Wien’s Displacement Law

Wien’s displacement law relates the observed wavelength of light from an object to its surface temperature, it states:

The black body radiation curve for different temperatures peaks at a wavelength that is inversely proportional to the temperature

This relation can be written as:

$λ_{m a x} \propto \frac{1}{T}$

Where:
- λ_max = the wavelength at which radiation is emitted at the greatest intensity (m)
- T = the surface temperature of an object (K)

Wien's Displacement Law Graph

Wien's Displacement Law Graph, for IB Physics Revision Notes

The intensity-wavelength graph shows how thermodynamic temperature links to the peak wavelength for four different bodies

Wien's Law equation is given by:

λ_maxT = 2.9 × 10⁻³ m K

This equation shows that the higher the temperature of a body, the shorter the wavelength at the peak intensity
- Additionally, as the object gains temperature, the intensity of radiation at every wavelength increases – this can be seen on the graph above
Consider an object being heated from room temperature:
- Initially, the object emits radiation with the greatest intensity in the infrared range
- As it gains heat, the peak of the curve shifts left to the red region of the visible spectrum (the object glows red)
- As it is heated further, the peak shifts left further until it is in the centre of the visible range, (the object glows white, an equal mix of wavelengths in the visible range)
- As it heats further still, the object will eventually glow blue and even emit UV radiation

Table to compare surface temperature and star colour

Star Colour	Temperature / K
blue	>33 000
blue-white	10 000 − 30 000
white	7500 − 10 000
yellow-white	6000 − 7500
yellow	5000 − 6000
orange	3500 − 5000
red	<3500

Worked example

The black-body radiation curve of an object at 900 K is shown in the diagram below.

2-1-15-wiens-displacement-law-worked-example-graph-ib-2025-physics

Which of the following shows the black-body radiation curve of an object at 650 K?

The dashed line represents the curve of the object at 900 K.

2-1-15-wiens-displacement-law-worked-example-graph-options

Answer: C

From Wien's displacement law: $λ_{m a x} \propto \frac{1}{T}$
Therefore, a curve with a longer peak wavelength will correspond to a lower temperature

Worked example

The spectrum of the star Rigel in the constellation of Orion peaks at a wavelength of 263 nm, while the spectrum of the star Betelgeuse peaks at a wavelength of 828 nm.

Determine which of these two stars, Betelgeuse or Rigel, is cooler.

Answer:

Step 1: List the known quantities

Maximum emission wavelength of Rigel = 263 nm = 263 × 10⁻⁹ m
Maximum emission wavelength of Betelgeuse = 828 × 10⁻⁹ m

Step 2: Write down Wien’s displacement law and rearrange for temperature T

$λ_{m a x} T = 2.9 \times 10^{- 3} m K$

$T = \frac{2.9 \times 10^{- 3}}{λ_{m a x}}$

Step 3: Calculate the surface temperature of each star

Rigel: $T = \frac{2.9 \times 10^{- 3}}{263 \times 10^{- 9}} = 11 027 = 11 000 K$

Betelgeuse: $T = \frac{2.9 \times 10^{- 3}}{828 \times 10^{- 9}} = 3502 = 3500 K$

Step 4: Write a concluding sentence

Betelgeuse has a surface temperature of 3500 K, therefore, it is much cooler than Rigel

Wien's law and stars in the Orion constellation

Wien's Law Orion and Betelguese Worked Example, for IB Physics Revision Notes

The Orion Constellation; cooler stars, such as Betelgeuse, appear red or yellow, while hotter stars, such as Rigel, appear white or blue

Exam Tip

Note that the temperature used in Wien’s Law is in Kelvin (K). Remember to convert from ^oC if the temperature is given in degrees in the question before using the Wien’s Law equation.

Wien’s Displacement Law (SL IB Physics)

Revision Note