Hydrogen-like Ions and Nuclear Charge

Introduction

Hydrogen is not the only system the exact solution describes. Remove all but one electron from any atom and what remains — He⁺, Li²⁺, Be³⁺, even U⁹¹⁺ — is a hydrogen-like ion : a single electron bound to a nucleus of charge +Ze. The mathematics is identical to hydrogen except that the nuclear charge is Z times larger. That single change produces simple, powerful scaling laws that let us predict energies, sizes and spectra of these ions without solving anything new.

Core explanation

The Hamiltonian. For one electron and a nucleus of charge Ze,

Ĥ = −(ħ²/2μ)∇² − Ze²/(4πε₀r)

Only the potential differs from hydrogen, by the factor Z. The angular parts of the solutions are unchanged, so the quantum numbers n, l and mₗ, the orbital shapes and the node counts are all exactly as in hydrogen.

Energy scales as Z². Solving the radial equation gives

Eₙ = −Z²μe⁴/(32π²ε₀²ħ²n²) ≈ −13.6 eV × Z²/n²

The factor Z² arises because a stronger nucleus both pulls the electron closer (by 1/Z) and attracts it more strongly at that closer distance (by Z). Kinetic energy rises as Z² too, because confining the electron to a smaller region raises its momentum.

Species Z Ground-state energy Ionisation energy --- --- --- --- H 1 −13.6 eV 13.6 eV He⁺ 2 −54.4 eV 54.4 eV Li²⁺ 3 −122.4 eV 122.4 eV Be³⁺ 4 −217.6 eV 217.6 eV

These predictions match the measured ionisation energies of the last electron closely; for example the second ionisation energy of helium is 54.4 eV.

Lengths scale as 1/Z. Every radial function depends on r only through ρ = Zr/a₀, so every characteristic distance shrinks by 1/Z: the 1s most probable radius is a₀/Z, ⟨r⟩₁ₛ = 1.5a₀/Z, and radial nodes move inwards proportionally. Li²⁺ is a third the size of hydrogen.

Spectra scale as Z². Transition wavenumbers follow

ṽ = R H Z² (1/n₁² − 1/n₂²)

so every line of He⁺ has four times the wavenumber of the corresponding hydrogen line, and a quarter of its wavelength. Hydrogen's Lyman-α line at 121.6 nm becomes 30.4 nm for He⁺, far in the ultraviolet.

An interesting coincidence. Because He⁺ energies are −54.4/n² eV, its even-n levels match hydrogen's: n = 2, 4, 6 of He⁺ lie at the same energies as n = 1, 2, 3 of hydrogen. As a result, some He⁺ lines, such as the n = 4 → 2 transition, lie almost on top of hydrogen lines, separated only by the small reduced-mass correction.

Formulae

Eₙ = −13.6 Z²/n² eV; ionisation energy from level n = 13.6 Z²/n² eV; r mp(1s) = a₀/Z; ⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)]; ṽ = R Z²(1/n₁² − 1/n₂²), with R ≈ 1.097 × 10⁷ m⁻¹.

Step-by-step reasoning

To predict a property of a hydrogen-like ion:

1. Identify Z (the atomic number, not the ion charge). 2. Find the corresponding hydrogen value. 3. Multiply energies and wavenumbers by Z². 4. Divide lengths by Z; divide wavelengths by Z². 5. For high precision, correct for the reduced mass of the heavier nucleus.

Visual explanation

Draw the energy-level ladders of H, He⁺ and Li²⁺ side by side on the same scale. The He⁺ ladder is stretched downwards by a factor of four and the Li²⁺ ladder by nine. The pattern of rungs — crowding together towards zero at high n — is identical; only the vertical scale changes.

Real-world analogy

Changing Z is like changing the strength of a spring holding a mass to a wall while keeping the same spring shape. A stiffer spring holds the mass closer and makes it harder to pull away, but the pattern of how it can vibrate is unchanged.

Real-world example

Astronomers use He⁺ lines to measure the temperature of hot stars and nebulae. The He II line at 468.6 nm (n = 4 → 3) appears only where there are photons energetic enough to remove helium's first electron, so it marks the hottest regions, such as the surroundings of Wolf–Rayet stars and the centres of planetary nebulae.

Why?

Why do lengths scale as exactly 1/Z and energies as exactly Z²? Substituting r = r′/Z into the Hamiltonian turns the kinetic term into Z² times its hydrogen form and the potential term into Z² times its hydrogen form. The whole equation is hydrogen's multiplied by Z², so the solutions are hydrogen's with lengths compressed by Z.

Common misconception

"For He⁺, Z = 1 because the ion has a charge of +1." Z is the nuclear charge number, 2 for helium, regardless of how many electrons remain. The ionic charge +1 reflects the difference between two protons and one electron.

Worked example

Question: Calculate the wavelength of the photon emitted when Li²⁺ drops from n = 2 to n = 1.

Reasoning: ΔE = 13.6 × 3² × (1 − 1/4) eV = 13.6 × 9 × 0.75 = 91.8 eV. λ = hc/ΔE = 1240 eV nm/91.8 eV.

Answer: λ ≈ 13.5 nm, in the extreme ultraviolet, nine times shorter than hydrogen's 121.6 nm.

Quick check

1. What is the energy of the n = 3 level of He⁺ in electronvolts, using the Z² scaling law? Answer: E₃ = −13.6 × 2²/3² = −13.6 × 4/9 ≈ −6.04 eV.

Exam focus

Remember the two scaling laws: energies as Z²/n², lengths as n²/Z (for the most probable radius of nodeless orbitals). Typical exam questions ask for an ionisation energy of He⁺ or Li²⁺ or the wavelength of a transition; show the Z² factor explicitly and use hc = 1240 eV nm for quick conversions.

Advanced insight

For heavy hydrogen-like ions the 1s electron moves at a speed of roughly Zαc, where α ≈ 1/137 is the fine-structure constant. For U⁹¹⁺, Zα ≈ 0.67, so relativistic effects are large and the non-relativistic formula fails badly; the Dirac equation is needed. Such highly charged ions, produced in storage rings, provide stringent tests of quantum electrodynamics in strong fields.

Summary

A hydrogen-like ion has one electron and a nucleus of charge Ze. Its orbitals have the same shapes and quantum numbers as hydrogen's, but energies scale as Z² (−54.4 eV for He⁺, −122.4 eV for Li²⁺) and distances as 1/Z. Spectral wavenumbers increase by Z², pushing the lines of He⁺ and Li²⁺ into the far ultraviolet. Z is the atomic number, not the ionic charge.

Practice questions

1. Calculate the ionisation energy of Be³⁺ in its ground state. Answer: 13.6 × 4² = 217.6 eV. 2. What is the most probable radius of the 1s electron in Be³⁺? Answer: a₀/4 ≈ 13.2 pm. 3. How does the wavelength of the He⁺ n = 3 → 2 line compare with hydrogen's 656 nm Balmer-α line? Answer: It is four times shorter, about 164 nm, because the wavenumber scales as Z² = 4. 4. Which level of He⁺ has the same energy as the hydrogen n = 2 level? Answer: The n = 4 level, since −54.4/16 = −3.4 eV, equal to hydrogen's E₂.