Understand the appearance and formation of emission and absorption line spectra.
Atoms possess a set of discrete energy states. An electron can only occupy these allowed levels, denoted by the principal quantum number $n = 1,2,3,\dots$.
When an electron moves between two levels $i$ and $j$ the atom either emits or absorbs a photon whose energy equals the difference between the two levels:
$$\Delta E = E_j - E_i = hu = \frac{hc}{\lambda}$$where $h$ is Planck’s constant, $u$ the frequency and $\lambda$ the wavelength of the photon.
Emission occurs when an electron in an excited state ($E_j$) drops to a lower state ($E_i$). The released photon produces a bright line at wavelength $\lambda$ in the spectrum.
If a beam of continuous radiation passes through a cool gas, photons whose energies match a possible transition are absorbed, creating dark lines (absorption lines) in the otherwise continuous spectrum.
For a given atom, transitions that share a common lower (or upper) level form a series of lines. In hydrogen the most important series are:
The Balmer series corresponds to transitions from $n \ge 3$ down to $n = 2$. Using the Rydberg formula:
$$\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right), \qquad n = 3,4,5,\dots$$| Transition | Upper level $n$ | Wavelength $\lambda$ (nm) | Colour (perceived) |
|---|---|---|---|
| H$_\alpha$ | 3 → 2 | 656.3 | Red |
| H$_\beta$ | 4 → 2 | 486.1 | Blue‑green |
| H$_\gamma$ | 5 → 2 | 434.0 | Violet |
| H$_\delta$ | 6 → 2 | 410.2 | Violet‑ultraviolet |
Atoms have quantised energy levels. Photons are emitted or absorbed when electrons transition between these levels, producing discrete lines in a spectrum. Emission lines appear bright against a dark background, while absorption lines appear dark against a continuous spectrum. The pattern of lines (spectral series) is characteristic of each element and forms the basis of spectroscopic identification.