In atoms containing more than one electron, most electronic states are strongly influenced by electron–electron repulsion. However, a special class of highly excited states—known as Rydberg states—can often be treated using ideas very similar to those developed for hydrogen.
Definition of a Rydberg state
A Rydberg state is an electronic state in which one electron is promoted to a very high principal quantum number \(n\), while the remaining electrons stay close to the nucleus in low-energy orbitals.
In such states, the excited electron spends most of its time far from the nucleus, where it experiences an almost Coulombic potential. As a result, the energy levels of Rydberg states form series that closely resemble those of hydrogen.
Effective nuclear charge
In a polyelectronic atom, inner electrons partially shield the nuclear charge from an outer electron. To account for this, Rydberg states are described using an effective nuclear charge
\[ Z^* = Z - \sigma, \]
where:
- \(Z\) is the atomic number,
- \(\sigma\) is the shielding constant, which depends on the electronic configuration and the orbital occupied by the Rydberg electron.
With this substitution, the energy of a Rydberg level can be approximated by a hydrogen-like expression:
\[ E_n \approx -\frac{hc\,R_M (Z^*)^2}{n^2}. \]
This approximation improves as \(n\) becomes large, because the Rydberg electron spends less time penetrating the inner electron cloud.
Estimating the shielding constant: Slater’s rules
The shielding constant \(\sigma\) can be estimated using Slater’s rules. These rules assign a shielding contribution to each electron based on its orbital relative to the electron of interest.
Electron group Shielding contribution to
- Electrons in higher shells 0.00 (no shielding)
- Other electrons in the same shell 0.35 each (0.30 for 1s)
- Electrons in the shell 0.85 each
- Electrons in shells or lower 1.00 each
To apply Slater’s rules:
- Write the electron configuration of the atom.
- Identify the electron of interest (here, the Rydberg electron).
- Sum the shielding contributions from all other electrons.
- Compute \(Z^* = Z - \sigma\).
Connection to ionization energies
Because Rydberg states lie very close to the ionization limit, the effective nuclear charge obtained from Slater’s rules can be used to estimate ionization energies of polyelectronic atoms.
In the next example, we will apply this approach to estimate the ionization energy of potassium by treating its outer electron as occupying a Rydberg state.
Big idea: Rydberg states of many-electron atoms behave like hydrogenic states when an effective nuclear charge is used, allowing atomic ionization energies and high-lying spectra to be understood with simple physical models.
Worked example: estimating the ionization potential of potassium using an effective nuclear charge
Potassium has one valence electron in the \(4s\) orbital. If we treat the ionization process as taking that electron from \(n=4\) to \(n=\infty\), then the ionization potential is approximately the magnitude of the hydrogen-like energy of the \(4s\) electron, but using an effective nuclear charge \(Z^*\).
Step 1: Electron configuration and identify the electron of interest
Potassium has \(Z=19\) and electron configuration \(\mathrm{K}: [\mathrm{Ar}]\,4s^1\). We want the effective nuclear charge felt by the \(4s\) electron.
Step 2: Use Slater’s rules to estimate the shielding constant \(\sigma\)
For a \(4s\) electron (\(n=4\)):
- Same shell (\(n=4\)): other electrons contribute \(0.35\) each. Here there are none (only one \(4s\) electron), so contribution = 0.
- \(n-1\) shell (\(n=3\)): electrons contribute \(0.85\) each. In \([\mathrm{Ar}]\), the \(3s^2 3p^6\) electrons give 8 electrons: \(8\times 0.85 = 6.80\).
- \(n-2\) or lower (shells \(n\le 2\)): electrons contribute \(1.00\) each. In \(1s^2 2s^2 2p^6\) there are 10 electrons: \(10\times 1.00 = 10.00\).
Therefore, \(\sigma = 0 + 6.80 + 10.00 = 16.80\), and
\[ Z^* = Z-\sigma = 19-16.80 = 2.20. \]
Step 3: Estimate the ionization potential from the hydrogen-like energy
Using the hydrogen-like energy model with an effective nuclear charge,
\[ E_n \approx -\frac{R\, (Z^*)^2}{n^2}, \]
where \(R\) is the Rydberg energy constant (for heavy nuclei, reduced-mass correction is negligible). Using \(R \approx 2.18\times 10^{-18}\ \text{J}\), the ionization potential (energy to reach \(n=\infty\)) is
\[ \mathrm{IP} \approx |E_{n=4}| = \frac{(2.18\times 10^{-18}\ \text{J})\,(2.20)^2}{4^2} = (2.18\times 10^{-18})\frac{4.84}{16}. \]
\[ \mathrm{IP} \approx 6.59\times 10^{-19}\ \text{J}. \]
Converting to electron-volts:
\[ \mathrm{IP}\;(\text{eV}) = \frac{6.59\times 10^{-19}\ \text{J}}{1.602\times 10^{-19}\ \text{J/eV}} \approx 4.12\ \text{eV}. \]
\[ \boxed{\mathrm{IP}(\mathrm{K}) \approx 4.12\ \text{eV}} \]
Interpretation
This estimate is close to the experimental ionization energy of potassium (about \(4.34\ \text{eV}\)), which is impressive given how simple the model is.
Why it works: the outer \(4s\) electron spends much of its time far from the nucleus, so it “feels” a Coulomb-like attraction from an effective charge \(Z^*\) rather than the full nuclear charge \(Z\).