Unit 15 · Topic 15.4

15.4 Blackbody Radiation

Every object turns some of its thermal energy into electromagnetic radiation. An ideal blackbody absorbs all the radiation that hits it and gives off a continuous spectrum that depends only on its temperature. Hotter objects peak at shorter wavelengths (Wien's law, $\lambda_{\max} = \frac{b}{T}$ ) and give off far more power ( $P = A\sigma T^4$ ). Explaining this spectrum required Planck's idea that light energy is quantized.

Key terms

blackbody
thermal radiation
peak wavelength
Wien's law
Stefan-Boltzmann law

Thermal radiation and blackbodies

Everything above absolute zero glows. You glow in the infrared; a stove burner glows red; the Sun glows white-yellow. This is thermal radiation: the jiggling charged particles in any warm object send out electromagnetic waves, and that energy comes out of the object's thermal energy.

A blackbody is an idealized object that absorbs all radiation falling on it, reflecting none. If it stays at a constant temperature, it must also give off energy as fast as it absorbs it. Its emitted spectrum is continuous (all wavelengths, no lines) and depends only on its temperature, not on what it's made of. Stars are close to blackbodies.

The blackbody curve

The spectrum is drawn as intensity per unit wavelength (vertical axis) against wavelength (horizontal axis). The curve starts near zero at very short wavelengths, rises steeply to a peak, then trails off slowly toward long wavelengths.

As temperature rises, the curve is higher at every wavelength, and its peak moves to shorter wavelengths. The curve for a hotter object completely encloses the curve for a cooler one; they never cross.

That's why a heated metal rod glows dull red, then orange, then yellow-white as it gets hotter: the peak slides from the infrared into the visible range.

Wien's law

The peak wavelength is inversely proportional to the absolute temperature: $\lambda_{\max} = \frac{b}{T}$ , where b = 2.90 × 10⁻³ m·K is Wien's constant. Double the kelvin temperature and the peak wavelength halves.

The Sun's surface (about 5800 K) peaks near 500 nm, in the visible range. Your body (about 310 K) peaks near 9.4 μm, in the infrared, which is why thermal cameras work. Infrared telescopes are cooled so that their own thermal glow doesn't swamp the faint infrared light they're trying to detect.

The Stefan-Boltzmann law

The total power radiated by a blackbody is $P = A\sigma T^4$ , where A is its surface area and σ = 5.67 × 10⁻⁸ W/(m²·K⁴). Power grows with area and very steeply with temperature: double the temperature and the power goes up 2⁴ = 16 times.

Temperatures in both laws must be in kelvin: T(K) = T(°C) + 273.

Why it needed quantum ideas

Classical physics predicted that a blackbody should give off more and more intensity at shorter and shorter wavelengths, without limit. That would mean infinite energy in the ultraviolet, a failure nicknamed the ultraviolet catastrophe. Real curves drop back to zero at short wavelengths.

In 1900, Max Planck fixed this by assuming that light energy comes only in chunks of size hf. High-frequency chunks are so large that a hot object rarely has enough energy to make them, so the short-wavelength end is cut off. The formula he got from that assumption, called Planck's law, matched the measured curves and started quantum theory.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Peak wavelength of the Sun
The Sun's surface temperature is about 5800 K. At what wavelength does its spectrum peak?
Show the solutionHide the solution
1. Step 1: $\lambda_{\max} = \dfrac{b}{T} = \dfrac{2.90 \times 10^{-3}\text{ m·K}}{5800\text{ K}} = 5.0 \times 10^{-7}$ m = 500 nm.
2. Step 2: That's in the visible range (blue-green), close to the middle of what our eyes detect.
Answer: 500 nm
Example 2
Factor of change for two stars
Star X has twice the surface temperature of star Y, and both have the same radius. Compare their peak wavelengths and their radiated powers. What if instead they had the same temperature but X had 3 times the radius?
Show the solutionHide the solution
1. Step 1: Peak wavelength: $\lambda_{\max} \propto \frac{1}{T}$ , so X's peak is half of Y's.
2. Step 2: Power: $P \propto T^4$ at equal area, so X radiates 2⁴ = 16 times as much power.
3. Step 3: Same temperature, 3 times the radius: area goes as r², so the area is 9 times bigger and the power is 9 times bigger. The peak wavelength is unchanged, because it depends only on temperature.
Answer: Twice the temperature: half the peak wavelength, 16 times the power. Three times the radius: same peak, 9 times the power.
Example 3Calculator allowed
Power from a hot sphere (kelvin trap)
A small blackbody sphere of radius 5.0 cm is held at 527°C. Find the power it radiates and its peak wavelength.
Show the solutionHide the solution
1. Step 1: Convert to kelvin first: T = 527 + 273 = 800 K. Using 527 in the T⁴ term would give an answer more than 5 times too small.
2. Step 2: Surface area: A = 4πr² = 4π(0.050 m)² = 0.0314 m².
3. Step 3: $P = A\sigma T^4 = (0.0314)(5.67 \times 10^{-8})(800)^4 = 730$ W.
4. Step 4: $\lambda_{\max} = \dfrac{2.90 \times 10^{-3}}{800} = 3.6 \times 10^{-6}$ m = 3.6 μm, in the infrared. It would barely glow visibly at all.
Answer: About 730 W; peak at 3.6 μm (infrared).

Common mistakes

Using Celsius in Wien's law or $P = A\sigma T^4$ . Always convert to kelvin.
Drawing hotter blackbody curves that cross cooler ones, or that peak at longer wavelengths. A hotter curve is higher everywhere and peaks farther left.
Treating power as proportional to T instead of T⁴. Doubling T multiplies the power by 16.
Saying a blackbody looks black. It absorbs everything, but a hot blackbody glows brightly (the Sun is a good example).

On the exam

Expect to sketch or interpret blackbody curves at two temperatures, and to justify which is hotter using the peak location and overall height.
Factor-of-change questions on power and peak wavelength are common. State the relationships ( $\lambda_{\max} \propto \frac{1}{T}$ , $P \propto AT^4$ ) as your reasoning.

Connected topics

Videos

AP Physics 2 - Unit 15 - Lesson 8 - Blackbody Radiation
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Blackbody radiation | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Quantization of Energy Part 1: Blackbody Radiation and the Ultraviolet Catastrophe
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The ULTRAVIOLET CATASTROPHE
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Max Planck Solves the Ultraviolet Catastrophe for Blackbody Radiation | Doc Physics
Doc SchusterWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 15.4 Blackbody Radiation. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A blackbody's absolute temperature is doubled while its size stays the same. By what factor does the power it radiates increase?

Question 2 of 4Calculator allowed

The spectrum of star X peaks at 400 nm, and the spectrum of star Y peaks at 800 nm. Treating both as blackbodies, how do their surface temperatures compare?

Question 3 of 4Calculator allowed

Wien's law is $\lambda_{\max} T = 2.90 \times 10^{-3}$ m·K. At about what wavelength does the thermal radiation from an object at 300 K (room temperature) peak?

Question 4 of 4Calculator allowed

Two stars have the same surface temperature, but star B has twice the radius of star A. How does the power radiated by star B compare with star A?

0 of 4 answered