Skip to main content
GIOPHYSICS
Menu
← University Physics IV

University Physics IV · Limits of Classical Physics · 1.4

Planck's quantised oscillator and the radiation law

This is the calculation that ends classical physics, and it is shorter than its reputation. Keep the classical mode count, restrict the wall oscillator to whole multiples of hν, and average with a sum instead of an integral. One denominator replaces kT — and everything measurable follows from it.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Planck changed exactly one thing and left the rest of the classical calculation standing. The cavity mode count 8πν²/c³ is wave geometry and survives untouched; what he replaced was the energy each mode is allowed to hold. Let a wall oscillator of frequency ν take only Eₙ = n hν, and the Boltzmann average becomes a geometric sum rather than an integral, giving ⟨E⟩ = hν/(e(hν/kT) − 1) in place of equipartition's flat kT.

Every consequence sits in that one denominator: for hν ≪ kT it expands to kT(1 − hν/2kT) and the Rayleigh–Jeans law returns; for hν ≫ kT it becomes hν e(−hν/kT) and Wien's separately fitted exponential returns; in between it interpolates, and the integral over all frequencies that used to diverge now converges, to (4σ/c)T⁴, so a black surface emits the σT⁴ Stefan had measured. The cost is a new constant of nature fitted to data, h = 6.63 × 10⁻³⁴ J s, and a premise Planck himself called an act of desperation — that energy exchange with the wall is grainy. What the fit buys is disproportionate: Wien's displacement constant, the Stefan–Boltzmann constant, and through k even Avogadro's number and the elementary charge, all from one curve.

What it does not buy is a quantised electromagnetic field. The graininess here sits in the material oscillators; the radiation is still classical modes.

Simple definition
The Planck distribution gives the energy per unit volume per unit frequency inside a cavity at temperature T, by multiplying the classical count of standing modes by the mean energy hν/(e(hν/kT) − 1) of an oscillator whose energies are restricted to whole multiples of hν.
Example
At T = 300 K a 10 µm mode has hν = 0.124 eV against kT = 0.0259 eV, so hν/kT = 4.79 and the mode holds ⟨n⟩ = 1/(e4.79 − 1) = 0.0083 quanta — a mean energy of 0.040 kT, not the kT equipartition promises.
Quantised ladder and its partition functionEₙ = n hν, n = 0, 1, 2, … Z = Σ e(−n hν/kT) = 1/(1 − e(−x))

The sum is geometric only because the rungs are evenly spaced — the one step a continuum of energies cannot take.

x = hν/kT is a pure number; hν and kT are in J. At 300 K, kT = 4.14 × 10⁻²¹ J = 25.9 meV.

Mean energy and mean occupancy of a mode⟨E⟩ = hν/(ex − 1) ⟨n⟩ = 1/(ex − 1) = Z − 1

Replaces equipartition's flat kT by a value that knows the frequency; the mode count is untouched, so this is the only door through which h enters.

⟨E⟩ in J, ⟨n⟩ a pure number. At x = 1, ⟨n⟩ = 0.582 and ⟨E⟩ = 0.582 hν, not kT.

Planck's spectral energy densityu(ν, T) = (8πν²/c³) · hν/(e(hν/kT) − 1)

Mode count × mean energy, nothing else. Multiply by c/4 for the spectral exitance, the power per unit area per unit frequency a black surface emits.

u in J m⁻³ Hz⁻¹, ν in Hz, c in m s⁻¹. The first bracket is the untouched classical mode count.

The two old laws as limits of one curvex ≪ 1: u → 8πν²kT/c³ x ≫ 1: u → (8πhν³/c³) e(−hν/kT)

Rayleigh–Jeans and Wien's fitted exponential stop being rival laws and become the two ends of this one.

At 300 K the first is within 1% only beyond 2.4 mm, the second only below 10.4 µm.

Wien's displacement constantλₘₐₓ T = hc/(4.9651 k) = 2.898 × 10⁻³ m K

Turns any measured peak into a temperature: 500 nm → 5800 K, 1.06 mm → 2.73 K.

4.9651 solves y = 5(1 − e(−y)); the frequency peak solves y = 3(1 − e(−y)), giving 2.8214 instead.

Stefan–Boltzmann constant from h, c and kσ = 2π⁵k⁴/(15h³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴

σ had been measured for two decades before it was derived; producing it from three constants is the law's hardest test.

Uses ∫₀^∞ x³dx/(ex − 1) = π⁴/15 = 6.4939, then M = cu/4. Energy density u = (4σ/c)T⁴.

01

Sum the ladder instead of integrating the continuum

Classically a wall oscillator of frequency ν may hold any energy, so its Boltzmann average is ⟨E⟩ = ∫E e(−E/kT)dE / ∫e(−E/kT)dE = kT — the same answer for every ν, which is exactly what makes the ultraviolet catastrophe unavoidable. Planck allowed only Eₙ = n hν. The two integrals become two sums, and with r = e(−hν/kT) both come from the geometric series: Σ rⁿ = 1/(1 − r) and Σ n rⁿ = r/(1 − r)². Their ratio is ⟨E⟩ = hν r/(1 − r) = hν/(e(hν/kT) − 1). Nothing else in the derivation moved. Notice where the frequency got in. In the classical integral the only energy scale is kT, set by the thermostat; in the sum the first available rung sits at hν, so the answer must depend on how that rung compares with kT. Write x = hν/kT and the result is ⟨E⟩ = kT · x/(ex − 1), one dimensionless function that equals 1 at x = 0 and dies exponentially after.

02

Why the exponential starves the high frequencies

The denominator is a probability statement. The chance of finding the oscillator on rung n is (1 − e(−x))e(−nx), so once hν exceeds kT even the first rung is expensive and every higher one is exponentially worse; the mean occupancy ⟨n⟩ = 1/(ex − 1) collapses. Numbers say it better. At 300 K, kT = 25.9 meV, while a green mode at 500 nm needs hν = 2.48 eV, so x = 95.9 and ⟨n⟩ ≈ e(−95.9) ≈ 2.3 × 10⁻⁴². A 30 m³ room holds about 6.0 × 10¹⁹ modes in a 1% frequency band around that line, so the expected number of green quanta in the room is around 1.4 × 10⁻²² — never, in any practical sense. Equipartition would have handed each of those modes a full kT and the room would glow. The classical mode count was never the problem: it still rises as ν², and it is the exponential that beats it down, which is why the integral over all frequencies now converges.

03

Mode count times mean energy is the whole law

Assembling is one multiplication: u(ν, T) = (8πν²/c³)⟨E⟩ = (8πhν³/c³)/(e(hν/kT) − 1), in J m⁻³ Hz⁻¹. Check it once against the coldest blackbody there is. The cosmic microwave background sits at T = 2.725 K, so kT = 3.763 × 10⁻²³ J, and at ν = 160 GHz a quantum is hν = 1.060 × 10⁻²² J: x = 2.817, ex − 1 = 15.73, and ⟨E⟩ = 6.74 × 10⁻²⁴ J, only 0.179 kT. The classical density 8πν²kT/c³ comes to 8.99 × 10⁻²⁵ J m⁻³ Hz⁻¹, while Planck's value is 1.61 × 10⁻²⁵, a factor of 5.58 lower — and this is the very frequency at which uν peaks, since x = 2.82 there. Two conversions are worth keeping. Multiply by c/4 to get the exitance a black surface radiates, and change variable with uλ dλ = uν dν, using |dν/dλ| = c/λ², to get uλ = (8πhc/λ⁵)/(e(hc/λkT) − 1).

04

Both old laws fall out, and so does the gap between them

Expand for small x: ex − 1 = x(1 + x/2 + …), so ⟨E⟩ = kT/(1 + x/2 + …) ≈ kT(1 − x/2) and u → 8πν²kT/c³. Rayleigh–Jeans is not overturned, it is demoted to a limit with a computable error. A 30 GHz mode at 300 K has x = 4.80 × 10⁻³, so equipartition is high by x/2 = 0.24% — fine for a microwave engineer. For large x the −1 is negligible and u → (8πhν³/c³)e(−hν/kT), which is Wien's exponential guess, its two fitted constants now identified as 8πh/c³ and h/k. The discipline worth having is to demand one percent rather than 'roughly'. At 300 K the Rayleigh–Jeans form holds to 1% only for λ > 2.4 mm (x < 0.02), and Wien's only for λ < 10.4 µm (x > 4.6). Between them lies a factor of 230 in wavelength, running from just past the 9.66 µm peak out to the millimetre band, where only the full expression will do.

05

Reading Wien's b and Stefan's σ off the curve

Two constants that had been measured now become predictions. Differentiate the wavelength form and set duλ/dλ = 0: the λ⁻⁵ prefactor and the exponential balance where y ey/(ey − 1) = 5, that is y = 5(1 − e(−y)), whose root is y = 4.9651. Since y = hc/λkT, this says λₘₐₓ T = hc/(4.9651 k) = 2.898 × 10⁻³ m K. Integrating instead over all frequencies needs ∫₀^∞ x³dx/(ex − 1) = π⁴/15 = 6.4939 and gives u = (8π⁵k⁴/15h³c³)T⁴; the emitted power is cu/4, so σ = 2π⁵k⁴/(15h³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴. Planck ran this backwards. From the measured σ and b he solved for h = 6.55 × 10⁻³⁴ J s and k = 1.346 × 10⁻²³ J K⁻¹, then took NA = R/k ≈ 6.2 × 10²³ and e = F/NA ≈ 1.56 × 10⁻¹⁹ C — both within a few percent of today's values, and in 1901 better than any direct measurement available.

06

What Planck quantised, and what he did not

The restriction Eₙ = n hν was imposed on the material oscillators in the cavity wall, not on the electromagnetic field. The field enters only through 8πν²/c³, which is classical standing-wave counting, and nothing in the derivation says a mode's energy travels in indivisible lumps. Planck treated ε = hν as a device for the entropy count and spent years trying to remove it; his 1912 'second theory' kept classical absorption and quantised emission alone. The modern derivation quantises the field mode itself, returns the same formula with an extra ½hν per mode, and the measured spectrum cannot see that term because it carries no temperature dependence. So take the modest claim: energy exchange between matter and radiation at frequency ν is grainy in units of hν. Whether the free field is itself quantised needs different evidence, and the experiments that settle it come later in this unit.

02

Change one variable at a time

Make the relationship visible.

Interactive model
5800 K
2.0 ×10¹⁴ Hz

Drag T from 8000 K down to 2000 K: the open circle slides left in proportion to T, and the dashed classical curve leaves the top of the frame ever earlier. Then, at 2000 K, walk the probe out to 5.8 and watch log₁₀⟨n⟩ fall past −6, long after the classical curve has run off the top.

Interactive physics modelSolid: Planck's u(ν) at T = 5800 K. Dashed: the classical 8πν²kT/c³. Both are scaled by the current Planck peak, which the open circle marks at hν/kT = 2.82. At the probe frequency hν/kT = 1.65, the mean occupancy is 0.236 and equipartition overstates the mode energy 2.56 times.Planck u(ν) at T = 5800 K, scaled to its own peaksolid: Planck dashed: Rayleigh–Jeans, 8πν²kT/c³open circle: peak, hν/kT = 2.821.00246ν / 10¹⁴ Hzprobe ν = 2.0, hν/kT = 1.65

hν / kT AT PROBE1.65

⟨E⟩ / kT AT PROBE0.391

log₁₀ OF ⟨n⟩-0.63

PEAK ν3.41 ×10¹⁴ Hz

Live interpretationhν / kT AT PROBE: 1.65. ⟨E⟩ / kT AT PROBE: 0.391. log₁₀ OF ⟨n⟩: −0.63. PEAK ν: 3.41 ×10¹⁴ Hz

03

Catch the common trap

Explain before calculating.

A cavity mode sits exactly at hν = kT. Using Planck's result for a set of levels Eₙ = n hν, what is the mean number of quanta ⟨n⟩ in that mode?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

EasyThe Sun's photosphere is at T = 5800 K. For a cavity mode at λ = 500 nm, find hν and kT in electronvolts, then the mean occupancy ⟨n⟩ and the mean energy ⟨E⟩ of that mode, and compare ⟨E⟩ with the kT equipartition would assign. Use h = 6.626 × 10⁻³⁴ J s, k = 1.381 × 10⁻²³ J K⁻¹, c = 2.998 × 10⁸ m s⁻¹.
  1. ν = c/λ = 2.998 × 10⁸ ÷ 5.00 × 10⁻⁷ = 5.996 × 10¹⁴ Hz, so hν = 6.626 × 10⁻³⁴ × 5.996 × 10¹⁴ = 3.973 × 10⁻¹⁹ J = 2.48 eV.
  2. kT = 1.381 × 10⁻²³ × 5800 = 8.010 × 10⁻²⁰ J = 0.500 eV, so x = hν/kT = 2.48/0.500 = 4.96.
  3. ⟨n⟩ = 1/(e4.96 − 1) = 1/(142.6 − 1) = 1/141.6 = 7.062 × 10⁻³ quanta.
  4. ⟨E⟩ = hν⟨n⟩ = 3.973 × 10⁻¹⁹ × 7.062 × 10⁻³ = 2.806 × 10⁻²¹ J.
  5. Compare: ⟨E⟩/kT = 2.806 × 10⁻²¹ ÷ 8.010 × 10⁻²⁰ = 0.0350, so equipartition is 28.5 times too generous — and this mode sits at the peak of the Sun's wavelength spectrum, not out on its tail.

Answerx = 4.96, ⟨n⟩ = 7.06 × 10⁻³, ⟨E⟩ = 2.81 × 10⁻²¹ J = 0.0350 kT; equipartition overstates it by 28.5 times.

MediumA cavity is held at 300 K. At ν = 3.0 × 10¹² Hz (λ ≈ 100 µm), find the Rayleigh–Jeans spectral energy density, the Planck value, and the factor by which the classical law overshoots. Take 8π = 25.13 and c³ = 2.695 × 10²⁵ m³ s⁻³.
  1. kT = 1.381 × 10⁻²³ × 300 = 4.143 × 10⁻²¹ J and hν = 6.626 × 10⁻³⁴ × 3.0 × 10¹² = 1.988 × 10⁻²¹ J, so x = hν/kT = 1.988/4.143 = 0.4798.
  2. Rayleigh–Jeans: u = 8πν²kT/c³ = 25.13 × (3.0 × 10¹²)² × 4.143 × 10⁻²¹ ÷ 2.695 × 10²⁵ = 3.48 × 10⁻²⁰ J m⁻³ Hz⁻¹.
  3. Planck replaces kT by hν/(ex − 1). With e0.4798 = 1.6158, ex − 1 = 0.6158, so ⟨E⟩ = 1.988 × 10⁻²¹ ÷ 0.6158 = 3.23 × 10⁻²¹ J = 0.779 kT.
  4. Hence uPlanck = 3.48 × 10⁻²⁰ × 0.779 = 2.71 × 10⁻²⁰ J m⁻³ Hz⁻¹.
  5. Overshoot factor = (ex − 1)/x = 0.6158/0.4798 = 1.28. Only 28% high at 100 µm — but hold 300 K and go to 500 nm, where x = 95.9, and the same ratio is 4.6 × 10³⁹.

AnsweruRJ = 3.48 × 10⁻²⁰ J m⁻³ Hz⁻¹ and uPlanck = 2.71 × 10⁻²⁰ J m⁻³ Hz⁻¹; the classical law is high by a factor of 1.28.

HardStarting from uλ = (8πhc/λ⁵)/(e(hc/λkT) − 1), show that the peak satisfies y = 5(1 − e(−y)) with y = hc/λkT, solve it by iteration, evaluate λₘₐₓ T, and apply it to a 5800 K photosphere. Then show that converting the frequency peak, at y = 2.8214, into a wavelength does not give the same answer.
  1. Write uλ = A λ⁻⁵(ey − 1)⁻¹ with y = hc/λkT, so dy/dλ = −y/λ. Then duλ/dλ = A[−5λ⁻⁶(ey − 1)⁻¹ + λ⁻⁶ y ey (ey − 1)⁻²].
  2. Set duλ/dλ = 0 and divide by A λ⁻⁶(ey − 1)⁻¹: 5 = y ey/(ey − 1), which rearranges to y = 5(1 − e(−y)).
  3. Iterate from y = 5: 5(1 − e⁻⁵) = 5(1 − 0.006738) = 4.9663; then 5(1 − e(−4.9663)) = 5(1 − 0.006969) = 4.9652; once more gives 4.9651. Take y = 4.9651.
  4. λₘₐₓ T = hc/(y k) = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (4.9651 × 1.381 × 10⁻²³) = 1.9865 × 10⁻²⁵ ÷ 6.857 × 10⁻²³ = 2.897 × 10⁻³ m K (2.898 × 10⁻³ m K with full-precision constants).
  5. At T = 5800 K, λₘₐₓ = 2.897 × 10⁻³ ÷ 5800 = 4.99 × 10⁻⁷ m = 499 nm: the Sun peaks in the green.
  6. The frequency peak sits at y = 2.8214, so νₘₐₓ = 2.8214 kT/h = 2.8214 × 8.010 × 10⁻²⁰ ÷ 6.626 × 10⁻³⁴ = 3.411 × 10¹⁴ Hz, and c/νₘₐₓ = 879 nm. The two peaks differ by 4.9651/2.8214 = 1.76 because uλ and uν are densities in different variables.

Answery = 4.9651 and λₘₐₓ T = 2.897 × 10⁻³ m K, so λₘₐₓ = 499 nm at 5800 K; the frequency peak converts to 879 nm, 1.76 times longer.