Spectral Series of Hydrogen

When hydrogen's electron drops from a higher energy level to a lower one it releases a photon. Because the energy levels are discrete — fixed by the principal quantum number $n$ — the emitted photons appear at discrete wavelengths, producing the sharp spectral lines that are a fingerprint of hydrogen.

Energy levels and photon energy

The energy of hydrogen's $n$ -th level is

$E_n = -\frac{13.6\,\text{eV}}{n^2}, \quad n = 1, 2, 3, \ldots$

A transition from $n_{\text{upper}}$ to $n_{\text{lower}}$ releases a photon of energy

$\Delta E = E_{n_{\text{lower}}} - E_{n_{\text{upper}}} = 13.6\,\text{eV}\!\left(\frac{1}{n_{\text{lower}}^2} - \frac{1}{n_{\text{upper}}^2}\right).$

Using $E = hc/\lambda$ , this gives the wavelength directly.

The Rydberg formula

Writing $\Delta E = hc/\lambda$ and defining the Rydberg constant $R_\infty = m_e e^4 / (8 \epsilon_0^2 h^3 c) \approx 1.0974 \times 10^7\,\text{m}^{-1}$ (the subscript $\infty$ denotes the limit of infinite nuclear mass; the hydrogen-specific value $R_H$ is slightly smaller because the proton recoils), the result is

$\frac{1}{\lambda} = R_H\!\left(\frac{1}{n_{\text{lower}}^2} - \frac{1}{n_{\text{upper}}^2}\right), \qquad R_H \approx 1.0968 \times 10^7\,\text{m}^{-1}.$

This is the Rydberg formula, first written empirically by Johann Balmer (1885) for the $n_{\text{lower}} = 2$ series and generalised by Johannes Rydberg (1888).

The named series

Each value of $n_{\text{lower}}$ defines a series:

| Series | $n_{\text{lower}}$ | Region | |---|---|---| | Lyman | 1 | ultraviolet | | Balmer | 2 | visible (partly) | | Paschen | 3 | near-infrared | | Brackett | 4 | infrared |

The Balmer series is the historically important one because its lines fall in or near the visible spectrum and were measurable with 19th-century spectroscopes.

H-alpha: a worked example

The first (longest-wavelength) Balmer line connects $n_{\text{upper}} = 3$ to $n_{\text{lower}} = 2$ . Substituting:

$\frac{1}{\lambda} = R_H\!\left(\frac{1}{4} - \frac{1}{9}\right) = R_H \cdot \frac{5}{36}$

$= 1.0968 \times 10^7 \times \frac{5}{36} \approx 1.524 \times 10^6\,\text{m}^{-1}.$

Inverting:

$\lambda = \frac{1}{1.524 \times 10^6} \approx 6.56 \times 10^{-7}\,\text{m} = 656\,\text{nm}.$

This deep-red line — H-alpha — is one of the most recognised lines in astronomy; it colours emission nebulae red in photographs.

Convergence limits

As $n_{\text{upper}} \to \infty$ the photon energy approaches $13.6\,\text{eV}/n_{\text{lower}}^2$ , the ionisation energy from level $n_{\text{lower}}$ . The spectral lines crowd together at the series limit and then there is a continuum of wavelengths for electrons freed with kinetic energy above zero.

Try it

This is a numerical exercise — return a number. Use the Rydberg formula to compute the wavelength (in nm) of the H-alpha line: the transition from $n_{\text{upper}} = 3$ down to $n_{\text{lower}} = 2$ in hydrogen. Use $R_H = 1.09678 \times 10^7\,\text{m}^{-1}$ .

Run your code to see the quantum state.