n + l = k runs over l = 0 to m − 1 with m = ⌈k/2⌉, and holds Σ 2(2l+1) = 2m² electrons.
- For k = 8 the group is 5f, 6d, 7p, 8s: m = 4 and the capacity is 32, which is the length of period 7.
- The next group, k = 9, is 5g, 6f, 7d, 8p, 9s: m = 5 and the capacity is 18 + 14 + 10 + 6 + 2 = 50. Period p is filled by group k = p + 1, so period 8 is 8s 5g 6f 7d 8p — fifty elements, Z = 119 to 168.
- Period 7 ends at oganesson, Z = 118, with 7p⁶ full, so element 119 opens period 8 on the next s subshell: [Og] 8s¹, an alkali metal in group 1.
- The caveat: the rule is a non-relativistic fit. At Z = 119 the 1s speed is of order Zαc ≈ 0.87c, spin–orbit splitting pushes 8p₁/₂ down until it competes with 8s, and Dirac–Fock calculations give a strongly contracted, much more tightly bound 8s than the Madelung picture implies, with the position of the 5g block unsettled.
AnswerPeriod 8 holds 50 elements (119 to 168) and element 119 is predicted [Og] 8s¹ — but the n + l rule is non-relativistic, and at this Z a Dirac treatment reorders 8s, 8p₁/₂ and 5g.