Atomic Spectra and the Rydberg Formula

In 1885, before quantum mechanics existed, Johann Balmer noticed that four visible lines in the hydrogen spectrum had wavelengths fitting a surprisingly simple pattern. Johann Rydberg later generalized it into the formula we use today.

The Rydberg formula

For hydrogen, the wavelength $\lambda$ of the light emitted (or absorbed) when an electron transitions between two energy levels $n_1$ and $n_2$ (with $n_2 > n_1 \geq 1$ ) obeys

$\frac{1}{\lambda} = R_H \!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right),$

where $R_H = 1.097 \times 10^7\,\text{m}^{-1}$ is the Rydberg constant for hydrogen. The integer $n_1$ labels the lower level; $n_2$ is the upper level the electron falls from.

The formula works because the energy of hydrogen level $n$ is

$E_n = -\frac{13.6\,\text{eV}}{n^2},$

so the photon energy released in a downward jump (the electron falling from the upper level $n_2$ to the lower level $n_1$ ) is

$\Delta E = E_{n_2} - E_{n_1} = 13.6\,\text{eV}\!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).$

Setting $\Delta E = hc/\lambda$ and using $hc = 1240\,\text{eV}\cdot\text{nm}$ , we can check that

$\frac{1}{\lambda} = \frac{13.6\,\text{eV}}{hc}\!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = \frac{13.6}{1240}\,\text{nm}^{-1}\!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right),$

which reproduces $R_H$ when converted to metres: $13.6/1240 \approx 0.01097\,\text{nm}^{-1} = 1.097 \times 10^7\,\text{m}^{-1}$ .

Named series

The series are named by their lower level $n_1$ :

| Series | $n_1$ | Spectral region | |--------|-------|-----------------| | Lyman | 1 | ultraviolet | | Balmer | 2 | visible | | Paschen| 3 | near-infrared |

The four lines Balmer originally catalogued correspond to $n_2 = 3, 4, 5, 6$ falling to $n_1 = 2$ . The reddest of these, the $n_2 = 3 \to n_1 = 2$ line, is called H-alpha and sits at $\lambda \approx 656\,\text{nm}$ — the distinctive crimson glow of hydrogen gas discharges and emission nebulae.

Deriving H-alpha step by step

For H-alpha, $n_1 = 2$ and $n_2 = 3$ :

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{5}{36} \approx 1.524 \times 10^6\,\text{m}^{-1}.$

Taking the reciprocal: $\lambda = 1/(1.524 \times 10^6) \approx 6.563 \times 10^{-7}\,\text{m} = 656.3\,\text{nm}$ .

Try it

This is a numerical exercise — your code should return a number, not a circuit. Apply the Rydberg formula to find the wavelength of the H-alpha line ( $n=3 \to n=2$ ) in nanometres.

Run your code to see the quantum state.