Wien's law states that the peak wavelength of the black body spectrum is inversely

Astrophysics & Cosmology - OCR A-Level Physics

$$\lambda_{max} T = 2.898 \times 10^{-3} \text{ m K}$$
  • Wien's lawThe peak wavelength of a black body spectrum is inversely proportional to temperature: λ_max = constant / T. states that the peak wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of the black body spectrum is inversely proportional to the surface temperature.
  • Hotter stars have shorter peak wavelengths: they appear blue/white.
  • Cooler stars have longer peak wavelengths: they appear red/orange.
  • The Sun's surface temperature (~5800 K) $gives \lambda_{max}$ $\approx 500$ nm (green/yellow light).
  • This law allows us to determine a star's surface temperature from its spectrum without visiting the star.
Worked Example
A star has a peak emission wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of 350 nm. Calculate its surface temperature.
Show Solution
1
Convert wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).

$\lambda_{max} = 350 \times 10^{-9} m.$

2
Apply Wien's law

$T = 2.898 \times 10^{-3} / \lambda_{max}.$

3

$T = 2.898 \times 10^{-3} / (350 \times 10^{-9}).$

4

$T = 2.898 \times 10^{-3} / (3.50 \times 10^{-7}).$

5

$T = 8280 K \approx 8300 K.$

Answer
$$T \approx 8300$ K (a hot blue-white star)$
Common Mistake MEDIUM
Wrong: Confusing \lambda_{max} with the maximum wavelength emitted. A star emits across a wide range of wavelengths.
Right: \lambda_{max} is the wavelength at which the intensityThe powerThe rate of energy transfer. Measured in watts (W). transmitted per unit area perpendicular to the wave direction. Measured in W m⁻². Proportional to amplitude squared. of emission is GREATEST (the peak of the black body curve), not the longest wavelength emitted.
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