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Section 5 of 9

Wien's Displacement Law

Part 5: Wien’s Displacement Law

The Peak Wavelength

The Planck function peaks at a wavelength that depends only on temperature:

Here $b = 0.2898\ \text{cm·K} = 2.898\times10^{6}\ \text{nm·K}$ is Wien’s displacement constant, and “peak wavelength” means the maximum of $B_\lambda$ (per unit wavelength). The story: $\lambda_{\text{peak}} \propto T^{-1}$ — hotter objects peak bluer, cooler objects redder. This is why color encodes temperature.

Problem

Using $\lambda_{\text{peak}} = bT^{-1}$ : (1) if $T$ doubles, $\lambda_{\text{peak}}$ does what? (2) If $T$ halves? (3) Star A at 6000 K versus Star B at 3000 K — how do their peaks compare?

Deep Dive: Where Does Wien's Law Come From? (Optional, Calculus)

Wien’s law comes from finding the maximum of the Planck function — setting $dB_\lambda/d\lambda = 0$ . The calculation (product and chain rule) is involved, but the result is elegant:

$\frac{hc}{\lambda_{\text{peak}} k_B T} \approx 4.965 \quad\Rightarrow\quad \lambda_{\text{peak}} = \frac{hc}{4.965\,k_B T} = \frac{b}{T},$

with $b = hc/(4.965\,k_B) = 0.2898\ \text{cm·K}$ . Wien’s law isn’t arbitrary — it follows directly from the Planck function. This is why calculus matters: it extracts simple, powerful results from complex equations.

Worked Example 1The Sun's Surface Temperature

Problem

The Sun’s spectrum peaks at approximately 500 nm (green-yellow). What is its surface temperature?

StepInvert Wien's law

$T = \frac{b}{\lambda_{\text{peak}}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{500\ \text{nm}} = 5796\ \text{K} \approx 5800\ \text{K}$

Dimensional check

$[T] = \dfrac{\text{cm·K}}{\text{cm}} = \text{K}$ ✓ (using nm·K / nm consistently).

Result

About 5800 K — determined from the color of sunlight alone, no thermometer and no spacecraft. This is astronomical inference in action.

Worked Example 2Comparing Betelgeuse and Rigel

Problem

Betelgeuse (a red supergiant) peaks near 830 nm; Rigel (a blue supergiant) peaks near 240 nm. Find their surface temperatures.

StepApply Wien's law to each

$T_{\text{Bet}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{830\ \text{nm}} \approx 3500\ \text{K}, \qquad T_{\text{Rigel}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{240\ \text{nm}} \approx 12{,}000\ \text{K}$

Dimensional check

nm·K divided by nm leaves K ✓ — a temperature, as required.

Result

Rigel is about $3.5\times$ hotter than Betelgeuse. In Orion you’re literally seeing temperature as color: the red shoulder (Betelgeuse) is cool, the blue foot (Rigel) is hot.

ESO image of Betelgeuse with solar system orbits overlaid for scale. The star's disk extends past Mars's orbit and approaches Jupiter's orbit. Inner planets (Mercury, Venus, Earth, Mars) would be inside the star. Angular scale bar shows 0.015 arcseconds. — Figure 9Betelgeuse is enormous — it would engulf Mercury through Mars and extend to about 4 AU. Red supergiants are cool (~3,500 K) but luminous because of their vast surface area.ESO/L. Calcada

Worked Example 3The Cosmic Microwave Background

Problem

The CMB — the afterglow of the Big Bang — peaks at about $0.11$ cm. What is its temperature?

StepInvert Wien's law (CGS)

$T_{\text{CMB}} = \frac{0.2898\ \text{cm·K}}{0.11\ \text{cm}} = 2.63\ \text{K} \approx 2.7\ \text{K}$

Dimensional check

cm·K divided by cm leaves K ✓.

Result

About 2.7 K — the temperature of the universe itself, cooling for 13.8 billion years as space expands. The CMB is the most perfect blackbody ever measured (COBE confirmed deviations $\lesssim 50$ ppm); Planck mapped its $\pm 0.0002$ K fluctuations, the seeds of all cosmic structure.

Planck satellite all-sky map of the Cosmic Microwave Background in Mollweide projection. Colors show tiny temperature fluctuations: blue regions are slightly cooler, red/orange regions slightly warmer than the 2.725 K average. — Figure 10The CMB is the most perfect blackbody ever measured (T = 2.725 K). These color variations show temperature fluctuations of only +/- 0.0002 K - the seeds of all cosmic structure.ESA/Planck Collaboration

Pause & Predict

A nebula glows bright red at one specific wavelength (656.3 nm); a nearby star shows a smooth rainbow with dark lines. Think about what's needed to produce each spectrum type.

Which object produces an emission spectrum, and which an absorption spectrum?

What does that tell you about the physical conditions in each object?

Part 6 develops Kirchhoff's laws to answer this.