University Physics IV · The Hydrogen Atom · 10.5

Hydrogen Spectrum, Bohr Radius & Rydberg Energy

The quantisation argument handed you an integer; here the integer becomes numbers a spectrometer can check. Build the two scales every hydrogenic quantity is priced in — one length, one energy — then spend them on series wavelengths, on one-electron ions, on isotope shifts, and on knowing where the discrete spectrum has to stop.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Truncating the radial series fixed the shape of the bound spectrum, Eₙ ∝ −1/n²; this topic is the prefactor, and the prefactor is the physics. The Coulomb problem contains three constants — ħ, the reduced mass μ, and the coupling e²/4πε₀ — and they combine into exactly one length and one energy: the Bohr radius a₀ ≈ 0.0529 nm and the Rydberg energy E_R ≈ 13.606 eV. The full solution is those two scales dressed with integers and Z: Eₙ = −Z²E_R/n², sizes near n²a₀/Z.

Because the two-body reduction put the recoil into μ, hydrogen's Rydberg sits one part in 1836 below the infinite-nuclear-mass value — a shift far above spectroscopic precision, and the lever that dragged deuterium into view. The model's cost is written at E = 0: normalisability quantises only decaying solutions, so above zero every energy is allowed. Discrete lines therefore crowd toward a series limit and hand over to a continuum — infinitely many bound levels accumulate below the edge, unbroken photoionisation runs beyond it.

And the whole account assumes a point nucleus, no relativity and no spin: the fine structure it omits enters (Zα)² smaller, which is precisely why this crude ladder works so well at Z = 1.

Simple definition: A hydrogenic atom's bound energies are Eₙ = −Z²E_R/n² and its sizes are near n²a₀/Z, where the Rydberg energy E_R ≈ 13.606 eV and the Bohr radius a₀ ≈ 0.0529 nm are built entirely from ħ, the reduced mass, the electron charge, and ε₀.
Example: For hydrogen's n = 3 → 2 drop the photon carries 13.606 × (1/4 − 1/9) = 1.89 eV, a wavelength near 1240/1.89 ≈ 656 nm — the red Balmer-α line any bench spectrometer resolves to a fraction of a nanometre.

Bohr radiusa₀ = 4πε₀ħ²/(μe²) = 0.05292 nm

The one length the constants can build: orbital sizes run as n²a₀/Z, so it fixes atomic dimensions before any equation is solved.

μ the reduced mass in kg, e in C, ε₀ in F m⁻¹; equivalently a₀ = ħ/(μcα)

Rydberg energyE_R = μe⁴/(2(4πε₀)²ħ²) = ħ²/(2μa₀²) = 13.606 eV

One number prices the atom: level depths, photon gaps and the ionisation edge are all rational multiples of it.

equivalently e²/(8πε₀a₀); as a wavenumber R∞ = E_R/hc = 1.09737 × 10⁷ m⁻¹

Bound spectrumEₙ = −Z²E_R/n², n = 1, 2, 3, …

Levels crowd as 1/n² — infinitely many bound states pile beneath E = 0 while the gaps between them shrink toward zero.

n = nᵣ + l + 1, radial nodes plus l plus one; only E < 0 is quantised

Line wavelengths1/λ = R Z² (1/n_f² − 1/nᵢ²)

Every observed line is one pair (nᵢ, n_f); letting nᵢ → ∞ gives the series limit, the last discrete mark before the continuum.

R in m⁻¹: use hydrogen's R_H = 1.09678 × 10⁷ m⁻¹, not R∞; hc = 1239.84 eV⋅nm

Reduced-mass correctionμ = mₑM/(mₑ + M) ⇒ R_M = R∞/(1 + mₑ/M)

The recoiling nucleus trims every binding energy by one part in 1836 — the isotope shift that separated deuterium from hydrogen.

M the nuclear mass; mₑ/Mₚ = 1/1836.15, so R_H = 1.09678 × 10⁷ m⁻¹

Hydrogenic scalingrₙ ≈ n²a₀/Z, Eₙ ∝ Z²/n², v/c ≈ Zα/n

He⁺ binds 54.4 eV; an n = 100 Rydberg atom spans 0.5 μm and holds on by 1.4 meV — one solution stretched five orders of magnitude.

α = e²/4πε₀ħc ≈ 1/137; covers He⁺, Li²⁺, inner shells, Rydberg states

Build the two scales before touching the equation

The Coulomb Hamiltonian holds three constants: ħ, the reduced mass μ, and the coupling e²/4πε₀, an energy-times-length worth 1.4400 eV⋅nm. There is exactly one way to build a length from them, a₀ = ħ²/(μ · e²/4πε₀), and evaluating it with ħc = 197.33 eV⋅nm and μc² ≈ 511 000 eV gives 197.33²/(511 000 × 1.4400) = 0.05292 nm. One natural energy follows, E_R = e²/(8πε₀a₀) = 1.4400/(2 × 0.05292) = 13.606 eV. So before any equation is solved you know atoms are ångström-sized and chemistry is electronvolt-priced; the solution only supplies the integers. Keep both faces of E_R at hand: ħ²/(2μa₀²) reads as the kinetic cost of confinement to a₀, and e²/(8πε₀a₀) as half the Coulomb energy at a₀. Their equality at the same a₀ is the virial theorem in disguise — the atom's size is a genuine compromise between confinement and attraction.

The ladder crowds as 1/n² — nothing like even rungs

For hydrogen Eₙ = −13.606/n² eV: −13.61, −3.40, −1.51, −0.85, −0.54, … The gaps collapse — 10.20 eV between the first pair, then 1.89 eV, then 0.66 eV — the opposite of the harmonic oscillator's equal spacing, because the Coulomb well keeps widening as it shallows. Two structural facts ride along. The energy depends on n alone, not on l or m, so level n holds n² orbital states (2n² with spin) — a degeneracy special to the 1/r potential. And the ladder has no top rung short of zero: however small ε is, infinitely many levels sit between −ε and 0, because the long 1/r tail always has room for one more distant, barely bound state. Contrast a finite square well, which holds a finite count. E = 0 is not a level but an accumulation point — a warning that something different starts there.

Lines are differences; series are choices of the floor

A photon carries a difference: hν = E_R(1/n_f² − 1/nᵢ²), or in wavenumbers 1/λ = R Z²(1/n_f² − 1/nᵢ²). Fixing the floor n_f defines a series. Drops to n_f = 1 form the ultraviolet Lyman series, opening at 13.606 × (1 − 1/4) = 10.20 eV — the 121.6 nm Lyman-α. Drops to n_f = 2 form the Balmer series, whose first four lines at 656, 486, 434 and 410 nm are the visible spectrum of a hydrogen discharge tube. As nᵢ → ∞ each series crowds toward its limit — for Balmer, E_R/4 = 3.40 eV, near 364.6 nm — where the lines pile up faster than any grating resolves them. One bookkeeping trap: the tabulated R∞ = 1.09737 × 10⁷ m⁻¹ belongs to a fictitious clamped nucleus. Hydrogen obeys R_H = 1.09678 × 10⁷ m⁻¹, and the difference displaces every line by one part in 1836 — about 0.36 nm at Balmer-α, enormous against spectroscopy that resolves parts per million.

The nucleus recoils: reduced mass and the isotope shift

The μ inside a₀ and E_R is mₑM/(mₑ + M), because the two-body reduction lodged the nuclear recoil there. Writing μ = mₑ/(1 + mₑ/M), hydrogen's levels are a factor 1/(1 + 1/1836.15) = 0.999456 of the clamped-nucleus values. A part-in-1836 correction sounds negligible; against instruments resolving parts per million it is a shout. It is also a discovery tool: the deuteron is twice the proton's mass, so deuterium's correction is half hydrogen's, and each of its lines sits slightly blue of its hydrogen partner — 0.18 nm at Balmer-α, the faint satellite from which Urey extracted deuterium in 1932. Pushed to extremes the 'correction' takes over: positronium's partner is a positron, μ = mₑ/2, and the whole ladder floats up to half depth with E_R = 6.80 eV; in muonic hydrogen μ ≈ 186 mₑ, the levels are keV-deep, and the orbit shrinks by the same factor to graze the nucleus — which is how that atom measures the proton's radius.

Dress it with Z²: ions, inner shells, Rydberg atoms

For a one-electron ion the coupling becomes Ze²/4πε₀. The coupling enters a₀ once, inversely, and E_R through its square, so a₀ → a₀/Z and E_R → Z²E_R: Eₙ = −13.606 Z²/n² eV. He⁺ therefore binds 54.4 eV and radiates its 2 → 1 line at 40.8 eV, 30.4 nm — the scaling that unmasked the Pickering series of hot stars, He⁺ drops to n_f = 4 masquerading as hydrogen lines with half-integer n. The same rule runs both directions. Inward: a K-shell electron at Z = 30 moves at v/c ≈ Zα ≈ 0.22, so the relativistic corrections, suppressed by (Zα)², stop being fine at high Z. Outward: a Rydberg atom at n = 100 spans n²a₀ ≈ 0.53 μm — virus-sized — and is bound by 13.606/10⁴ eV = 1.36 meV, more than an order of magnitude below room temperature's kT ≈ 25 meV, which is why such atoms survive only cold, dilute, and shielded from stray fields.

Above zero, nothing is quantised — and you can see it

The truncation argument had a single premise: a bound state must decay at infinity to be normalisable, and only E < 0 solutions can. For E > 0 the radial solution oscillates for ever, is admissible at every energy, and describes an electron that arrives, deflects, and leaves. So the spectrum is discrete only below zero — and observation says exactly that. In absorption, series lines march toward the limit, and there the spectrum changes character rather than stopping: any photon above 13.6 eV ionises ground-state hydrogen and gives the surplus to the electron as kinetic energy, so a continuous photoionisation edge extends beyond the last resolvable line. Run backwards, a free electron of any energy can be captured, emitting a continuum photon — the Balmer continuum shortward of 364.6 nm that makes the 'Balmer jump' in nebular and stellar spectra. Bound ladder plus continuum form the complete set of states; drop the continuum and even formal completeness sums come out wrong.

Change one variable at a time

Make the relationship visible.

Interactive model

Lower level n_f2

Jump Δn = nᵢ − n_f1

Partner mass M1836 mₑ

Set n_f = 2 and Δn = 1 to land on Balmer-α near 656 nm, then drag the partner mass down from 1836: almost nothing moves until M falls below a few hundred, and at M = 1 — positronium — the whole ladder sits at half depth. That long flat stretch is why hydrogen's shift is only one part in 1836.

RYDBERG E_R13.599 eV

PHOTON ΔE1.89 eV

PHOTON λ656 nm

SERIES-LIMIT λ365 nm

Live interpretationRYDBERG E_R: 13.599 eV. PHOTON ΔE: 1.89 eV. PHOTON λ: 656 nm. SERIES-LIMIT λ: 365 nm

Catch the common trap

Explain before calculating.

Misconception check

Students read E = 0 as the atom's topmost level and the series limit as where hydrogen's spectrum ends. Both miss what the truncation argument actually proved. Normalisability bites only for E < 0, so zero is not a level: it is an accumulation point, with infinitely many bound states crowding beneath it and an unquantised continuum starting immediately above. The observed spectrum therefore does not stop at the 364.6 nm Balmer limit — it switches from discrete lines to the continuous absorption of photoionisation. And a 15 eV photon is not 'off resonance because no two levels differ by 15 eV': it ionises ground-state hydrogen and hands the spare 1.4 eV to the freed electron as kinetic energy, which is unquantised precisely because the truncation argument never applied above zero.

Hydrogen's Lyman-α transition, n = 2 → 1, emits a 10.2 eV photon. Ignoring the small reduced-mass difference, what photon energy does the same n = 2 → 1 transition give in the one-electron ion He⁺, where Z = 2?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe Balmer-β line of hydrogen is the n = 4 → 2 transition. Find its photon energy and wavelength, then find the wavelength of the Balmer series limit. Use E_R = 13.606 eV and hc = 1239.8 eV⋅nm.

The photon carries the level difference: ΔE = E_R(1/n_f² − 1/nᵢ²) = 13.606 × (1/4 − 1/16) = 13.606 × 0.1875 = 2.551 eV.
Convert with hc: λ = 1239.8/2.551 = 486.0 nm — the blue-green line that sits beside the red 656 nm Balmer-α in any discharge tube.
The series limit is nᵢ → ∞, so the 1/nᵢ² term vanishes: ΔEₗᵢₘᵢₜ = E_R/4 = 3.402 eV.
λₗᵢₘᵢₜ = 1239.8/3.402 = 364.5 nm. Every Balmer line lies between 656 nm and this ultraviolet edge, crowding toward it as nᵢ grows — and beyond the edge the photoionisation continuum takes over.

AnswerΔE = 2.55 eV, λ = 486 nm; the Balmer series limit sits at 364.5 nm.

MediumHydrogen's Balmer-α line sits near 656.3 nm. The deuteron's mass is 3670.5 mₑ and the proton's is 1836.2 mₑ. Find the reduced-mass factor μ/mₑ for each isotope, the wavelength splitting between the two Balmer-α lines, and which isotope's line is bluer.

Levels scale with μ/mₑ = 1/(1 + mₑ/M). Hydrogen: 1/(1 + 1/1836.2) = 0.999456. Deuterium: 1/(1 + 1/3670.5) = 0.999728.
Wavelength scales inversely with the level factor, λ ∝ 1/μ, so the fractional split is Δλ/λ ≈ mₑ/Mₚ − mₑ/M_D = 5.446 × 10⁻⁴ − 2.724 × 10⁻⁴ = 2.722 × 10⁻⁴.
Δλ = 656.3 nm × 2.722 × 10⁻⁴ = 0.179 nm.
Deuterium's larger μ binds every level more deeply, so its photons are more energetic: the deuterium line sits 0.18 nm to the blue. Urey resolved exactly this faint satellite in 1932 — deuterium discovered one part in 3700 from line centre.

Answerμ_H/mₑ = 0.999456, μ_D/mₑ = 0.999728; Δλ ≈ 0.18 nm, with the deuterium line on the blue side.

HardIn a nebula, a free electron with kinetic energy 0.85 eV is captured by a proton directly into the n = 2 level of hydrogen. (a) Find the emitted photon's energy and wavelength. (b) Show the photon lies beyond the Balmer series limit and explain why no discrete line can sit there. (c) Find the photon emitted when the atom then completes its cascade to the ground state.

(a) Energy conservation across the capture: the electron falls from E = +0.85 eV in the continuum to E₂ = −13.606/4 = −3.40 eV, so the photon carries 0.85 + 3.40 = 4.25 eV, and λ = 1239.8/4.25 = 291.7 nm.
(b) The Balmer limit is the nᵢ → ∞ edge at E_R/4 = 3.40 eV, i.e. 364.5 nm. Bound-bound drops into n = 2 are capped at that 3.40 eV, but this electron starts above E = 0, where energy is unquantised — so capture photons fill a continuum shortward of the limit instead of a line. Ours lands 73 nm inside it.
(c) The cascade finishes with n = 2 → 1: ΔE = 13.606 × (1 − 1/4) = 10.20 eV, so λ = 1239.8/10.20 = 121.5 nm — Lyman-α (measured at 121.6 nm; the reduced-mass E_R restores the difference these rounded inputs drop).
Together the two signatures — a continuous 'Balmer jump' starting at 364.5 nm plus strong Lyman-α — are precisely what recombining nebulae show, and the jump's shape thermometers the free-electron energies.

Answer(a) 4.25 eV at 291.7 nm. (b) It sits 73 nm shortward of the 364.5 nm Balmer limit — allowed only because the initial state is unquantised continuum. (c) 10.2 eV at 121.5 nm, Lyman-α.