The Rydberg Formula
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You should be comfortable with the concept of atomic line spectra and the idea that atoms absorb or emit light at specific, discrete wavelengths. Basic algebraic manipulation of fractions and scientific notation is required.
Why this matters
The Rydberg formula provides the first precise mathematical link between the observed spectral lines of hydrogen and the underlying quantization of energy.
Before the development of quantum mechanics, the spectrum of hydrogen was a mystery. While chemists knew that hydrogen produced specific lines of light, they could not explain why those lines appeared at such specific intervals. The Rydberg formula, derived empirically from experimental data, revealed a simple mathematical pattern hidden within the chaos of spectral lines. This formula was the crucial bridge that allowed physicists to propose the Bohr model, marking the transition from classical physics to the quantum understanding of the atom that defines modern chemistry.
The Pattern in the Lines
Imagine looking at the spectrum of hydrogen. You see distinct lines of colour, but they are not evenly spaced. Some are close together, others far apart. In the 19th century, scientists noticed that if you calculated the reciprocal of the wavelength (often called the wavenumber) for each line, the numbers fit a surprisingly simple pattern. The difference between any two lines could be explained by subtracting two specific fractions from each other. These fractions always involved the squares of whole numbers. This observation suggested that the energy levels of the electron in a hydrogen atom are not continuous, like a ramp, but discrete, like the steps of a staircase. The Rydberg formula captures this "staircase" nature, showing that light is emitted or absorbed only when an electron jumps between these fixed steps.
The Mathematical Formulation
The Rydberg formula expresses the wavenumber \(\tilde{\nu}\) (which is \(1/\lambda\)) of the emitted or absorbed light. It relates the wavelength to two integers, \(n_1\) and \(n_2\), which represent the lower and higher energy levels involved in the transition.
Here, \(R_H\) is the Rydberg constant for hydrogen, approximately \(1.097\times10^7\,\mathrm{m^{-1}}\). The integers \(n_1\) and \(n_2\) must satisfy the condition \(n_2 > n_1 \ge 1\). The level energies are \(E_n = -hcR_H/n^2\), so the terms \(1/n_1^2\) and \(1/n_2^2\) measure how deeply the lower and upper levels are bound. The difference between these terms determines the energy of the photon involved in the transition. Because the energy of a photon is inversely proportional to its wavelength, a larger difference in the squared integers results in a shorter wavelength (higher energy) photon.
Generalization to Hydrogen-like Ions
The formula above is specific to hydrogen, which has a single proton in its nucleus. However, the same mathematical structure applies to any ion that contains only one electron, such as helium ion (\(\mathrm{He^+}\)) or lithium ion (\(\mathrm{Li^{2+}}\)). These are known as hydrogen-like ions. The key difference is the nuclear charge, denoted by \(Z\). In hydrogen, \(Z=1\). In helium, \(Z=2\). The electrostatic attraction between the nucleus and the electron is stronger in ions with higher \(Z\), which compresses the energy levels and increases the energy differences between them.
Notice the factor \(Z^2\) multiplying the Rydberg constant. This means that for a given transition (same \(n_1\) and \(n_2\)), the wavenumber of the light emitted by a hydrogen-like ion is \(Z^2\) times larger than that of hydrogen. Consequently, the wavelength is shorter, and the energy is higher. This scaling law allows us to predict the spectra of these ions with high precision using the same fundamental constant \(R_H\).
Spectral Series and Regions
Depending on the value of the lower energy level \(n_1\), the Rydberg formula generates different series of spectral lines. Each series corresponds to a specific region of the electromagnetic spectrum. The most famous series is the Lyman series, where \(n_1=1\). Transitions to this ground state involve large energy changes, resulting in ultraviolet radiation. The Balmer series, where \(n_1=2\), involves transitions to the first excited state. These transitions have smaller energy differences and fall within the visible region of the spectrum, which is why they were the first to be observed and studied. The Paschen series (\(n_1=3\)) and Brackett series (\(n_1=4\)) fall into the infrared region. As \(n_1\) increases, the energy differences between levels decrease, shifting the spectral lines to longer wavelengths.
The result
The Rydberg formula
For a hydrogen atom, the wavelength \(\lambda\) of light emitted or absorbed as the electron transitions between two allowed energy levels \(n_1\) and \(n_2\) (integers, \(n_2>n_1\ge1\)) satisfies \(\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\), where \(R_H\approx1.097\times10^7\,\text{m}^{-1}\) is the Rydberg constant for hydrogen. For a hydrogen-like ion of nuclear charge \(Z\), the same formula holds with \(R_H\) replaced by \(Z^2R_H\).
A common misconception is that the Rydberg constant \(R_H\) is a universal constant that does not change with the element. While \(R_\infty\) (the Rydberg constant for an infinitely heavy nucleus) is a fundamental physical constant, the value used for hydrogen, \(R_H\), is slightly smaller due to the finite mass of the proton. The electron and proton orbit their common centre of mass, not a fixed nucleus. This "reduced mass" effect means \(R_H\) is slightly less than \(R_\infty\). For most introductory calculations, using \(1.097\times10^7\,\mathrm{m^{-1}}\) is sufficient, but for high-precision spectroscopy, the distinction between \(R_H\) and \(R_\infty\) is critical. C
Worked example
Problem. Find the shortest wavelength in the Balmer series of hydrogen (the series limit, \(n_2 \to \infty\), \(n_1 = 2\)) and the energy needed to ionise a hydrogen atom from \(n = 2\). Use \(R_H = 1.097\times10^7\,\mathrm{m^{-1}}\), \(h = 6.626\times10^{-34}\,\mathrm{J\,s}\), \(c = 3.00\times10^8\,\mathrm{m\,s^{-1}}\).
- As \(n_2 \to \infty\), \(1/n_2^2 \to 0\), so \(1/\lambda = R_H/2^2 = 1.097\times10^7/4 = 2.743\times10^6\,\mathrm{m^{-1}}\).
- Invert: \(\lambda = 1/(2.743\times10^6) = 3.646\times10^{-7}\,\mathrm{m} = 364.6\,\mathrm{nm}\), just into the ultraviolet.
- Photon energy at the limit: \(E = hc/\lambda = (6.626\times10^{-34})(3.00\times10^8)/(3.646\times10^{-7}) = 5.45\times10^{-19}\,\mathrm{J}\).
- In electronvolts: \(5.45\times10^{-19}/1.602\times10^{-19} = 3.40\,\mathrm{eV}\), which is \(13.6\,\mathrm{eV}/4\), the binding energy of the \(n = 2\) level.
Units check. \(R_H\) is in \(\mathrm{m^{-1}}\), so its reciprocal is a length; \(\mathrm{J\,s} \times \mathrm{m\,s^{-1}} / \mathrm{m} = \mathrm{J}\).
Three quick tests
- What is the value of \(n_1\) for the Balmer series?
Answer
2 - If \(Z=2\) (helium ion), how does the wavenumber of a transition compare to hydrogen for the same \(n_1, n_2\)?
Answer
It is 4 times larger. - Which series corresponds to transitions ending at the ground state?
Answer
Lyman series
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.