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Planck’s radiation law and Wien’s displacement law

Derive Planck’s radiation law and Wien’s displacement law, then use them to analyse spectra and radiative exchange.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Planck radiation law and Wien displacement law. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

At a glance

  • Prerequisites: blackbody idea, exponentials, differentiation of a single-variable function.
  • Outcomes: use Planck spectrum, identify peak wavelength, and apply Wien’s law.
  • Key result: λₚₑₐₖT = b with b ≈ 2.898 × 10⁻³ m K.
  • Common trap: confusing spectral density per wavelength with total emitted power.

Setup

For a blackbody, spectral emissive power per wavelength interval is

Eλ(T) = (2π hc²)/(λ⁵(e^(hc/(λ kBT))-1)).

Definitions:

  • Eλ dλ is power per area emitted in [λ,λ + dλ].
  • Total emission is ∫₀^∞ Eλ dλ.

Physical interpretation:

  • Planck’s law is the equilibrium spectrum you get when electromagnetic modes exchange energy with matter with quantized energies (the resolution of the classical ultraviolet catastrophe).
  • The dimensionless combination
    x = hc/(λ kBT)
    compares photon energy scale hc/λ to thermal energy kBT. Large x means “photon energy is expensive” so emission is strongly suppressed.
Pitfalls (Planck spectrum + peaks)
  • Be clear whether you are using the spectrum per wavelength (Eλ) or per frequency (Eν). They are related, but their peaks occur at different numerical values. - Use kelvin in x and in Wien’s law. Celsius in the fourth-power or exponential context gives large errors. - “Spectral density” is not “total power”: you only get total emission after integrating over wavelength (Stefan-Boltzmann law).

Core derivation / explanation

Set derivative to zero for the peak:

dEλ/dλ = 0.

With substitution x = hc/(λ kBT), the condition becomes

5(1-e^(-x)) = x.

Numerical root: x ≈ 4.965. Therefore

λₚₑₐₖT = hc/4.965kB = 2.898 × 10⁻³ m K.

This is Wien’s displacement law.

Related lessons:

  • This topic upgrades the basic radiation law from UY1: radiation by resolving where in wavelength the energy is emitted.
  • Integrating Planck’s law over all wavelengths gives Stefan-Boltzmann total-power scaling ∝ T⁴.
  • The need for Planck’s quantization is one of the classic failures of naive equipartition (see UY1: crisis for equipartition theorem).

Units check:

  • Eλ has units W m⁻³ (power per area per wavelength).
  • In Wien’s law, [λ T] = m K.

Physical limit checks:

  • Short wavelength limit (x≫1): Wien tail Eλ∼ λ⁻⁵e^(-x).
  • Long wavelength limit (x≪1): Rayleigh-Jeans behavior Eλ ∝ T/λ⁴.

Peak wavelength of sunlight

Worked example 1

Solar blackbody approximation

Problem

Approximate the Sun’s photosphere as a blackbody at 5800 K. Estimate the peak wavelength of its spectrum per wavelength interval, using b = 2.898 × 10⁻³ m K.
Show full solution

  1. Apply Wien's displacement law

    Method

    Rearrange λₚₑₐₖT = b to λₚₑₐₖ = b/T.

    Reason

    This law locates the peak of the wavelength spectrum, and the supplied temperature is in kelvin.

    Working

    Answer

    λₚₑₐₖ = (2.898 × 10⁻³)/5800 ≈ 5.0 × 10⁻⁷ m = 500 nm.

  2. Locate the peak in the spectrum

    Working

    Answer

    λₚₑₐₖ ≈ 500 nm, in visible light. This locates a spectral maximum; it does not mean all sunlight has this wavelength.

Practice set

Check your understanding 1

Find the peak wavelength of a 300 K blackbody’s spectrum per wavelength interval. Use b = 2.898 × 10⁻³ m K.

Show hint
Divide the displacement constant by absolute temperature.
Show answer
λₚₑₐₖ = b/T = 2.898 × 10⁻³/300 = 9.66 × 10⁻⁶ m = 9.66 μm, in the infrared.

Check your understanding 2

What happens to a blackbody’s peak wavelength when its absolute temperature doubles?

Show hint
Keep the product λₚₑₐₖT constant.
Show answer
It halves: λₚₑₐₖ = b/T, so b/(2T) is half of b/T. The peak shifts toward shorter wavelengths.

Check your understanding 3

Why does Planck’s law avoid the ultraviolet divergence of the classical radiation spectrum?

Show hint
Examine e^(-hc/(λ kBT)) as wavelength tends to zero.
Show answer
At short wavelengths the photon energy greatly exceeds kBT, so the spectrum is suppressed by e^(-hc/(λ kBT)). This exponential tends to zero faster than the λ⁻⁵ factor grows, preventing a divergent ultraviolet contribution.

Summary + next steps

  • Planck law gives the full blackbody spectrum.
  • Wien’s law gives peak shift: hotter bodies peak at shorter wavelengths.
  • Spectral and total-radiation views are complementary.

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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