Black-body radiationEdexcel A-Level Physics: Revision notes

A black body is an idealised object that absorbs all electromagnetic radiation incident on it, at every wavelength, and at constant temperature emits a continuous spectrum that depends only on its temperature. Stars, and the interior of a furnace with a small opening, are good approximations.

A hot black body emits across a continuous range of wavelengths, so its radiation is not confined to one colour.

A radiation curve plots the intensity (power per unit area) against wavelength for a black body at a fixed temperature. Every curve rises to a single peak at the wavelength $\lambda_{max}$ and then falls.

As the temperature increases:

• the peak moves to a shorter wavelength
• the intensity increases at every wavelength, so curves for hotter bodies lie above those for cooler ones
• the area under the curve, the total power per unit area, increases rapidly

The total power radiated by a black body, its luminosity $L$, depends on its surface area $A$ and thermodynamic temperature $T$:

$L = \sigma A T^4$

where $\sigma = 5.67 \times 10^{-8}$ W m⁻² K⁻⁴ is the Stefan-Boltzmann constant. For a spherical star $A = 4\pi r^2$, so $L = 4\pi r^2 \sigma T^4$.

Worked example. For a radius of $7.0 \times 10^{8}$ m and temperature $5.8 \times 10^{3}$ K, $A = 6.2 \times 10^{18}$ m², so $L = 5.67 \times 10^{-8} \times 6.2 \times 10^{18} \times (5.8 \times 10^{3})^4 = 4.0 \times 10^{26}$ W.

Doubling $T$ increases $L$ by $2^4 = 16$.

The peak wavelength of a black body is inversely proportional to its thermodynamic temperature:

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

Worked example. The Sun's peak wavelength is $5.0 \times 10^{-7}$ m, so $T = 2.898 \times 10^{-3} / 5.0 \times 10^{-7} = 5.8 \times 10^{3}$ K.

Hotter stars peak at shorter wavelengths, so a star at $10\,000$ K (peak in the ultraviolet) looks blue-white, and one at $3000$ K (peak in the infrared) looks red.

Combining the laws lets you compare stars. Because $\sigma$ and $4\pi$ cancel,

$\dfrac{L_1}{L_2} = \left(\dfrac{r_1}{r_2}\right)^2 \left(\dfrac{T_1}{T_2}\right)^4$

A cool star can be extremely luminous if its radius is large. For example, with $r_1 = 100\,r_2$ and $T_1 = \frac{1}{2}T_2$, $L_1/L_2 = 100^2 \times 0.5^4 = 625$.

Wien's law gives the temperature from the colour, the Stefan-Boltzmann law then gives the luminosity from the temperature and radius, and either can be rearranged to find the radius.