Effective Nuclear Charge from the Variation Method

Introduction

Keeping helium's electrons in hydrogen-like orbitals with the full nuclear charge Z = 2 gives an energy of −2.75 Eₕ, about 0.15 Eₕ above the experimental value. The orbitals are too compact: each electron actually spends part of its time shielded from the nucleus by the other. The variation method lets the atom decide for itself how much shielding there is. We replace Z in the orbital by an adjustable charge ζ and find the value that gives the lowest energy.

Core explanation

The trial function. Take

ψ(1,2) = (ζ³/π) e^(−ζr₁) e^(−ζr₂)

This is a product of two normalised 1s-type orbitals with exponent ζ. The Hamiltonian is still the true one, with nuclear charge Z = 2:

Ĥ = −½∇₁² − ½∇₂² − Z/r₁ − Z/r₂ + 1/r₁₂

The three energy contributions. Using standard hydrogen-like expectation values, scaled for exponent ζ:

- kinetic energy of each electron: ζ²/2, so both together give ζ² - attraction to the actual nucleus: each electron has ⟨1/r⟩ = ζ, so the total is −2Zζ - electron repulsion: ⟨1/r₁₂⟩ = 5ζ/8

Therefore

E(ζ) = ζ² − 2Zζ + 5ζ/8

Minimising. dE/dζ = 2ζ − 2Z + 5/8 = 0, which gives

ζ = Z − 5/16

For helium, ζ = 2 − 0.3125 = 1.6875. Substituting back, E min = −ζ² = −(27/16)² ≈ −2.848 Eₕ ≈ −77.5 eV.

How good is it? The experimental energy is −2.904 Eₕ (−79.0 eV). The error has fallen from 0.154 Eₕ with ζ = 2 to 0.056 Eₕ, about 1.9%. One adjustable number has removed almost two-thirds of the error.

The meaning of screening. The optimum ζ is smaller than Z by σ = 5/16 ≈ 0.31. Each electron behaves as though it orbits a nucleus of charge about 1.69 rather than 2. The other electron's charge cloud partly surrounds the nucleus and cancels some of its attraction. The value 0.31 is close to the 0.30 used in Slater's empirical rules for another 1s electron — the variation method gives a theoretical basis for those rules.

Orbital expansion. Because the orbital radius scales as 1/ζ, the optimised helium orbital is larger than the ζ = 2 orbital by a factor 2/1.6875 ≈ 1.19. Screening makes the atom bigger, which reduces the electron repulsion at a modest cost in nuclear attraction.

The virial theorem again. At the optimum, ⟨T⟩ = ζ² and ⟨V⟩ = −2ζ², so ⟨V⟩ = −2⟨T⟩, exactly as required for Coulomb systems.

Formulae

E(ζ) = ζ² − 2Zζ + (5/8)ζ (atomic units, two 1s electrons)

ζ opt = Z − 5/16; E min = −(Z − 5/16)²

σ = Z − ζ = 5/16 ≈ 0.31

Step-by-step reasoning

1. Replace Z in the orbital exponent by a parameter ζ. 2. Evaluate kinetic, attraction and repulsion terms as functions of ζ, keeping the real Z in the attraction. 3. Add them to form E(ζ). 4. Differentiate and set the derivative equal to zero. 5. Solve for ζ and substitute to get E min. 6. Interpret Z − ζ as the screening constant.

Visual explanation

Picture the radial distribution function of one helium electron for three cases: ζ = 2 (too tight), ζ = 1.69 (optimised) and the exact result. The optimised curve peaks at r = 1/ζ ≈ 0.59 a₀ instead of 0.50 a₀, shifting outwards to sit very close to the true distribution.

Real-world analogy

Standing behind a friend near a campfire: you still feel warmth, but your friend blocks part of it. You feel an "effective" fire that is weaker than the real one. Each helium electron is partly blocked from the nucleus by the other electron in the same way.

Real-world example

Effective nuclear charges explain periodic trends in atomic radii and ionisation energies across the periodic table. Computational chemists also optimise orbital exponents for molecules: in H₂ the best single 1s exponent is about 1.17 to 1.2, larger than 1, because the orbitals contract towards the bond.

Why?

Why is ζ smaller than Z but not much smaller? Reducing ζ lowers the repulsion (which scales as ζ) but also weakens the nuclear attraction (which scales as Zζ with the larger coefficient 2Z). The attraction dominates, so only a small reduction pays off.

Common misconception

"ζ is the charge of the nucleus seen by an outside observer." It is a parameter describing the shape of each electron's orbital. The actual nuclear charge is still +2, and it is the real Z that appears in the attraction term of the Hamiltonian.

Worked example

Question: Use the optimised energy to estimate the first ionisation energy of helium.

Reasoning: E(He) = −2.848 Eₕ. E(He⁺) = −Z²/2 = −2.000 Eₕ exactly. Ionisation energy = −2.000 − (−2.848) = 0.848 Eₕ. Converting, 0.848 × 27.21 ≈ 23.1 eV.

Answer: About 23.1 eV, compared with the experimental 24.6 eV — much better than the 20.4 eV from the unscreened estimate.

Quick check

1. What is the optimum effective nuclear charge for helium and what screening constant does it imply? Answer: ζ = 27/16 = 1.6875, so the screening constant is σ = 2 − 1.6875 = 0.3125.

Exam focus

A classic derivation question: build E(ζ) term by term, minimise, and state ζ, E and the percentage error. Mark schemes insist on keeping the true Z in the nuclear attraction while using ζ in the orbital. Always convert the final energy to eV and comment on the physical meaning of screening.

Advanced insight

The same formula applies to any two-electron ion: for Li⁺, ζ = 3 − 5/16 = 2.6875 and E = −7.223 Eₕ against the exact −7.280 Eₕ. As Z grows the fractional error shrinks, because repulsion becomes a smaller part of the total. Better results need different exponents for the two electrons or explicit r₁₂ terms.

Summary

Replacing Z by a variational parameter ζ in the helium 1s orbitals gives E(ζ) = ζ² − 2Zζ + 5ζ/8. Minimising gives ζ = Z − 5/16 = 1.6875 and E = −2.848 Eₕ, only 1.9% above experiment. The reduction of ζ below Z measures screening: each electron partly shields the nucleus from the other, so the orbitals expand.

Practice questions

1. Derive the optimum ζ from E(ζ) = ζ² − 2Zζ + 5ζ/8. Answer: dE/dζ = 2ζ − 2Z + 5/8 = 0 gives ζ = Z − 5/16. 2. Calculate the optimised energy of the two-electron ion Be²⁺ (Z = 4) in hartrees. Answer: ζ = 4 − 0.3125 = 3.6875, so E = −(3.6875)² ≈ −13.60 Eₕ. 3. By what factor is the optimised helium orbital larger than an orbital with ζ = 2? Answer: Orbital size scales as 1/ζ, so the factor is 2/1.6875 ≈ 1.19. 4. Why is the variational energy with ζ = 1.6875 guaranteed to be at least as low as the energy with ζ = 2? Answer: ζ = 2 is one member of the same family of trial functions, and minimising over ζ can only find an energy equal to or lower than any particular member.