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Physics V course sequence · Interactive lesson

Screening and 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.

UPV-12.06 · Lesson 6 of 9 in this unit. Position is not a completion score.
Starting model: Screening and shell structureEffective radial potential in hartree against r in bohr radii. Solid: V_eff = l(l+1)/2r² − Z_eff(r)/r with Z_eff = 1 + (Z−1)e^(−r/d), falling from Z = 11 at the nucleus to 1 far out. Dashed: the same l with Z_eff = 1 everywhere. The vertical marks the core radius 0.50 a₀, where the barrier is 0.0 Ha and the extra core pull 7.4 Ha.Veff(r) / hartree · l = 0core charge Z = 11dashed: unscreened, Zeff = 1core radius 0.50 a₀+10−1−2246r / a₀
Start with a picture. Change the model in the lab below.
1

Build the model

Break the exact 1/r law and n stops being the only label.

The key idea

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.

The complete physical model and assumptions

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.

1The 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.

2Penetration, 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(rc) 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.

3Reading 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.

4From 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.

5Where 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.

6Valence 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Quantum-defect term formula

Eₙₗ = − Zc² R∞hc / (n − δₗ)²

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

Effective central potential

Veff(r) = − e²Zeff(r)/4πε₀r + ħ²l(l+1)/2mₑᵣ²

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

Penetration of the core

uₙₗ ∝ r(l+1) as r → 0∫₀(rc)|u|² dr ∝ rc(2l+3)

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

Subshell capacity and period length

2(2l+1)Σ(l=0)(m−1) 2(2l+1) = 2m²

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

Madelung (n + l) filling order

fill by increasing n + l, ties by increasing n

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

Effective charge from a measured defect

Zeff = n/(n − δₗ), Eₙₗ = − 13.606 Zeff²/n² eV

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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

11 e
3 to 20 e; steps of 1.
0.50 a₀
0.2 to 1.5 a₀; steps of 0.05.
0
0 to 3; steps of 1.

Try this. 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.

Live results

Zeff 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

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelEffective radial potential in hartree against r in bohr radii. Solid: V_eff = l(l+1)/2r² − Z_eff(r)/r with Z_eff = 1 + (Z−1)e^(−r/d), falling from Z = 11 at the nucleus to 1 far out. Dashed: the same l with Z_eff = 1 everywhere. The vertical marks the core radius 0.50 a₀, where the barrier is 0.0 Ha and the extra core pull 7.4 Ha.Veff(r) / hartree · l = 0core charge Z = 11dashed: unscreened, Zeff = 1core radius 0.50 a₀+10−1−2246r / a₀
Effective radial potential in hartree against r in bohr radii. Solid: Veff = l(l+1)/2r² − Zeff(r)/r with Zeff = 1 + (Z−1)e(−r/d), falling from Z = 11 at the nucleus to 1 far out. Dashed: the same l with Zeff = 1 everywhere. The vertical marks the core radius 0.50 a₀, where the barrier is 0.0 Ha and the extra core pull 7.4 Ha.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Core charge at the nucleus Z: 11 e; Core screening radius d: 0.50 a₀; Orbital quantum number l: 0

Baseline 11; choose 3 to 20, in steps of 1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

Because 4s fills before 3d, students read the Madelung sequence as a fixed energy ladder and conclude that 4s lies below 3d inside every atom. The prediction fails at the first test. On such a ladder iron's outermost electrons would be the 3d ones, so Fe⁺ ought to be [Ar]3d⁵4s²; it is not, Fe⁺ is [Ar]3d⁶4s¹ and Fe²⁺ is [Ar]3d⁶, and the 4s electrons are the ones that leave. From scandium onwards the 3d orbital energy already lies below 4s in the neutral atom, and 4s is occupied anyway because what an atom minimises is the total energy of the whole determinant, not the sum of the one-electron energies. Those are different objects: each εᵢ already contains the repulsion with every other electron, so Σᵢ εᵢ counts every pair twice. The rule orders configurations along the sequence of neutral atoms — one more proton and one more electron at every step — and says nothing about taking an electron away from a single fixed atom.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Sodium'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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

Sodium'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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Use 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.