Many atomic spectra can be understood using the same framework developed for the hydrogen atom, provided the system contains only one electron. Such systems are known as hydrogen-like atoms or ions.
Hydrogen-like atomic systems
A hydrogen-like system consists of a single electron bound to a nucleus of charge \( +Ze \), where \(Z\) is the atomic number.
Common examples include:
- Hydrogen: \( \mathrm{H} \) (\(Z=1\))
- Singly ionized helium: \( \mathrm{He}^+ \) (\(Z=2\))
- Doubly ionized lithium: \( \mathrm{Li}^{2+} \) (\(Z=3\))
Because there is only one electron, electron–electron repulsion is absent, and the Schrödinger equation can be solved exactly using the same methods as for hydrogen.
Effect of nuclear charge
For a hydrogen-like system, the Coulomb potential is
\[ V(r) = -\frac{Ze^2}{4\pi\varepsilon_0 r}. \]
Increasing the nuclear charge \(Z\) strengthens the attraction between the electron and the nucleus, leading to more tightly bound electronic states and larger energy level spacings.
The Rydberg equation for hydrogen-like systems
The wavenumber of radiation associated with a transition between two energy levels \(n_i\) and \(n_f\) in a hydrogen-like system is given by the Rydberg equation:
\[ \tilde{\nu} = R_M\,Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right), \]
where:
- \(Z\) is the nuclear charge,
- \(R_M\) is the mass-corrected Rydberg constant appropriate for the specific nucleus,
- \(n_i > n_f\) for emission (and \(n_i < n_f\) for absorption).
The factor \(Z^2\) reflects the stronger Coulomb attraction in systems with larger nuclear charge.
Physical interpretation
Compared to hydrogen, spectral lines for hydrogen-like ions with \(Z > 1\) appear at much higher wavenumbers (shorter wavelengths). For example, transitions in \( \mathrm{He}^+ \) occur at four times the wavenumber of the corresponding hydrogen transitions.
Big idea: hydrogen-like systems share the same underlying quantum structure as hydrogen, but their spectra scale simply with nuclear charge and reduced mass, allowing a wide class of atomic spectra to be understood with a single equation.
Worked example: wavelength of the \(4 \rightarrow 2\) transition in \(\mathrm{^4He}^+\)
\(\mathrm{He}^+\) is a hydrogen-like ion (one electron) with nuclear charge \(Z=2\). Its spectral lines are described by the Rydberg equation with a mass-corrected Rydberg constant \(R_M\).
Step 1: Mass-corrected Rydberg constant for helium-4
The reduced-mass correction replaces the electron mass with the reduced mass of the electron–nucleus system. In practice, this is often written as
\[ R_M \;=\; \frac{R_\infty}{1+\dfrac{m_e}{M}} \]
For an \(\alpha\) particle (the \(^4\mathrm{He}\) nucleus), \(M \approx 6.644657\times 10^{-27}\,\text{kg}\), and \(m_e = 9.109384\times 10^{-31}\,\text{kg}\), so \(\dfrac{m_e}{M}\approx 1.3709\times 10^{-4}\). Using \(R_\infty = 109737.3157\ \text{cm}^{-1}\),
\[ R_M \;=\; \frac{109737.3157\ \text{cm}^{-1}}{1+1.3709\times 10^{-4}} \;\approx\; 109722.27\ \text{cm}^{-1}. \]
Step 2: Apply the hydrogen-like Rydberg equation
For a transition \(n_i \rightarrow n_f\),
\[ \tilde{\nu} = R_M Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right). \]
Here \(n_i=4\), \(n_f=2\), and \(Z=2\), so
\[ \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) = \left(\frac{1}{2^2}-\frac{1}{4^2}\right) = \left(\frac{1}{4}-\frac{1}{16}\right) = \frac{3}{16}. \]
\[ \tilde{\nu} = (109722.27\ \text{cm}^{-1})\,(2^2)\left(\frac{3}{16}\right) = (109722.27\ \text{cm}^{-1})\left(\frac{12}{16}\right) = 0.75\,R_M \approx 8.22917\times 10^{4}\ \text{cm}^{-1}. \]
Step 3: Convert wavenumber to wavelength
Using \(\lambda = 1/\tilde{\nu}\) and \(1\ \text{cm} = 10^{7}\ \text{nm}\),
\[ \lambda(\text{nm}) = \frac{10^{7}}{\tilde{\nu}(\text{cm}^{-1})} = \frac{10^{7}}{8.22917\times 10^{4}} \approx 1.21519\times 10^{2}\ \text{nm}. \]
\[ \boxed{\lambda \approx 121.52\ \text{nm}} \]
Interpretation: \(121.5\ \text{nm}\) is in the vacuum ultraviolet (VUV)