Blackbody radiation

Builds on: Sections 4.1, 4.3 and S4
Reading time: 40 min
Exercises: 4
Olympiad problems: 2

In Orion, Betelgeuse at the top left shines orange and Rigel at the bottom right blue-white. The difference in colour is a difference in temperature. The surface of Betelgeuse is at about 3600 K and that of Rigel at about 12 000 K. A piece of iron in a fire shows the same thing, glowing dull red, then orange, then yellow-white as it grows hotter. In this section we describe the light given out by a hot body, find where its spectrum peaks and how much light it gives out in all, and use the two rules to find the sizes of stars we can never see as discs.

4.6.1The spectrum of a hot body

Before you readMake a guess first

A star has twice the surface temperature of the Sun and the same radius. How many times as much light does it give out?

A blackbody is an ideal body that absorbs all the light falling on it. When it is hot, the light it gives out has a spectrum that depends on its temperature alone, not on what it is made of. The inside of a star is very close to this ideal, since light there is absorbed and given out again many times before it escapes, and the spectra of stars are close to those of blackbodies. Max Planck found the shape of the spectrum in 1900, and to explain it he had to assume that energy is given out and absorbed in packets of $h c / λ$ , which later became the photons of Section 4.5.

ResultThe Planck law(4.10)

B_{λ} (T) = \frac{2 h c^{2}}{λ^{5}} \frac{1}{e^{h c / λ k T} - 1}

$B_{λ} (T)$	the power given out by each square metre of the surface, for each metre of wavelength, into each unit of solid angle, in W m⁻² m⁻¹ sr⁻¹
$λ$	the wavelength
$T$	the temperature of the body, in kelvin
$h, c, k$	the Planck constant, the speed of light and the Boltzmann constant, $1.381 \times 10^{- 23}$ J K⁻¹

At short wavelengths, where

h c / λ k T

is much more than 1, the

- 1

can be dropped, and

B_{λ}

falls off steeply as

e^{- h c / λ k T}

. At long wavelengths, where

h c / λ k T

is much less than 1,

e^{x} - 1 \approx x

and

B_{λ} \approx 2 c k T / λ^{4}

. Radio astronomers use this second form, which is proportional to the temperature.

Figure 4.3Blackbody spectra at different temperaturesinteractive

Temperature

Peak wavelength
Light from each square metre

The figure shows the power given out by each square metre of a blackbody for each nanometre of wavelength, against wavelength, for the chosen temperature as a heavy curve and for the Sun, at 5772 K, as a dotted curve. P is the peak of the heavy curve and S the peak of the Sun's. The two thin lines bound the visible light, from 400 to 700 nm. The slider sets the temperature, and the vertical scale changes to fit the taller curve. A body as cool as , peaks in the infrared and gives out little blue light. , peaks in the ultraviolet, and , gives out 19 times as much light from each square metre as the Sun.

Two features of the curves matter most. As the temperature rises, the peak moves to shorter wavelengths, and the whole curve rises at every wavelength, so the area under it grows quickly. Each feature has its own rule.

ResultWien's displacement law(4.11)

λ_{max} T = b = 2.898 \times 10^{- 3} m K

$λ_{max}$	the wavelength at which $B_{λ}$ is greatest
$T$	the temperature, in kelvin
$b$	Wien's constant

The law follows from setting the slope of the Planck curve to zero. It refers to the peak of the power for each unit of wavelength. A spectrum drawn for each unit of frequency peaks at a different place.

Example 4.11Where the Sun's light peaks

The Sun's surface has a temperature of 5772 K. At what wavelength does its spectrum peak?

Solution

By (4.11),

λ_{max} = \frac{2.898 \times 10^{- 3}}{5772} = 5.02 \times 10^{- 7} m = 502 nm .

AnswerThe Sun's light peaks at 502 nm, in the blue-green, near the middle of the band the eye is most sensitive to.

4.6.2The total light

The area under a Planck curve is the total power given out by each square metre of the surface, at all wavelengths. Adding it up gives a simple result, found by Josef Stefan from measurements in 1879 and derived by Ludwig Boltzmann. The power from each square metre grows as the fourth power of the temperature. A star of radius $R$ has a surface of area $4 π R^{2}$ , which gives its luminosity.

ResultThe Stefan-Boltzmann law(4.12)

\begin{aligned} F_{surf} & = σ T^{4} \\ L & = 4 π R^{2} σ T_{eff}^{4} \end{aligned}

$F_{surf}$	the power given out by each square metre of a blackbody's surface, in W m⁻²
$σ$	the Stefan-Boltzmann constant, $5.670 \times 10^{- 8}$ W m⁻² K⁻⁴
$L, R$	the luminosity and radius of a star
$T_{eff}$	the effective temperature of the star

A star is not exactly a blackbody, so the second line defines its effective temperature, the temperature of a blackbody of the same size and the same luminosity. For the Sun it is 5772 K.

Quick checkThe Sun's luminosity again

The Sun has radius $6.957 \times 10^{8}$ m and effective temperature 5772 K. What is its luminosity?

Comparing a star with the Sun is easier than putting in the constants, since $4 π σ$ cancels and $L / L_{⊙} = (R / R_{⊙})^{2} (T / T_{⊙})^{4}$ . A luminosity from the distance modulus of Section 4.3 and a temperature from the colour or spectrum together give the radius of a star, though even the largest stars are too small in angle to be seen as discs by most telescopes.

Example 4.12The size of a red supergiantfinish it

A red supergiant has a luminosity of $1.0 \times 10^{5} L_{⊙}$ and an effective temperature of 3600 K. What is its radius, in units of the Sun's radius?

The start

From (4.12), $R / R_{⊙} = \sqrt{L / L_{⊙}} (T_{⊙} / T)^{2}$ , and $\sqrt{1.0 \times 10^{5}} = 316$ .

Your stepNow multiply by

(T_{⊙} / T)^{2}

, with

T_{⊙} = 5772

solar radii

Common errorsSection 4.6

Temperatures in Wien's law and the Stefan-Boltzmann law must be in kelvin. A temperature in degrees Celsius gives nonsense, since the laws need the temperature counted from absolute zero.
Twice the temperature gives 16 times the light for each square metre, because the power from each square metre goes as $T^{4}$ .
Luminosity grows as $R^{2}$ , not as $R$ . A star twice the Sun's radius at the Sun's temperature gives out four times the Sun's light.

ExercisesSection 4.64 questions

4.6.1
At what wavelength does the spectrum of a star at 10 000 K peak?
nm
4.6.2
A star has the Sun's effective temperature and 100 times its luminosity. What is its radius in units of the Sun's?
solar radii
4.6.3
One star is three times as hot as another and has the same radius. How many times as luminous is it?
times
4.6.4
Two stars of the same size are seen, one blue-white and one orange. Which gives out more light?

Olympiad problemsPast IOAA problems, restated2 problems

2007A star at 4000 K2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.8. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

At what wavelength does a star with a surface temperature of 4000 K give out most strongly?

I solved it without the solution

AnswerThe star gives out most strongly at about 724 nm, at the red edge of visible light.

2013Distance, luminosity and radius of a star10 points

Adapted from IOAA 2013 (Volos), theory problem S6. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A star has effective temperature 8700 K, absolute magnitude 1.6 and apparent magnitude 7.2. Ignoring extinction, find (a) its distance, (b) its luminosity, and (c) its radius. Take the absolute magnitude of the Sun as 4.83 and its luminosity as $3.828 \times 10^{26}$ W.

I solved it without the solution

AnswerThe star is 132 pc away, about 20 times as luminous as the Sun at $7.5 \times 10^{27}$ W, and $1.36 \times 10^{9}$ m in radius, about twice the Sun's.