Wien's displacement law describes how the peak wavelength of light emitted by a hot object depends on its temperature. It is especially important for understanding blackbody radiation, which is the ideal pattern of electromagnetic radiation from an object that absorbs and emits perfectly. As temperature increases, the brightest part of the emitted spectrum moves toward shorter wavelengths.
This helps explain why cool stars look red, hotter stars look white or blue, and very hot objects can glow beyond the visible range.
Key Facts
- Wien's displacement law: lambda_max = b / T
- Wien's constant: b = 2.898 x 10^-3 m K
- lambda_max is the wavelength where the blackbody spectrum has maximum intensity.
- Higher temperature means smaller lambda_max because lambda_max is inversely proportional to T.
- Lower temperature means larger lambda_max, so cooler objects peak farther into the infrared or red part of the spectrum.
- Temperature from peak wavelength: T = b / lambda_max
Vocabulary
- Blackbody
- An ideal object that absorbs all incoming radiation and emits radiation with a spectrum determined only by its temperature.
- Peak wavelength
- The wavelength at which an object emits the greatest intensity of radiation.
- Wien's displacement law
- A law stating that the peak wavelength of blackbody radiation is inversely proportional to the object's absolute temperature.
- Kelvin
- The SI unit of absolute temperature, where 0 K represents absolute zero.
- Spectrum
- The range of wavelengths or frequencies of electromagnetic radiation emitted or absorbed by an object.
Common Mistakes to Avoid
- Using Celsius instead of Kelvin, which is wrong because Wien's law requires absolute temperature in kelvin.
- Thinking hotter objects peak at longer wavelengths, which is wrong because lambda_max = b / T shows that peak wavelength decreases as temperature increases.
- Confusing peak wavelength with total brightness, which is wrong because Wien's law gives the location of the maximum, not the total emitted power.
- Assuming a star's visible color exactly equals its peak wavelength, which is wrong because stars emit a broad spectrum and human color perception combines many wavelengths.
Practice Questions
- 1 A star has a surface temperature of 5800 K. Use lambda_max = b / T with b = 2.898 x 10^-3 m K to find its peak wavelength in meters and nanometers.
- 2 An object has peak emission at 1.0 x 10^-6 m. What is its temperature in kelvin?
- 3 Two stars have temperatures of 3500 K and 12000 K. Explain which star would appear redder, which would appear bluer, and why Wien's displacement law predicts this.