Stefan-Boltzmann Law & Stellar Radii (CIE A Level Physics)

Revision Note

Test yourself

Author

Ashika

Last updated

29 October 2024

Stefan-Boltzmann law

A star’s luminosity depends on its:
- surface temperature
- radius

The relationship between these is known as the Stefan-Boltzmann law, which states:

The total energy emitted by a black body per unit area per second is proportional to the fourth power of the absolute temperature of the body

It is equal to:

$L = 4 π r^{2} σ T^{4}$

Where:
- L = luminosity of the star (W)
- r = radius of the star (m)
- σ = the Stefan-Boltzmann constant
- T = surface temperature of the star (K)

Estimating the radius of stars

The radius of a star can be estimated by combining Wien’s displacement law and the Stefan–Boltzmann law
The procedure for this is as follows:
- Use Wien’s displacement law to find the surface temperature of the star
- Use the inverse square law of flux equation to find the luminosity of the star (if given the radiant flux and stellar distance)
- Then, use the Stefan-Boltzmann law to calculate the radius of the star

Summary of Stellar Equations (1), downloadable AS & A Level Physics revision notes

Worked example

Betelgeuse is our nearest red giant star.

It has a luminosity of 4.49 × 10³¹ W and emits radiation with a peak wavelength of 850 nm.

Calculate the ratio of the radius of Betelgeuse r_B to the radius of the Sun r_s.

Radius of the sun r_s = 6.95 × 10⁸ m.

Answer:

Step 1: Write down Wien’s displacement law

$λ_{m a x} T = 2.9 \times 10^{- 3} m K$

Step 2: Rearrange Wien’s displacement law to find the surface temperature of Betelgeuse

$T = \frac{2.9 \times 10^{- 3}}{λ_{m a x}} = \frac{2.9 \times 10^{- 3}}{850 \times 10^{- 9}} = 3410 K$

Step 3: Write down the Stefan-Boltzmann law

$L = 4 π r^{2} σ T^{4}$

Step 4: Rearrange for r and calculate the stellar radius of Betelgeuse

$r_{B} = \sqrt{\frac{L}{4 π σ T^{4}}} = \sqrt{\frac{4.49 \times 10^{31}}{4 π \times (5.67 \times 10^{- 8}) \times {(3410)}^{4}}} = 6.83 \times 10^{11} m$

Step 5: Calculate the ratio $\frac{r_{B}}{r_{s}}$

$\frac{r_{B}}{r_{s}} = \frac{6.83 \times 10^{11}}{6.95 \times 10^{8}} = 983$

Therefore, the radius of Betelgeuse is about 1000 times larger than the Sun’s radius

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Test yourself Next topic

Did this page help you?

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.