Black-Body Radiation Laws: Stefan's and Wien's Revision Notes | A Level AQA Physics

1. Why Stars Glow Different Colours

Look up at the night sky and you will notice that stars are not all the same colour. Rigel appears blue-white; Betelgeuse glows distinctly orange-red. That colour difference is not a quirk of the atmosphere — it is a direct readout of each star's surface temperature. Two laws, Stefan's law and Wien's displacement law, turn that colour information into precise, quantitative physics.

Before we can apply those laws, we need one crucial assumption: we treat a star as a black body. A black body absorbs all radiation that falls on it and re-emits radiation whose spectrum depends only on temperature — not on the material it is made from. Stars are not perfect black bodies, but they are close enough that the model gives excellent agreement with observation, and AQA explicitly requires you to state this assumption.

The black-body assumption is powerful because it means we can predict the full emission spectrum of a star from a single number: its surface temperature $T$. That is the foundation on which both Wien's law and Stefan's law rest.

2. The Shape of Black-Body Radiation Curves

Picture a graph with wavelength on the horizontal axis and intensity (power emitted per unit wavelength) on the vertical axis. For any black body at a fixed temperature, the curve rises steeply from zero at very short wavelengths, reaches a single sharp peak at a wavelength called $\lambda_{\text{max}}$, then falls away gradually at longer wavelengths. This characteristic shape is the same for every black body — only the position of the peak and the overall height of the curve change with temperature.

Two trends are immediately visible from the curves. First, as temperature increases, the peak shifts to shorter wavelengths — hotter stars emit most strongly in the blue or ultraviolet region. Second, the area under the curve (total power emitted) grows dramatically with temperature — hotter stars are far more luminous. These two trends are captured precisely by Wien's displacement law and Stefan's law respectively.

Notice also that each curve is asymmetric: the rise on the short-wavelength side is steeper than the fall on the long-wavelength side. Examiners sometimes ask you to describe the shape — use the words "rises steeply", "single peak at $\lambda_{\text{max}}$", and "falls gradually" to secure those description marks.

3. Wien's Displacement Law

Wien's displacement law gives us the quantitative link between peak wavelength and temperature. The relationship is elegantly simple: the product of the peak wavelength $\lambda_{\text{max}}$ and the absolute temperature $T$ is a constant.

Because $\lambda_{\text{max}}$ and $T$ are inversely proportional, doubling the temperature halves the peak wavelength. That is why a star at 12 000 K peaks in the ultraviolet while a star at 3 000 K peaks in the infrared. The rearrangement you will use most often in calculations is $T = \frac{2.9 \times 10^{-3}}{\lambda_{\text{max}}}$.

Worked example: Estimating Stellar Temperature Using Wien's Displacement Law

The star Sirius A has a peak emission wavelength of $2.9 \times 10^{-7}$ m. Estimate its surface temperature.

1. $\lambda_{\text{max}} T = 2.9 \times 10^{-3}$ m K — This is the starting equation. Both sides must have consistent units: metres for wavelength, kelvin for temperature. [M1]
2. $T = \frac{2.9 \times 10^{-3}}{\lambda_{\text{max}}}$ — Divide both sides by $\lambda_{\text{max}}$. This rearrangement is the one you will use in virtually every Wien's law question. [M1]
3. $T = \frac{2.9 \times 10^{-3}}{2.9 \times 10^{-7}}$ — The wavelength is already in metres, so no conversion is needed. Carry units through explicitly.
4. $T = \frac{2.9}{2.9} \times \frac{10^{-3}}{10^{-7}} = 1 \times 10^{-3-(-7)} = 1 \times 10^{4}$ K — Divide the coefficients: $2.9 \div 2.9 = 1$. Subtract the exponents: $-3 - (-7) = +4$. So $T = 10^{4}$ K = 10 000 K. [A1]
5. $T \approx 10\,000$ K — Always include kelvin. The word 'estimate' in the question signals that the answer is approximate — this is expected and acceptable.

Answer: $T \approx 10\,000$ K

Notice the word "estimate" in the question. AQA uses this deliberately because the black-body model is an approximation for real stars. Your answer does not need to be exact to the nearest kelvin — the method and correct order of magnitude are what earn the marks.

4. Stefan's Law and Stellar Luminosity

Wien's law tells us where the peak of the spectrum sits. Stefan's law tells us how much total power a black body radiates across all wavelengths. This total power output is what astronomers call luminosity, $L$, measured in watts.

For a spherical star of radius $R$, the total surface area is $4\pi R^2$. Multiplying the power per unit area by the full surface area gives the luminosity: $L = 4\pi R^2 \sigma T^4$. This single equation connects three observable or inferable quantities — luminosity, radius, and surface temperature — and it is one of the most powerful tools in stellar astrophysics.

The $T^4$ dependence is the most important feature of Stefan's law. Doubling the temperature does not double the luminosity — it increases it by a factor of $2^4 = 16$. This is why a relatively small increase in surface temperature produces an enormous increase in a star's brightness. A star at 10 000 K radiates $2^4 = 16$ times more power per unit area than a star at 5 000 K, even if both stars are the same size.

Worked example: Calculating Stellar Luminosity Using Stefan's Law

A star has a surface temperature of $6000$ K and a radius of $7.0 \times 10^{8}$ m. Calculate its luminosity. ($\sigma = 5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$)

1. $L = 4\pi R^2 \sigma T^4$ — This combines the power per unit area ($\sigma T^4$) with the full surface area of a sphere ($4\pi R^2$). [M1]
2. $R^2 = (7.0 \times 10^{8})^2 = 7.0 \times 7.0 \times 10^{8 \times 2} = 49 \times 10^{16} = 4.9 \times 10^{17}$ m$^2$ — Square the coefficient and double the exponent separately. $7.0 \times 7.0 = 49$, and $10^8 \times 10^8 = 10^{16}$. Then write in standard form: $4.9 \times 10^{17}$.
3. $T^4 = 6000^4 = (6.0 \times 10^3)^4 = 6.0^4 \times 10^{12}$ $6.0^4 = 6.0 \times 6.0 \times 6.0 \times 6.0 = 36 \times 36 = 1296$ $T^4 = 1296 \times 10^{12} = 1.296 \times 10^{15}$ K$^4$ — Work through the fourth power step by step: $6 \times 6 = 36$, then $36 \times 36 = 1296$. Never skip this — a common error is writing $6^4 = 24$ (multiplying instead of raising to a power).
4. $L = 4\pi \times (4.9 \times 10^{17}) \times (5.67 \times 10^{-8}) \times (1.296 \times 10^{15})$ — Gather all terms before multiplying. Keep track of powers of ten separately from the numerical coefficients. [M1]
5. $4 \times \pi \times 4.9 \times 5.67 \times 1.296$ $= 4 \times 3.1416 \times 4.9 \times 5.67 \times 1.296$ $\approx 4 \times 3.1416 \times 36.03$ $\approx 4 \times 113.2 \approx 452.8$ — Work in stages: $4.9 \times 5.67 \approx 27.78$; $27.78 \times 1.296 \approx 36.0$; $36.0 \times \pi \approx 113.1$; $113.1 \times 4 \approx 452$.
6. $10^{17} \times 10^{-8} \times 10^{15} = 10^{17+(-8)+15} = 10^{24}$ — Add exponents when multiplying powers of ten: $17 - 8 + 15 = 24$.
7. $L \approx 452 \times 10^{24}$ W $= 4.5 \times 10^{26}$ W — Convert to standard form. The Sun's actual luminosity is approximately $3.8 \times 10^{26}$ W, so this answer is physically reasonable for a Sun-like star. [A1]

Answer: $L \approx 4.5 \times 10^{26}$ W

That sanity check in the final step is worth doing in an exam. If your luminosity comes out as $10^{5}$ W or $10^{50}$ W, something has gone wrong with the powers of ten. The Sun's luminosity ($\approx 3.8 \times 10^{26}$ W) is a useful benchmark to keep in mind.

5. Comparing Stars: Applying Both Laws Together

In practice, exam questions often ask you to compare two stars using both laws. Earlier we used Wien's law to find temperature from peak wavelength; now we can feed that temperature into Stefan's law to compare luminosities. The two laws work as a pair.

A particularly common question type asks you to find the ratio of luminosities of two stars. If both stars have the same radius, the ratio simplifies to $\frac{L_1}{L_2} = \left(\frac{T_1}{T_2}\right)^4$. If the radii differ as well, you must include the $R^2$ terms: $\frac{L_1}{L_2} = \frac{R_1^2 T_1^4}{R_2^2 T_2^4}$. Working with ratios avoids having to substitute the Stefan-Boltzmann constant at all.

Worked example: Comparing Luminosities of Two Stars

Star X has surface temperature $T_X = 12\,000$ K and radius $R_X = 2.0 \times 10^{9}$ m. Star Y has surface temperature $T_Y = 6\,000$ K and radius $R_Y = 2.0 \times 10^{9}$ m. Calculate the ratio $L_X / L_Y$.

1. $\frac{L_X}{L_Y} = \frac{4\pi R_X^2 \sigma T_X^4}{4\pi R_Y^2 \sigma T_Y^4}$ — Write Stefan's law for each star, then form the ratio. The constants $4\pi$ and $\sigma$ appear in both numerator and denominator.
2. $\frac{L_X}{L_Y} = \frac{R_X^2 T_X^4}{R_Y^2 T_Y^4}$ — $4\pi$ and $\sigma$ cancel completely. This is why ratio questions are efficient — you do not need the value of $\sigma$.
3. $R_X = R_Y = 2.0 \times 10^{9}$ m, so $\frac{R_X^2}{R_Y^2} = 1$ — Both stars have the same radius, so the radius ratio equals 1 and drops out.
4. $\frac{T_X}{T_Y} = \frac{12\,000}{6\,000} = 2$ — Star X is twice as hot as Star Y.
5. $\left(\frac{T_X}{T_Y}\right)^4 = 2^4 = 2 \times 2 \times 2 \times 2 = 16$ — $2^4$: first $2 \times 2 = 4$, then $4 \times 2 = 8$, then $8 \times 2 = 16$. Never write $2^4 = 8$ — that is $2^3$.
6. $\frac{L_X}{L_Y} = 16$ — Star X is 16 times more luminous than Star Y, despite being only twice as hot. This illustrates the dramatic effect of the fourth-power dependence in Stefan's law. [A1]

Answer: $L_X / L_Y = 16$

6. The Black-Body Assumption: Validity and Limitations

Recall from the opening section that treating a star as a black body is an assumption, not a fact. AQA expects you to be able to state this assumption clearly and to understand what it implies. A real stellar spectrum shows absorption lines — dark gaps where specific wavelengths have been absorbed by cooler gas in the star's outer atmosphere. These lines sit on top of an underlying continuous spectrum that closely follows the black-body curve.

The assumption is valid because the continuous component of the spectrum — the part we use to identify $\lambda_{\text{max}}$ — does match the black-body prediction well. The absorption lines affect only narrow wavelength ranges and do not significantly shift the peak. For the purposes of applying Wien's law and Stefan's law, the black-body model gives results that agree with independent measurements of stellar temperatures and luminosities.

7. Exam Technique and Common Mistakes

These two laws generate a predictable set of exam questions, and examiners use the same traps year after year. Knowing where students lose marks is as valuable as knowing the physics itself.

For extended-response questions that ask you to 'explain how astronomers determine the surface temperature of a star', a full-mark answer needs four elements: state the black-body assumption, state Wien's displacement law with the correct formula, explain that $\lambda_{\text{max}}$ is identified from the observed spectrum, and show how $T$ is calculated by rearranging the law. Omitting the assumption is the most common reason students drop a mark on this type of question.

Stefan's law and Wien's displacement law together form the quantitative backbone of stellar classification. Wien's law reads the star's colour and converts it to temperature; Stefan's law uses that temperature alongside the star's size to give its total power output. Both laws rest on the black-body assumption — state it, use it, and you have the tools to analyse any star the examiner places in front of you.