AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Stefan’s law and Wien’s displacement law.
Full specification wording
Classification by temperature, black-body radiation (A-level only): Stefan’s law and Wien’s displacement law.
A black body absorbs and emits all wavelengths, and its spectrum depends only on temperature.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 1 of 4
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
A black body absorbs and emits all wavelengths, and its spectrum depends only on temperature. Stefan’s law gives total power per unit area as P/A = σT⁴, so a star of radius r radiates L = 4πr²σT⁴. Wien’s displacement law gives the peak wavelength as λmax = 2.9 × 10⁻³ / T, with T in kelvin and λmax in metres. Together they classify stars: a hotter star is more luminous at every wavelength and peaks at a shorter wavelength, which is why blue stars are hotter than red ones. For a star of known luminosity and temperature, Stefan’s law gives its radius; for a known temperature, Wien’s law gives the peak wavelength. Both laws assume ideal black-body behaviour, so real spectra may show absorption lines and departures from the ideal curve.
What earns the marks
- States Stefan’s law as P = σAT⁴ or L = 4πr²σT⁴ and uses it to compare luminosities or radii of stars at different temperatures.
- States Wien’s displacement law as λmaxT = 2.9 × 10⁻³ m K and uses it to find peak wavelength or surface temperature.
- Explains that a black body absorbs and emits all wavelengths and that its spectrum depends only on temperature.
- Uses the two laws together to classify a star by temperature and to relate luminosity, radius and peak wavelength.
- Applies the inverse-square law for intensity where relevant, distinguishing luminosity from observed intensity at a distance.
Common mistakes and how to put them right
Using degrees Celsius in either law.
Put it right by converting to kelvin by adding 273.
Forgetting to square the radius or to include 4π in L = 4πr²σT⁴.
Put it right by writing the full surface-area term.
Assuming a cooler star is always less luminous.
Put it right by noting that luminosity also depends on radius, so a large cool star can outshine a small hot one.
Exam tips
- Quote the constant with its unit, 2.9 × 10⁻³ m K, and keep λmax in metres before converting to nm.
- Show the rearrangement before substituting values, especially when finding T from λmax or r from L and T.
- Use ratios when comparing two stars, since constants and 4π cancel and the arithmetic is simpler.
Try the lesson
The start of the Classification by temperature, black-body radiation (A-level only): Stefan’s law and Wien’s displacement law. lesson, made from checked teaching notes for this specification point.
Have a go first
According to Wien’s displacement law, what happens to the peak wavelength of a black-body spectrum as the temperature of the body increases?
Listen
A black body absorbs and emits all wavelengths, and its spectrum depends only on temperature. Stefan’s law gives total power per unit area as P/A = σT⁴, so a star of radius r radiates L = 4πr²σT⁴. Wien’s displacement law gives the peak wavelength as λmax = 2.9 × 10⁻³ / T, with T in kelvin and λmax in metres. Together they classify stars: a hotter star is more luminous at every wavelength and peaks at a shorter wavelength, which is why blue stars are hotter than red ones. For a star of known luminosity and temperature, Stefan’s law gives its radius; for a known temperature, Wien’s law gives the peak wavelength. Both laws assume ideal black-body behaviour, so real spectra may show absorption lines and departures from the ideal curve. A strong answer earns marks for these points. States Stefan’s law as P = σAT⁴ or L = 4πr²σT⁴ and uses it to compare luminosities or radii of stars at different temperatures. States Wien’s displacement law as λmaxT = 2.9 × 10⁻³ m K and uses it to find peak wavelength or surface temperature. Explains that a black body absorbs and emits all wavelengths and that its spectrum depends only on temperature. Uses the two laws together to classify a star by temperature and to relate luminosity, radius and peak wavelength. A common slip: Using degrees Celsius in either law. Instead, converting to kelvin by adding 273. Exam tip: Quote the constant with its unit, 2.9 × 10⁻³ m K, and keep λmax in metres before converting to nm.
Key words
Say what each word means before you turn the card.
Black body
Try answering aloud or in your head before turning the card.
Stefan’s law
Try answering aloud or in your head before turning the card.
Wien’s displacement law
Try answering aloud or in your head before turning the card.
Luminosity and radius
Try answering aloud or in your head before turning the card.
See every definition
- Black body
- Absorbs and emits all wavelengths, and its spectrum depends only on temperature.
- Stefan’s law
- Gives total power per unit area as P/A = σT⁴, so a star of radius r radiates L = 4πr²σT⁴.
- Wien’s displacement law
- Gives the peak wavelength as λmax = 2.9 × 10⁻³ / T, with T in kelvin and λmax in metres.
- Luminosity and radius
- For a star of known luminosity and temperature, Stefan’s law gives its radius.
There is more to this lesson
The full lesson goes on to check your understanding and help you earn the marks. Create a free account to carry on, with your progress saved.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Test your understanding of Classification by temperature, black-body radiation (A-level only): Stefan’s law and Wien’s displacement law.
Active recall beats re-reading. Take an instant quiz now to test what you’ve learned and lock it into long-term memory.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge