Blackbody Radiation &
the Planck Hypothesis
Change one rule. Recover the spectrum.
Count the cavity modes, see why classical equipartition fails, and derive the energy rule that opened the quantum world.
Same physics. A deeper understanding.
Build the model — one change, a new prediction
The geometry survives. The classical allocation of energy does not.
1Cavity radiation
A small aperture samples an equilibrium radiation field. The ideal spectrum depends on temperature, not the wall material.
2Count the modes
Standing electromagnetic waves give the number of modes per volume and frequency.
3Classical payment
Equipartition assigns a mean thermal energy kBT to every mode.
The high-frequency integral diverges: the ultraviolet catastrophe.
See the failure →4Planck’s energy steps
Use the ladder En = nhν and sum its normalised Boltzmann probabilities.
High-frequency modes are suppressed. The total thermal energy is finite.
Derive the average →A hole in a hot box
An almost perfect absorber is also an almost perfect emitter.
- Repeated reflections make absorption likely.
- Walls and radiation share an equilibrium temperature.
- A small hole disturbs the internal field only weakly.
From energy in the cavity to emitted power
If cavities at equal temperature had different equilibrium spectra, frequency-selective exchange would contradict thermal equilibrium. That is why the ideal spectrum is universal.
For isotropic radiation, the hemisphere average of the normal velocity gives Mν = cuν/4. Spectral radiance instead satisfies Bν = cuν/(4π), and Mν = πBν. Do not confuse radiance per steradian with hemispherical exitance.
Real surfaces can have frequency-dependent emissivity. A well-designed equilibrium cavity approximates a blackbody; an arbitrary glowing object need not.
Counting the standing waves
A large-cavity result, with two transverse polarisations.
g has units m⁻³ Hz⁻¹ = s m⁻³. At ν = 10¹⁴ Hz, g ≈ 9.33 × 10³ m⁻³ Hz⁻¹.
Derive the count and check the finite-box limit
For a cube of side L, k components are integer multiples of π/L and ν = c|k|/(2π). In integer-index space the sphere has radius R = 2Lν/c. Count one octant, multiply by two polarisations, then differentiate: N ≈ 2 × (1/8) × (4πR³/3), and g = (1/V)dN/dν.
Boundary-index corrections are negligible only when many modes are available. A 1 cm conducting cube has its lowest electromagnetic mode near 21.2 GHz and five modes below 30 GHz; the continuum formula predicts about 8.4. Do not treat g as an exact count for a tiny cavity at its lowest frequencies.
A 1 litre cavity contains approximately 2.34 × 10¹⁴ modes between 500 and 501 THz. The count is large; the correct mean energy per mode is the crucial second ingredient.
Why classical physics fails
Rayleigh–Jeans is a low-frequency approximation, not a law for every frequency.
Two quadratic terms → kBT per mode
The electric and magnetic contributions together give kBT by equipartition. Multiply by g(ν) to obtain uν,RJ.
The result grows without bound as the cutoff rises. A frequency cutoff chosen only to remove the infinity does not explain the observed shape.
The failure is at high ν / short λ. “Ultraviolet catastrophe” refers to this unbounded classical prediction, not a physical catastrophe in the apparatus.
A numerical scale check
At 1500 K, integrating Rayleigh–Jeans to 10¹⁵ Hz gives about 6.44 J m⁻³. The entire Planck spectrum gives about 0.00383 J m⁻³, a factor of roughly 1680 smaller. These are calculated predictions, not a new experimental dataset.
At 2 THz and 1500 K, x = hν/(kBT) ≈ 0.064, so Rayleigh–Jeans exceeds Planck by about 3.3%. It is useful when the energy step is small relative to the thermal scale.
Planck’s ladder
A geometric series replaces the classical continuous-energy average.
Normalise first, then average
Z = Σqⁿ = 1/(1 − q), while Σnqⁿ = q/(1 − q)². Divide the weighted energy sum by Z. Omitting this normalisation gives a wrong mean energy.
At hν = kBT: P₀ ≈ 0.632, P₁ ≈ 0.233, P₂ ≈ 0.0855 and ⟨E⟩ ≈ 0.582kBT. At hν = 5kBT, P₀ ≈ 0.993 and ⟨E⟩ ≈ 0.0339kBT.
The equally spaced ladder is a modern teaching derivation of Planck’s distribution. A quantum oscillator also has a zero-point offset ½hν; the expressions here concern thermal energy above that offset. Temperature-independent zero-point energy is not part of the emitted thermal spectrum.
Wien scaling uν = ν³f(ν/T) is consistent with steps proportional to frequency. A constant energy step independent of ν would not reproduce that scaling.
Two limits, two peaks, one finite power
Use x = hν/(kBT) to choose the approximation.
Low frequency · x ≪ 1
Leading term: Rayleigh–Jeans. The same limit follows as trial h → 0 at fixed ν and T.
High frequency · x ≫ 1
Wien’s exponential tail suppresses the high-frequency energy.
Peak positions are density-dependent
For Bν: hνpeak = 2.82144kBT.
For Bλ: λpeakT = 2.89777 × 10⁻³ m K.
At 5800 K: νpeak ≈ 341 THz and λpeak ≈ 500 nm, but c/νpeak ≈ 879 nm. These are not contradictory.
Integrated power · Stefan–Boltzmann
σ = 2π5kB415c2h3
σ ≈ 5.670374419 × 10⁻⁸ W m⁻² K⁻⁴. M is hemispherical exitance, not spectral radiance.
Peak and area derivations
Differentiate x³/(eˣ − 1): the nonzero root of 3(1 − e⁻ˣ) = x is 2.82144. For the wavelength spectrum, the numerator is x⁵ and 5(1 − e⁻ˣ) = x gives 4.96511.
For the area, substitute x = hν/(kBT) into M = π∫Bνdν. The dimensionless integral ∫₀∞x³/(eˣ − 1)dx = π⁴/15 leaves the T⁴ law and the stated σ.
Fitting h and k from a spectrum
Shape and total power constrain different combinations.
What the observations can identify
- The calibrated peak position and temperature determine h/kB.
- Total exitance gives kB⁴/h³ through σ.
- Together they can determine both parameters in the historical formulation.
- The low-frequency radiance loses its h dependence; use the turnover and tail for sensitivity to h.
Modern SI values are exact
kB = 1.380649 × 10⁻²³ J K⁻¹
Today these are defining constants. The lab’s trial-h control is a model-fitting exercise, not evidence that h varies.
Real fitting requires measured spectral radiance, calibration, temperature uncertainty and an emissivity model. The reference below is calculated.
Interactive spectrum lab
Explore temperature, energy-step size and the choice of spectral variable.
The reference uses the exact SI value of h. Its dots are computed, not measurements. Changing trial h explores model sensitivity; it does not change a physical constant.
Classical vs quantum comparison
Same cavity geometry. Different thermal energy per mode.
| Feature | Rayleigh–Jeans | Planck |
|---|---|---|
| Mode density | 8πν²/c³ | 8πν²/c³ (unchanged) |
| Mean thermal energy | kBT | hν / (exp(hν/kBT) − 1) |
| Spectral radiance Bν | 2ν²kBT/c² | (2hν³/c²) / (exp(hν/kBT) − 1) |
| Integrated thermal energy | Diverges without a cutoff | Finite; energy density ∝ T⁴ |
| Scope | hν ≪ kBT | Equilibrium ideal blackbody spectrum |
Catch the common trap
Planck’s 1900 proposal is not Einstein’s light-quantum hypothesis.
Planck introduced energy elements in modelling resonators and their exchange with radiation. Einstein’s 1905 proposal attributed energy quanta to light itself. Do not collapse these distinct historical steps into “Planck proposed photons”.
The ladder derivation above is the standard modern statistical treatment. The blackbody spectrum alone does not reproduce all the separate evidence for photon behaviour.
Practice & worked examples
The lesson’s original three problems, with full solutions.
Easy · Mode probabilities
A cavity is held at 600 K. One of its modes has frequency 25.0 THz (λ = 12.0 μm). Using Planck's ladder Eₙ = nhν with Boltzmann weights, find the probability that the mode sits in each of its three lowest levels, check that the probabilities are normalised, and give the mean energy as a fraction of kT and in joules.
View solution →
- kT = (1.3806 × 10⁻²³ J K⁻¹)(600 K) = 8.284 × 10⁻²¹ J and hν = (6.626 × 10⁻³⁴ J s)(2.50 × 10¹³ Hz) = 1.657 × 10⁻²⁰ J, so hν/kT = 2.00: one rung of the ladder costs twice the thermal energy scale.
- The ratio of successive weights is q = e(−hν/kT) = e(−2.00) = 0.1353, and Pₙ = (1 − q)qⁿ. So P₀ = 0.865, P₁ = 0.865 × 0.1353 = 0.117 and P₂ = 0.117 × 0.1353 = 0.0158.
- Normalisation: the three lowest levels hold 0.9975 of the probability, and all the levels from n = 3 upward share q³ = 0.0025. The total is (1 − q)/(1 − q) = 1 exactly, as a probability distribution must be.
- Mean energy: ⟨E⟩/kT = (hν/kT)/(e(hν/kT) − 1) = 2.00/(7.389 − 1) = 0.313, so ⟨E⟩ = 0.313 × 8.284 × 10⁻²¹ J = 2.59 × 10⁻²¹ J. The mean number of quanta is 1/(e² − 1) = 0.157.
- Checks: ⟨E⟩ carries the units of hν, joules. It lies between the two limits, kT for hν ≪ kT and hν e(−hν/kT) = 0.271 kT for hν ≫ kT, and nearer the second, because at hν = 2kT the mode already spends 86% of its time in the ground level.
P₀ = 0.865, P₁ = 0.117, P₂ = 0.0158, with 0.0025 in all higher levels together, summing to 1; ⟨E⟩ = 0.313 kT = 2.59 × 10⁻²¹ J, under a third of the classical kT.
Medium · Energy in a frequency band
A 1.5 L cavity is held at 900 K. For the narrow band from 44.75 THz to 45.25 THz (λ ≈ 6.7 μm), find how many modes it contains, the energy those modes hold according to Rayleigh–Jeans and according to Planck, and the power in that band escaping through a 1.0 mm² hole in the wall.
View solution →
- Mode density at the band centre, ν = 4.50 × 10¹³ Hz: g = 8πν²/c³ = 8π(4.50 × 10¹³ Hz)² ÷ (2.998 × 10⁸ m s⁻¹)³ = 1.889 × 10³ s m⁻³.
- Number of modes: N = gVΔν = (1.889 × 10³ s m⁻³)(1.5 × 10⁻³ m³)(5.0 × 10¹¹ Hz) = 1.417 × 10¹² modes. The band is only 1.1% wide, so evaluating g at its centre is accurate to about 1 part in 10⁵.
- Classical payment: kT = (1.381 × 10⁻²³ J K⁻¹)(900 K) = 1.243 × 10⁻²⁰ J per mode, so ERJ = NkT = 1.76 × 10⁻⁸ J.
- Planck payment: hν = 2.982 × 10⁻²⁰ J, so hν/kT = 2.399 and ⟨E⟩ = hν/(e(2.399) − 1) = 2.982 × 10⁻²⁰ J ÷ 10.01 = 2.98 × 10⁻²¹ J = 0.240 kT. Then EP = N⟨E⟩ = 4.22 × 10⁻⁹ J.
- Ratio: ERJ/EP = (e(2.399) − 1)/2.399 = 4.17. It depends on hν/kT alone, not on V or Δν, which confirms that the two theories differ only in the energy per mode, never in the count.
- Power through the hole: the band's energy density is EP/V = 2.81 × 10⁻⁶ J m⁻³, and a hole of area A passes (c/4)(EP/V)A = (7.49 × 10⁷ m s⁻¹)(2.81 × 10⁻⁶ J m⁻³)(1.0 × 10⁻⁶ m²) = 2.1 × 10⁻⁴ W. Rayleigh–Jeans would promise 8.8 × 10⁻⁴ W.
1.42 × 10¹² modes; Rayleigh–Jeans gives 1.76 × 10⁻⁸ J and Planck 4.22 × 10⁻⁹ J, a factor of 4.17 apart; about 0.21 mW leaves through the hole in this band.
Hard · Extracting h from a ratio
A cavity at 1500 K, its temperature known from a calibrated thermocouple, is viewed by a spectrometer whose relative spectral response is calibrated but whose absolute sensitivity is not. It finds the spectral radiance at 120 THz to be 1.023 ± 0.005 times that at 60.0 THz. Taking k = 1.380649 × 10⁻²³ J K⁻¹, extract h with its uncertainty, and show what the Rayleigh–Jeans and Wien forms make of the same measurement.
View solution →
- A ratio is the right observable: the unknown absolute sensitivity, frequency-independent throughput and the solid angle multiply both readings and cancel. What remains is r = Bν(2ν₁)/Bν(ν₁) with ν₁ = 6.00 × 10¹³ Hz.
- Write y = hν₁/kT. Planck's law gives r = 8(ey − 1)/(e(2y) − 1), and because e(2y) − 1 = (ey − 1)(ey + 1), the ratio collapses exactly to r = 8/(ey + 1). No approximation has been made.
- Invert: ey = 8/r − 1 = 8/1.023 − 1 = 6.820, so y = ln 6.820 = 1.920. With kT = (1.380649 × 10⁻²³ J K⁻¹)(1500 K) = 2.0710 × 10⁻²⁰ J, h = ykT/ν₁ = 1.920 × 2.0710 × 10⁻²⁰ J ÷ (6.00 × 10¹³ Hz) = 6.627 × 10⁻³⁴ J s.
- Uncertainty: dy/dr = −8/[r(8 − r)] = −1.121, so the uncertainty in y is 1.121 × 0.005 = 0.0056, a relative 0.0056/1.920 = 0.29% that h inherits: ±0.019 × 10⁻³⁴ J s. A 0.5% error in the ratio becomes 0.3% in h.
- Rayleigh–Jeans, with Bν ∝ ν², predicts r = 4 for every h: it cannot measure h at all, and the observed 1.023 rules it out. Wien's form, Bν ∝ ν³e(−hν/kT), gives r = 8e(−y), so y = ln(8/1.023) = 2.057 and h = 7.10 × 10⁻³⁴ J s, 7% too high, because at y ≈ 2 the −1 in Planck's denominator is not negligible.
h = (6.63 ± 0.02) × 10⁻³⁴ J s, consistent with 6.626 × 10⁻³⁴ J s; the Wien form would return 7.10 × 10⁻³⁴ J s, and Rayleigh–Jeans predicts a ratio of 4 whatever h is.
Illustrative problem inputs, not a newly collected dataset. The uncertainty shown propagates the ratio uncertainty only; uncertainty in T or relative spectral calibration would also contribute in a real experiment.
Quick check
Five checks, including the original ladder problem · this visit only
Summary & next steps
Explain the count, the energy rule and the evidence separately.
- Large-cavity mode density is g(ν) = 8πν²/c³, including both polarisations.
- Equipartition gives a divergent Rayleigh–Jeans spectrum.
- Normalised Boltzmann weights on nhν give Planck’s mean thermal energy.
- Keep Bν, Bλ, uν and integrated exitance M distinct, with their units.
- The photoelectric effect is the next step in examining the light-quantum idea.
Prerequisites, sources and modelling notes
Prerequisites: waves and standing-wave boundary conditions, basic thermal physics, exponentials, geometric series, differentiation and integration. This is university-level teaching; it is not a claim of complete alignment to every school course.
All diagrams are schematics or calculated models. The curve markers are not laboratory observations. The cavity is assumed large compared with relevant wavelengths, the field is at thermal equilibrium, and the aperture samples isotropic radiation.
- NIST · Defining constants of the SI
- University of Texas · Planck radiation law
- University of Texas · Cavity energy and emitted power
- University of Texas · Stefan–Boltzmann derivation
- Nobel archives · Planck, Einstein and light quanta
Lesson calculations and references checked 29 September 2026. The actual Quantum Mechanics I course sequence is retained.