Atomic models and line spectra

What is it about?

A gas of single atoms does not glow in all colours but only in very specific ones: behind a prism you see separate bright lines, a line spectrum. Every element has its own pattern, like a fingerprint.

The electrons of an atom can only have certain energies, the energy levels. When the electron jumps from a higher level $E_\text{u}$ (upper) to a lower level $E_\text{l}$ , the atom emits exactly one photon with

\Delta E = E_\text{u} - E_\text{l} = h \cdot f, \qquad \lambda = \frac{c}{f} = \frac{h \cdot c}{\Delta E},

with $h = 6.626 \cdot 10^{-34}\,\text{J}\cdot\text{s}$ and $h \cdot c \approx 1240\,\text{eV}\cdot\text{nm}$ . For hydrogen the Bohr model gives $E_n = -\frac{13.6\,\text{eV}}{n^2}$ . Jumps ending on the same level form a series: Lyman ( $n = 1$ , UV), Balmer ( $n = 2$ , visible lines at $656$ , $486$ , $434$ and $410\,\text{nm}$ ), Paschen ( $n = 3$ , infrared).

Absorption: white light passing through a cool gas loses exactly the photons whose energy fits a jump upwards – dark lines appear at the same places as the bright emission lines (Fraunhofer lines).

The orbits in the picture are a model and not to scale. For helium, sodium and mercury the levels are taken from tables. The simulation uses vacuum wavelengths; tables for air give slightly smaller values, e.g. the yellow helium line at $587.7\,\text{nm}$ (vacuum) instead of $587.6\,\text{nm}$ (air).

Try it

Watch a jump from $n = 3$ to $n = 2$ in slow motion. Load example
Tap the other arrows of the Balmer series. How do $\Delta E$ , $f$ and $\lambda$ change? Balmer series
Choose a jump that ends at $n = 1$ . Why is no colour visible? Lyman-α
Let the gas glow and compare the four elements. All lines
Switch to absorption. Sodium in absorption

Tasks

Task 1: The blue-green hydrogen line

Calculate energy, frequency and wavelength of the photon for the jump from $n = 4$ to $n = 2$ . Load task

Show solution

$\Delta E = -0.85\,\text{eV} - (-3.40\,\text{eV}) = 2.55\,\text{eV}$ , $f = \frac{\Delta E}{h} \approx 6.17 \cdot 10^{14}\,\text{Hz}$ , $\lambda \approx 486\,\text{nm}$ . Show in the simulation

Task 2: Invisible light

Calculate the wavelength for the jump from $n = 2$ to $n = 1$ . Why can’t you see it? Load task

Show solution

$\Delta E = 10.2\,\text{eV}$ , $\lambda = \frac{1240\,\text{eV}\cdot\text{nm}}{10.2\,\text{eV}} \approx 122\,\text{nm}$ – ultraviolet, outside the visible range of about $380$ to $780\,\text{nm}$ . Show in the simulation

Task 3: The yellow sodium line

Sodium street lamps glow at $589\,\text{nm}$ . What is the energy difference of the levels involved? Load task

Show solution

$\Delta E = \frac{1240\,\text{eV}\cdot\text{nm}}{589\,\text{nm}} \approx 2.10\,\text{eV}$ (jump $3\text{p} \to 3\text{s}$ ). Show in the simulation

Task 4: Series limit

What is the shortest wavelength of the Balmer series (jump from $E = 0$ to $n = 2$ )?

Show solution

$\Delta E = 3.40\,\text{eV}$ , $\lambda \approx 365\,\text{nm}$ (just in the UV). The levels crowd together near $E = 0$ , so the lines crowd together too.

Ideas for the classroom

Combine with real spectral tubes and a hand spectroscope: look at the real spectrum first, then find the origin of each line here.
With val=0 the values in the picture are hidden for calculation exercises.
Common stumbling blocks:
- “The electron emits light while orbiting.” – Light is only emitted during a jump, one photon per jump.
- “Higher levels give red light.” – What matters is the energy difference.
- “Negative energies make no sense.” – $E = 0$ means the electron has just become free; bound states lie below.