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Physics (IAL)

Black Body Radiation

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Edexcel IAL Physics · Thermal Radiation

Black Body Radiation

🔑 The Big Idea: Every object radiates energy as electromagnetic waves just because it has a temperature — and the hotter it is, the more energy it pumps out, and the shorter (bluer) the wavelength of its peak radiation becomes.
Summary — What This Chapter Covers
  • All objects, no matter their temperature, emit black body radiation — thermal energy in the form of EM waves.
  • A perfect black body absorbs 100% of radiation hitting it (reflects/transmits none) — and is therefore also the best possible emitter.
  • Black bodies produce a characteristic intensity vs wavelength curve that depends only on temperature.
  • The Stefan-Boltzmann Law tells us the total power radiated: L = σAT⁴.
  • Wien's Law tells us where the peak of that curve sits: λ_max T = 2.9 × 10⁻³ m K.
  • Hotter objects emit shorter, more energetic wavelengths (shift towards blue/UV); cooler objects emit longer wavelengths (shift towards red/IR).
  • Stars are the closest real-world approximation to perfect black bodies — this is how astronomers estimate stellar surface temperatures from colour alone.
Core Topics

1What Is Black Body Radiation?

Here's the surprising bit: you don't need to set something on fire for it to give off radiation. Anything with a temperature above absolute zero radiates energy — your body, a cup of tea, a brick wall, the Sun, all of it. This radiation is called black body radiation, and it's just thermal energy escaping as electromagnetic waves.

Most everyday objects (room temperature, your body, a warm mug) radiate mostly in the infrared part of the spectrum — invisible to our eyes but detectable with a thermal camera. But as an object gets hotter, two things happen simultaneously:

  • It radiates more total energy per second (it gets "louder" overall).
  • The peak wavelength shifts shorter — towards visible light, then towards blue/UV if it's hot enough (like a star).
Analogy: Think of a metal poker heated in a fire. At first it just feels warm (infrared only, invisible). Heat it more and it starts to glow dull red. Heat it further and it glows orange, then yellow-white. You're literally watching the peak wavelength of its black body spectrum slide down through the visible spectrum as temperature rises — this is black body radiation in action, right in front of you.

The Perfect Black Body — Definition

Definition to memorise word-for-word

A perfect black body is an object that absorbs all the radiation incident on it, and does not reflect or transmit any radiation.

Here's the clever logic chain examiners love to test: a good absorber is also a good emitter. So a perfect black body — being the best possible absorber — is automatically the best possible emitter too. And since black objects are what you get when all visible colours are absorbed (no light bounces back to your eye), a perfect black body would visually appear black at low temperatures. That's where the name comes from — it's not about being literally black in colour at all temperatures, it's about being a perfect absorber/emitter.

ColourAbsorbingEmitting
BlackGood absorberGood emitter
Dull / DarkReasonable absorberReasonable emitter
WhitePoor absorberPoor emitter
ShinyVery poor absorber (good reflector)Very poor emitter
Quick check
Why do shiny/silver surfaces feel like they stay cool for longer and also keep food warm in foil? Because shiny = poor absorber AND poor emitter — it reflects radiation away (stops it getting in) and struggles to radiate energy away (stops it getting out). Same property, two useful jobs!
Practice Question
Q1. Explain why a matte black kettle would lose heat faster (once boiled and left to cool) than a shiny, polished silver kettle of the same size and starting temperature.

2Black Body Radiation Curves

If you plot intensity (y-axis) against wavelength (x-axis) for the radiation coming off an object at a fixed temperature, you get a smooth, skewed hump-shaped curve. This is the black body spectrum, and its exact shape depends on one thing only: temperature.

As temperature increases: the curve gets taller (more total energy) AND the peak shifts to a shorter wavelength.

Two separate effects are packed into this single curve, and exam questions love to test whether you can spot both:

  • Effect 1 — Total area under curve increases with T. This is captured by the Stefan-Boltzmann Law (total power radiated).
  • Effect 2 — Peak wavelength shifts shorter as T increases. This is captured by Wien's Law (colour/peak position).

Recall from the EM spectrum: shorter wavelength = higher energy (this is why UV and X-rays are dangerous, but radio waves aren't). So as an object heats up and its peak wavelength shrinks, it's not just glowing brighter — the individual photons it's giving off are carrying more energy too.

Analogy: Picture a crowd doing "the wave" at a stadium. A cold, sluggish crowd (low T) makes a slow, low wave that barely rises — long "wavelength" between peaks, low energy. A hyped-up crowd (high T) makes the wave ripple through fast and violently — short "wavelength," high energy, and the whole stadium (total intensity) is roaring louder overall. Same idea: heat something up, and both the "loudness" (total power) and the "speed of the ripple" (shortness of peak wavelength) increase together.
Real-world connection: stars
No object is a truly perfect black body — but stars are the closest real approximation we have. This is exactly why astronomers can point a telescope at a star, measure the wavelength where its spectrum peaks, and calculate its surface temperature using nothing but Wien's Law. Blue stars are scorching hot; red stars are (relatively) cool.
Practice Question
Q2. Sketch (in words) what happens to a black body radiation curve if the temperature of the object is doubled. Describe two separate changes.

3The Stefan-Boltzmann Law

This law answers the question: "How much total power does an object radiate?" It turns out this depends on just two things — how hot the object is, and how much surface area it has to radiate from (makes sense: more surface = more "exits" for the energy to escape through).

Stefan-Boltzmann Law
L = σAT⁴

In plain words: the total energy radiated by a black body, per unit area, per second, is proportional to the fourth power of its absolute temperature.

L = luminosity / total power radiated (W)
σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
A = surface area of the object (m²)
T = absolute (Kelvin) surface temperature (K)
Surface area of a sphere (stars, planets)
A = 4πr²
r = radius of the sphere (m)

The T⁴ is the part that trips people up because it's so extreme. If you double an object's absolute temperature, you don't double its power output — you multiply it by 2⁴ = 16. Triple the temperature and power goes up by 3⁴ = 81 times! This is why the Sun's surface, at ~5800 K compared to Earth's ~290 K (roughly 20× hotter), radiates roughly 20⁴ ≈ 160,000 times more power per square metre than the Earth's surface does.

Common mistake
Students often plug in temperature in °C by accident. The Stefan-Boltzmann Law (and Wien's Law) only works with absolute temperature in Kelvin. Always convert first: T(K) = T(°C) + 273.

Worked Example

Worked Example (from the textbook)
A camel has a body temperature of 40°C and a surface area of 16 m². Calculate the total power radiated by the camel.
Practice Question
Q3. A star has a surface temperature of 6000 K and a radius of 7.0 × 10⁸ m. Calculate its luminosity. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

4Wien's Law

While the Stefan-Boltzmann Law tells us how much energy is radiated in total, Wien's Law tells us where the peak of the black body curve sits — in other words, which wavelength carries the most intensity, which roughly tells us what colour the object glows.

Wien's Displacement Law
λmax T = 2.9 × 10⁻³ m K

In plain words: the peak wavelength of a black body's spectrum is inversely proportional to its absolute temperature — hotter objects peak at shorter wavelengths.

λmax = wavelength at peak intensity (m)
T = absolute surface temperature (K)

Because it's an inverse relationship (λ_max ∝ 1/T), as T goes up, λ_max must come down — and vice versa. This single equation is the reason we can look at a star's colour and immediately estimate its temperature without ever touching it.

Colour of starApprox. Temperature / K
Blue> 33,000
Blue-white10,000 – 30,000
White7,500 – 10,000
Yellow-white6,000 – 7,500
Yellow (like our Sun)5,000 – 6,000
Orange3,500 – 5,000
Red< 3,500
Analogy: Think of tuning a radio dial. As you heat an object up, it's like the "broadcast frequency" of its peak radiation is being tuned steadily from long-wave (infrared, radio-ish) toward short-wave (blue, then UV). A red star is broadcasting on a lower "channel" than a blue star — the blue star is simply hotter and pumping its peak energy out at shorter wavelengths.

Worked Example

Worked Example (from the textbook)
The spectrum of the star Rigel peaks at 263 nm, while Betelgeuse peaks at 828 nm. Which star is cooler?
Practice Question
Q4. A light bulb filament peaks in intensity at a wavelength of 1160 nm. Calculate the filament's surface temperature.
Watch the units!
Wavelengths in these problems are usually given in nm or μm — always convert to metres before substituting into Wien's Law (1 nm = 1×10⁻⁹ m, 1 μm = 1×10⁻⁶ m). This is one of the most common places marks get dropped in exams.
What to Memorise

Perfect Black Body

Absorbs all incident radiation; reflects/transmits none. Best possible absorber = best possible emitter.

Stefan-Boltzmann Law

L = σAT⁴ — total power radiated. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Power ∝ T⁴ (extremely sensitive to temperature).

Sphere Surface Area

A = 4πr² — needed to find A for stars/planets before using Stefan-Boltzmann.

Wien's Displacement Law

λ_max T = 2.9 × 10⁻³ m K — peak wavelength is inversely proportional to temperature.

Always Use Kelvin

T(K) = T(°C) + 273 — both laws require absolute temperature, never Celsius.

Colour-Temperature Link

Hotter → shorter peak wavelength → white/blue. Cooler → longer peak wavelength → red/yellow.

Concepts Checklist
Exam Tips & Common Mistakes

Forgetting to convert to Kelvin

The single most common lost mark. Both L = σAT⁴ and λ_max T = 2.9×10⁻³ m K require absolute temperature. If a question gives °C, add 273 first — every time, no exceptions.

Wavelength unit errors

Values are often given in nm or μm but the constant 2.9×10⁻³ m K requires λ_max in metres. Always convert before substituting, and double check your final answer's order of magnitude makes sense.

Underestimating the power of T⁴

Students often treat the Stefan-Boltzmann Law as if power scales linearly with temperature. Remember: doubling T means multiplying power by 16, not 2. Examiners love asking "by what factor does power change if T doubles/triples?" — always raise the temperature ratio to the 4th power.

Confusing "black" with the colour black

A "black body" isn't necessarily black in appearance — the Sun is a near-perfect black body but obviously isn't black! The term refers to its perfect absorption/emission property, not its visual colour at all temperatures.

Mixing up which law answers which question

If the question asks for total power/energy/luminosity → use Stefan-Boltzmann (L = σAT⁴). If the question asks about peak wavelength or colour → use Wien's Law (λ_max T = 2.9×10⁻³ m K). Read the question carefully to spot which quantity is actually being asked for.

Examiner's favourite phrasing
Watch for questions that give you two stars' peak wavelengths and ask you to compare their temperatures, colours, or which is more luminous — these often combine Wien's Law AND Stefan-Boltzmann Law in a single multi-step question. Work out T first using Wien's Law, then plug that T into Stefan-Boltzmann if luminosity or power is also asked for.
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