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.

Helpful background. Coulomb Potential to identify Coulomb length and energy scales.

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 an allowed radiative transition from $n_i$ to $n_f$, neglecting photon recoil, the photon energy in the ideal model is

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

The energy difference does not by itself guarantee an allowed transition: angular-momentum and parity selection rules determine the relevant matrix element. 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.

Exercises

1. Ignoring reduced-mass corrections, estimate the $n=1$ and $n=2$ energies and most probable ground-state radius of $\mathrm{He}^+$. Compare the orbital length scale with hydrogen.

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 $a_0/2$. At fixed $n,\ell$, the orbital length scale is half the hydrogen value. For the first excited shell,

$$E_2^{(2)}=\frac{E_1^{(2)}}4\simeq-13.6\,\mathrm{eV}.$$

Thus the $\mathrm{He}^+$ $n=2$ binding energy matches the hydrogen $n=1$ value within the stated mass approximation, although their quantum numbers and states differ.

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. Does finite nuclear mass increase or decrease the binding magnitude relative to an infinitely heavy nucleus?

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.

For a finite nuclear mass,

$$\frac{\mu_Z}{m_e}=\frac{M_N}{m_e+M_N}=\frac1{1+m_e/M_N}<1.$$

The finite-mass binding magnitude is therefore smaller than the infinite-nuclear-mass value by this factor, and the orbital length is larger by its reciprocal. This separates the direction of the reduced-mass correction from the additional nuclear and radiative effects required for precision isotope comparisons.

References

• H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Springer, 1957.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• 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.