University Physics II · Introduction to Modern Physics · 15.1

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.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

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.

Simple definition: 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.

Rayleigh–Jeans lawu(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 lawu(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 meanEₙ = 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 ratiox = 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 lawλmax 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 lawM = 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

A 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.

Count 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.

One 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 is the next topic's business.

The 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.

Wien 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.

Bounded 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Cavity temperature T1500 K

Probe q, where λ = 10^q μm0.00

Drag the probe right to long wavelengths and the two curves merge into one line — x falls below 1, the steps stop showing, and the classical law is exact again. Drag it left and they part by whole decades. Raise T and the crossover slides left, carrying λmax = b/T with it; the classical line never turns over at all.

PROBE λ1.00 μm

x = hc/λkT9.592

PLANCK ÷ RAYLEIGH–JEANS0.0007

MEAN QUANTA ⟨n⟩ = 1/(eˣ − 1)0.000

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

Catch the common trap

Explain before calculating.

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.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 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.

Wien's law: λmax = b/T = 2.898 × 10⁻³ m K ÷ 2800 K = 1.035 × 10⁻⁶ m = 1.035 μm.
That peak sits in the near infrared, past the 700 nm red edge of vision — most of the filament's radiated power never becomes light at all.
At 550 nm: λT = 550 × 10⁻⁹ m × 2800 K = 1.54 × 10⁻³ m K, so x = 1.4388 × 10⁻² m K ÷ 1.54 × 10⁻³ m K = 9.34.
A quantum there costs nine kT, so Planck ÷ Rayleigh–Jeans = x/(eˣ − 1) = 9.34 ÷ (1.14 × 10⁴ − 1) = 8.2 × 10⁻⁴ — classical mode counting overstates the green output by a factor of 1.2 × 10³, so it cannot be used.

Answerλmax = 1.035 μm, in the near infrared; at 550 nm x = 9.34, and Rayleigh–Jeans runs high by a factor of 1.2 × 10³.

MediumA 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.

At 10 μm: λT = 1.00 × 10⁻⁵ m × 1500 K = 1.50 × 10⁻² m K, so x = 1.4388 × 10⁻² m K ÷ 1.50 × 10⁻² m K = 0.959.
eˣ − 1 = 2.610 − 1 = 1.610, so the ratio is 0.959 ÷ 1.610 = 0.596 and ⟨n⟩ = 1 ÷ 1.610 = 0.621 — roughly one quantum for every second mode, and the classical law already 68% high.
At 600 nm: λT = 6.00 × 10⁻⁷ m × 1500 K = 9.00 × 10⁻⁴ m K, so x = 1.4388 × 10⁻² m K ÷ 9.00 × 10⁻⁴ m K = 16.0.
eˣ − 1 = 8.77 × 10⁶, so the ratio is 16.0 ÷ 8.77 × 10⁶ = 1.8 × 10⁻⁶ and ⟨n⟩ = 1.1 × 10⁻⁷ — that mode is in its ground state essentially all the time.
Same cavity, same mode count 8πν²/c³, same T: the only thing separating a usable classical answer from one wrong by 5.5 × 10⁵ is the value of x.

Answer10 μm: x = 0.959, ratio 0.596, ⟨n⟩ = 0.621. 600 nm: x = 16.0, ratio 1.8 × 10⁻⁶, ⟨n⟩ = 1.1 × 10⁻⁷.

HardRadio 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.

Planck ÷ Rayleigh–Jeans = x/(eˣ − 1), so "within 1%" is the condition x/(eˣ − 1) ≥ 0.99.
For small x, eˣ − 1 = x(1 + x/2 + x²/6 + …), so x/(eˣ − 1) ≈ 1 − x/2 + x²/12; setting that to 0.99 gives x ≈ 0.0201, and checking it exactly, e^0.0201 − 1 = 0.020303 and 0.0201 ÷ 0.020303 = 0.990.
Turn the bound into a wavelength: λ = (hc/k)/(xT) = 1.4388 × 10⁻² m K ÷ (0.0201 × 2.725 K) = 0.263 m. Shorter than about 26 cm and the classical law is more than 1% high.
The 21 cm line: x = 1.4388 × 10⁻² m K ÷ (0.210 m × 2.725 K) = 0.025143, and eˣ − 1 = 0.025462, so x/(eˣ − 1) = 0.025143 ÷ 0.025462 = 0.9875 — Planck sits 1.25% below the classical value, just outside the 1% band.
So brightness temperatures at 21 cm carry a small Planck correction that grows as λ shortens. The classical law is not wrong here; it is bounded, and the bound is a number you can evaluate.

Answerλ ≳ 0.26 m (x ≲ 0.020) for 1% accuracy at 2.725 K; the 21 cm line sits at x = 0.0251 with a ratio of 0.9875, so Rayleigh–Jeans runs about 1.3% high there.