Syllabus Edition

First teaching 2023

First exams 2025

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Wien’s Displacement Law (HL IB Physics)

Revision Note

Katie M

Author

Katie M

Expertise

Physics

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:

lambda subscript m a x end subscript space proportional to space 1 over T

  • Where:
    • λmax = the wavelength at which radiation is emitted at the greatest intensity (m)
    • = 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:

λmaxT = 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:  lambda subscript m a x end subscript space proportional to space 1 over T
  • Therefore, a curve with a longer peak wavelength will correspond to a lower temperature

2-1-15-wiens-displacement-law-worked-example-solution

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

lambda subscript m a x end subscript T space equals space 2.9 space cross times space 10 to the power of negative 3 end exponent space straight m space straight K

T space equals space fraction numerator 2.9 space cross times space 10 to the power of negative 3 end exponent over denominator lambda subscript m a x end subscript end fraction

Step 3: Calculate the surface temperature of each star

Rigel:  T space equals space fraction numerator 2.9 space cross times space 10 to the power of negative 3 end exponent over denominator 263 space cross times space 10 to the power of negative 9 end exponent end fraction space equals space 11 space 027 space equals space 11 space 000 space straight K

Betelgeuse:  T space equals space fraction numerator 2.9 space cross times space 10 to the power of negative 3 end exponent over denominator 828 space cross times space 10 to the power of negative 9 end exponent end fraction space equals space 3502 space equals space 3500 space straight 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.

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.