University Physics V · The Hydrogen Atom · 11.5

Energy Levels, Scales & the Continuum

Where −13.6 eV actually comes from, why it is α²/2 of an electron's rest energy, and what the reduced mass does to it. Then the half of the spectrum the level diagram hides: above E = 0 the eigenvalues run continuously, and no sum over bound states is complete without them.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Solving the radial equation produced a truncation condition; this topic reads what the eigenvalue it produced actually says. Written in natural constants it is Eₙ = −½ μc²α²Z²/n², so the entire atomic energy scale is an electron's rest energy demoted by α²/2 — about one part in 37 600 — which is both why a bound electron is a non-relativistic problem and why relativistic corrections arrive one further factor of α² down. The same two constants fix the length: a₀ = ħ/(mₑ c α), the reduced Compton wavelength stretched by 1/α, with n²a₀/Z the scale 1/⟨1/r⟩ rather than any orbit.

The reduced mass is honest bookkeeping rather than polish: replacing mₑ by μ = mₑ M/(mₑ + M) lifts every hydrogen level by 5.44 parts in 10⁴ and every deuterium level by 2.72 parts in 10⁴, and the difference between those two is the whole of the H–D splitting Urey used to find deuterium. What it costs is the illusion that the ladder is the spectrum. Because the Coulomb tail dies only as 1/r, the bound levels are infinite in number and accumulate at E = 0 rather than stopping; above threshold the eigenvalues run continuously, their eigenfunctions are δ-normalised rather than square-integrable, and completeness on each (l, m) subspace reads Σₙ|n⟩⟨n| + ∫₀^∞ dE |E⟩⟨E| = Î.

Drop the integral and the resolution of the identity is simply false — which is exactly the error a finite numerical box commits when it replaces the continuum with pseudo-states.

Simple definition: The hydrogenic energy levels are the negative eigenvalues of the Coulomb radial Hamiltonian, Eₙ = −½μc²α²Z²/n²: a discrete ladder that accumulates at E = 0, above which the spectrum is continuous and unbound.
Example: With μ/mₑ = 0.999456, hydrogen has E₁ = −13.598 eV and E₂ = −3.400 eV, so the 1→2 line costs 10.199 eV (121.6 nm), 13.598 eV ionises, and ⟨1/r⟩ = 1/(n²a₀) puts the n = 1 length scale at 0.0529 nm.

Bound eigenvalues in natural constantsEₙ = −½ μ c² α² Z² / n², n = 1, 2, 3, …

Shows the atomic scale as α²/2 of a rest energy, so hydrogen is non-relativistic by construction.

μ the reduced mass (use μc² in eV), α = 1/137.036 a pure number, Z the nuclear charge number

The Rydberg energy, and the number it givesE_R = ½ mₑ c² α² = 13.6057 eV → Eₙ = −13.598 eV / n² for H

The gap between the two is 5.44 × 10⁻⁴ and sets the scale of every isotope shift; the H–D shift is the difference of two such corrections, 2.72 × 10⁻⁴.

13.598 eV carries the μ correction for hydrogen; 13.6057 eV is the infinite-nuclear-mass limit

The atomic length, and what n²a₀/Z meansa₀ = ħ/(mₑ c α) = 0.052918 nm, ⟨1/r⟩ = Z/(n² a₀)

Puts n = 12 near 7.6 nm and n = 50 near 132 nm — Rydberg atoms are very nearly macroscopic.

n²a₀/Z is the scale 1/⟨1/r⟩, exact for every l; the mean radius is ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)], and ⟨v²⟩^½/c = Zα/n

Reduced mass and the isotope shiftμ = mₑ M/(mₑ + M), Eₙ ∝ μ, lengths ∝ 1/μ, Δλ/λ = −Δμ/μ

Puts Hα at 656.46 nm and Dα 0.179 nm to the blue — the shift that revealed deuterium in 1931.

M is the nuclear mass; μ/mₑ = 0.999456 for H, 0.999728 for D, and exactly 0.5 for positronium

Series limits from the Rydberg formula1/λ = R_H Z² (1/n_f² − 1/nᵢ²), R_H = 1.09678 × 10⁷ m⁻¹

Lyman limit 91.18 nm, Balmer limit 364.7 nm in vacuum — where lines stop and continuous absorption begins.

R_H = E_R(μ/mₑ)/hc in m⁻¹; letting nᵢ → ∞ gives the series limit rather than another line

Completeness across the thresholdΣₙ |n l m⟩⟨n l m| + ∫₀^∞ dE |E l m⟩⟨E l m| = Î

The bound states alone are not a basis: expanding a kicked or ionised state needs the integral too.

Bound kets are unit-normalised; continuum kets obey ⟨E|E′⟩ = δ(E − E′), so they carry units of energy^(−1/2)

Reading the eigenvalue in natural constants

Series truncation gave n = nᵣ + l + 1 and an energy; the useful form strips the SI clutter away. Write Eₙ = −½ μc²α²Z²/n², with α = e²/4πε₀ħc = 1/137.036. Everything on the right is either a rest energy or a pure number, so the arithmetic is one line: ½ × 510 999 eV × (1/137.036)² = ½ × 27.211 eV = 13.606 eV. That number, the Rydberg energy E_R, is the whole atomic energy scale, and the formula says where it comes from — a rest energy demoted by α²/2 ≈ 2.66 × 10⁻⁵. Two consequences follow at once. By the virial theorem ⟨T⟩ = −E₁ = 13.6 eV, so ⟨v²⟩^½/c = √(2 × 13.598/511 000) = 0.00730 ≈ α, and the non-relativistic p²/2μ we assumed is self-consistent rather than merely convenient. And the leading relativistic correction, one further factor of α² down, lands near 10⁻⁴ eV — invisible without an etalon, and the business of the next unit.

One length, set by the same two constants

The energy scale fixes the length scale. Combining a₀ = ħ/(mₑ c α) with the reduced Compton wavelength ƛ_C = ħ/(mₑ c) = 386.16 fm gives a₀ = ƛ_C/α = 386.16 fm × 137.036 = 52 918 fm = 0.052918 nm, and E_R = ħ²/(2mₑ a₀²) closes the loop between them. Notice which mass sits where: the atomic length goes as 1/μ while Eₙ goes as μ, so the reduced mass moves length and energy in opposite directions and leaves the product E_R a₀ = ½ αħc = 0.7200 eV⋅nm alone — the product E_R a₀², by contrast, is ħ²/2μ and does depend on the mass. The n and Z scaling is what to carry, and it belongs to an expectation value rather than to an orbit: ⟨1/r⟩ = Z/(n²a₀) exactly, for every l, so n²a₀/Z is the scale 1/⟨1/r⟩, while the mean radius itself is l-dependent, ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)], and only the circular state l = n − 1 has its radial density peaking at n²a₀/Z. The virial theorem turns the same scale into a speed, ⟨v²⟩^½/c = Zα/n. So He⁺ in n = 3 has 1/⟨1/r⟩ = 0.238 nm and E = −6.05 eV, while a hydrogen Rydberg state at n = 50 sits on a scale of 2500a₀ = 132 nm and binds by only 5.44 meV — a fifth of room-temperature kT, which is why such atoms survive only in cold, dilute beams.

The reduced mass is a measurable 5-in-10⁴

The Coulomb problem is two-body, and reducing it to one body replaces mₑ by μ = mₑ M/(mₑ + M). For hydrogen M/mₑ = 1836.15, so μ/mₑ = 1836.15/1837.15 = 0.999456: every level is shallower by 5.44 parts in 10⁴, and E₁ is −13.598 eV, not the infinite-mass −13.6057 eV. That is not a rounding detail. Deuterium has μ/mₑ = 3670.48/3671.48 = 0.999728, larger by 2.72 × 10⁻⁴, so every deuterium line sits that fraction to the blue: Balmer-α at 656.46 nm in hydrogen appears at 656.28 nm in deuterium, a 0.179 nm doublet and the signature Urey used in 1931. Push μ further and the whole atom rescales. Positronium has μ = mₑ/2 exactly, so E₁ = −6.80 eV and its length scale is 2a₀. Muonic hydrogen has μ = 185.8 mₑ, giving E₁ = −2.53 keV and a scale of 285 fm — small enough that the proton's finite size shifts the levels measurably.

Why the ladder never ends

The levels do not run out; they crowd. E_{n+1} − Eₙ = E_R(1/n² − 1/(n+1)²) ≈ 2E_R/n³, so the 1→2 gap is 10.20 eV while the 50→51 gap is 0.211 meV, and the density of levels, |dn/dE| = n³/(2E_R), diverges as E → 0⁻. There are therefore infinitely many bound states, and they accumulate at threshold rather than stopping at some deepest-but-one. The reason is the range of the potential, not the depth of the well: −Ze²/4πε₀r falls off too slowly for the centrifugal term to win at large r, so there is always another shallow state further out. Replace the tail by a screened Yukawa form e^(−r/λ)/r and the count becomes finite — for l = 0 the last bound state vanishes once λ drops below about 0.84 a₀. That is why H⁻, whose departing electron sees a neutral atom and only a short-range polarisation tail, has exactly one bound state while neutral hydrogen has an endless series.

Above threshold, and what changes there

At E = 0 the character of the spectrum changes. For E > 0 every value is an eigenvalue: write E = ħ²k²/2μ with k continuous, and the radial solutions oscillate all the way out, so they are not square-integrable and cannot be normalised to one. They are normalised to a delta instead, ⟨E|E′⟩ = δ(E − E′), whose argument carries units of energy, so the kets carry energy^(−1/2) — the same move made for plane waves in the momentum basis. What is not the same is the phase. The 1/r tail never stops acting, so the asymptotic form is sin(kr − lπ/2 − η ln 2kr + σₗ), with the Sommerfeld parameter η = −Z/ka₀ supplying a logarithmic distortion no finite-range potential produces. Physically this is photoionisation. A 20.00 eV photon on ground-state hydrogen leaves E = −13.598 + 20.00 = +6.40 eV of kinetic energy, and because the eigenvalues are continuous, so is the absorption: the Lyman lines converge on 91.18 nm and are followed by a continuous edge, not by nothing.

Completeness, and what a finite box does to it

The resolution of the identity on a fixed (l, m) subspace is Σₙ |n l m⟩⟨n l m| + ∫₀^∞ dE |E l m⟩⟨E l m| = Î — bound sum plus continuum integral — and the continuum half is not a small correction. Change the nuclear charge suddenly, as the beta decay of tritium does, and the overlap of the old ground state with the new bound states does not reach one; the missing weight is real ionisation, called shake-off. A numerical diagonalisation on a grid of radius R never sees the integral. It returns discrete positive eigenvalues instead, spaced as the box modes kⱼ ≈ jπ/R allow. Those pseudo-states are not states of the atom, and the test is to double R: a true bound level moves by a fraction of a per cent, a pseudo-state moves a long way and multiplies in number. They still earn their place, though — as a quadrature rule for the continuum integral they restore the sum rule Σ|c|² = 1 that the bound states alone can never reach.

Change one variable at a time

Make the relationship visible.

Interactive model

Level highlighted, n2

Photon energy hν10.2 eV

Set hν to 10.2 eV and the arrow lands exactly on n = 2; nudge it past 13.60 eV and the tip crosses into the dashed continuum, where the surplus is kinetic and may take any value. Then run n from 1 to 12 and watch the levels pile up against E = 0 instead of spreading out.

LEVEL ENERGY Eₙ-3.400 eV

1→n LINE PHOTON10.199 eV

FINAL ENERGY E₁+hν-3.40 eV

FREED ELECTRON KE0.00 eV

Live interpretationLEVEL ENERGY Eₙ: −3.400 eV. 1→n LINE PHOTON: 10.199 eV. FINAL ENERGY E₁+hν: −3.40 eV. FREED ELECTRON KE: 0.00 eV

Catch the common trap

Explain before calculating.

A ground-state hydrogen atom (E₁ = −13.598 eV) absorbs a photon of energy 16.00 eV. What is the final state?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA He⁺ ion (Z = 2) sits in the n = 3 level. Taking the Rydberg energy as 13.606 eV and ignoring the reduced-mass correction, find the level energy, the photon energy needed to ionise from there, and the length scale 1/⟨1/r⟩. Then compare that scale with ⟨r⟩ for the circular state l = 2.

The hydrogenic eigenvalue is Eₙ = −E_R Z²/n², so the Z² in the numerator and the n² in the denominator do all the work: E₃ = −13.606 × 2²/3² eV.
13.606 × 4 = 54.424 eV, and 54.424/9 = 6.047 eV, so E₃ = −6.047 eV.
Ionising means reaching E = 0, the bottom of the continuum, not some higher line. The photon must supply the binding energy, 6.047 eV — and any larger photon also works, with the surplus appearing as kinetic energy.
The length scales the other way, and it comes from an expectation value: ⟨1/r⟩ = Z/(n²a₀) exactly, so 1/⟨1/r⟩ = n²a₀/Z = 9 × 0.052918 nm / 2 = 0.47626/2 = 0.2381 nm.
That is a scale, not an orbit, and not the mean radius: ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)] = (0.052918/4)(27 − 6) nm = 0.013230 × 21 = 0.2778 nm for l = 2. The two differ by 17 per cent, and only ⟨r⟩ depends on l.

AnswerE₃ = −6.047 eV, ionisation from n = 3 costs 6.047 eV, and 1/⟨1/r⟩ = 0.2381 nm — four times the binding energy and half the length of hydrogen's n = 3 level, exactly the Z² and 1/Z scalings — while the mean radius of the l = 2 state is larger, 0.2778 nm.

MediumBalmer-α (n = 3 → 2) lies at 656.46 nm in vacuum for hydrogen. Using mₚ/mₑ = 1836.15 and m_d/mₑ = 3670.48, find the wavelength of the same line in deuterium and the splitting between the two.

Every level carries one factor of μ, so every transition energy does, and λ = hc/ΔE is proportional to 1/μ. Only the ratio μ_H/μ_D is needed; the 656.46 nm anchor supplies everything else.
μ/mₑ = M/(M + mₑ). Hydrogen: 1836.15/1837.15 = 0.9994557. Deuterium: 3670.48/3671.48 = 0.9997276.
λ_D/λ_H = μ_H/μ_D = 0.9994557/0.9997276 = 0.9997280, a fractional shift of 2.720 × 10⁻⁴.
Δλ = 656.46 nm × 2.720 × 10⁻⁴ = 0.1786 nm, so λ_D = 656.46 − 0.179 = 656.28 nm. The deuterium line lies to the blue because the heavier nucleus gives a larger μ and therefore deeper levels.

Answerλ_D = 656.28 nm, 0.179 nm to the blue of hydrogen's 656.46 nm. A 5-parts-in-10⁴ mass correction shows up as a resolvable doublet — this is the measurement that identified deuterium.

HardA radial l = 0 diagonalisation for hydrogen uses a hard-wall box of radius R = 60a₀. It returns bound levels down to −13.60 eV, one level at −0.520 eV, and about fifteen positive eigenvalues below +8 eV. Say which levels you can trust, identify the −0.520 eV level, and explain what the positive eigenvalues are.

A level is resolved only if uₙ has died well inside the wall. For l = 0 the classical turning point sits at 2n²a₀ and uₙ only decays beyond it, so the box is comfortable while 2n² ≪ 60, i.e. n ≪ 5.5: trust n ≤ 4, whose turning points reach at most 32a₀, and treat n = 5, whose turning point is 50a₀ against a wall at 60a₀, as box-contaminated.
Test the suspect against the exact value: E₅ = −13.598/25 = −0.5439 eV. The returned −0.520 eV is 0.024 eV high, a 4.4% error, and the sign is right — a hard wall compresses the state and pushes every eigenvalue up.
Confirm by doubling the box. At R = 120a₀ a genuine bound eigenvalue shifts by a fraction of a per cent and settles onto −0.5439 eV; anything that moves a long way was never a physical state.
The positive eigenvalues are exactly that. The true spectrum above E = 0 is continuous with ⟨E|E′⟩ = δ(E − E′); the wall replaces it with box modes kⱼ ≈ jπ/R. Count them: at E = 8 eV, k = √(2 × 511 000 × 8) eV / 197.33 eV⋅nm = 2859 eV / 197.33 eV⋅nm = 14.5 nm⁻¹, and R = 60a₀ = 3.175 nm, so j ≈ kR/π ≈ 14.6 — about the fifteen the run returned.
They are still worth keeping. No single one is a state of the atom, but together they act as a quadrature rule for ∫dE |E⟩⟨E|, restoring the Σ|c|² = 1 that the bound states alone can never reach.

AnswerTrust n ≤ 4; −0.520 eV is a box-shifted n = 5, whose true value is −0.5439 eV; the fifteen positive eigenvalues are discretised continuum pseudo-states, exposed by doubling R and useful only as a quadrature for the continuum integral.