25.2 Stellar radii
- Syllabus
- 9702–2028–2029
- Topic
- 25.2
- Level
- A2
Learning objectives
25.2.1Wien’s displacement law λmax ∝ 1 / T to estimate the peak surface• recall and use Wien’s displacement law λmax ∝ 1 / T to estimate the peak surface temperature of a star25.2.2The Stefan–Boltzmann law L = 4 πσr 2 T 4• use the Stefan–Boltzmann law L = 4 πσr 2 T 425.2.3Wien’s displacement law and the Stefan–Boltzmann law to estimate the radius• use Wien’s displacement law and the Stefan–Boltzmann law to estimate the radius of a star
Wien's law uses the emission peak to estimate stellar surface temperature
λmaxT=b≈2.90×10−3mKT=b/λmax
T is the thermodynamic surface temperature of the star and λmax is the wavelength at which its emitted spectral intensity/rate is maximum.
For λmax=624 nm, T=(2.90×10⁻³)/(624×10⁻⁹)=4.65×10³ K.
A cooler star peaks at a longer wavelength and has a lower thermal-emission curve; a hotter star peaks at a shorter wavelength.
Use the emitted/rest-frame peak wavelength. If redshift makes the observed λmax too large and is ignored, Wien's law gives a temperature that is too low. Convert nm to m and use kelvin.
Stefan-Boltzmann links luminosity to stellar area and T⁴
L=4πr2σT4σ=5.67×10−8Wm−2K−4
A star's surface emits power per unit area σT⁴; multiplying by spherical surface area 4πr² gives total luminosity L.
r=√[L/(4πσT4)]T=[L/(4πσr2)](1/4)
For the Sun L=3.85×10²⁶ W and T=5780 K, r=√{3.85×10²⁶/[4π(5.67×10⁻⁸)(5780)⁴]}=6.96×10⁸ m.
At fixed radius, L∝T⁴; at fixed temperature, L∝r². Doubling T multiplies L by 16.
Use radius, not diameter, and thermodynamic temperature in kelvin. Keep 4π and apply the square or fourth root only after isolating r² or T⁴.
Estimate stellar radius by combining spectrum temperature with luminosity
- Read the star's emitted peak wavelength λmax and use Wien's law for T. 2. Obtain luminosity L independently, for example from measured flux and known distance. 3. Use Stefan-Boltzmann to solve r.
T=b/λmaxL=4πd2Fr=√[L/(4πσT4)]
If λmax=415 nm, T=(2.90×10⁻³)/(415×10⁻⁹)=6.99×10³ K.
With L=5.48×10²⁶ W, r=√{5.48×10²⁶/[4π(5.67×10⁻⁸)(6.99×10³)⁴]}=5.67×10⁸ m.
For equal luminosity, a hotter star must have a smaller radius because each square metre emits much more power (∝T⁴).
Peak wavelength determines T, not r by itself. Do not substitute radiant flux F at Earth where luminosity L is required; first convert using distance if needed.