Hydrogenic Ions

Hydrogenic ions are one-electron Coulomb systems with nuclear charge $+Ze$. They include hydrogen, $\mathrm{He}^+$, $\mathrm{Li}^{2+}$, $\mathrm{Be}^{3+}$, and any idealized one-electron ion with a point nucleus. They are not neutral many-electron atoms; the word hydrogenic means “hydrogen-like one-electron Coulomb problem.”

The main result is simple: after the reduced mass and nuclear charge are inserted, the dimensionless Schrödinger problem is the same as hydrogen. Lengths scale like $1/Z$, energies scale like $Z^2$, and finite nuclear mass replaces $m_e$ by the reduced mass.

Hamiltonian

For an electron of mass $m_e$ bound to a nucleus of charge $+Ze$ and mass $M_N$, the relative-coordinate Hamiltonian is

$$\hat H = -\frac{\hbar^2}{2\mu_Z}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r},$$

where

$$\mu_Z = \frac{m_eM_N}{m_e+M_N}$$

is the electron-nucleus reduced mass. The subscript $Z$ is a reminder that different nuclei have different masses, even when the electron is the same.

This is the same attractive Coulomb potential problem as hydrogen, with $Z$ and $\mu_Z$ left explicit. Spin, relativistic corrections, finite nuclear size, and quantum-electrodynamic effects are not part of this nonrelativistic model.

Length Scale

Define the charge-$Z$, reduced-mass Bohr scale

$$a_Z = \frac{4\pi\epsilon_0\hbar^2}{\mu_ZZe^2}.$$

If $a_0$ denotes the infinite-nuclear-mass hydrogen Bohr radius,

$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_ee^2},$$

then

$$a_Z = \frac{m_e}{\mu_Z}\frac{a_0}{Z}.$$

For heavy nuclei, $\mu_Z$ is close to $m_e$, so the dominant scaling is

$$a_Z\approx\frac{a_0}{Z}.$$

Thus $\mathrm{He}^+$ orbitals are roughly half the size of hydrogen orbitals with the same quantum numbers, and $\mathrm{Li}^{2+}$ orbitals are roughly one third the size. The reduced-mass factor gives a smaller isotope-dependent correction.

Energy Scale

The corresponding energy scale is

$$E_Z = \frac{\hbar^2}{2\mu_Za_Z^2} = \frac{\mu_ZZ^2e^4}{2(4\pi\epsilon_0)^2\hbar^2}.$$

Using the infinite-nuclear-mass Rydberg energy $\mathrm{Ry}_\infty$, this can be written as

$$E_Z = \frac{\mu_Z}{m_e}Z^2\mathrm{Ry}_\infty.$$

The ideal nonrelativistic bound-state energies are

$$E_n^{(Z)} = - \frac{E_Z}{n^2} = - \frac{\mu_Z}{m_e} \frac{Z^2\mathrm{Ry}_\infty}{n^2}, \qquad n=1,2,\ldots .$$

The leading scaling is

$$\lvert E_n^{(Z)}\rvert\propto \mu_ZZ^2.$$

This is why one-electron ions become much more tightly bound as $Z$ increases.

Wavefunction Scaling

The angular functions are unchanged:

$$Y_\ell^m(\theta,\phi).$$

The radial functions have the same dimensionless shape as hydrogen, but their radial argument is measured in units of $a_Z$. If $\rho=r/a_Z$, then a normalized radial wavefunction has the schematic scaling

$$R_{n\ell}^{(Z)}(r) = a_Z^{-3/2} \mathcal R_{n\ell}\!\left(\frac{r}{a_Z}\right),$$

where $\mathcal R_{n\ell}$ is the same dimensionless function for every point-Coulomb one-electron ion.

For the ground state,

$$R_{10}^{(Z)}(r) = 2a_Z^{-3/2}e^{-r/a_Z}.$$

The radial probability density

$$P_{10}^{(Z)}(r) = \lvert R_{10}^{(Z)}(r)\rvert^2r^2$$

is maximal at

$$r=a_Z.$$

Thus the most probable ground-state radius scales approximately as $a_0/Z$ for heavy nuclei.

Examples

Ignoring small reduced-mass corrections, the ground-state binding energies and length scales are:

Ion	Nuclear charge	Ground-state energy	Ground-state radius scale
$\mathrm{H}$	$Z=1$	$-13.6\,\mathrm{eV}$	$a_0$
$\mathrm{He}^+$	$Z=2$	$-54.4\,\mathrm{eV}$	$a_0/2$
$\mathrm{Li}^{2+}$	$Z=3$	$-122.4\,\mathrm{eV}$	$a_0/3$

These are nonrelativistic point-nucleus estimates. Real spectroscopic values include reduced-mass corrections and, at higher precision, relativistic, radiative, and nuclear-size effects.

Transition Scaling

For a transition from $n_i$ to $n_f$, the photon energy in the ideal model is

$$\Delta E = E_Z \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right), \qquad n_i\gt n_f.$$

At fixed quantum numbers, transition frequencies scale as

$$\nu\propto \mu_ZZ^2,$$

and wavelengths scale approximately as

$$\lambda\propto\frac{1}{\mu_ZZ^2}.$$

For example, the $\mathrm{He}^+$ transition with the same $n_i\to n_f$ labels as a hydrogen Lyman transition has roughly four times the photon energy and one quarter the wavelength, before reduced-mass corrections.

Degeneracy and Corrections

The ideal spinless point-Coulomb spectrum still depends only on $n$, not on $\ell$ or $m$:

$$E_n^{(Z)} = - \frac{E_Z}{n^2}.$$

For each fixed $n$, the spatial degeneracy remains

$$n^2.$$

This is the same special Coulomb degeneracy described in Degeneracy of the Hydrogen Atom. It is not removed by changing $Z$ or by replacing $m_e$ with $\mu_Z$.

Corrections grow in importance as $Z$ increases. Relativistic effects scale roughly with powers of $Z\alpha$, where $\alpha$ is the fine-structure constant. For high-$Z$ hydrogenic ions, the nonrelativistic Schrödinger model is no longer enough for precision work; one must use relativistic and finite-nuclear-size treatments. The nonrelativistic formulas here remain the clean scaling baseline.

Common Mistakes

• Calling neutral helium a hydrogenic ion. $\mathrm{He}^+$ is hydrogenic; neutral helium has two electrons.
• Forgetting that $\mathrm{Li}^{2+}$ is one-electron, while $\mathrm{Li}^+$ is two-electron.
• Scaling energies as $Z$ instead of $Z^2$.
• Scaling lengths as $1/Z^2$ instead of $1/Z$.
• Ignoring reduced mass when comparing isotopes or precision spectra.
• Assuming the nonrelativistic point-nucleus model remains quantitatively accurate for very large $Z$.
• Treating the $n^2$ degeneracy as a generic central-potential result rather than a special Coulomb result.

Where This Is Used

• Coulomb Potential defines the $Z$-dependent length and energy scales used here.
• Hydrogen Atom gives the $Z=1$ solution whose dimensionless form is being rescaled.
• Radial Wavefunctions gives the dimensionless radial shapes and node structure that scale with $a_Z$.
• Atomic Orbitals explains why the same $s,p,d,\ldots$ orbital notation applies after the hydrogenic length rescaling.
• Continuum States of the Coulomb Problem explains how positive-energy Coulomb states scale with $Z$ and reduced mass.
• Degeneracy of the Hydrogen Atom explains why the ideal one-electron Coulomb spectrum remains independent of $\ell$.
• Units and Constants states the charge convention $e\gt 0$ used in the nuclear potential.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Springer, 1957.

Exercises

1. Ignoring reduced-mass corrections, estimate the ground-state energy and most probable ground-state radius of $\mathrm{He}^+$.

Solution

For $\mathrm{He}^+$, $Z=2$. The ground-state energy scales as

$$E_1^{(Z)} = -Z^2(13.6\,\mathrm{eV}).$$

Therefore

$$E_1^{(2)} \approx -4(13.6\,\mathrm{eV}) = -54.4\,\mathrm{eV}.$$

The most probable ground-state radius scales as

$$a_Z\approx\frac{a_0}{Z},$$

so for $\mathrm{He}^+$ it is approximately $a_0/2$.

2. Compare a hydrogen and $\mathrm{Li}^{2+}$ transition with the same $n_i\to n_f$ labels, ignoring reduced-mass corrections.

Solution

For $\mathrm{Li}^{2+}$, $Z=3$. Transition energies scale as $Z^2$, so the $\mathrm{Li}^{2+}$ photon energy is approximately

$$3^2=9$$

times the corresponding hydrogen photon energy. Since wavelength is inversely proportional to photon energy,

$$\lambda_{\mathrm{Li}^{2+}} \approx \frac{1}{9}\lambda_{\mathrm{H}}.$$

3. Show that reduced mass changes both the length and energy scales.

Solution

The length scale is

$$a_Z = \frac{4\pi\epsilon_0\hbar^2}{\mu_ZZe^2},$$

so increasing $\mu_Z$ decreases $a_Z$. The energy scale is

$$E_Z = \frac{\mu_ZZ^2e^4}{2(4\pi\epsilon_0)^2\hbar^2},$$

so increasing $\mu_Z$ increases the binding-energy scale. Thus heavier reduced mass makes a hydrogenic ion smaller and more tightly bound, all else equal.