The Lyman Series: Ultraviolet Hydrogen Lines

1 · Predict

Electrons dropping all the way down to the ground state (n = 1) emit the Lyman series. Compared to the Balmer series (dropping to n = 2), would you expect Lyman photons to have more or less energy?

2 · Set Up

  1. Open the lyman-alpha preset and press Reset. This preset's default transition is n = 2 → n = 1 (Lyman-alpha).
  2. Enable the emitted-wavelength readout.
  3. Set the starting energy level n_initial for each trial (dropping down to n_final = 1) and record the emitted wavelength.

3 · Collect Data

Starting level n_iEmitted wavelength λ (nm)
2
3
4

Plot emitted wavelength λ (y-axis) against starting level n_i (x-axis) for your three trials.

4 · Analyze

  1. For one trial, compute 1/λ = R(1/n_f² − 1/n_i²) using n_f = 1. Compare to the table. Confirm all three wavelengths fall below 122 nm — deep ultraviolet, invisible to the eye.
  2. Compare your Lyman wavelengths to the balmer-alpha experiment's visible-light wavelengths. Explain why dropping to n = 1 (a much bigger energy gap than dropping to n = 2) always produces higher-energy, shorter-wavelength photons.

5 · Extend

  1. Lyman-series ultraviolet light is almost entirely absorbed by Earth's atmosphere before reaching the ground. Explain why space telescopes (rather than ground-based ones) are needed to observe hydrogen's Lyman-alpha emission from distant astronomical objects.
  2. As n_i → ∞, 1/λ approaches R exactly (the series limit), corresponding to an electron barely escaping the atom entirely (ionization) rather than a specific transition. Using your formula, explain why the series limit represents the ionization energy from the ground state.

The Physics Behind This Experiment

Lyman Series (Ground-State Transitions)

Transitions ending at n_f = 1 release the largest possible energy gaps in hydrogen, since the ground state sits far below every excited level — producing the shortest-wavelength (highest-energy) spectral series, entirely in the ultraviolet.

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