25.2.1Medical physics and astronomy (A Level)
Wien’s Displacement Law
5 min read 10 parts Lesson 16 of 22 1 quiz
Lesson summary
25.2.1 Wien's Displacement Law — AS PHYSICS 9702 notes: State Wien's displacement law and use it to find the surface temperature of a star from its peak wavelength.Black body = an object which absorbs all the radiation that falls on it, is also a good emitter and does not reflect…
- Theoretical
- Intensity-wavelength graph
- Proportionality
- Doubling T
- Peak position
- Seen colour
Section 1 of 10Black-body radiation
Black-body radiation
Black body
Black body an object which absorbs all the radiation that falls on it, is also a good emitter and does not reflect or transmit radiation.
Theoretical
- A black body is theoretical, and stars are the best approximation.
Spectrum
- Its emitted spectrum depends on temperature alone.
Intensity-wavelength graph
- A smooth hump that rises from zero at short wavelengths, reaches one maximum and falls slowly at long wavelengths.
Peak wavelength
- The wavelength at the top of the hump is λmax.
Section 2 of 10Wien's displacement law
Statement
- The black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the thermodynamic temperature.
Proportionality
- λmax ∝ 1/T.
Constant
- λmaxT = 2.9 × 10−3 m K.
- λmax is in metres and T is in kelvin.
Doubling T
- The peak wavelength halves.
Section 3 of 10Effect of temperature on the curve
Peak position
- A higher temperature moves the peak to a shorter wavelength.
Intensity
- A hotter body emits greater intensity at every wavelength, so its curve is higher.
Area
- The area under the curve rises as T4, which is why hotter bodies are brighter overall.
Curves
- Curves for different temperatures do not cross.
Examples
- The Sun's peak is in the green at about 500 nm, and a red giant peaks in the infrared.
Section 4 of 10Star colour and temperature
| Colour of star | Temperature / K |
|---|---|
| Blue | above 33 000 |
| Blue-white | 10 000 to 30 000 |
| White | 7500 to 10 000 |
| Yellow-white | 6000 to 7500 |
| Yellow | 5000 to 6000 |
| Orange | 3500 to 5000 |
| Red | below 3500 |
Pattern
- Hotter stars tend to be white or blue and cooler stars tend to be red or yellow.
Seen colour
- The peak wavelength is not simply the colour seen, since a 3000 K star peaks in the infrared but still looks red.
Section 5 of 10Using the law on a star
1
Find the peak wavelength from the star's spectrum.
2
Convert the wavelength to metres (1 nm = 10−9 m).
3
Divide 2.9 × 10−3 m K by λmax to get T in kelvin.
4
Check the result is sensible, since main-sequence stars range from about 3000 K to over 30 000 K.
Surface temperature
- The law gives the photosphere temperature, not the core temperature, which is millions of kelvin.
Celsius
- Convert from °C to K before using the law.
Real spectra
- Absorption lines from cooler outer layers are present, but the overall envelope follows the black-body curve.
Section 6 of 10Reading intensity graphs
Hottest curve
- The curve with its peak furthest to the left is the hottest object.
Height
- The tallest curve is also the hottest, since peak intensity rises steeply with temperature.
Sketching
- A hotter body has a peak moved left and higher, and a cooler body has a peak moved right and lower.
Logarithmic axes
- Read the peak value carefully from the scale before applying the law.
Section 7 of 10Links
Radius of a star
- Combined with the Stefan-Boltzmann law and the luminosity, Wien's law allows the radius to be found.
Glowing metal
- A heated metal glows red, then orange, then white as its temperature rises.
Microwave background
- A peak near 1 mm corresponds to a temperature of about 2.7 K.
Section 8 of 10More from the specification
1Wien's displacement law
- Wien's displacement law states that the black-body radiation curve peaks at a wavelength lambda_max inversely proportional to thermodynamic temperature T: lambda_max * T = b ~ 2.898 x 10-3 m K.
2Stellar surface temperature and appearance
- In the constellation Orion, Betelgeuse appears distinctly red because its peak emission is at longer wavelengths (lambda_max) corresponding to a cooler surface temperature (~3500 K),
- whereas Rigel appears blue-white due to its much hotter surface temperature (~12 000 K).
3Stefan-Boltzmann law and stellar radii
- Combining Wien's law with the Stefan-Boltzmann equation, astronomers rearrange equations to determine stellar radii R = sqrt(L / (4 pi sigma T4) from measured luminosity and peak wavelength.
- (L = 4 pi R2 sigma T4
4Calculating temperature and radius ratios
- Wien's law equation can be expressed as a ratio T1 / T2 = lambda_max,2 / lambda_max,1,
- allowing astronomers to compare stellar surface temperatures and radii without knowing exact distances.
5Wien's displacement law
- Wien's displacement law states that the black-body radiation curve peaks at a wavelength lambda_max inversely proportional to thermodynamic temperature T: lambda_max * T = b ~ 2.898 x 10-3 m K.
6Stellar surface temperature and appearance
- In the constellation Orion, Betelgeuse appears distinctly red because its peak emission is at longer wavelengths (lambda_max) corresponding to a cooler surface temperature (~3500 K),
- whereas Rigel appears blue-white due to its much hotter surface temperature (~12 000 K).
7Stefan-Boltzmann law and stellar radii
- Combining Wien's law with the Stefan-Boltzmann equation, astronomers rearrange equations to determine stellar radii R = sqrt(L / (4 pi sigma T4) from measured luminosity and peak wavelength.
- (L = 4 pi R2 sigma T4
8Calculating temperature and radius ratios
- Wien's law equation can be expressed as a ratio T1 / T2 = lambda_max,2 / lambda_max,1,
- allowing astronomers to compare stellar surface temperatures and radii without knowing exact distances.
Section 9 of 10Key terms
| Term | Meaning |
|---|---|
| Black body | An object that absorbs all radiation falling on it and emits a spectrum determined by its temperature alone. |
| Peak wavelength (λmax) | The wavelength at which a black body emits its maximum intensity. |
| Wien's displacement law | The peak wavelength emitted by a black body is inversely proportional to its thermodynamic temperature. |
| Thermodynamic temperature | The absolute temperature in kelvin. |
| Photosphere | The visible surface layer of a star whose temperature the spectrum gives. |
Section 10 of 10Exam technique
1
Using the peak wavelength in nanometres without converting — the constant is in m K, so λ must be in metres.
2
Using temperature in degrees Celsius — Wien's law needs kelvin.
3
Thinking a hotter star has a longer peak wavelength — the relation is inverse, so hot means short wavelength.
4
Believing the peak wavelength is the colour we see — a 3000 K star peaks in the infrared but still looks red.
5
Treating the law as giving the core temperature — it gives the surface temperature.