University Physics V · Foundations of Quantum Mechanics · 2.1

Four quantitative failures of the classical account

You cannot rebuild mechanics until you know what the old mechanics owes. This lesson collects the four debts with numbers attached — an atom dead in 16 ps, spectra that subtract where orbits would multiply, heat capacities that switch off on schedule, an entropy constant classical theory cannot name — and finds one h behind all four.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Around 1900 classical physics made four sharp, quantitative predictions about ordinary matter and lost all four by measurable margins. Maxwell plus Newton say a hydrogen electron radiates at the Larmor rate and spirals into the nucleus in 16 ps, chirping continuously on the way down — atoms should neither persist nor emit lines. The lines that do appear obey Ritz's rule: every wavenumber is a difference T(m) − T(n) drawn from one list of terms, where a periodic orbit would deliver a fundamental and its harmonics.

Equipartition fixes a diatomic C_V at 7/2 R at every temperature; hydrogen pays 5/2 R at 300 K and is back to 3/2 R below about 80 K, as if modes switched off below private thresholds. And classical statistical mechanics cannot state argon's absolute entropy at all: the phase-space integral leaves an arbitrary cell size that calorimetry pins at h³ per particle. Taken singly, each failure can be patched inside its own domain.

What no patchwork explains is that the fits share a part: the same 6.63×10⁻³⁴ J s sizes the stable atom, spaces the term ladder, sets every freeze-out threshold and cuts the entropy cell. Four unrelated laboratories returning one constant — that convergence, not any single failure, is the case for one new theory.

Simple definition: The classical failure inventory is the list of sharp wrong predictions — atomic collapse in picoseconds, harmonic spectra, temperature-blind heat capacities, an undefined entropy constant — that classical mechanics, electrodynamics and statistical mechanics make about ordinary matter.
Example: At the Bohr radius the classical electron radiates 4.7×10⁻⁸ W and spirals in within 1.6×10⁻¹¹ s; the age of the universe, 4.4×10¹⁷ s, exceeds that lifetime by a factor of nearly 3×10²⁸.

Larmor radiated powerP = e²a²/(6πε₀c³)

Prices acceleration in watts: a circling electron must radiate, so the planetary atom's collapse is a definite classical prediction — and a false one.

e = 1.60×10⁻¹⁹ C, a in m s⁻², ε₀ = 8.85×10⁻¹² F m⁻¹; at the Bohr radius P = 4.7×10⁻⁸ W

Classical infall timet = r₀³/(4rₑ²c), rₑ = e²/(4πε₀m_ec²)

Turns 'unstable' into 16 ps and about 10⁵ orbits. The r₀³ means no plausible starting radius rescues the atom.

r₀ = 5.29×10⁻¹¹ m, rₑ = 2.82×10⁻¹⁵ m, c = 3.00×10⁸ m s⁻¹ → t = 1.6×10⁻¹¹ s

Ritz combination principleν̃ = T(m) − T(n), T(n) = R_H/n², n > m

Differences of observed lines are themselves lines: subtract Lyman-α from Lyman-β and Balmer-α at 656.46 nm (vacuum) appears. Orbits would emit harmonics nf, which are absent.

wavenumbers ν̃ and term values T in m⁻¹; for hydrogen R_H = 1.0968×10⁷ m⁻¹

Equipartition and freeze-outC_V = (f/2)R classicallya mode counts only while k_BT ≳ ΔE, with ΔE = ℏω (vibration) or ℏ²/I (rotation)

Classical C_V is temperature-blind. The measured staircase 3/2 → 5/2 → 7/2 R assigns each mode a price set by ℏ — a per-mode energy scale classical mechanics does not own.

f = quadratic energy terms; for H₂, θᵣₒₜ = ℏ²/(2Ik_B) ≈ 85 K and θ_vib = ℏω/k_B ≈ 6300 K; R = 8.314 J K⁻¹ mol⁻¹

Sackur–Tetrode entropyS = Nk_B[ln(V/(Nλ³)) + 5/2], λ = h/√(2πmk_BT)

Absolute entropy exists only once phase space is cut into cells of h³ per particle — and calorimetry picks out the same h that radiation does.

λ = thermal de Broglie wavelength in m; argon at 87.3 K, 1 atm: λ = 29.6 pm, S = 129.2 J K⁻¹ mol⁻¹

The shared constantR_H = μe⁴/(8ε₀²h³c)θ_vib = ℏω/k_Bθᵣₒₜ = ℏ²/(2Ik_B)cell = h³

The inventory's verdict: unrelated experiments keep fitting the same constant, which no set of independent patches explains.

h = 6.626×10⁻³⁴ J s, ℏ = h/2π, μ = mₑₘₚ/(mₑ + mₚ) the reduced mass; one h in the term ladder, the freeze-outs and the entropy cell

The Larmor clock gives the planetary atom 16 picoseconds

Take Rutherford's atom at face value. An electron on a circular Coulomb orbit of radius r has centripetal acceleration a = e²/4πε₀mₑᵣ², and Larmor prices that acceleration at P = e²a²/6πε₀c³. At the Bohr radius the bill is 4.7×10⁻⁸ W against a stored binding energy of 2.2×10⁻¹⁸ J. The loss per orbit is tiny, so the orbit stays nearly circular while it shrinks: setting dE/dt = −P with E = −e²/8πε₀r gives dr/dt = −(4/3)rₑ²c/r², and integrating from r₀ = 5.29×10⁻¹¹ m to zero yields t = r₀³/4rₑ²c = 1.6×10⁻¹¹ s — about 10⁵ turns. The prediction fails twice over: atoms visibly last, and the radiation is wrong too, since the emitted frequency is the orbital one, 6.6×10¹⁵ Hz at r₀ and rising as r⁻³/², a continuous upward chirp where hydrogen actually emits sharp lines. Every input is a measured constant, so there is nothing to adjust — 16 ps is not a model parameter but a verdict.

Spectral lines combine as differences, never as harmonics

A classical bound charge is a periodic system, and periodic motion Fourier-decomposes into a fundamental f and integer harmonics 2f, 3f, … — so a classical atom should emit evenly spaced multiples of one frequency. Hydrogen does not. Rydberg and Ritz found that every observed wavenumber is the difference of two entries in a single list of term values, T(n) = R_H/n² with R_H = 1.0968×10⁷ m⁻¹. The rule is generative: subtract the Lyman-α wavenumber from Lyman-β and the result, 1.5233×10⁶ m⁻¹, is the observed Balmer-α line at 656.46 nm in vacuum (656.28 nm in air) — a prediction in a different series, made by arithmetic on known lines. Doubling a known line's frequency, by contrast, predicts lines that are never there. Data organised as one ledger of terms whose differences are the observables reads like a set of energy eigenvalues with transitions between them: the shape of the spectrum votes for states over orbits.

Heat capacities switch off on a schedule

Equipartition hands every quadratic term in the energy ½k_BT regardless of temperature, so H₂ — three translations, two rotations, one vibration counting kinetic and potential parts — should hold C_V = 7/2 R from freezing point to furnace. The measurement says 5/2 R at 300 K, falling to 3/2 R below about 80 K, with vibration only beginning to stir above roughly 1000 K. Solids repeat the pattern: Dulong–Petit's 3R fits lead at room temperature, while diamond sits at 6.1 J K⁻¹ mol⁻¹, a quarter of it. The empirical rule is that a mode contributes only while k_BT is at least comparable to its level spacing — ℏω for a vibration, ℏ²/I for a rotor — and the characteristic temperatures θᵣₒₜ = ℏ²/2Ik_B ≈ 85 K and θ_vib = ℏω/k_B ≈ 6300 K do the bookkeeping for hydrogen, whose rotational step is centred near 150 K rather than at θᵣₒₜ because ortho and para H₂ must each jump ΔJ = 2, gaps of 6θᵣₒₜ and 10θᵣₒₜ. Classical mechanics cannot reproduce any of this, because equipartition's derivation never consults ω or I at all — it owns no dial by which a stiff mode could cost more to excite than a soft one. Every measured freeze-out temperature is therefore a measurement of a level spacing over k_B, and the ℏ inside that spacing is Planck's.

Sackur–Tetrode: a gas in equilibrium measures h

Boltzmann's S = k_B ln W needs W to be a pure count, but the classical phase-space volume ∫d³x d³p per particle carries dimensions of (J s)³: it counts nothing until divided by a cell volume h₀³ (with an N! for indistinguishability). Classically h₀ is arbitrary, so absolute entropies — and with them vapour-pressure constants and chemical equilibria — are undefined; only entropy differences survive. Sackur and Tetrode closed the gap in 1912 by fitting h₀ to data. For argon vapour at its 87.3 K boiling point the formula S = Nk_B[ln(V/(Nλ³)) + 5/2], with thermal wavelength λ = h/√(2πmk_BT) = 29.6 pm, gives 129.2 J K⁻¹ mol⁻¹ when h₀ = h — matching the third-law value assembled from ∫Cₚ dT/T and latent heats. A cell of (2h)³ would fall 3R ln 2 = 17.3 J K⁻¹ mol⁻¹ short. Nothing in the experiment radiates or shows a spectrum, yet the equilibrium thermodynamics of a monatomic gas that neither emits nor absorbs a photon returns Planck's constant.

Four patches, or one constant

None of the four failures was unanswerable alone, and contemporaries patched each locally: perhaps Maxwell's equations fail inside atoms; perhaps Ritz combinations emerge from cunningly coupled oscillators; perhaps hidden constraints immobilise the frozen modes; perhaps the entropy cell is a fitted constant with no meaning. Each patch is cheap in its own domain, and no experiment inside that domain refutes it. The inventory's force is that the patches share no parts, but the fits do. The entropy cell is h³. The freeze-out thresholds are ℏω/k_B and ℏ²/2Ik_B. The term ladder is R_H/n² with R_H = μe⁴/8ε₀²h³c, built from the same h. Even the surviving atom votes: the only action that e, mₑ and a 10⁻¹⁰ m atomic radius can form is √(r mₑₑ²/4πε₀) ≈ 10⁻³⁴ J s, the size of ℏ, so whatever halts the Larmor spiral must carry Planck's constant too. Electrodynamic, spectroscopic, thermal and thermodynamic measurements all point at 6.63×10⁻³⁴ J s. Independent patches have no reason to agree on anything; one theory with one new constant must. The rest of this unit builds that theory, and the inventory fixes its acceptance test: discrete states whose differences are the lines, level spacings per mode set by ℏ, and h³ cells in phase space.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotation quantum θᵣₒₜ85 K

Vibration quantum θ_vib6300 K

Gas temperature T300 K

Keep θ_vib at hydrogen's 6300 K and walk T down from 10⁴ K: vibration is half frozen by about 2100 K and C_V settles on 5/2 R, then rotation fades below θᵣₒₜ ≈ 85 K, half gone near 30 K, leaving 3/2 R. Equipartition has no ℏω to compare k_BT with, so its dashed line cannot step at any temperature.

C_V AT T2.49 R

ROTATION SHARE0.99 R

VIBRATION SHARE0.00 R

GAP BELOW 7/2 R1.01 R

Live interpretationC_V AT T: 2.49 R. ROTATION SHARE: 0.99 R. VIBRATION SHARE: 0.00 R. GAP BELOW 7/2 R: 1.01 R

Catch the common trap

Explain before calculating.

Misconception check

The mistake is to point at one entry — usually the 16 ps collapse — and call it the proof of quantum mechanics. On its own it proves only that something stops the spiral, and the cheapest stopper is a postulate: Bohr's first postulate of 1913 simply declares certain orbits non-radiating, which is exactly the kind of local patch this inventory warns about. What promoted that postulate to a theory was not the collapse it cured but the debt it paid elsewhere — quantising action in units of h reproduced the Rydberg constant, a number from spectroscopy, using the h that Planck had already fitted to cavity radiation and that Sackur and Tetrode were fitting to argon's entropy. A patch earns the name theory only when it settles a failure in a different laboratory with a constant it did not get to choose. Judge every proposed fix by that test, not by whether it makes one anomaly go away.

Classical electrodynamics is applied without amendment to a hydrogen electron circling at the Bohr radius. What does the theory predict for the atom?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe classical hydrogen atom starts with its electron on a circular orbit at r₀ = 5.29×10⁻¹¹ m. Radiation shrinks the orbit at dr/dt = −(4/3)rₑ²c/r², with rₑ = 2.82×10⁻¹⁵ m and c = 3.00×10⁸ m s⁻¹. Find the collapse time, and estimate the number of orbits completed if the initial orbital frequency is 6.6×10¹⁵ Hz.

Separate variables: r² dr = −(4/3)rₑ²c dt. Integrating r from r₀ down to 0 gives r₀³/3 = (4/3)rₑ²c⋅t, so t = r₀³/(4rₑ²c) — and the r₀³ means halving the starting radius cuts the lifetime eightfold.
Numerator: r₀³ = (5.29×10⁻¹¹ m)³ = 1.48×10⁻³¹ m³.
Denominator: 4rₑ²c = 4 × (2.82×10⁻¹⁵ m)² × 3.00×10⁸ m s⁻¹ = 4 × 7.95×10⁻³⁰ × 3.00×10⁸ = 9.54×10⁻²¹ m³ s⁻¹.
t = 1.48×10⁻³¹ / 9.54×10⁻²¹ = 1.55×10⁻¹¹ s ≈ 16 ps (1.56×10⁻¹¹ s with CODATA inputs).
Orbits: N ≈ f₀t = 6.6×10¹⁵ Hz × 1.55×10⁻¹¹ s ≈ 1.0×10⁵ — an underestimate, since the frequency rises as r⁻³/² on the way down; integrating f dt with that weighting exactly doubles it, to 2×10⁵.

Answert = r₀³/(4rₑ²c) ≈ 1.6×10⁻¹¹ s — the classical atom survives about 16 ps and of order 10⁵ orbits.

MediumHydrogen's Lyman-α and Lyman-β lines are measured at vacuum wavelengths 121.567 nm and 102.572 nm. Use the Ritz combination principle to predict a third line from their difference, identify it, and state what a classical oscillator at the Lyman-α frequency would predict instead.

Convert to wavenumbers, the additive currency of the combination principle: ν̃_α = 1/(121.567 nm) = 8.22592×10⁶ m⁻¹ and ν̃_β = 1/(102.572 nm) = 9.74925×10⁶ m⁻¹.
Both are differences against the same ground term: ν̃_α = T(1) − T(2) and ν̃_β = T(1) − T(3), with T(n) = R_H/n². Subtracting cancels the shared T(1): ν̃_β − ν̃_α = T(2) − T(3).
ν̃_β − ν̃_α = 9.74925×10⁶ − 8.22592×10⁶ = 1.52333×10⁶ m⁻¹, so λ = 1/(1.52333×10⁶ m⁻¹) = 656.46 nm. As a check, T(2) − T(3) = 5R_H/36 = 1.5233×10⁶ m⁻¹ with R_H = 1.0968×10⁷ m⁻¹.
That is Balmer-α, the n = 3 → 2 red line, observed at 656.46 nm in vacuum (656.28 nm in air) — a line in a different series, predicted purely by subtracting two known lines.
A classical periodic charge would instead radiate harmonics of one fundamental: 2ν̃_α = 1.64518×10⁷ m⁻¹, i.e. 60.78 nm — below the 91.2 nm Lyman limit 1/R_H, where hydrogen shows no line at all.

AnswerThe difference predicts 656.46 nm — Balmer-α, exactly where it is observed in vacuum. The classical harmonic at 60.8 nm does not exist.

HardArgon vapour at its normal boiling point, T = 87.3 K and p = 1.013×10⁵ Pa, has a measured third-law molar entropy close to 129 J K⁻¹ mol⁻¹. Evaluate the Sackur–Tetrode entropy using a phase-space cell of h³ (m_Ar = 6.63×10⁻²⁶ kg), and find how far the prediction moves if the cell were (2h)³ instead.

Thermal wavelength: 2πm k_BT = 2π × 6.63×10⁻²⁶ × 1.381×10⁻²³ × 87.3 = 5.02×10⁻⁴⁶ kg² m² s⁻²; its square root is 2.24×10⁻²³ kg m s⁻¹, so λ = h/√(2πmk_BT) = 6.626×10⁻³⁴/2.24×10⁻²³ = 2.96×10⁻¹¹ m.
Volume per atom from the ideal-gas law: V/N = k_BT/p = 1.381×10⁻²³ × 87.3/1.013×10⁵ = 1.19×10⁻²⁶ m³.
Occupation ratio: λ³ = 2.59×10⁻³² m³, so V/(Nλ³) = 1.19×10⁻²⁶/2.59×10⁻³² = 4.60×10⁵ — about half a million cells per atom, which is why the gas moves classically yet still counts cells.
S = R[ln(4.60×10⁵) + 5/2] = 8.314 × (13.04 + 2.50) = 8.314 × 15.54 = 129.2 J K⁻¹ mol⁻¹, matching the calorimetric value built from ∫Cₚ dT/T plus latent heats.
A cell of (2h)³ is 8 times larger, dividing the count per atom by 8: ΔS = −R ln 8 = −3R ln 2 = −17.3 J K⁻¹ mol⁻¹, giving 111.9 — a 13% miss, far outside calorimetric uncertainty. The gas measures h.

AnswerS(h³) = 129.2 J K⁻¹ mol⁻¹, in agreement with experiment; the (2h)³ cell predicts 111.9, wrong by 17.3 J K⁻¹ mol⁻¹.