What is Black Body Radiation in the context of Edexcel International A Levels Physics?

Black Body Radiation refers to the theoretical concept of an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. In the context of Edexcel International A Levels Physics, it is studied under Thermodynamics and Radiation, where students learn how such a body emits radiation at a characteristic spectrum that depends solely on its temperature, known as Planck's law.

How does Planck's Law relate to Black Body Radiation?

Planck's Law is fundamental to understanding Black Body Radiation. It describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature. This law is crucial in Edexcel International A Levels Physics as it explains how the intensity of radiation emitted varies with wavelength and temperature, providing insights into the distribution of energy across different wavelengths.

Why is the concept of a black body important in Thermodynamics and Cosmology?

The concept of a black body is important because it serves as a model for understanding how objects emit radiation. In Thermodynamics, it helps explain energy transfer processes, while in Cosmology, it is used to study cosmic microwave background radiation, which is a remnant from the early universe. This concept is integral to the Edexcel International A Levels curriculum as it bridges the gap between theoretical physics and real-world applications.

What is the significance of Wien's Displacement Law in Black Body Radiation?

Wien's Displacement Law states that the wavelength at which the emission of a black body spectrum is maximized is inversely proportional to its temperature. This law is significant because it allows us to determine the temperature of stars and other celestial bodies by analyzing their emitted radiation. In the Edexcel International A Levels Physics curriculum, this law is crucial for understanding the relationship between temperature and radiation wavelength.

How does the Stefan-Boltzmann Law apply to Black Body Radiation?

The Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature. This law is pivotal in the study of Black Body Radiation as it quantifies the power emitted by a black body, which is essential for calculations in Thermodynamics and Cosmology. In Edexcel International A Levels Physics, understanding this law helps students analyze energy emission and absorption processes.

Why is "Substituting temperature in degrees Celsius instead of kelvin" a common mistake in Black Body Radiation?

Why it happens: Data are sometimes quoted in °C, and students substitute the number directly without converting. How to avoid it: Both Wien's law and the Stefan–Boltzmann law require the **absolute temperature in kelvin**. Convert with T/K = /°C + 273 before substituting. Using °C is disastrous in T^4 because of the fourth power. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Treating $\lambda_{max}$ as directly proportional to $T$ (multiplying instead of dividing)" a common mistake in Black Body Radiation?

Why it happens: Students remember ‘\lambda_max and T are related’ but forget it is the *product* that is constant. How to avoid it: Wien's law is \lambda_maxT = constant, so \lambda_max 1/T — a **hotter** body peaks at a **shorter** wavelength. To find \lambda_max, *divide* the constant by T; to find T, divide the constant by \lambda_max. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Forgetting the fourth power of $T$ in the Stefan–Boltzmann law (or the square in $4\pi r^2$)" a common mistake in Black Body Radiation?

Why it happens: The exponents are easy to drop when copying the formula, and calculators hide the slip. How to avoid it: Write L = A T^4 in full and raise T to the **fourth** power; use A = 4 r^2 (the radius is **squared**). A quick sanity check: doubling T should multiply the power by 2^4 = 16. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Using the wrong surface area for a star — e.g. $\pi r^2$ (a disc) instead of $4\pi r^2$ (a sphere)" a common mistake in Black Body Radiation?

Why it happens: The circular cross-sectional area r^2 is more familiar, so it is used by mistake. How to avoid it: A star radiates from its **whole spherical surface**, so A = 4 r^2. The disc area r^2 is for the light *intercepted* by a target, not the area emitting; for luminosity always use 4 r^2. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Confusing luminosity (total power radiated) with the brightness observed from Earth" a common mistake in Black Body Radiation?

Why it happens: Both are loosely called ‘brightness’, so students assume the brightest-looking star is the most luminous. How to avoid it: **Luminosity** is the *total* power a star radiates (an intrinsic property, L = A T^4). **Apparent brightness** is the intensity that reaches us, which also depends on distance ( 1/d^2). A near, faint star can outshine a distant, luminous one. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Leaving the wavelength in nanometres (or micrometres) when using Wien's law" a common mistake in Black Body Radiation?

Why it happens: Peak wavelengths are naturally quoted in nm, and the number is substituted as it appears. How to avoid it: Convert to metres before substituting: 1\ nm = 10^-9\ m, 1\ m = 10^-6\ m. Since Wien's constant is 2.90×10^-3\ m K, a wavelength left in nm gives a temperature 10^9 times too large. Source: Edexcel IAL Physics Unit 5 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Black Body Radiation

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Short Notes - Black Body Radiation

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Detailed Study Notes

Detailed notes on Thermodynamics, Radiation, Oscillations and Cosmology for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Black Body Radiation — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 5)

What a black body is and why stars behave like one; how the intensity–wavelength curves change with temperature; Wien's displacement law ( $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ ) for finding peak wavelengths and stellar surface temperatures; the Stefan–Boltzmann law ( $L = \sigma A T^4$ , with $A = 4\pi r^2$ for a star) for luminosity; and how the two laws combine to estimate a star's radius — the foundation for Astronomy and Cosmology.

At a glance

A black body absorbs (and emits) radiation of every wavelength perfectly. Stars are close approximations, so their spectra follow the ideal black-body curve set by surface temperature.
Intensity–wavelength curves: each has one peak; higher temperature → higher peak AND peak at shorter wavelength; the curves never cross (a hotter body emits more at every wavelength).
Wien's displacement law: $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ . The product is constant, so $\lambda_{max}\propto 1/T$ — divide the constant by $T$ (or by $\lambda_{max}$ ) to find the other.
Stefan–Boltzmann law: $L = \sigma A T^4$ with $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . Strong $T^4$ dependence — doubling $T$ multiplies power by $16$ .
A star is a sphere: its radiating area is $A = 4\pi r^2$ , so $L = 4\pi r^2 \sigma T^4$ .
Estimate a star's radius: Wien's law gives $T$ from $\lambda_{max}$ , then $r = \sqrt{L/(4\pi\sigma T^4)}$ gives $r$ from $L$ and $T$ .
Always use $T$ in kelvin, convert wavelengths to metres, keep full precision through the fourth power, and quote answers to 3 s.f.
Luminosity ≠ apparent brightness: luminosity is the total power radiated; observed brightness also depends on distance ( $\propto 1/d^2$ ).

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Understand that a black body is a perfect absorber and emitter of radiation at all wavelengths, and that stars approximate black bodies.
Interpret black-body intensity–wavelength curves and how they change with temperature (peak height, peak position, non-crossing).
Use Wien's displacement law $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ to find the peak wavelength from temperature, or a star's surface temperature from $\lambda_{max}$ .
Use the Stefan–Boltzmann law $L = \sigma A T^4$ (with $A = 4\pi r^2$ for a star) to calculate and compare luminosities, temperatures and radii.
Combine Wien's law and the Stefan–Boltzmann law to estimate the radius of a star.
Distinguish a star's luminosity (total power radiated) from its apparent brightness observed at Earth.

What is a black body, and why stars behave like one

A perfect absorber is a perfect emitter — and stars come remarkably close.

A black body is an idealised object that absorbs all the electromagnetic radiation incident on it, at every wavelength — it reflects and transmits none. A key consequence of being a perfect absorber is that it is also a perfect emitter: at any given temperature it radiates the maximum possible power at every wavelength.

The name is a little misleading. A black body is not necessarily "black" to look at — the Sun is a good black body yet is dazzlingly bright. The word "black" refers only to the absorption: nothing incident is reflected back. The classic laboratory model is a small hole in the wall of a heated cavity: radiation entering the hole bounces around inside and is almost entirely absorbed, so the hole behaves as a near-perfect black body, and the radiation streaming out of it is ideal black-body radiation.

Why stars count as black bodies. A star's outer layers are dense, hot gas that absorbs and re-emits radiation across the whole spectrum extremely efficiently. To a good approximation the light leaving the star follows the ideal black-body intensity–wavelength curve fixed by its surface temperature alone. This single idea — that a star's spectrum depends only on its surface temperature — is what lets us read a star's temperature, and then its size, straight from the radiation it sends us. It is the launch-pad for the Astronomy and Cosmology topics.

The distribution of that emitted energy across wavelengths is governed by two laws you must be able to use:

Wien's displacement law — where (at which wavelength) the emission peaks.
The Stefan–Boltzmann law — how much total power is radiated.

Black body = perfect absorber of all wavelengths ⇒ perfect emitter of all wavelengths.
‘Black’ refers to absorption, not appearance — the Sun is a bright black body.
A star's spectrum depends only on its surface temperature (to a good approximation).
Two laws describe it: Wien (peak wavelength) and Stefan–Boltzmann (total power).

The intensity–wavelength curves and how temperature changes them

One peak each; hotter means higher and further left; curves never cross.

Plotting the intensity of the radiation against wavelength for a black body gives a smooth curve that rises from zero, reaches a single peak at a wavelength $\lambda_{max}$ , then tails off at longer wavelengths. The shape is fixed entirely by the temperature. There are three features the examiners test again and again.

Higher temperature → higher peak intensity. The whole curve lifts, so at every wavelength a hotter body radiates more strongly.
Higher temperature → peak at a shorter wavelength. The maximum shifts towards the blue/ultraviolet end. This is Wien's law in graphical form.
The curves never cross. Because a hotter body emits more at every wavelength, its curve lies entirely above a cooler body's curve.

As temperature rises, the peak grows taller and moves to a shorter wavelength (dashed arrow). A hotter body radiates more at every wavelength, so the curves never cross — the 7000 K curve lies entirely above the 5000 K curve, which lies entirely above the 3500 K curve.

The position of that peak, $\lambda_{max}$ , is the single most useful reading you can take from the curve: feed it into Wien's law and you have the surface temperature. The total area under a curve is proportional to the total power radiated per unit area, and it grows extremely rapidly with temperature — that rapid growth is quantified by the Stefan–Boltzmann law.

Reading the peak wavelength λ_max off the curve and substituting into Wien's law gives the surface temperature. The area under the whole curve represents the total power radiated per unit area (Stefan–Boltzmann).

Each curve has one peak; the area under it ∝ total power radiated per unit area.
Higher T ⇒ higher peak AND shorter peak wavelength.
Curves never cross — a hotter body emits more at every wavelength.
Read λ_max off the curve → Wien's law → surface temperature.

Wien's displacement law — finding temperature from the peak

λ_max T is a fixed constant, so λ_max ∝ 1/T.

Wien's displacement law links the peak wavelength to the absolute temperature: $\lambda_{max}\,T = 2.90\times10^{-3}\ \text{m K}.$

Because the right-hand side is a constant, the peak wavelength is inversely proportional to the absolute temperature, $\lambda_{max}\propto 1/T$ . The two everyday uses are:

Find the peak wavelength from a known temperature: $\lambda_{max}=\dfrac{2.90\times10^{-3}}{T}$ .
Find a star's surface temperature from its observed peak: $T=\dfrac{2.90\times10^{-3}}{\lambda_{max}}$ .

Worked check (the Sun). The Sun's surface is about $5800\ \text{K}$ , so $\lambda_{max}=\frac{2.90\times10^{-3}}{5800}=5.00\times10^{-7}\ \text{m}=500\ \text{nm},$ right in the visible (green-yellow) — which is why the Sun looks yellow-white and why our eyes are most sensitive near this wavelength.

Two habits that protect the marks.

Keep $T$ in kelvin. If a temperature is given in °C, convert with $T/\text{K}=\theta/\text{°C}+273$ before substituting.
Convert wavelengths to metres. A peak quoted as $500\ \text{nm}$ is $5.00\times10^{-7}\ \text{m}$ ; the constant is in $\text{m K}$ , so a wavelength left in nanometres gives a temperature $10^{9}$ times too large.

Direction of the shift. Hotter → shorter peak wavelength (bluer); cooler → longer peak wavelength (redder). A blue star like Rigel is far hotter than a red star like Betelgeuse. If you ever find a hotter body giving a longer peak wavelength, you have divided the wrong way round.

$\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ (a constant), so $\lambda_{max}\propto 1/T$ .
$\lambda_{max}=2.90\times10^{-3}/T$ ; or $T=2.90\times10^{-3}/\lambda_{max}$ .
Hotter ⇒ shorter peak wavelength (bluer); cooler ⇒ longer (redder).
Use $T$ in kelvin and $\lambda$ in metres before substituting.

See the full worked example for black body radiation →

The Stefan–Boltzmann law — total power and a star as a sphere

L = σAT⁴, and for a star A = 4πr², so L = 4πr²σT⁴.

The Stefan–Boltzmann law gives the total power (the luminosity) radiated by a black body across all wavelengths: $L = \sigma A T^4,$ where $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ is the Stefan–Boltzmann constant, $A$ is the radiating surface area, and $T$ is the absolute surface temperature. Two features dominate every calculation:

The $T^4$ dependence is fierce. Doubling the temperature multiplies the power by $2^4=16$ ; tripling it multiplies by $3^4=81$ . Small temperature differences make huge luminosity differences.
A star is a sphere, so its whole surface radiates and the area is $A=4\pi r^2$ . Substituting, $\boxed{L = 4\pi r^2\,\sigma T^4}.$

A star radiates from its entire spherical surface, so its area is A = 4πr² (not the disc area πr²). Its luminosity is L = σAT⁴ = 4πr²σT⁴.

Worked check (the Sun). With $r=6.96\times10^{8}\ \text{m}$ and $T=5800\ \text{K}$ : $A=4\pi r^2=6.09\times10^{18}\ \text{m}^2$ , and $T^4=1.13\times10^{15}\ \text{K}^4$ , so $L=(5.67\times10^{-8})(6.09\times10^{18})(1.13\times10^{15})\approx 3.9\times10^{26}\ \text{W},$ matching the Sun's accepted luminosity. If you only want the surface flux (power per unit area) rather than the total, use $L/A=\sigma T^4$ on its own.

Luminosity is not apparent brightness. Luminosity is the total power a star pours out — an intrinsic property. What we see from Earth is the intensity that actually reaches us, which is diluted by distance as $1/d^2$ . A nearby dim star can look brighter than a distant brilliant one, so never assume the brightest-looking star is the most luminous.

$L = \sigma A T^4$ , $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ .
For a star $A = 4\pi r^2$ , so $L = 4\pi r^2 \sigma T^4$ .
$T^4$ : doubling $T$ multiplies power by 16 — a very strong dependence.
Surface flux (power per unit area) = $\sigma T^4$ ; luminosity ≠ observed brightness ( $\propto 1/d^2$ ).

See the full worked example for black body radiation →

Combining the two laws to estimate a star's radius

Wien gives T from λ_max; Stefan then gives r from L and T.

The real power of this topic — and the direct feed into Astronomy and Cosmology — is combining the two laws to work out how big a star is when you can measure only its peak wavelength and its luminosity. The chain has three links.

Wien's law → surface temperature. From the observed $\lambda_{max}$ , $\displaystyle T=\frac{2.90\times10^{-3}}{\lambda_{max}}$ .
Stefan–Boltzmann law → radius. Write $L=4\pi r^2\sigma T^4$ and rearrange for $r$ : $r=\sqrt{\dfrac{L}{4\pi\sigma T^4}}.$
Interpret the size (e.g. compare with the Sun's radius $\approx 7.0\times10^{8}\ \text{m}$ ).

Worked check. A star with $L=4.5\times10^{28}\ \text{W}$ peaks at $\lambda_{max}=350\ \text{nm}=3.50\times10^{-7}\ \text{m}$ .

Wien: $T=\dfrac{2.90\times10^{-3}}{3.50\times10^{-7}}=8.29\times10^{3}\ \text{K}$ .
Denominator: $4\pi\sigma T^4=4\pi(5.67\times10^{-8})(8.29\times10^{3})^4=3.36\times10^{9}\ \text{W m}^{-2}$ .
Radius: $r=\sqrt{\dfrac{4.5\times10^{28}}{3.36\times10^{9}}}=3.66\times10^{9}\ \text{m}$ — about $5.3$ solar radii.

Two competing effects decide a star's size. A red giant has a cool surface (low $\sigma T^4$ , so low flux) yet an enormous luminosity — the only way to reconcile them is a huge radius. A hot but faint white dwarf is the reverse: high flux, low luminosity, so it must be tiny. The maths of $r=\sqrt{L/(4\pi\sigma T^4)}$ captures both cases automatically.

Exam discipline for this chain.

Carry full calculator precision through $T$ before raising it to the fourth power — rounding $T$ early can shift $T^4$ enough to lose the accuracy mark.
Take the square root at the end (a very common slip is to quote $r^2$ as $r$ ).
Keep every quantity in SI units: $\lambda$ in metres, $T$ in kelvin, $L$ in watts.

Step 1 — Wien: $T = 2.90\times10^{-3}/\lambda_{max}$ .
Step 2 — Stefan rearranged: $r = \sqrt{L/(4\pi\sigma T^4)}$ .
Red giant = cool but luminous ⇒ huge radius; white dwarf = hot but faint ⇒ tiny radius.
Keep full precision through $T$ , remember the square root, use SI units.

See the full worked example for black body radiation →

Quick recap

A black body absorbs and emits all wavelengths perfectly; stars approximate black bodies, so their spectra depend only on surface temperature.
Intensity–wavelength curves: higher T gives a higher peak at a shorter wavelength, and the curves never cross.
Wien's displacement law: $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ , so $\lambda_{max}\propto 1/T$ — hotter bodies peak at shorter (bluer) wavelengths.
Stefan–Boltzmann law: $L = \sigma A T^4$ ( $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ ); for a star $A = 4\pi r^2$ , so $L = 4\pi r^2 \sigma T^4$ , with a strong $T^4$ dependence.
Estimate a star's radius: Wien gives $T$ from $\lambda_{max}$ , then $r = \sqrt{L/(4\pi\sigma T^4)}$ .
Luminosity is the total power radiated (intrinsic); apparent brightness also depends on distance ( $\propto 1/d^2$ ).
Use $T$ in kelvin, wavelengths in metres, keep full precision through the fourth power, and quote 3 s.f.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Wien's displacement law: $\lambda_{max}\,T = 2.90\times10^{-3}\ \text{m K}$ — so $\lambda_{max}\propto 1/T$ (hotter ⇒ shorter peak wavelength).
Stefan–Boltzmann law: $L = \sigma A T^4$ , with $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ .
A star is a sphere: $A = 4\pi r^2$ , so $L = 4\pi r^2 \sigma T^4$ .
Radius from $L$ and $T$ : $r = \sqrt{\dfrac{L}{4\pi\sigma T^4}}$ .
Surface flux (power per unit area) $= \sigma T^4$ .
Curves never cross; higher $T$ ⇒ higher peak AND shorter peak wavelength.
Doubling $T$ multiplies radiated power by $2^4 = 16$ (strong $T^4$ dependence).
Luminosity = total power radiated; apparent brightness also depends on distance ( $\propto 1/d^2$ ).
Always use $T$ in kelvin and $\lambda$ in metres.

How it’s examined

Black body radiation is Unit 5 content and appears on the Unit 5 paper (WPH15, Thermodynamics, Radiation, Oscillations and Cosmology), a structured IA2 paper with short calculations, ‘state/explain’ items and multi-step astrophysical problems. Typical formats: define a black body or state Wien's/Stefan's law (1–2 marks, AO1); describe how the intensity–wavelength curves change with temperature — peak height, peak position, non-crossing (2–3 marks, AO1/AO2); a single-step Wien calculation for $\lambda_{max}$ or surface temperature (2 marks, M and A marks); a Stefan–Boltzmann luminosity calculation requiring $A = 4\pi r^2$ (3–4 marks); a ratio/comparison of two stars using $(T_1/T_2)^4$ or $(r_1/r_2)^2$ (2–4 marks); and the flagship multi-step item combining both laws to estimate a star's radius via $r = \sqrt{L/(4\pi\sigma T^4)}$ (5–6 marks, AO2/AO3). ALWAYS convert temperatures to kelvin and wavelengths to metres, show the equation and substitution, keep full precision through the fourth power, and quote a 3 s.f. answer with units. Watch the luminosity-versus-apparent-brightness distinction — examiners reward candidates who note that observed brightness depends on distance ( $1/d^2$ ) as well as luminosity.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 5 (WPH15) examiner reports; Edexcel IAL Physics Unit 5 sample assessment materials and mark schemes. Last reviewed 2026-07-06.

Take this whole topic with you

Step-by-step worked examples — Black Body Radiation

Step-by-step solutions to past-paper-style questions on black body radiation, written exactly the way a tutor would explain them at the board.

Question type:

1Peak wavelength from temperature (2 marks)
Getting startedDirect calculation• Wien's law, AO2, Unit 5
Question
The surface of a star behaves as a black body at a temperature of $6000\ \text{K}$ . Calculate the wavelength at which the star emits the greatest intensity of radiation. Take Wien's constant as $2.90\times10^{-3}\ \text{m K}$ . (2 marks)
Step-by-step solution
1. Step 1
  Use Wien's displacement law, $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ , rearranged for the peak wavelength. Keep $T$ in kelvin.
  $\lambda_{max} = \frac{2.90\times10^{-3}}{T}$
2. Step 2
  Substitute $T = 6000\ \text{K}$ and evaluate to 3 s.f.
  $\lambda_{max} = \frac{2.90\times10^{-3}}{6000} = 4.83\times10^{-7}\ \text{m}$
Answer
$\lambda_{max} = 4.83\times10^{-7}\ \text{m}$ ( $483\ \text{nm}$ ), in the visible (blue-green) region.
Examiner tip
Mark scheme: (1) use of $\lambda_{max} = 2.90\times10^{-3}/T$ ; (2) $4.83\times10^{-7}\ \text{m}$ to 3 s.f. with unit. The commonest slip is multiplying instead of dividing — Wien's law is $\lambda_{max}T =$ constant, so a hotter body has a shorter peak wavelength.
2Surface temperature from peak wavelength (2 marks)
Getting startedDirect calculation• Wien's law, AO2, Unit 5
Question
A star emits radiation with a peak intensity at a wavelength of $480\ \text{nm}$ . Estimate the surface temperature of the star. Take Wien's constant as $2.90\times10^{-3}\ \text{m K}$ . (2 marks)
Step-by-step solution
1. Step 1
  Rearrange Wien's displacement law for temperature. Convert the wavelength to metres: $480\ \text{nm} = 4.80\times10^{-7}\ \text{m}$ .
  $T = \frac{2.90\times10^{-3}}{\lambda_{max}}$
2. Step 2
  Substitute and evaluate to 3 s.f.
  $T = \frac{2.90\times10^{-3}}{4.80\times10^{-7}} = 6.04\times10^{3}\ \text{K}$
Answer
$T = 6.04\times10^{3}\ \text{K}$ (about $6040\ \text{K}$ ).
Examiner tip
Mark scheme: (1) use of $T = 2.90\times10^{-3}/\lambda_{max}$ ; (2) $6.04\times10^{3}\ \text{K}$ to 3 s.f. Leaving $\lambda_{max}$ in nanometres gives an answer $10^{9}$ times too large — always convert to metres first.
3Luminosity from the Stefan–Boltzmann law (2 marks)
Getting startedDirect calculation• Stefan–Boltzmann law, AO2, Unit 5
Question
A black-body surface of area $2.0\times10^{18}\ \text{m}^2$ is at a temperature of $5200\ \text{K}$ . Calculate the total power it radiates. Take $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . (2 marks)
Step-by-step solution
1. Step 1
  Use the Stefan–Boltzmann law with the surface area given. $T$ must be in kelvin and is raised to the fourth power.
  $L = \sigma A T^4$
2. Step 2
  Substitute $A = 2.0\times10^{18}\ \text{m}^2$ and $T = 5200\ \text{K}$ ( $T^4 = 7.31\times10^{14}\ \text{K}^4$ ) and evaluate to 3 s.f.
  $L = (5.67\times10^{-8})(2.0\times10^{18})(5200)^4 = 8.29\times10^{25}\ \text{W}$
Answer
$L = 8.29\times10^{25}\ \text{W}$ .
Examiner tip
Mark scheme: (1) use of $L = \sigma A T^4$ ; (2) $8.29\times10^{25}\ \text{W}$ to 3 s.f. The classic error is forgetting the fourth power (using $T$ or $T^2$ ). Because of the $T^4$ dependence, a small temperature change makes a huge luminosity change.
4Luminosity of a star from its radius (4 marks)
Building confidenceMulti-step problem• Stefan–Boltzmann law, surface area, AO2, Unit 5
Question
A star has a radius of $8.0\times10^{8}\ \text{m}$ and a surface temperature of $6500\ \text{K}$ , and behaves as a black body. Calculate its luminosity (total power radiated). Take $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . (4 marks)
Step-by-step solution
1. Step 1
  The star is a sphere, so first find its surface area from $A = 4\pi r^2$ .
  $A = 4\pi r^2 = 4\pi (8.0\times10^{8})^2$
2. Step 2
  Evaluate the surface area.
  $A = 8.04\times10^{18}\ \text{m}^2$
3. Step 3
  Apply the Stefan–Boltzmann law $L = \sigma A T^4$ with the area just found.
  $L = \sigma A T^4 = (5.67\times10^{-8})(8.04\times10^{18})(6500)^4$
4. Step 4
  Evaluate ( $6500^4 = 1.785\times10^{15}\ \text{K}^4$ ) to 3 s.f.
  $L = 8.14\times10^{26}\ \text{W}$
Answer
$L = 8.14\times10^{26}\ \text{W}$ (about twice the Sun's luminosity of $3.9\times10^{26}\ \text{W}$ ).
Examiner tip
Mark scheme: (1) $A = 4\pi r^2$ ; (2) $8.04\times10^{18}\ \text{m}^2$ ; (3) use of $L = \sigma A T^4$ ; (4) $8.14\times10^{26}\ \text{W}$ to 3 s.f. The most-penalised error is using $A = \pi r^2$ or $A = 2\pi r$ — a star's radiating area is its full spherical surface $4\pi r^2$ .
5Effect of doubling the temperature on power (4 marks)
Building confidenceMulti-step problem• Stefan–Boltzmann law, ratio, AO2, AO3, Unit 5
Question
Two stars have the same radius, but star X has a surface temperature exactly twice that of star Y. Both behave as black bodies. (a) Show that star X radiates 16 times as much power as star Y. (b) Calculate the ratio of their peak wavelengths $\lambda_{max,X} : \lambda_{max,Y}$ . (4 marks)
Step-by-step solution
1. Step 1
  (a) For equal radii the areas are equal, so from $L = \sigma A T^4$ the luminosity ratio depends only on $T^4$ .
  $\frac{L_X}{L_Y} = \frac{\sigma A T_X^4}{\sigma A T_Y^4} = \left(\frac{T_X}{T_Y}\right)^4$
2. Step 2
  Substitute $T_X = 2T_Y$ .
  $\frac{L_X}{L_Y} = (2)^4 = 16$
3. Step 3
  (b) From Wien's law $\lambda_{max} \propto 1/T$ , so a higher temperature gives a shorter peak wavelength.
  $\frac{\lambda_{max,X}}{\lambda_{max,Y}} = \frac{T_Y}{T_X} = \frac{1}{2}$
Answer
(a) $L_X/L_Y = (T_X/T_Y)^4 = 2^4 = 16$ , so star X radiates 16 times as much power. (b) $\lambda_{max,X} : \lambda_{max,Y} = 1 : 2$ — star X's peak is at half the wavelength (it looks bluer).
Examiner tip
Mark scheme: (a) recognise areas equal so ratio $= (T_X/T_Y)^4$ (1); $= 16$ (1). (b) $\lambda_{max} \propto 1/T$ (1); ratio $1:2$ (1). This tests both laws at once — Stefan gives the $T^4$ dependence of power, Wien gives the $1/T$ dependence of peak wavelength.
6Peak wavelength and surface flux together (4 marks)
Building confidenceMulti-step problem• Wien's law, Stefan–Boltzmann law, AO2, Unit 5
Question
A black body radiates with peak intensity at a wavelength of $1.0\times10^{-6}\ \text{m}$ . (a) Calculate its surface temperature. (b) Calculate the power radiated per unit surface area (the surface flux). Take $2.90\times10^{-3}\ \text{m K}$ and $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . (4 marks)
Step-by-step solution
1. Step 1
  (a) Use Wien's law to find the temperature from the peak wavelength.
  $T = \frac{2.90\times10^{-3}}{\lambda_{max}} = \frac{2.90\times10^{-3}}{1.0\times10^{-6}}$
2. Step 2
  Evaluate.
  $T = 2.90\times10^{3}\ \text{K}$
3. Step 3
  (b) The power per unit area is $L/A = \sigma T^4$ (the Stefan–Boltzmann law without the area factor).
  $\frac{L}{A} = \sigma T^4 = (5.67\times10^{-8})(2900)^4$
4. Step 4
  Evaluate ( $2900^4 = 7.07\times10^{13}\ \text{K}^4$ ) to 3 s.f.
  $\frac{L}{A} = 4.01\times10^{6}\ \text{W m}^{-2}$
Answer
(a) $T = 2.90\times10^{3}\ \text{K}$ . (b) Surface flux $= 4.01\times10^{6}\ \text{W m}^{-2}$ .
Examiner tip
Mark scheme: (a) $T = 2.90\times10^{-3}/\lambda_{max}$ (1); $2900\ \text{K}$ (1). (b) use of $\sigma T^4$ (1); $4.01\times10^{6}\ \text{W m}^{-2}$ (1). Note that $\sigma T^4$ alone gives power per unit area; multiply by $4\pi r^2$ only when the total luminosity is wanted.
7Estimating a star's radius from L and λmax (6 marks)
StretchMulti-step problem• Wien's law, Stefan–Boltzmann law, radius, AO2, AO3, Unit 5
Question
A distant star has a luminosity of $4.5\times10^{28}\ \text{W}$ and its black-body radiation peaks at a wavelength of $350\ \text{nm}$ . Estimate the radius of the star. Take $2.90\times10^{-3}\ \text{m K}$ and $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . (6 marks)
Step-by-step solution
1. Step 1
  First find the surface temperature from Wien's law. Convert $350\ \text{nm} = 3.50\times10^{-7}\ \text{m}$ .
  $T = \frac{2.90\times10^{-3}}{\lambda_{max}} = \frac{2.90\times10^{-3}}{3.50\times10^{-7}} = 8.29\times10^{3}\ \text{K}$
2. Step 2
  Now use the Stefan–Boltzmann law with $A = 4\pi r^2$ , and rearrange for $r$ .
  $L = \sigma (4\pi r^2) T^4 \;\Rightarrow\; r = \sqrt{\frac{L}{4\pi\sigma T^4}}$
3. Step 3
  Evaluate $T^4$ using the full-precision temperature: $(8286)^4 = 4.71\times10^{15}\ \text{K}^4$ .
  $T^4 = 4.71\times10^{15}\ \text{K}^4$
4. Step 4
  Evaluate the denominator $4\pi\sigma T^4$ .
  $4\pi\sigma T^4 = 4\pi(5.67\times10^{-8})(4.71\times10^{15}) = 3.36\times10^{9}\ \text{W m}^{-2}$
5. Step 5
  Substitute into the expression for $r$ .
  $r = \sqrt{\frac{4.5\times10^{28}}{3.36\times10^{9}}} = \sqrt{1.34\times10^{19}}$
6. Step 6
  Evaluate the square root to 3 s.f.
  $r = 3.66\times10^{9}\ \text{m}$
Answer
$r = 3.66\times10^{9}\ \text{m}$ (about $5.3$ solar radii). The method is: Wien's law gives $T$ from $\lambda_{max}$ , then the Stefan–Boltzmann law with $A = 4\pi r^2$ gives $r$ from $L$ and $T$ .
Examiner tip
Mark scheme: (1) $T$ from Wien's law; (2) $8.29\times10^{3}\ \text{K}$ ; (3) rearrange $L = 4\pi r^2 \sigma T^4$ for $r$ ; (4) correct substitution; (5)/(6) $r = 3.66\times10^{9}\ \text{m}$ to 3 s.f. The two-law link is the whole point of this Astrophysics-feeding question; carry full precision through $T$ before the fourth power, and remember to take the square root at the end.
8Comparing two stars' luminosities (5 marks)
StretchMulti-step problem• Stefan–Boltzmann law, ratio, AO2, AO3, Unit 5
Question
Star A and star B have the same radius. Star A has a surface temperature of $10\,000\ \text{K}$ and star B has a surface temperature of $5000\ \text{K}$ . (a) Calculate the ratio of their luminosities $L_A : L_B$ . (b) Star B is observed to appear far brighter in the night sky than star A. State one reason why apparent brightness need not match luminosity. (5 marks)
Step-by-step solution
1. Step 1
  (a) With equal radii the areas cancel, so the luminosity ratio depends only on $T^4$ .
  $\frac{L_A}{L_B} = \left(\frac{T_A}{T_B}\right)^4$
2. Step 2
  Substitute the temperatures.
  $\frac{L_A}{L_B} = \left(\frac{10\,000}{5000}\right)^4 = 2^4 = 16$
3. Step 3
  So star A is 16 times as luminous as star B, even though B looks brighter.
  $L_A : L_B = 16 : 1$
4. Step 4
  (b) Luminosity is the total power the star radiates; apparent brightness (observed intensity) depends also on how far away the star is — intensity falls as $1/d^2$ .
5. Step 5
  So star B can look brighter simply because it is much closer to us, even though star A radiates far more power in total.
Answer
(a) $L_A : L_B = (10\,000/5000)^4 = 16 : 1$ . (b) Apparent brightness depends on distance as well as luminosity (observed intensity $\propto 1/d^2$ ), so a closer, less luminous star can appear brighter than a distant, more luminous one.
Examiner tip
Mark scheme: (a) ratio $= (T_A/T_B)^4$ (1); $= 16$ (1); state A is more luminous (1). (b) luminosity = total power radiated (1); apparent brightness also depends on distance / inverse-square (1). The key discriminator is distinguishing luminosity (intrinsic total power) from observed intensity (what reaches us, diluted by distance).
9Radius of a cool, luminous red giant (6 marks)
StretchMulti-step problem• Stefan–Boltzmann law, Wien's law, radius, AO2, AO3, Unit 5
Question
A red giant has a luminosity of $1.2\times10^{31}\ \text{W}$ and a surface temperature of $3600\ \text{K}$ , and behaves as a black body. (a) Calculate its radius. (b) Comment on the size of your answer compared with the Sun (radius $7.0\times10^{8}\ \text{m}$ ). Take $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ . (6 marks)
Step-by-step solution
1. Step 1
  (a) Use the Stefan–Boltzmann law with $A = 4\pi r^2$ and rearrange for the radius.
  $L = \sigma (4\pi r^2) T^4 \;\Rightarrow\; r = \sqrt{\frac{L}{4\pi\sigma T^4}}$
2. Step 2
  Evaluate $T^4$ for the cool surface: $(3600)^4 = 1.68\times10^{14}\ \text{K}^4$ .
  $T^4 = 1.68\times10^{14}\ \text{K}^4$
3. Step 3
  Evaluate the denominator $4\pi\sigma T^4$ .
  $4\pi\sigma T^4 = 4\pi(5.67\times10^{-8})(1.68\times10^{14}) = 1.20\times10^{8}\ \text{W m}^{-2}$
4. Step 4
  Substitute into the expression for $r$ .
  $r = \sqrt{\frac{1.2\times10^{31}}{1.20\times10^{8}}} = \sqrt{1.00\times10^{23}}$
5. Step 5
  Evaluate the square root to 3 s.f.
  $r = 3.16\times10^{11}\ \text{m}$
6. Step 6
  (b) Compare with the Sun: $r / r_{\odot} = (3.16\times10^{11})/(7.0\times10^{8}) \approx 450$ .
  $\frac{r}{r_{\odot}} \approx 4.5\times10^{2}$
Answer
(a) $r = 3.16\times10^{11}\ \text{m}$ . (b) This is roughly $450$ times the Sun's radius — despite the cool surface ( $3600\ \text{K}$ , so a low flux $\sigma T^4$ ), the enormous luminosity can only be produced by a vast surface area, which is characteristic of a red giant.
Examiner tip
Mark scheme: (a) rearrange $L = 4\pi r^2 \sigma T^4$ (1); $T^4$ (1); correct substitution (1); $r = 3.16\times10^{11}\ \text{m}$ (1). (b) ratio $\approx 450$ (1); a large cool star must be huge to be so luminous (1). This shows how the two competing effects — low temperature (low $\sigma T^4$ ) but high luminosity — force a large radius.

Model Answers — Black Body Radiation

High-scoring sample answers for black body radiation on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by a black body, and explain why stars are treated as black bodies.
Model answer
A black body is an idealised object that absorbs all the electromagnetic radiation of every wavelength that falls on it (reflecting and transmitting none), and — being a perfect absorber — is also a perfect emitter at every wavelength.

Stars are treated as black bodies because a star's dense, hot gas absorbs and re-emits radiation extremely efficiently across the whole spectrum, so its emission closely follows the ideal black-body intensity–wavelength curve set by its surface temperature.
Why this scores
Why this scores 2/2. (1) absorbs (and emits) all wavelengths perfectly; (2) stars approximate this so their spectra follow the black-body curve. Saying only ‘it looks black’ misses the mark — the defining property is perfect absorption/emission at all wavelengths.
Question 2
2 marks
Q (2 marks). State Wien's displacement law and describe what it tells you about the peak wavelength as temperature increases.
Model answer
Wien's displacement law states that the wavelength of peak emission of a black body is inversely proportional to its absolute temperature: $\lambda_{max}\,T = 2.90\times10^{-3}\ \text{m K} \quad (\text{a constant}).$ Because the product is constant, as the temperature increases the peak wavelength decreases — a hotter body radiates most strongly at a shorter wavelength (the peak shifts towards the blue/ultraviolet).
Why this scores
Why this scores 2/2. (1) $\lambda_{max}T =$ constant, i.e. $\lambda_{max} \propto 1/T$ ; (2) higher $T$ → shorter peak wavelength. The frequent error is stating the relationship as direct proportion — it is inverse.
Question 3
2 marks
Q (2 marks). State the Stefan–Boltzmann law for the power radiated by a black body, defining each symbol, and give the surface area of a star of radius $r$ .
Model answer
The Stefan–Boltzmann law gives the total power (luminosity) radiated by a black body as $L = \sigma A T^4,$ where $L$ is the luminosity (W), $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ is the Stefan–Boltzmann constant, $A$ is the radiating surface area (m²) and $T$ is the absolute surface temperature (K). For a spherical star the surface area is $A = 4\pi r^2,$ so $L = 4\pi r^2 \sigma T^4$ .
Why this scores
Why this scores 2/2. (1) $L = \sigma A T^4$ with symbols defined and $T$ to the fourth power; (2) $A = 4\pi r^2$ for a star. Writing $T^2$ , or using $\pi r^2$ for the area, are the two mark-losing slips.
Question 4
3 marks
Q (3 marks). Sketch, in words, how the black-body intensity–wavelength curve changes as the temperature of the body is increased. Refer to the peak height, the peak position and whether the curves cross.
Model answer
As the temperature increases:

The peak (maximum) intensity is higher — the whole curve rises, so more power is radiated at every wavelength.

The peak shifts to a shorter wavelength (towards the blue/UV), in line with Wien's law $\lambda_{max} \propto 1/T$ .

The curves never cross — a hotter body emits more intensity at every wavelength than a cooler one, so its curve lies entirely above the cooler body's curve.

The total area under each curve (proportional to the total power radiated per unit area) grows very rapidly, consistent with the $\sigma T^4$ dependence.
Why this scores
Why this scores 3/3. (1) higher peak intensity; (2) peak at shorter wavelength; (3) curves do not cross (hotter emits more at all wavelengths). Many candidates correctly shift the peak but wrongly draw the curves crossing at long wavelengths — the higher- $T$ curve is always above.
Question 5
3 marks
Q (3 marks). The Sun's surface temperature is $5800\ \text{K}$ . (a) Calculate the wavelength at which the Sun's radiation peaks. (b) State in which region of the spectrum this lies. Take $2.90\times10^{-3}\ \text{m K}$ .
Model answer
(a) Using Wien's displacement law: $\lambda_{max} = \frac{2.90\times10^{-3}}{T} = \frac{2.90\times10^{-3}}{5800} = 5.00\times10^{-7}\ \text{m}.$

(b) This is $500\ \text{nm}$ , which lies in the visible region of the spectrum (green-yellow light) — consistent with the Sun appearing as a yellow-white star and with our eyes being most sensitive near this wavelength.
Why this scores
Why this scores 3/3. (a) use of $\lambda_{max} = 2.90\times10^{-3}/T$ (1) and $5.00\times10^{-7}\ \text{m}$ (1); (b) visible region (1). Quoting the answer as $500\ \text{nm}$ and naming the spectral region secures both parts.
Question 6
3 marks
Q (3 marks). The black-body radiation from a star peaks at a wavelength of $580\ \text{nm}$ . (a) Calculate the star's surface temperature. (b) A second star peaks at a shorter wavelength. State, with a reason, whether the second star is hotter or cooler. Take $2.90\times10^{-3}\ \text{m K}$ .
Model answer
(a) Convert $580\ \text{nm} = 5.80\times10^{-7}\ \text{m}$ and use Wien's law: $T = \frac{2.90\times10^{-3}}{\lambda_{max}} = \frac{2.90\times10^{-3}}{5.80\times10^{-7}} = 5.00\times10^{3}\ \text{K}.$

(b) The second star is hotter. Because $\lambda_{max} \propto 1/T$ , a shorter peak wavelength corresponds to a higher temperature.
Why this scores
Why this scores 3/3. (a) use of $T = 2.90\times10^{-3}/\lambda_{max}$ (1) and $5.00\times10^{3}\ \text{K}$ (1); (b) hotter, because $\lambda_{max} \propto 1/T$ (1). The reasoning mark needs the explicit inverse relationship, not just ‘hotter’.
Question 7
4 marks
Q (4 marks). A star of radius $1.4\times10^{9}\ \text{m}$ has a surface temperature of $4800\ \text{K}$ and behaves as a black body. Calculate its luminosity. Take $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ .
Model answer
First find the surface area of the star (a sphere): $A = 4\pi r^2 = 4\pi (1.4\times10^{9})^2 = 2.463\times10^{19}\ \text{m}^2.$ Then apply the Stefan–Boltzmann law with $T^4 = (4800)^4 = 5.308\times10^{14}\ \text{K}^4$ : $L = \sigma A T^4 = (5.67\times10^{-8})(2.463\times10^{19})(5.308\times10^{14}).$ $L = 7.41\times10^{26}\ \text{W}.$
Why this scores
Why this scores 4/4. (1) $A = 4\pi r^2$ ; (2) $2.46\times10^{19}\ \text{m}^2$ ; (3) use of $L = \sigma A T^4$ ; (4) $7.41\times10^{26}\ \text{W}$ to 3 s.f. Forgetting the fourth power, or using $A = \pi r^2$ , are the standard errors — a star radiates from its whole spherical surface.
Question 8
6 marks
Q (6 marks). A star has a luminosity of $8.0\times10^{28}\ \text{W}$ and its radiation peaks at $290\ \text{nm}$ . (a) Find its surface temperature. (b) Estimate its radius. Take $2.90\times10^{-3}\ \text{m K}$ and $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ .
Model answer
(a) Convert $290\ \text{nm} = 2.90\times10^{-7}\ \text{m}$ and use Wien's law: $T = \frac{2.90\times10^{-3}}{2.90\times10^{-7}} = 1.00\times10^{4}\ \text{K}.$

(b) Rearrange the Stefan–Boltzmann law with $A = 4\pi r^2$ for the radius: $r = \sqrt{\frac{L}{4\pi\sigma T^4}}.$ With $T^4 = (1.00\times10^{4})^4 = 1.00\times10^{16}\ \text{K}^4$ , the denominator is $4\pi\sigma T^4 = 4\pi(5.67\times10^{-8})(1.00\times10^{16}) = 7.12\times10^{9}\ \text{W m}^{-2}.$ Therefore $r = \sqrt{\frac{8.0\times10^{28}}{7.12\times10^{9}}} = \sqrt{1.12\times10^{19}} = 3.35\times10^{9}\ \text{m}.$
Why this scores
Why this scores 6/6. (a) use of Wien's law (1) and $1.00\times10^{4}\ \text{K}$ (1). (b) rearrange $L = 4\pi r^2 \sigma T^4$ for $r$ (1), $T^4$ (1), correct substitution (1), $r = 3.35\times10^{9}\ \text{m}$ (1). The full two-law chain (Wien → $T$ , Stefan → $r$ ) is exactly the technique that feeds the Astronomy/Cosmology topic; keep full precision in $T$ before the fourth power.
Question 9
6 marks
Q (6 marks). Two stars, P and Q, have the same surface temperature, but star P has three times the radius of star Q. (a) Show that star P radiates nine times the power of star Q. (b) Both stars have the same surface temperature yet star Q appears brighter from Earth. Explain how this is possible.
Model answer
(a) From the Stefan–Boltzmann law $L = \sigma A T^4 = \sigma (4\pi r^2) T^4$ . With the same temperature $T$ , the constant factors and $T^4$ cancel in the ratio, leaving only the areas (i.e. the radii squared): $\frac{L_P}{L_Q} = \frac{r_P^2}{r_Q^2} = \left(\frac{r_P}{r_Q}\right)^2.$ Substituting $r_P = 3\,r_Q$ : $\frac{L_P}{L_Q} = 3^2 = 9,$ so star P radiates nine times the power of star Q.

(b) Luminosity is the total power radiated, but the apparent brightness seen from Earth is the radiation intensity that actually reaches us, which falls off with distance as $1/d^2$ . Star Q can appear brighter if it is much closer to Earth than star P, so that its radiation is spread over a smaller area by the time it reaches us — despite P radiating nine times as much power in total.
Why this scores
Why this scores 6/6. (a) $L \propto r^2$ at fixed $T$ (1); ratio $= (r_P/r_Q)^2$ (1); $= 9$ (1). (b) luminosity = total power radiated (1); apparent brightness depends on distance / inverse-square (1); Q closer so appears brighter (1). The examined distinction is luminosity (intrinsic) versus observed intensity/apparent brightness (distance-dependent).
Question 10
6 marks
Q (6 marks). A red giant has a luminosity of $1.0\times10^{30}\ \text{W}$ and a surface temperature of $3000\ \text{K}$ . (a) Calculate its radius. (b) Its peak wavelength is $967\ \text{nm}$ ; state which spectral region this is in and comment on why a red giant appears red. Take $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ (Sun radius $\approx 7.0\times10^{8}\ \text{m}$ ).
Model answer
(a) Rearrange $L = 4\pi r^2 \sigma T^4$ for the radius: $r = \sqrt{\frac{L}{4\pi\sigma T^4}}.$ With $T^4 = (3000)^4 = 8.10\times10^{13}\ \text{K}^4$ , the denominator is $4\pi\sigma T^4 = 4\pi(5.67\times10^{-8})(8.10\times10^{13}) = 5.77\times10^{7}\ \text{W m}^{-2}.$ Therefore $r = \sqrt{\frac{1.0\times10^{30}}{5.77\times10^{7}}} = \sqrt{1.73\times10^{22}} = 1.32\times10^{11}\ \text{m}$ — about $190$ times the Sun's radius, as expected for a giant.

(b) A peak at $967\ \text{nm}$ lies just beyond the visible, in the near-infrared. With the peak at long wavelengths, the star emits relatively more red light than blue within the visible band, so its overall colour appears red — consistent with its low surface temperature (Wien's law: cool → long $\lambda_{max}$ ).
Why this scores
Why this scores 6/6. (a) rearrange for $r$ (1); $T^4$ (1); correct substitution (1); $r = 1.32\times10^{11}\ \text{m}$ (1). (b) near-infrared (1); cool star → long $\lambda_{max}$ → looks red (1). The physics insight rewarded is that a cool surface (low $\sigma T^4$ ) combined with a high luminosity forces an enormous radius — the signature of a red giant.

Key Formulae — Black Body Radiation

The formulae you need to memorise for black body radiation on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Wien's displacement law
$\lambda_{max}\,T = 2.90\times10^{-3}\ \text{m K}$
$\lambda_{max}$
wavelength of peak (maximum-intensity) emission / $\text{m}$
$T$
absolute (thermodynamic) temperature of the surface / $\text{K}$
$2.90\times10^{-3}$
Wien's constant / $\text{m K}$
When to use
To find the peak wavelength of a black body from its temperature, or (rearranged) a star's surface temperature from its observed $\lambda_{max}$ . Because the product is constant, $\lambda_{max} \propto 1/T$ : hotter bodies peak at shorter wavelengths.
Example
A $6000\ \text{K}$ star peaks at $\lambda_{max} = 2.90\times10^{-3}/6000 = 4.83\times10^{-7}\ \text{m}$ (483 nm).
Stefan–Boltzmann law (luminosity)
$L = \sigma A T^4$
$L$
luminosity — total power radiated / $\text{W}$
$\sigma$
Stefan–Boltzmann constant $= 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$
$A$
surface area of the radiating body / $\text{m}^2$
$T$
absolute surface temperature / $\text{K}$
When to use
To find the total power a black body radiates. The strong $T^4$ dependence means doubling $T$ multiplies the power by 16. Use $L/A = \sigma T^4$ alone for the power per unit surface area (surface flux).
Example
A surface of $2.0\times10^{18}\ \text{m}^2$ at $5200\ \text{K}$ : $L = (5.67\times10^{-8})(2.0\times10^{18})(5200)^4 = 8.29\times10^{25}\ \text{W}$ .
Surface area of a star (sphere)
$A = 4\pi r^2$
$A$
surface area of the sphere / $\text{m}^2$
$r$
radius of the star / $\text{m}$
When to use
To get the radiating surface area of a star for the Stefan–Boltzmann law. Combining gives $L = 4\pi r^2 \sigma T^4$ ; rearranged, $r = \sqrt{L/(4\pi\sigma T^4)}$ estimates a star's radius from $L$ and $T$ .
Example
A star of radius $8.0\times10^{8}\ \text{m}$ has $A = 4\pi(8.0\times10^{8})^2 = 8.04\times10^{18}\ \text{m}^2$ .

Key Definitions and Keywords — Black Body Radiation

Definitions to memorise and the exact keywords mark schemes credit for black body radiation answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Black body
Examiner keyword
An idealised object that absorbs all the electromagnetic radiation of every wavelength incident on it (and, being a perfect absorber, is also a perfect emitter at every wavelength). Stars are close approximations.
Black-body radiation
Examiner keyword
The characteristic continuous spectrum of electromagnetic radiation emitted by a black body, whose intensity–wavelength distribution depends only on the body's absolute temperature.
Intensity–wavelength curve
Examiner keyword
A graph of radiated intensity against wavelength for a black body. It rises to a single peak at $\lambda_{max}$ then falls; higher temperatures give a higher peak at a shorter wavelength, and the curves never cross.
Peak wavelength ( $\lambda_{max}$ )
Examiner keyword
The wavelength at which a black body emits the greatest intensity of radiation — the position of the maximum of its intensity–wavelength curve.
Wien's displacement law
Examiner keyword
The peak wavelength of a black body is inversely proportional to its absolute temperature: $\lambda_{max}T = 2.90\times10^{-3}\ \text{m K}$ . Hotter bodies peak at shorter wavelengths.
Wien's constant
The constant of proportionality in Wien's displacement law, $2.90\times10^{-3}\ \text{m K}$ , equal to the product $\lambda_{max}T$ for any black body.
Stefan–Boltzmann law
Examiner keyword
The total power radiated by a black body is $L = \sigma A T^4$ — proportional to the surface area and to the fourth power of the absolute temperature.
Stefan–Boltzmann constant ( $\sigma$ )
Examiner keyword
The constant in the Stefan–Boltzmann law, $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ , giving the power radiated per unit area per $\text{K}^4$ .
Luminosity
Examiner keyword
The total power radiated by a star (or black body) across all wavelengths, in watts. An intrinsic property of the star, given by $L = \sigma A T^4$ ; not the same as the brightness observed from Earth.
Surface flux
The power radiated per unit surface area of a black body, $L/A = \sigma T^4$ . Depends only on the absolute temperature (to the fourth power).
Apparent brightness (observed intensity)
Examiner keyword
The radiation intensity received from a star at Earth. It depends on both the star's luminosity and its distance, falling as $1/d^2$ , so it is not the same as luminosity.
Thermodynamic (absolute) temperature
Examiner keyword
Temperature measured in kelvin from absolute zero. Both Wien's law and the Stefan–Boltzmann law require $T$ in kelvin, never degrees Celsius.
Perfect emitter
A body that emits the maximum possible radiation at every wavelength for its temperature. A black body is a perfect emitter as well as a perfect absorber.
Stellar radius from L and T
A star's radius estimated by combining the two laws: Wien's law gives $T$ from $\lambda_{max}$ , then $r = \sqrt{L/(4\pi\sigma T^4)}$ gives $r$ from $L$ and $T$ (using $A = 4\pi r^2$ ).

Common Mistakes and Misconceptions — Black Body Radiation

The traps other students keep falling into on black body radiation questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Substituting temperature in degrees Celsius instead of kelvin
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Data are sometimes quoted in °C, and students substitute the number directly without converting.
How to avoid it
Both Wien's law and the Stefan–Boltzmann law require the absolute temperature in kelvin. Convert with $T/\text{K} = \theta/\text{°C} + 273$ before substituting. Using °C is disastrous in $\sigma T^4$ because of the fourth power.
✕Treating $\lambda_{max}$ as directly proportional to $T$ (multiplying instead of dividing)
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students remember ‘ $\lambda_{max}$ and $T$ are related’ but forget it is the product that is constant.
How to avoid it
Wien's law is $\lambda_{max}T =$ constant, so $\lambda_{max} \propto 1/T$ — a hotter body peaks at a shorter wavelength. To find $\lambda_{max}$ , divide the constant by $T$ ; to find $T$ , divide the constant by $\lambda_{max}$ .
✕Forgetting the fourth power of $T$ in the Stefan–Boltzmann law (or the square in $4\pi r^2$ )
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
The exponents are easy to drop when copying the formula, and calculators hide the slip.
How to avoid it
Write $L = \sigma A T^4$ in full and raise $T$ to the fourth power; use $A = 4\pi r^2$ (the radius is squared). A quick sanity check: doubling $T$ should multiply the power by $2^4 = 16$ .
✕Using the wrong surface area for a star — e.g. $\pi r^2$ (a disc) instead of $4\pi r^2$ (a sphere)
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
The circular cross-sectional area $\pi r^2$ is more familiar, so it is used by mistake.
How to avoid it
A star radiates from its whole spherical surface, so $A = 4\pi r^2$ . The disc area $\pi r^2$ is for the light intercepted by a target, not the area emitting; for luminosity always use $4\pi r^2$ .
✕Confusing luminosity (total power radiated) with the brightness observed from Earth
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Both are loosely called ‘brightness’, so students assume the brightest-looking star is the most luminous.
How to avoid it
Luminosity is the total power a star radiates (an intrinsic property, $L = \sigma A T^4$ ). Apparent brightness is the intensity that reaches us, which also depends on distance ( $\propto 1/d^2$ ). A near, faint star can outshine a distant, luminous one.
✕Leaving the wavelength in nanometres (or micrometres) when using Wien's law
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Peak wavelengths are naturally quoted in nm, and the number is substituted as it appears.
How to avoid it
Convert to metres before substituting: $1\ \text{nm} = 10^{-9}\ \text{m}$ , $1\ \mu\text{m} = 10^{-6}\ \text{m}$ . Since Wien's constant is $2.90\times10^{-3}\ \text{m K}$ , a wavelength left in nm gives a temperature $10^{9}$ times too large.

Black Body Radiation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Black Body Radiation

Short Notes - Black Body Radiation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Peak wavelength from temperature (2 marks)

2Surface temperature from peak wavelength (2 marks)

3Luminosity from the Stefan–Boltzmann law (2 marks)

4Luminosity of a star from its radius (4 marks)

5Effect of doubling the temperature on power (4 marks)

6Peak wavelength and surface flux together (4 marks)

7Estimating a star's radius from L and λmax (6 marks)

8Comparing two stars' luminosities (5 marks)

9Radius of a cool, luminous red giant (6 marks)

Wien's displacement law

Stefan–Boltzmann law (luminosity)

Surface area of a star (sphere)

Black body

Black-body radiation

Intensity–wavelength curve

Peak wavelength (λmax\lambda_{max}λmax​)

Wien's displacement law

Wien's constant

Stefan–Boltzmann law

Stefan–Boltzmann constant (σ\sigmaσ)

Luminosity

Surface flux

Apparent brightness (observed intensity)

Thermodynamic (absolute) temperature

Perfect emitter

Stellar radius from L and T

✕Substituting temperature in degrees Celsius instead of kelvin

✕Treating λmax\lambda_{max}λmax​ as directly proportional to TTT (multiplying instead of dividing)

✕Forgetting the fourth power of TTT in the Stefan–Boltzmann law (or the square in 4πr24\pi r^24πr2)

✕Using the wrong surface area for a star — e.g. πr2\pi r^2πr2 (a disc) instead of 4πr24\pi r^24πr2 (a sphere)

✕Confusing luminosity (total power radiated) with the brightness observed from Earth

✕Leaving the wavelength in nanometres (or micrometres) when using Wien's law

Black Body Radiation — frequently asked questions

Peak wavelength ( $\lambda_{max}$ )

Stefan–Boltzmann constant ( $\sigma$ )

✕Treating $\lambda_{max}$ as directly proportional to $T$ (multiplying instead of dividing)

✕Forgetting the fourth power of $T$ in the Stefan–Boltzmann law (or the square in $4\pi r^2$ )

✕Using the wrong surface area for a star — e.g. $\pi r^2$ (a disc) instead of $4\pi r^2$ (a sphere)