P26: Black Body Radiation
Black body radiation, absorption and emission of radiation, the relationship between temperature and peak emission wavelength, and Earth's radiation balance.
What is a Black Body?
A perfect black body is an object that absorbs all the radiation incident upon it. No radiation is reflected or transmitted. A perfect black body is also the best possible emitter of radiation.
- A perfect black body absorbs all wavelengths of EM radiation that fall on it
- Since no radiation is reflected or transmitted, it appears completely black at low temperatures
- A perfect black body is the best possible emitter because good absorbers are also good emitters
- Stars can be approximated as black bodies
No real object is a perfect black body, but some materials come close. Black velvet absorbs about 99% of visible light. A small hole in a heated cavity behaves almost exactly like a perfect black body.
Absorption and Emission of Radiation
All objects continually absorb and emit infrared radiation. The temperature of an object depends on the balance between the radiation it absorbs and the radiation it emits.
- If an object absorbs more radiation than it emits, its temperature increases
- If an object emits more radiation than it absorbs, its temperature decreases
- If an object absorbs radiation at the same rate as it emits, its temperature stays constant (thermal equilibrium)
The type and intensity of radiation an object emits depends on its temperature. As temperature increases:
- The total power of radiation emitted across all wavelengths increases
- The peak wavelength of the emitted radiation decreases (shifts towards shorter wavelengths)
- The colour of a heated object changes from red to orange to yellow to white as temperature increases
Temperature and Peak Wavelength
As the temperature of a black body increases, the peak wavelength of the emitted radiation becomes shorter. This is known as Wien's displacement law.
- A cool object (e.g. a human body at 37 C) emits mainly long-wavelength infrared
- A hotter object (e.g. a hot poker at 800 C) emits visible light as well as infrared, appearing red
- At even higher temperatures (e.g. the Sun at 5800 K), the peak emission is in the visible range
- At extremely high temperatures (e.g. hot blue stars), peak emission shifts towards ultraviolet
| Object | Approximate Temperature | Peak Emission | Appearance |
|---|---|---|---|
| Human body | 37 C (310 K) | Far infrared | Invisible (IR only) |
| Hot iron | 800 C (1073 K) | Infrared / red light | Dull red glow |
| Light bulb filament | 2500 C (2773 K) | Near infrared | Orange-yellow glow |
| Sun (surface) | 5800 K | Visible (green-yellow) | White-yellow |
| Hot blue star | 20,000 K | Ultraviolet | Blue-white |
Black Body Radiation Curves
A black body radiation curve shows how the intensity of radiation varies with wavelength. The curve has a characteristic shape with a peak at the wavelength of maximum emission.
- At higher temperatures, the entire curve is higher (more intense radiation at all wavelengths)
- At higher temperatures, the peak shifts to shorter wavelengths
- The area under the curve represents the total power emitted per unit surface area
- At room temperature, the peak is in the infrared region and the object does not glow visibly
Stefan-Boltzmann Law
Total Power Emitted
P = sigma x A x T to the power 4
P = total power emitted (W)
sigma = Stefan-Boltzmann constant (5.67 x 10^-8 W/m2K4)
A = surface area of the object (m2)
T = absolute temperature (K)
The total power radiated by a black body is proportional to the fourth power of its absolute temperature. A small increase in temperature causes a large increase in the power radiated.
Worked Example
Calculate the total power radiated by a sphere of radius 0.1 m at a temperature of 1000 K.
Solution:
Surface area A = 4 pi r squared = 4 x 3.14 x 0.01 = 0.126 m2
P = sigma x A x T^4
P = 5.67 x 10^-8 x 0.126 x (1000)^4
P = 5.67 x 10^-8 x 0.126 x 10^12
P = 5.67 x 0.126 x 10^4 = 7144 W
Earth's Radiation Balance
The temperature of the Earth depends on the balance between the radiation it receives from the Sun and the radiation it emits back into space.
- The Sun emits short-wavelength radiation (visible light and UV) which passes through the atmosphere
- The Earth absorbs this radiation and warms up
- The Earth emits long-wavelength infrared radiation back towards space
- Some of this infrared radiation is absorbed by greenhouse gases in the atmosphere (CO2, methane, water vapour)
- The atmosphere then re-radiates some of this energy back towards the Earth's surface (the greenhouse effect)
If the balance is disturbed (e.g. by increasing greenhouse gas concentrations), the Earth's average temperature changes. More greenhouse gases mean more infrared is absorbed and re-emitted back to the surface, causing global warming.
Factors Affecting Earth's Temperature
| Factor | Effect on Temperature | Mechanism |
|---|---|---|
| Greenhouse gases increasing | Temperature rises | More infrared absorbed and re-emitted to surface |
| Increased cloud cover | Can increase or decrease | Clouds reflect sunlight (cooling) but also trap IR (warming) |
| Ice/albedo change | Feedback loop | Less ice means less reflection, more absorption, further warming |
| Volcanic eruptions | Temporary cooling | Aerosols reflect sunlight back to space |
| Solar activity changes | Varies with output | More solar radiation means more energy absorbed |
Practice Questions
1. Define what is meant by a perfect black body. [2 marks]
A perfect black body is an object that absorbs all the radiation incident upon it. No radiation is reflected or transmitted. It is also the best possible emitter of radiation.
2. Explain why a hot star appears blue-white while a cooler star appears red. [3 marks]
Hotter objects emit radiation with a shorter peak wavelength. A very hot star has its peak emission in the ultraviolet or blue end of the visible spectrum, so it appears blue-white. A cooler star has its peak emission at longer wavelengths in the red part of the visible spectrum.
3. Describe how the black body radiation curve changes when the temperature of an object increases. [3 marks]
As temperature increases, the intensity of radiation at all wavelengths increases (the entire curve is higher). The peak of the curve shifts to a shorter wavelength. The total area under the curve (total power emitted) increases.
4. Explain how increasing the concentration of greenhouse gases in the atmosphere affects the Earth's temperature. [3 marks]
Greenhouse gases absorb the long-wavelength infrared radiation emitted by the Earth. They then re-emit some of this radiation back towards the Earth's surface. With higher concentrations of greenhouse gases, more infrared is trapped and re-emitted, so the Earth's surface temperature increases.
5. An object absorbs 200 W of radiation and emits 150 W. Will its temperature increase, decrease, or stay the same? Explain your answer. [2 marks]
Its temperature will increase because it is absorbing more radiation (200 W) than it is emitting (150 W). The net energy gain causes the temperature to rise until absorption and emission are balanced.
6. The Sun emits radiation with a peak wavelength in the visible region. If the Sun's temperature were to decrease, what would happen to the peak wavelength and the total power emitted? [2 marks]
The peak wavelength would increase (shift towards longer wavelengths, into the infrared) and the total power emitted would decrease because power is proportional to T to the power 4.
Maths Skills
Interpreting Radiation Curves
- The x-axis of a black body radiation curve shows wavelength; the y-axis shows intensity (power per unit area per unit wavelength).
- The peak of the curve gives the wavelength of maximum emission.
- The area under the curve represents the total power radiated per unit surface area.
- When comparing curves at different temperatures, both the height and the peak position change.
Reading a Radiation Curve
A star has a radiation curve that peaks at a wavelength of 500 nm. Estimate its approximate temperature using the fact that the Sun (5800 K) peaks at about 500 nm.
Since this star has the same peak wavelength as the Sun, its temperature is approximately 5800 K. If the peak were at 250 nm (half the wavelength), the temperature would be roughly double (about 11,600 K), because peak wavelength is inversely proportional to temperature.
Wavelength-Temperature Relationship
Wien's Displacement Law
λmax ∝ 1 / T
If temperature doubles, the peak wavelength halves. If temperature triples, the peak wavelength becomes one third.
Worked Example
The Sun has a surface temperature of 5800 K and emits radiation with a peak wavelength of about 500 nm. A hot blue star has a surface temperature of 11,600 K. Calculate its peak wavelength.
λmax ∝ 1 / T, so if T doubles from 5800 K to 11,600 K, the peak wavelength halves.
λmax = 500 / 2 = 250 nm (in the ultraviolet region)
Stefan-Boltzmann Law Calculations
Comparing Power Output
Star A has a temperature of 4000 K and Star B has a temperature of 8000 K. Both have the same surface area. How many times more power does Star B emit?
Power ∝ T⁴. Ratio = (8000/4000)⁴ = 2⁴ = 16 times more power.
A relatively small change in temperature produces a very large change in power output because of the T⁴ relationship.
Common Misconceptions
"A black body is always black in colour." A black body is defined by its ability to absorb and emit all radiation perfectly, not by its colour. The Sun is approximately a black body — it absorbs virtually all radiation incident upon it and emits radiation according to its temperature. At high temperatures, a black body glows brightly (the Sun appears white-yellow, not black). The term 'black body' refers to its perfect absorption properties at low temperatures, when it would indeed appear black because it reflects nothing.
"The Earth is cooling down because it radiates energy into space." The Earth is approximately in thermal equilibrium. It absorbs energy from the Sun (mainly as visible light and UV) and emits roughly the same amount of energy back into space (as infrared). The Earth's average temperature remains roughly constant over time. However, if the balance is disturbed — for example, by increased greenhouse gas concentrations trapping more infrared — the Earth warms up until a new equilibrium is reached at a higher temperature.
"Objects only emit radiation when they are hot enough to glow." All objects above absolute zero (0 K) emit electromagnetic radiation. At room temperature, objects emit infrared radiation that we cannot see. An object only begins to glow visibly when its temperature exceeds about 800 K (dull red), but it was already emitting infrared radiation at lower temperatures. A human body at 37°C emits infrared radiation continuously, which is how thermal imaging cameras detect people in the dark.
6-Mark Extended Question
Explain how the balance between absorbed and emitted radiation determines Earth's temperature. Discuss how human activities affect this balance. [6 marks]
The Earth's temperature is determined by the balance between the radiation it absorbs and the radiation it emits. The Sun emits short-wavelength radiation (visible light and ultraviolet) which passes through the atmosphere and is absorbed by the Earth's surface, causing it to warm up. The Earth then emits long-wavelength infrared radiation back towards space. However, greenhouse gases in the atmosphere — including carbon dioxide, methane and water vapour — absorb some of this infrared radiation and re-emit it in all directions, including back towards the Earth's surface. This is the greenhouse effect and it keeps the Earth warm enough to sustain life. When the Earth is in equilibrium, the total power absorbed equals the total power emitted, and the temperature remains stable. Human activities are disturbing this balance by increasing the concentration of greenhouse gases in the atmosphere. Burning fossil fuels releases CO2; agriculture and landfill produce methane; and deforestation reduces the absorption of CO2 by plants. More greenhouse gases mean more infrared radiation is absorbed and re-emitted back to the surface, so the Earth emits slightly less radiation than it absorbs and its temperature rises. This is global warming. As the Earth warms, ice caps melt, reducing the albedo (reflectivity) of the surface, meaning more solar radiation is absorbed — a positive feedback loop that amplifies the warming. The Earth will only reach a new equilibrium at a higher temperature.
AO3: Analyse and Evaluate
The diagram below shows black body radiation curves for two stars. Star A has a surface temperature of 3000 K. Star B has a surface temperature of 6000 K.
- Star A: peak wavelength = 966 nm (infrared), low overall intensity
- Star B: peak wavelength = 483 nm (visible blue-green), much higher overall intensity
(a) Explain why Star B's peak wavelength is approximately half that of Star A.
(b) The area under Star B's curve is 16 times larger than under Star A's curve. Show that this is consistent with the Stefan-Boltzmann law, assuming both stars have the same surface area.
(c) A student claims: "Star A emits no visible light so it would be invisible to the human eye." Evaluate this claim using the radiation curve data.
(a) Wien's displacement law states that peak wavelength is inversely proportional to absolute temperature: λmax ∝ 1/T. Star B's temperature is twice that of Star A (6000 K / 3000 K = 2). Therefore, Star B's peak wavelength should be half that of Star A: 966 / 2 = 483 nm, which matches the data.
(b) The Stefan-Boltzmann law states that total power ∝ T⁴. The ratio of temperatures is 6000/3000 = 2. The ratio of powers should be 2⁴ = 16. This matches the observation that the area under Star B's curve is 16 times larger, confirming the Stefan-Boltzmann law.
(c) The student's claim is not entirely correct. While Star A's peak emission is in the infrared (966 nm), the radiation curve shows that it still emits some radiation across all wavelengths, including the visible range (roughly 400–700 nm). The tail of the curve extends into the visible region. Star A would emit some red light, so it would appear as a dim, red-coloured star to the human eye. However, it would be much dimmer in visible light than Star B because its peak is in the infrared. The claim that it is "invisible" is incorrect — it would simply appear very red and relatively dim compared to hotter stars.
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