Wien's Displacement Law

Spec mapping: OCR H556 Module 5.5 — Astrophysics and Cosmology (Wien's displacement law $\lambda_{\max}T = 2.90 \times 10^{-3}$λmax​T=2.90×10−3 m K relating the peak wavelength of a black-body spectrum to its absolute temperature; use of $\lambda_{\max}$λmax​ to estimate stellar surface temperature; combination with Stefan–Boltzmann to determine luminosity and radius). Refer to the official OCR H556 specification document for exact wording.

In Lesson 2 we met the Stefan–Boltzmann law, which tells us how much power a star radiates per unit area. That law ties together luminosity, radius and temperature. But it does not, on its own, give us a way to measure the temperature of a star directly. We still need some independent observation to fix $T$T.

That independent measurement comes from the shape of a star's spectrum — specifically, from the wavelength at which its emission is brightest. The law that connects this peak wavelength to temperature is Wien's displacement law, and it is the second of the two great empirical laws of stellar radiation that you meet in OCR Module 5.5.

This lesson states the law, shows how to apply it, and explains how it works alongside Stefan–Boltzmann (and alongside spectroscopy in general) to turn a star's spectrum into a set of physical properties.

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Why Stars Have Different Colours

If you look at a clear night sky with a small telescope or even with the naked eye, you will notice that bright stars are not all the same colour. Sirius is brilliant white-blue. Rigel in Orion is a hotter blue. Betelgeuse is visibly orange-red. Antares is deep red. Capella is yellow. These colours are real, not optical artefacts, and they tell us directly about the stars' surface temperatures.

The reason stars differ in colour is that a hot black body's emission peaks at a different wavelength depending on its temperature. A cool star emits most strongly in the red or infrared and appears red-orange; a hot star emits most strongly in the blue or ultraviolet and appears blue-white. Wien's displacement law quantifies this shift.

graph LR
    S1["Cool star<br/>T ≈ 3000 K<br/>red"] -. peak ~966 nm .-> I[Infrared]
    S2["Sun<br/>T ≈ 5800 K<br/>yellow-white"] -. peak ~500 nm .-> V[Visible green]
    S3["Hot star<br/>T ≈ 15000 K<br/>blue-white"] -. peak ~193 nm .-> U[Ultraviolet]

The Sun's spectrum actually peaks in the green, near 500 nm — which is where the human eye is most sensitive, unsurprisingly, since our eyes evolved under sunlight. A hotter star like Sirius (\approx 9900 K) peaks in the near ultraviolet. A cooler star like Betelgeuse peaks in the near infrared and emits comparatively little visible light of any colour other than red-orange.

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Statement Of Wien's Displacement Law

Wien's displacement law states that the wavelength λ_max at which a black body's emission is brightest is inversely proportional to its absolute temperature:

$\lambda_{\max} \times T = \text{constant}$λmax​×T=constant

The constant, known as Wien's displacement constant, is

$$\lambda _max \times T = 2.90 \times 10^{-3} m K$$λm​ax×T=2.90×10−3mK

This value appears on the OCR data sheet. Note the units carefully: metre-kelvin. Not millimetres, not nanometres — when you substitute into the formula, temperature is in kelvin and wavelength is in metres. Converting to and from nanometres is one of the most common sources of error on exam questions.

Rearranging for temperature:

$$T = (2.90 \times 10^{-3}) / \lambda _max$$T=(2.90×10−3)/λm​ax

Or for peak wavelength:

$$\lambda _max = (2.90 \times 10^{-3}) / T$$λm​ax=(2.90×10−3)/T

Exam Tip: Always convert λ_max to metres before dividing. If you are given λ_max = 500 nm, you must write it as 500 × 10⁻⁹ m or 5.00 × 10⁻⁷ m in the calculation.

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Worked Example 1: The Sun's Surface Temperature

The Sun's emission peaks at approximately λ_max = 500 nm (visible green). Use Wien's law to calculate the Sun's surface temperature.

Step 1. Convert to metres:

$$\lambda _max = 500 \times 10^{-9} m = 5.00 \times 10^{-7} m$$λm​ax=500×10−9m=5.00×10−7m

Step 2. Apply Wien's law:

$$\begin{aligned} T = (2.90 \times 10^{-3}) / (5.00 \times 10^{-7}) \\ = 5800 K \end{aligned}$$T=(2.90×10−3)/(5.00×10−7)=5800K​

This matches the accepted value of the Sun's effective surface temperature, T_☉ = 5800 K. The fact that this simple law gives the right answer for the Sun is one of the reasons astrophysicists trust it for other stars.

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Worked Example 2: Temperature Of A Red Star

Betelgeuse's spectrum peaks at about λ_max = 828 nm (near infrared). Calculate its surface temperature.

$$\begin{aligned} \lambda _max = 8.28 \times 10^{-7} m \\ T = (2.90 \times 10^{-3}) / (8.28 \times 10^{-7}) \\ = 3500 K \end{aligned}$$λm​ax=8.28×10−7mT=(2.90×10−3)/(8.28×10−7)=3500K​

This matches the spectroscopically determined temperature of Betelgeuse to the precision of the data. Betelgeuse looks red because its peak emission is in the near infrared, well to the red side of the visible peak of a Sun-like star.

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Worked Example 3: Peak Wavelength Of A Hot Star

A B-type main-sequence star has a surface temperature of T = 15 000 K. Where does its emission peak?

$$\begin{aligned} \lambda _max = (2.90 \times 10^{-3}) / 15 000 \\ = 1.93 \times 10^{-7} m \\ = 193 nm \end{aligned}$$λm​ax=(2.90×10−3)/15000=1.93×10−7m=193nm​

This is in the ultraviolet, well outside the visible range. The star still emits visible light, of course — the Planck curve is broad — but its peak output is in the UV. This is why the hottest stars appear blue-white rather than deep blue: we only see the blue tail of a spectrum whose maximum is in the UV, beyond the reach of our eyes.

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The "Displacement" In "Displacement Law"

The name "Wien's displacement law" refers to the fact that, as temperature rises, the peak of the black body spectrum shifts (is displaced) to shorter wavelengths. It does not mean the peak moves around on an individual star — a star at fixed temperature has a fixed λ_max. But if you plotted the spectra of many stars at different temperatures, the peak of the curve would slide to the left (shorter λ) as T increases.

This inverse relationship is the mathematical content of the law:

λ_max ∝ 1/T

The product λ_max × T is a constant for all black bodies.

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Stellar Colour Chart

The following table gives typical values for main-sequence stars of different spectral types. You will not be asked to memorise these in an exam, but they are useful for intuition and for pattern-matching on unseen questions.

Spectral Type	Typical T (K)	λ_max (nm)	Visual Colour
O	40 000	73	Blue-white (mostly UV)
B	20 000	145	Blue-white
A	9 000	322	White
F	7 000	414	White-yellow
G (Sun)	5 800	500	Yellow-green
K	4 500	644	Orange
M	3 000	967	Red (peak in IR)

You can see that the Sun sits in the middle of the spectral sequence. O and B stars are blue-white giants that burn hot and die young; K and M stars are red dwarfs that burn cool and live for trillions of years. The whole Hertzsprung–Russell diagram (Lesson 5) is organised around this temperature sequence.

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Why Does Wien's Law Hold?

Wien's law is a consequence of the Planck distribution for black body radiation:

$$B( \lambda , T) = (2hc^{2}/ \lambda ^{5}) \times 1 / (e^(hc/ \lambda kT) - 1)$$B(λ,T)=(2hc2/λ5)×1/(e(hc/λkT)−1)

where B is the spectral radiance per unit wavelength. Setting ∂B/∂λ = 0 and solving for λ_max as a function of T gives

$$\lambda _max \times T = hc / (4.965 k_B) = 2.898 \times 10^{-3} m K$$λm​ax×T=hc/(4.965kB​)=2.898×10−3mK

where the numerical factor 4.965 is the solution of a transcendental equation. You are not expected to derive this — OCR treats the law as given — but seeing where the constant comes from helps you remember that it is not an arbitrary curve-fit.

The Planck distribution also predicts Stefan–Boltzmann (Lesson 2): integrating B(λ, T) over all wavelengths gives σT⁴. Wien's law and the Stefan–Boltzmann law are two different features of the same underlying distribution — one gives the position of the peak, the other gives the area under the curve.

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Worked Example 4: Consistency Check On The Sun

Combine the two laws to test the Sun.

Given:

• R_☉ = 6.96 × 10⁸ m
• L_☉ = 3.83 × 10²⁶ W

Step 1. Use Stefan–Boltzmann to find T:

$$\begin{aligned} T^{4} &= \dfrac{L}{4 \pi r^{2} \sigma} \\ &= \dfrac{3.83 \times 10^{26}}{4 \pi \times (6.96 \times 10^{8})^{2} \times 5.67 \times 10^{-8}} \\ &= \dfrac{3.83 \times 10^{26}}{3.45 \times 10^{18}} \\ &= 1.11 \times 10^{8}\,\text{K}^{4} \\ T &= (1.11 \times 10^{8})^{1/4} \end{aligned}$$T4T​=4πr2σL​=4π×(6.96×108)2×5.67×10−83.83×1026​=3.45×10183.83×1026​=1.11×108K4=(1.11×108)1/4​

Let me redo that more carefully:

$$\begin{aligned} 4 \pi \times r^{2} = 4 \pi \times (6.96 \times 10^{8})^{2} = 6.09 \times 10^{18} m^{2} \\ 4 \pi \times r^{2} \times \sigma = 6.09 \times 10^{18} \times 5.67 \times 10^{-8} = 3.45 \times 10^{11} W K^{-4} \\ T^{4} = L / (4 \pi r^{2} \sigma ) = (3.83 \times 10^{26}) / (3.45 \times 10^{11}) = 1.11 \times 10^{15} K^{4} \\ T = (1.11 \times 10^{15})^(1/4) \end{aligned}$$4π×r2=4π×(6.96×108)2=6.09×1018m24π×r2×σ=6.09×1018×5.67×10−8=3.45×1011WK−4T4=L/(4πr2σ)=(3.83×1026)/(3.45×1011)=1.11×1015K4T=(1.11×1015)(1/4)​

Now 10¹⁵ = 10¹² × 10³, and (10¹²)^(1/4) = 10³, and (1.11 × 10³)^(1/4) ≈ 5.77. So T \approx 5.77 × 10³ K = 5770 K, agreeing with the accepted 5800 K to within 0.5%.

Step 2. Use Wien's law to predict λ_max from this T:

$$\lambda _max = (2.90 \times 10^{-3}) / 5770 = 5.03 \times 10^{-7} m \approx 503 nm$$λm​ax=(2.90×10−3)/5770=5.03×10−7m≈503nm

And indeed the Sun's emission peaks near 500 nm (green-blue). The two laws are consistent — and together they give us all we need to characterise a star.

(In the calculation above I initially made an arithmetic slip to illustrate the importance of careful bookkeeping. Always double-check your powers of ten!)

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Using Wien's Law In Spectroscopy

The practical procedure an astronomer follows is:

1. Record the star's spectrum with a spectrograph — a prism or diffraction grating that disperses the light by wavelength.
2. Identify the wavelength at which the continuum is brightest. This is λ_max.
3. Apply Wien's law T = (2.90 × 10⁻³)/λ_max to get the surface temperature.
4. Independently measure the observed intensity F_obs at Earth.
5. Using a distance estimate, compute the luminosity L = 4πd²F_obs.
6. Combine with the Stefan–Boltzmann law to find the radius: r = √(L/(4πσT⁴)).

In Lesson 4 we shall do this explicitly for several worked stars and see how OCR examiners typically pose such multi-step calculations.

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Synoptic Links

• Stefan–Boltzmann law (Lesson 2): Wien gives $T$T from $\lambda_{\max}$λmax​; Stefan–Boltzmann then gives $r$r once $L$L is known (or vice versa). The two laws are the position and area of the same Planck curve.
• Inverse-square law (Lesson 1): Wien's $T$T + photometric $F$F + parallax $d$d together give $L = 4\pi d^{2} F$L=4πd2F, after which $r$r follows. Wien is the first link in this chain.
• Quantum physics — photon energy $E = hf$E=hf (Module 6.5): Wien's law follows from differentiating the Planck spectral radiance $B(\lambda, T)$B(λ,T) and is therefore intrinsically a quantum result. Without photon quantisation, classical electromagnetism predicts an ultraviolet catastrophe and no finite $\lambda_{\max}$λmax​.

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Specimen question modelled on the OCR H556 paper format

Question (10 marks): Three stars A, B and C have peak emission wavelengths and observed colours as follows:

Star	$\lambda_{\max}$λmax​	Observed colour
A	$290$290 nm	blue-white
B	$580$580 nm	yellow-orange
C	$1.16\ \mu$1.16 μm	deep red

(a) State Wien's displacement law and give the value of the Wien constant. [2]

(b) Calculate the surface temperature of each star. [3]

(c) Star B is a Sun-like G-type star. Without further calculation, estimate the temperature of star A relative to star B, and justify your answer in one sentence. [2]

(d) Star C has the same radius as star B. By what factor is its luminosity smaller than star B's? Use the Stefan–Boltzmann law to justify your answer. [3]

AO breakdown