Preview

Models of the Hydrogen Atom

Q: What wavelength photon is needed to excite hydrogen from n=1 to n=2?

ΔE = 13.6(1/1² − 1/2²) = 10.2 eV; λ = hc/ΔE ≈ 121 nm (UV).

Q: Why is the Balmer series visible light while Lyman series is UV?

Balmer ends at n=2 (lower energy transitions); Lyman ends at n=1 (higher energy = shorter λ).

Q: What was wrong with the Bohr model that the quantum model fixed?

Bohr can't explain multi-electron atoms, fine structure, or the uncertainty principle.

Explore quantum models from Bohr to Schrödinger

1885, Basel. A schoolteacher named Johann Balmer stares at four colored lines in hydrogen's glow — red, blue-green, violet, deep violet — and discovers their wavelengths fit one eerily simple formula. Nobody knows why. The formula is a recipe without a kitchen. Twenty-eight years later the mystery turns into a crisis: Rutherford's atom has electrons circling a tiny nucleus, and classical physics says every orbiting electron must radiate, spiral inward, and crash in a nanosecond. Atoms should collapse; matter should not exist. In Copenhagen in 1913, Niels Bohr answers with a audacious bargain: the electron may occupy only certain orbits — angular momentum quantized in units of ℏ — and in those orbits it simply does not radiate. It shines only when it falls from one allowed rung to a lower one, and the photon it releases carries exactly the difference: hf = Eᵢ − E_f. Out of that single postulate comes Eₙ = −13.6/n² eV, and with it the radius ladder rₙ = n²a₀, the size of the atom, the ionization energy, and — to gasps — Balmer's recipe, derived from first principles to the last digit of the Rydberg constant. The model cannot handle helium, and quantum mechanics will later trade the tidy circles for probability clouds. But for hydrogen, Bohr's numbers are exact. In this lab you stand on the ladder yourself: six glowing rings, one electron, one repeating fall. Park the electron on any rung, aim its drop, and read the photon the atom pays each time it falls — red H-alpha at 656.5 nm, or the ultraviolet edge of the Lyman abyss. The ladder is crowded at the top and bottomless at n=1. Choose the drop.

What you'll be able to do

Price any transition with hf = 13.6(1/nf² − 1/ni²) eV and λ = 1240/ΔE nm: at the default state (n=3 → n=2) the panel reads ΔE = 1.889 eV and λ = 656.5 nm — the red H-alpha line, first rung of the Balmer series ('part of the Balmer series is visible', OpenStax CP2e §30.3, verbatim) — while dropping the target to n=1 prices 12.089 eV / 102.6 nm, exactly 6.4× the energy in the ultraviolet ('The Lyman series is entirely in the UV', verbatim). The same recipe derives the Rydberg constant R = 1.097×10⁷ /m from first principles (CP2e Eq. 30.32) and prices the balmer preset's 4→2 photon at 486.3 nm — Example 30.1's own 486 nm second Balmer line
Read the two Bohr scaling laws off the panel's live rows: sweep Principal level n 1 → 6 and the Orbital radius row climbs 0.0529 → 0.2116 → 0.4761 → 0.8464 → 1.3225 → 1.9044 nm (rₙ = n²a₀ with a₀ = 0.0529 nm = 0.529×10⁻¹⁰ m, CP2e Eq. 30.23 — the n=3 row's 9a₀ = 0.476 nm is the HTML quiz Q2's keyed answer) while the Energy row rises −13.600 → −3.400 → −1.511 → −0.850 → −0.544 → −0.378 eV with the steps shrinking 10.2 → 1.89 → 0.66 → 0.31 → 0.17 eV — the 1/n² crowd toward zero (Chem 2e Fig 6.14; E(n=3) = −1.511 eV is Example 6.4's own −2.421×10⁻¹⁹ J)
Keep the ionization ledger straight: '13.6 eV is needed to ionize hydrogen' from the ground state (CP2e §30.3, verbatim — 'an experimentally verified number'), but the binding weakens as 13.6/n² eV, so at n=6 only 0.378 eV holds the electron — 'as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases and it is held less tightly in the atom' and 'E = 0 corresponds to the ionization limit' (Chem 2e §6.2, verbatim). Higher n means EASIER to strip — the HTML quiz Q3's keyed answer

Formulas

r_n = n^2 a_0,\quad a_0 = 0.0529\text{ nm}

Bohr orbital radius for level n (HTML sidebar f1)

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

Bohr energy levels

\Delta E = hf = \frac{hc}{\lambda}

Photon energy for transition

\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

Rydberg formula

Make a prediction

Default state: the electron sits at n = 3 and the transition target is n = 2 — the panel pre-computes the famous red H-alpha photon, λ = 656.5 nm (Balmer series). Now drag the Transition to level slider from 2 down to 1. Same starting level n = 3, deeper fall. What does the Wavelength row do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Explore

Read the pre-computed H-alpha rows at the default state (ΔE = 1.889 eV, λ = 656.5 nm, Balmer), then watch the demo cycle run the transition on its own clock: the electron slides from the n=3 ring down to n=2 with a burst of red photons, dwells on the lower rung, and climbs back to repeat — the red line fires again every cycle, same λ every time.
Drag Principal level n from 1 to 6 and watch two rows at once: Orbital radius climbs 0.0529 → 1.9044 nm with GROWING gaps (the n² law — each ring farther out than the last step) while Energy rises −13.600 → −0.378 eV with SHRINKING gaps (10.2 → 1.89 → 0.66 → 0.31 → 0.17 eV — the 1/n² crowd toward zero). The electron visibly loiters on the outer rings (ω ∝ 1/n³ visual pacing).
Park n = 6 and sweep Transition to level from 5 down to 1 — the whole series ladder on one lever: ΔE climbs 0.166 → 0.472 → 1.133 → 3.022 → 13.222 eV while λ dives 7460 → 2626 → 1094 → 410 → 93.8 nm, walking Pfund+ → Brackett → Paschen → Balmer → Lyman without ever touching the starting level. Five series, one gap formula.
Click the three preset buttons in turn and read one atom in three bands: Ground State (r = 0.0529 nm, E = −13.600 eV — the bottom rung, 13.6 eV to ionize) → Balmer (4→2, 486.3 nm blue-green, the line OpenStax Example 30.1 prices) → Paschen (5→3, 1282.2 nm — infrared your eyes can't see but the panel prices anyway).

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It dives to 102.6 nm in the ultraviolet — ΔE jumps from 1.889 eV to 12.089 eV (exactly 6.4×), and the Series row flips from Balmer to Lyman: the photon prices the GAP, and the fall to n=1 is far deeper

The photon is paid out of the DIFFERENCE between two levels, never out of one level alone: hf = 13.6(1/nf² − 1/ni²) eV (OpenStax CP2e §30.3, Eq. 30.30). From n=3, targeting n=2 prices ΔE = 13.6×(1/4 − 1/9) = 1.8889 eV → λ = 1240/1.8889 = 656.5 nm, red, Balmer. Retargeting to n=1 prices ΔE = 13.6×(1/1 − 1/9) = 12.0889 eV → λ = 1240/12.0889 = 102.6 nm, ultraviolet, Lyman ('The Lyman series is entirely in the UV', §30.3 verbatim) — exactly 6.4× the energy, because the gap (1/nf² − 1/ni²) grew from 5/36 to 8/9. Option B is the misconception this gate targets: the intuition that the starting rung owns the photon — the sim's own quiz Q1 ships it as the 'same wavelength' trap, and its keyed feedback gives the refutation numbers (12.09 eV vs 1.89 eV). Option C runs the energy bookkeeping backwards — and it's quiz Q1's other trap ('longer wavelength (lower energy)'): the electron that lands at n=1 ends LOWER (−13.6 eV vs −3.4 eV), so it gave UP more, not less — 'if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns' (Chem 2e §6.2, verbatim). Bigger drop, bigger photon, shorter wavelength.

Quiz (0/3)

What wavelength photon is needed to excite hydrogen from n=1 to n=2?

Why is the Balmer series visible light while Lyman series is UV?

What was wrong with the Bohr model that the quantum model fixed?

You can now

Price any transition with hf = 13.6(1/nf² − 1/ni²) eV and λ = 1240/ΔE nm: at the default state (n=3 → n=2) the panel reads ΔE = 1.889 eV and λ = 656.5 nm — the red H-alpha line, first rung of the Balmer series ('part of the Balmer series is visible', OpenStax CP2e §30.3, verbatim) — while dropping the target to n=1 prices 12.089 eV / 102.6 nm, exactly 6.4× the energy in the ultraviolet ('The Lyman series is entirely in the UV', verbatim). The same recipe derives the Rydberg constant R = 1.097×10⁷ /m from first principles (CP2e Eq. 30.32) and prices the balmer preset's 4→2 photon at 486.3 nm — Example 30.1's own 486 nm second Balmer line
Read the two Bohr scaling laws off the panel's live rows: sweep Principal level n 1 → 6 and the Orbital radius row climbs 0.0529 → 0.2116 → 0.4761 → 0.8464 → 1.3225 → 1.9044 nm (rₙ = n²a₀ with a₀ = 0.0529 nm = 0.529×10⁻¹⁰ m, CP2e Eq. 30.23 — the n=3 row's 9a₀ = 0.476 nm is the HTML quiz Q2's keyed answer) while the Energy row rises −13.600 → −3.400 → −1.511 → −0.850 → −0.544 → −0.378 eV with the steps shrinking 10.2 → 1.89 → 0.66 → 0.31 → 0.17 eV — the 1/n² crowd toward zero (Chem 2e Fig 6.14; E(n=3) = −1.511 eV is Example 6.4's own −2.421×10⁻¹⁹ J)
Keep the ionization ledger straight: '13.6 eV is needed to ionize hydrogen' from the ground state (CP2e §30.3, verbatim — 'an experimentally verified number'), but the binding weakens as 13.6/n² eV, so at n=6 only 0.378 eV holds the electron — 'as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases and it is held less tightly in the atom' and 'E = 0 corresponds to the ionization limit' (Chem 2e §6.2, verbatim). Higher n means EASIER to strip — the HTML quiz Q3's keyed answer