Planck’s Hypothesis

Key idea: Use quantised oscillator energies to explain the Planck blackbody distribution, its classical limit, and high-frequency suppression.

  • Advanced Physics
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Learning objectives

  • Connect blackbody evidence, Planck quantisation, photon momentum, and Compton scattering.

Planck resolved the ultraviolet catastrophe by changing how matter exchanges energy with the electromagnetic field. Oscillators of frequency f can occupy only discrete energies separated by hf, so high-frequency excitation becomes unlikely when hf greatly exceeds the thermal energy scale kT.

1. Hypothesis and assumptions

For an oscillator of frequency f,

Eₙ = nhf, n = 0,1,2,…

The allowed energies are discrete, with adjacent levels separated by

Δ E = hf

Planck originally quantised the energy exchange of material oscillators in the cavity walls. The later photon model interprets a quantum of electromagnetic radiation as carrying energy hf. Keep those historical steps distinct even though modern calculations use the photon language naturally.

2. Mean energy and spectrum

Applying Boltzmann weighting to the allowed oscillator energies gives the mean thermal energy per mode

⟨E⟩ = hf/(e^(hf/(kT))-1)

Multiplying by the electromagnetic mode density gives the spectral energy density per unit frequency:

u_f(f,T) = ((8π hf³)/c³)1/(e^(hf/(kT))-1)

Here u_f df is the energy per unit volume in the frequency interval [f,f + df].

The wavelength form is not obtained by merely substituting f = c/λ; the interval also transforms. The result is

u_λ(λ,T) = ((8π hc)/λ⁵)1/(e^(hc/(λ kT))-1)

3. Why quantisation fixes the spectrum

Define the dimensionless ratio

x = hf/kT

  • If x≪1, then e^x-1 ≈ x. Hence ⟨E⟩ ≈ kT, recovering classical equipartition and the Rayleigh–Jeans limit.
  • If x≫1, then ⟨E⟩ ≈ hf e^(-x). The exponential factor overwhelms the increasing number of modes and suppresses the high-frequency spectrum.

Thus Planck’s law agrees with classical physics where classical physics works, but remains finite where the classical model diverges.

4. Photon relations

For a photon in vacuum,

E = hf = hc/λ

and the relativistic massless-particle relation E = pc gives

p = E/c = h/λ

This photon momentum is the key input to Compton scattering.

5. Common mistakes

  • Starting n at one and excluding the zero-energy state in Planck’s original oscillator model.
  • Saying all energies in nature must be integer multiples of one universal amount. The spacing is hf and depends on oscillator frequency.
  • Treating hf as the energy of the oscillator at every temperature. It is the spacing between adjacent levels; the mean energy depends on T.
  • Substituting f = c/λ into a spectral density without transforming the interval.
  • Describing quantisation as an arbitrary high-frequency cutoff. The suppression follows continuously from the Boltzmann factor.

6. Worked Examples

Modelled example 1

Spacing of oscillator energies

Core

Problem

Find the energy spacing for an oscillator at 6.0 × 10¹⁴ Hz.
Study the worked solution
  1. Apply Planck's quantum

    Method

    Δ E = 4.0 × 10⁻¹⁹ J = 2.5 eV.

    Reason

    Adjacent oscillator energies are separated by hf.

    Working

    Δ E = hf = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴) = 4.0 × 10⁻¹⁹ J

Guided practice 2

Testing the classical limit

About 5 min

Problem

For f = 1.0 × 10¹¹ Hz and T = 300 K, calculate x = hf/(kT) and decide whether equipartition is reasonable.

Try this before viewing the solution

Hints

Hint 1: compare with unity
The classical limit requires x≪1.
View solution step by step
  1. Calculate the ratio

    Method

    x = 0.016.

    Reason

    Use absolute temperature and consistent SI constants.

    Working

    x = ((6.63 × 10⁻³⁴)(1.0 × 10¹¹))/((1.38 × 10⁻²³)(300)) = 0.016
  2. Select the limit

    Method

    Equipartition is a good approximation.

    Reason

    x≪1 makes e^x-1 ≈ x and hence ⟨E⟩ ≈ kT.

    Working

    x≪1

Challenge 3

High-frequency suppression

Minimal support

Independent transfer

At a particular f and T, hf/(kT) = 10. Estimate ⟨E⟩/(hf) and interpret it.

Try this before viewing the solution

Hints

Hint 1: use the mean-energy factor
Divide Planck’s mean energy by hf.
View solution step by step
  1. Evaluate the factor

    Method

    ⟨E⟩/(hf) = 4.54 × 10⁻⁵.

    Reason

    The ratio is 1/(e¹⁰-1).

    Working

    (⟨E⟩)/hf = 1/(e¹⁰-1) = 4.54 × 10⁻⁵
  2. Interpret

    Method

    The mode is thermally populated only very weakly.

    Reason

    The quantum spacing is ten times the thermal scale.

    Working

    hf≫ kT

7. Model check

Mind stretcher 1: Planck distribution limit checkExtension

The Planck distribution must approach Rayleigh–Jeans at low frequency and fall exponentially at high frequency. Checking both limits is more reliable than memorising the full expression without its assumptions.

Next: Compton Shift.

Continue with the next resource in this course.

Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics