Wien's Displacement Law (OCR A Level Physics): Revision Note

Exam code: H556

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Katie M

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Wien's Law

Black Body Radiator

  • An ideal black body radiator is one that absorbs and emits all wavelengths. 

    • A black body is a theoretical object, however, stars are the best approximation there is

  • The radiation emitted from a black body has a characteristic spectrum that is determined by the temperature alone

Wiens Law Graph

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

Wien's Displacement Law

  • Wien’s displacement law relates the observed wavelength of light from an object to its surface temperature, and can be summarised as:

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 maximum wavelength emitted by an object at the peak intensity (m) 

    • = the surface temperature of an object (K) 

  • This equation shows that the higher the temperature of a body:

    • the shorter the wavelength at the peak intensity, so hotter objects tend to be white or blue, and cooler objects tend to be red or yellow

    • the greater the intensity of the radiation at each wavelength

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 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, lambda subscript R = 263 nm

  • Maximum emission wavelength of Betelgeuse, lambda subscript B = 828 nm

Step 2: Recall Wien’s displacement law

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

Step 3: Compare the relative surface temperature of each star

lambda subscript B over lambda subscript R space equals space T subscript R over T subscript B

T subscript R over T subscript B space equals space 828 over 263 space almost equal to space 3.1

Step 4: Write a concluding sentence

  • The surface temperature of Rigel is about 3 times hotter than that of Betelgeuse

  • Therefore, Betelgeuse is much cooler than Rigel

Wiens Law Orion, downloadable AS & A Level 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

Examiner Tips and Tricks

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 Wien’s law equation.

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Katie M

Author: Katie M

Expertise: Physics Content Creator

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.