What Is Wien's Displacement Law and How Does It Find a Star's Temperature?

Wien's displacement law states that the wavelength at which a black body emits its greatest intensity is inversely proportional to its absolute temperature. By measuring a star's peak wavelength, astronomers can estimate its surface temperature.

This is part of B.1 Thermal energy transfers and is required for both SL and HL Physics.

The Mechanism

A star can be approximated as a black body, meaning its emission spectrum depends mainly on its temperature. Wien's displacement law is

$$\lambda_{\text{max}}T = 2.9 \times 10^{-3}\ \text{m K}$$

where:

• $\lambda_{\text{max}}$ is the wavelength of maximum intensity in metres
• $T$ is the absolute surface temperature in kelvin

Rearranging gives

$$T = \frac{2.9 \times 10^{-3}}{\lambda_{\text{max}}}.$$

A hotter star has a shorter peak wavelength, while a cooler star has a longer peak wavelength. This explains why hotter stars appear bluer and cooler stars appear redder, although stars emit a continuous range of wavelengths rather than only one colour.

Because real stars are only approximate black bodies, the temperature obtained is an estimate. Absorption lines alter narrow parts of the spectrum, but the overall continuous curve still has a measurable maximum. The law concerns surface temperature, not the much higher temperature in the stellar core.

Worked Example

Suppose a star's emission spectrum has its maximum intensity at $500\ \text{nm}$. First convert nanometres to metres:

$$500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.00 \times 10^{-7}\ \text{m}.$$

Then apply Wien's law:

$$T = \frac{2.9 \times 10^{-3}}{5.00 \times 10^{-7}} = 5.8 \times 10^3\ \text{K}.$$

The star's estimated surface temperature is therefore $5800\ \text{K}$.

A common misconception is that $\lambda_{\text{max}}$ is the only wavelength emitted. It is actually the wavelength at which the intensity is greatest.

Exam Technique

For a Calculate question, show the equation, convert the wavelength into metres, substitute, and give the temperature in kelvin. Check the inverse relationship: a shorter $\lambda_{\text{max}}$ must produce a higher temperature.

In an exam, a graph may provide the peak wavelength rather than stating it directly. Read the horizontal axis and its prefix carefully. If the axis is in nanometres, use $1\ \text{nm}=10^{-9}\ \text{m}$ before substituting into the data-booklet equation. This avoids unit errors.