Black-body radiation and stellar luminosityEdexcel International A Level Physics: Revision notes

Section 1

Black-body radiators

A black body radiator is an object that absorbs all the electromagnetic radiation that falls on it at every wavelength, and emits a continuous spectrum that depends only on its temperature, not on its material. Stars are close to black bodies.

A radiation curve plots the intensity (power per unit area) emitted at each wavelength against wavelength. Each curve rises to a single peak and falls again. As the temperature rises, the curve moves to shorter wavelengths and the intensity increases at every wavelength: the peak becomes higher and shifts towards the blue end, and the area under the curve (total power per unit area) increases greatly.

Key termsblack body radiatorradiation curve

Common mistake

Saying that a black body is always black. A hot black body glows; it is 'black' because it reflects nothing.

Section 2

Wien's law

Wien's law relates the wavelength of maximum intensity $\lambda_{max}$ to the thermodynamic temperature $T$ of the black body:

$\lambda_{max}T = 2.898 \times 10^{-3}$ m K

So $\lambda_{max} \propto 1/T$ : doubling the temperature halves the peak wavelength. Cooler stars (about 3500 K) peak in the red or infrared; hotter stars (about 10 000 K) peak in the blue or ultraviolet, so they look blue-white. Wien's law lets astronomers find the surface temperature of a star from its spectrum.

Worked example. A star's peak wavelength is 293 nm, so $T = 2.898 \times 10^{-3} / 2.93 \times 10^{-7} = 9.9 \times 10^3$ K. The Sun at 5800 K peaks at about 500 nm.

Key termsWien's lawpeak wavelength

Common mistake

Using temperatures in degrees Celsius. Wien's law and the Stefan–Boltzmann law need kelvin.

Section 3

The Stefan–Boltzmann law and luminosity

The luminosity $L$ of a star is the total power it radiates, in watts. For a black body of surface area $A$ at temperature $T$ :

$L = \sigma A T^4$

where $\sigma = 5.67 \times 10^{-8}$ W m⁻² K⁻⁴. For a spherical star $A = 4\pi R^2$ , so $L = 4\pi R^2\sigma T^4$ . Doubling the temperature increases the luminosity by a factor of $2^4 = 16$ , while doubling the radius increases it by a factor of 4.

Worked example. For $R = 6.96 \times 10^8$ m and $T = 5.80 \times 10^3$ K: $L = 5.67 \times 10^{-8} \times 4\pi \times (6.96 \times 10^8)^2 \times (5.80 \times 10^3)^4 = 3.9 \times 10^{26}$ W.

Key termsluminosityStefan–Boltzmann law

Exam tip

To compare two stars, use ratios: L₂/L₁ = (R₂/R₁)² × (T₂/T₁)⁴, so you do not need to calculate σ.

Section 4

Intensity at a distance

Radiation spreads out uniformly over a sphere, so the intensity $I$ (power per unit area, W m⁻²) at a distance $d$ from a star of luminosity $L$ is:

$I = \frac{L}{4\pi d^2}$

Intensity follows an inverse square law with distance. At the Earth, $d = 1.50 \times 10^{11}$ m, so for the Sun $I = 3.9 \times 10^{26} / (4\pi \times (1.50 \times 10^{11})^2) = 1.4 \times 10^3$ W m⁻². If the intensity and distance of a star are measured, its luminosity is $L = 4\pi d^2 I$ .

Key termsintensity

Section 5

Putting the laws together

Combining the laws lets astronomers find the radius of a star: use Wien's law to find $T$ from the peak wavelength, find $L$ from the measured intensity and distance ( $L = 4\pi d^2 I$ ), then $R = \sqrt{L / (4\pi\sigma T^4)}$ .

A cool star can be very luminous if it is huge. A red supergiant at 3500 K with $R = 5.5 \times 10^{11}$ m has $L = 3.2 \times 10^{31}$ W, about $8 \times 10^4$ times the Sun's, because its surface area is $6 \times 10^5$ times larger, which outweighs the $T^4$ factor of $0.13$ .

Key termsradius of a star

Exam tip

When both T and R differ, give the two factors separately: the area effect (R²) and the temperature effect (T⁴).

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Black-body radiation and stellar luminosity

A star of radius 6.96 × 10⁸ m has a surface temperature of 5.80 × 10³ K and may be treated as a black body radiator. A planet orbits it at a distance of 1.50 × 10¹¹ m. Use: Stefan–Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; Wien's law constant = 2.898 × 10⁻³ m K.
Calculate the intensity of the radiation from the star at the planet.2 marks
Two stars, A and B, have the same radius and may both be treated as black body radiators. The surface temperature of star A is 3500 K and that of star B is 7000 K. Use: Stefan–Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; Wien's law constant = 2.898 × 10⁻³ m K.
Explain why star B appears bluer than star A.2 marks
An astronomer studies a star that radiates as a black body with a luminosity of 2.5 × 10²⁸ W. The star is at a distance of 8.1 × 10¹⁶ m from the Earth. The wavelength at which the radiation from the star has maximum intensity is 293 nm. Use: Stefan–Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; Wien's law constant = 2.898 × 10⁻³ m K.
Calculate the intensity of the radiation from the star at the Earth.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).