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Optional unit · institution-dependent · Interactive lesson

Blackbody Radiation & Planck's Hypothesis

Count cavity modes classically and the energy runs away at short wavelengths. Let each mode take energy only in whole steps of hν and the measured curve falls out, along with Wien's law, Stefan's law, and a boundary for classical physics.

UPII-15.01 · Lesson 1 of 8 in this unit. Position is not a completion score.
Starting model: Blackbody Radiation & Planck's HypothesisSpectral energy density against wavelength, both axes logarithmic, for a cavity at 1500 K. The dashed Rayleigh–Jeans line 8πkT/λ⁴ climbs without limit towards short wavelengths; the solid Planck curve tracks it at long wavelengths, turns over at λmax = 1.93 μm and then collapses. At the probe wavelength 1.00 μm the ratio x = hc/λkT is 9.59, so Planck ÷ Rayleigh–Jeans = x/(eˣ − 1) = 0.0007 and the mean number of quanta in that mode is 0.000.u(λ, T) log scalePlanck solid · Rayleigh–Jeans dashedλ = 1.00 μm · x = hc/λkT = 9.59Rayleigh–Jeans high by 3.2 decadesλmax = 1.93 μm0.1 μmwavelength λ1 μm10 μm
Start with a picture. Change the model in the lab below.

Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.

1

Build the model

The steps are hν. The ratio hν/kT decides whether you can see them.

The key idea

A blackbody is an ideal absorber whose thermal radiation spectrum depends only on temperature, and Planck's hypothesis is that a radiation mode of frequency ν exchanges energy only in whole multiples of hν.

Example. An electric hob at 1000 K glows dull red and peaks at 2.9 μm, deep in the infrared. Heat the same metal to 3000 K and the peak moves to 0.97 μm and it looks white. Temperature alone moved the curve.

The complete physical model and assumptions

A cavity in thermal equilibrium emits a spectrum fixed by its temperature alone — not by what the walls are made of — which makes it a clean test of any theory of matter and radiation. Classical physics fails that test in one specific way. Count the standing-wave modes in the cavity, give each the kT that equipartition demands, and you get u(λ) = 8πkT/λ⁴: correct at long wavelengths, unbounded at short ones, with infinite total energy. Planck's repair was a single restriction — a mode of frequency ν exchanges energy only in whole steps hν — and its consequence is a Boltzmann factor that starves any mode whose step costs more than kT. Everything quantitative then lives in one dimensionless ratio, x = hν/kT. Where x ≪ 1 the steps are invisible and the classical result comes back exactly; where x ≫ 1 the mode is frozen out. Classical physics was not deleted here. It was given a boundary, and told where it is allowed to be right.

1A spectrum that depends on nothing but temperature

Cut a small hole in a closed cavity whose walls sit at temperature T, and the radiation leaking out has a spectrum fixed by T alone. Kirchhoff established this in 1860 from thermodynamics: at every wavelength a good absorber must be an equally good emitter, or you could drive heat from a cold body to a hot one. So the curve carries no fingerprint of iron, or soot, or ceramic. That universality is exactly what makes it a hard test — there is nothing left to tune. By the 1890s the curve was measured well enough to argue over: intensity falling to zero at short wavelengths, one peak that moves left as T rises, a long tail, and a total emitted flux going as T⁴. A hob at 1000 K peaks at 2.9 μm; the Sun's photosphere at 5772 K peaks at 502 nm. Wien's displacement law and Stefan's fourth power were both fitted to data before anyone could derive the shape lying between them.

2Count the modes, hand each one kT, watch it diverge

The classical calculation has two honest steps. First, count the standing electromagnetic waves that fit inside the cavity: between ν and ν + dν there are 8πν²/c³ per unit volume, the 8π already carrying both polarizations. Second, apply equipartition. Each mode is a harmonic oscillator with two quadratic degrees of freedom, so its average energy is kT, the same for every frequency. Multiply: u(ν) = 8πν²kT/c³, or in wavelength u(λ) = 8πkT/λ⁴. At long wavelengths that matches the measurements. Everywhere else it is a disaster. The mode count keeps climbing as ν², so the predicted density has no peak at all, and ∫u dν diverges — a cavity at any temperature would hold infinite energy and drain itself into the ultraviolet. Ehrenfest named this the ultraviolet catastrophe in 1911. There is no arithmetic slip to blame: the mode count is geometry, and equipartition is standard statistical mechanics.

3One restriction: energy in whole steps of hν

Planck's move was to deny each oscillator a continuum of energies. It may hold only Eₙ = nhν with n = 0, 1, 2, …, where h is a new constant carrying units of J s. Redo the average with Boltzmann weights exp(−nhν/kT). Both sums are geometric, and writing x = hν/kT they collapse to ⟨E⟩ = hν/(eˣ − 1). Read the two limits. When x ≪ 1 the rungs sit far closer together than kT, the staircase looks smooth, and ⟨E⟩ → kT: equipartition restored, not contradicted. When x ≫ 1 even the first rung costs more than a thermal kick usually delivers, e⁻ˣ buries the mode in its ground state, and ⟨E⟩ collapses. Multiply the mode count by ⟨E⟩ rather than kT and you have u(λ, T) = (8πhc/λ⁵)/(exp(hc/λkT) − 1). Note carefully what was quantized: the energy exchanged by the cavity's material oscillators, not the electromagnetic field itself. Treating the quantum as a particle of light was Einstein's step in 1905, and it waits for the photoelectric effect later in this unit.

4The ratio hν/kT does all the deciding

Everything quantitative follows from x = hν/kT = hc/(λkT), and hc/k = 1.4388 × 10⁻² m K makes it a one-line calculation: x = (1.4388 × 10⁻² m K)/(λT). Planck's prediction divided by the classical one is exactly x/(eˣ − 1). Three cases. Microwaves in a 300 K room at λ = 1.0 mm: x = 0.048, ratio 0.976, so the classical law is 2.4% high and entirely usable. Visible light at the Sun's surface, λ = 500 nm and T = 5800 K: x = 4.96, ratio 0.035, and the mean occupancy ⟨n⟩ = 1/(eˣ − 1) is 0.0071, so such a mode is empty 99.3% of the time even there. Visible light from a 300 K wall: x = 95.9, ratio 2 × 10⁻⁴⁰. That last number is why a room-temperature object is not merely dim to the eye but utterly dark, and why the classical prediction of a fierce ultraviolet glow is not a small correction to be patched.

5Wien and Stefan–Boltzmann drop out, with h attached

Two laws that had been fitted to data now become consequences. Differentiate u(λ, T) with respect to λ and set the result to zero: the condition reduces to x = 5(1 − e⁻ˣ), whose non-zero root is 4.9651, so λmax T = hc/(4.9651k) = 2.898 × 10⁻³ m K. Wien's constant stops being measured and becomes h, c and k. Integrate u over all wavelengths and the divergence is gone: the total is finite and proportional to T⁴, with emitted flux M = σT⁴ and σ = 2π⁵k⁴/(15h³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴. One new constant, fitted once, delivers both. Planck's own 1901 fit to the measured curves returned h = 6.55 × 10⁻³⁴ J s — 1.1% below the value that now defines the kilogram — and k = 1.346 × 10⁻²³ J K⁻¹, 2.5% low; dividing the gas constant by that k handed him Avogadro's number to within 3%, out of a radiation spectrum.

6Bounded domains, not deleted laws

The classical result was never erased. It is the x → 0 limit of the correct one, and radio astronomers still use it: the cosmic microwave background is a 2.725 K blackbody peaking at 1.06 mm, but at the 21 cm line x = 0.025, the Rayleigh–Jeans form is within 1.3%, and that is why microwave sky maps are quoted as brightness temperatures at all. This is the pattern the rest of the unit repeats. Newtonian mechanics is the v/c → 0 limit of relativity; ray optics is the λ → 0 limit of wave optics; classical trajectories are the ħ → 0 limit of quantum mechanics. A theory that survived real measurement keeps its domain, and any replacement is obliged to reproduce it there — the correspondence requirement, which is a constraint on the new theory rather than a concession to the old one. The working question is never whether classical physics is true. Identify the characteristic ratio, evaluate it, then choose the model.

2

Key equations

Read the units, assumptions and limits before substituting.

Rayleigh–Jeans law

u(λ, T) = 8πkT/λ⁴

Mode count times kT per mode. Integrating it over all λ diverges — the ultraviolet catastrophe.

Spectral energy density in J m⁻³ per metre of wavelength; k = 1.381 × 10⁻²³ J K⁻¹

Planck radiation law

u(λ, T) = (8πhc/λ⁵)/(exp(hc/λkT) − 1)

The Boltzmann factor shuts down every mode whose quantum hc/λ costs more than kT.

h = 6.626 × 10⁻³⁴ J s; per unit frequency it reads (8πhν³/c³)/(exp(hν/kT) − 1)

Quantized energy and its mean

Eₙ = n h ν, n = 0, 1, 2, … · ⟨E⟩ = hν/(exp(hν/kT) − 1)

A Boltzmann-weighted average over the allowed rungs; classical kT is its closely spaced limit.

⟨E⟩ → kT when hν ≪ kT · mean quanta per mode ⟨n⟩ = 1/(exp(hν/kT) − 1)

Quantum-to-thermal ratio

x = hν/kT = hc/(λkT) = (1.4388 × 10⁻² m K)/(λT)

One number sets the regime: Planck divided by Rayleigh–Jeans is exactly x/(eˣ − 1).

Dimensionless · kT = 25.9 meV at 300 K, 0.50 eV at 5800 K

Wien displacement law

λmax T = b = 2.898 × 10⁻³ m K

The Sun at 5772 K peaks at 502 nm; skin at 306 K peaks at 9.5 μm.

b = hc/(4.9651 k), where 4.9651 solves x = 5(1 − e⁻ˣ)

Stefan–Boltzmann law

M = σT⁴ · σ = 2π⁵k⁴/(15h³c²)

Integrate Planck's curve over all λ: finite, fourth power, and σ built from h, c and k.

σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴ · M is emitted power per unit area, W m⁻²
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1500 K
300 to 6000 K; steps of 100.
0.00
−1 to 1; steps of 0.05.

Try this. Drag the probe right and the two curves close up without quite meeting: at 1500 K the 10 μm end still has x = 0.96 and Planck ÷ Rayleigh–Jeans = 0.60, and only near 6000 K does x fall to 0.24 with the ratio at 0.88. Drag it left and they part by whole decades — about 40 of them at 0.1 μm and 1500 K. Raise T and the crossover slides left, carrying λmax = b/T with it; the classical line never turns over at all.

Live results

PROBE λ
1.00 μm
x = hc/λkT
9.592
PLANCK ÷ RAYLEIGH–JEANS
0.0007
MEAN QUANTA ⟨n⟩ = 1/(eˣ − 1)
0.000

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics II modelSpectral energy density against wavelength, both axes logarithmic, for a cavity at 1500 K. The dashed Rayleigh–Jeans line 8πkT/λ⁴ climbs without limit towards short wavelengths; the solid Planck curve tracks it at long wavelengths, turns over at λmax = 1.93 μm and then collapses. At the probe wavelength 1.00 μm the ratio x = hc/λkT is 9.59, so Planck ÷ Rayleigh–Jeans = x/(eˣ − 1) = 0.0007 and the mean number of quanta in that mode is 0.000.u(λ, T) log scalePlanck solid · Rayleigh–Jeans dashedλ = 1.00 μm · x = hc/λkT = 9.59Rayleigh–Jeans high by 3.2 decadesλmax = 1.93 μm0.1 μmwavelength λ1 μm10 μm
Spectral energy density against wavelength, both axes logarithmic, for a cavity at 1500 K. The dashed Rayleigh–Jeans line 8πkT/λ⁴ climbs without limit towards short wavelengths; the solid Planck curve tracks it at long wavelengths, turns over at λmax = 1.93 μm and then collapses. At the probe wavelength 1.00 μm the ratio x = hc/λkT is 9.59, so Planck ÷ Rayleigh–Jeans = x/(eˣ − 1) = 0.0007 and the mean number of quanta in that mode is 0.000.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Cavity temperature T: 1500 K; Probe q, where λ = 10q μm: 0.00

Baseline 1500; choose 300 to 6000, in steps of 100.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

"The peak of the spectrum" is not a property of the light alone — it depends on what you plot it against. For T = 5772 K, Wien's law gives λmax = 502 nm, while the per-unit-frequency curve peaks at νmax = 2.8214kT/h = 3.39 × 10¹⁴ Hz, whose wavelength is 883 nm. Both describe the same Sun. The fixed ratio 4.9651/2.8214 = 1.76 comes from converting a density per dλ into one per dν with |dν/dλ| = c/λ². Never feed νmax into λ = c/ν and call the result Wien's law.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A tungsten lamp filament runs at 2800 K. Find the wavelength its spectrum peaks at, then evaluate x = hc/(λkT) at 550 nm, in the middle of the visible band, and say whether classical mode counting can be trusted there. Take b = 2.898 × 10⁻³ m K and hc/k = 1.4388 × 10⁻² m K.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A furnace cavity is held at 1500 K. At λ = 10 μm and at λ = 600 nm, evaluate x, the ratio x/(eˣ − 1) of Planck to Rayleigh–Jeans, and the mean occupancy ⟨n⟩ = 1/(eˣ − 1). Use hc/k = 1.4388 × 10⁻² m K.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Radio astronomers still quote cosmic microwave background maps as Rayleigh–Jeans brightness temperatures. For the 2.725 K background, find the shortest wavelength at which the classical law is within 1% of Planck, then test the 21 cm line against that boundary. Use hc/k = 1.4388 × 10⁻² m K.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A cavity is held at 300 K. Using x = hc/(λkT), with hc/k = 1.4388 × 10⁻² m K, compare the Rayleigh–Jeans and Planck spectral energy densities at λ = 1.0 mm and at λ = 500 nm.
7

Review & continue

A blackbody is an ideal absorber whose thermal radiation spectrum depends only on temperature, and Planck's hypothesis is that a radiation mode of frequency ν exchanges energy only in whole multiples of hν.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.