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AP Physics 2: Algebra Notes

7.4.4 Wien’s Law and Peak Wavelength

AP Syllabus focus: 'The peak wavelength emitted by a blackbody decreases as temperature increases, as described by Wien’s law.'

Thermal radiation gives a direct clue about temperature. For an ideal blackbody, the wavelength of strongest emission shifts in a predictable way, so Wien’s law is a powerful way to connect spectra to temperature.

Peak Wavelength in a Thermal Spectrum

A blackbody is an ideal object that emits electromagnetic radiation based only on its temperature. It does not emit just one wavelength. Instead, it produces a continuous spectrum with some wavelengths emitted more strongly than others.

In that spectrum, the peak wavelength is especially important.

Peak wavelength: The wavelength at which a blackbody emits the greatest intensity of radiation for a given temperature.

The peak wavelength is found at the highest point of an intensity-versus-wavelength graph. This does not mean the object emits radiation only at that wavelength. It means that wavelength corresponds to the maximum emission within the full spread of emitted wavelengths.

Wien’s Law

Wien’s law describes how the peak wavelength depends on temperature. As temperature increases, the peak wavelength becomes shorter.

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Energy emitted vs. wavelength curves for blackbodies at 300 K, 400 K, and 500 K, with an annotation showing the peak occurs at approximately 0.0029/T0.0029/T. The peak shifting to shorter wavelengths as temperature rises is the graphical signature of Wien’s displacement law and helps you read λmax\lambda_{max} directly from a spectrum plot. Source

As temperature decreases, the peak wavelength becomes longer.

λmaxT=b \lambda_{max}T=b

λmax \lambda_{max} = peak wavelength, in meters

T T = absolute temperature, in kelvin

b b = Wien’s constant, approximately 2.90×103 mK2.90\times10^{-3}\ m\cdot K

This equation shows an inverse relationship between peak wavelength and temperature. If one quantity goes up, the other goes down in the same proportion. For example, if the temperature doubles, the peak wavelength becomes half as large.

Because the law uses absolute temperature, temperature must be in kelvin, not degrees Celsius. A temperature in Celsius must be converted before the equation is used.

What the Law Means Physically

Hotter objects tend to have their strongest emission at shorter wavelengths. Cooler objects tend to have their strongest emission at longer wavelengths. This is why heating an object can change the part of the electromagnetic spectrum where its radiation is most intense.

A useful way to think about this is by comparing broad regions of the spectrum:

  • Cooler objects often have peak wavelengths in the infrared.

  • Moderately hot objects may have peak wavelengths in the visible range.

  • Very hot objects can have peak wavelengths in the ultraviolet.

This shifting peak helps explain why temperature affects the apparent color of glowing objects. A relatively cooler glowing object may appear reddish, while a hotter one may appear whiter or bluish because the strongest emission moves toward shorter wavelengths.

Wien’s law does not describe only visible light. It applies to the full electromagnetic spectrum, including infrared and ultraviolet wavelengths.

Reading Blackbody Graphs

A blackbody spectrum is often shown as a smooth curve on a graph of intensity versus wavelength.

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Blackbody emission curves plotted against wavelength for several temperatures (e.g., 3500 K, 6000 K, 10,000 K). The figure makes the Wien’s-law trend visually obvious: as TT increases, the peak of the spectrum shifts to shorter wavelengths (left) and the emitted intensity increases across the curve. Source

The position of the highest point on the curve gives the peak wavelength.

When temperature increases:

  • the peak moves left on a wavelength graph, toward shorter wavelengths

  • the curve usually becomes taller as well

  • the overall distribution shifts so shorter wavelengths become more significant

For this subtopic, the key idea is the horizontal shift of the peak. Wien’s law tells you where the peak occurs, not the total amount of power emitted.

A common misunderstanding is to assume that the peak wavelength is the “only” wavelength the object emits. In reality, a blackbody emits a range of wavelengths on both sides of the peak. The peak simply identifies the wavelength of maximum intensity.

Proportional Reasoning with Wien’s Law

You do not always need a full numerical calculation. Since λmaxT \lambda_{max}T is constant, proportional reasoning is often enough.

If temperature changes by some factor, peak wavelength changes by the inverse factor:

  • if T T increases by a factor of 2, then λmax \lambda_{max} decreases by a factor of 2

  • if T T increases by a factor of 3, then λmax \lambda_{max} decreases by a factor of 3

  • if λmax \lambda_{max} becomes 4 times larger, then T T becomes 4 times smaller

This relationship is often tested conceptually, especially when comparing two blackbodies.

Units and Measurement

The SI unit for wavelength in Wien’s law is the meter. However, spectra are often described using nanometers because visible and near-visible wavelengths are very small.

When using the equation, keep units consistent:

  • convert nanometers to meters if needed

  • use kelvin for temperature

  • make sure the constant and the wavelength use compatible units

Careless unit handling can give an answer that is numerically incorrect even if the physics idea is right.

Common Errors to Avoid

  • Using Celsius instead of kelvin

  • Thinking a higher temperature means a longer peak wavelength

  • Confusing peak wavelength with the full range of emitted wavelengths

  • Assuming Wien’s law gives total emitted power

  • Forgetting that the law applies to an ideal blackbody spectrum

Understanding Wien’s law means recognizing that temperature determines where the emission curve reaches its maximum, and that hotter blackbodies peak at shorter wavelengths.

FAQ

Room-temperature objects do emit thermal radiation, but their peak wavelength is usually in the infrared, not the visible range.

That means:

  • they are radiating energy

  • the radiation is mostly invisible to human eyes

  • infrared cameras can detect it even when we cannot

Visible glowing usually requires a much higher temperature so that enough emission reaches the visible part of the spectrum.

A blackbody emits a wide range of wavelengths, not just one.

When the object is very hot, the visible part of its spectrum can become broad and intense across many colors at once. Your eye then combines that mixture and may perceive it as white rather than as a pure red or blue color.

This is one reason very hot stars can appear white.

No. The location of the peak depends on how the spectrum is graphed.

A graph of intensity versus wavelength and a graph of intensity versus frequency do not have peaks that correspond by a simple inverse conversion. Because of this, the “peak” for wavelength and the “peak” for frequency are not just related by $f=c/\lambda$ using the same numerical maximum.

For AP Physics 2 Algebra, Wien’s law is used with wavelength.

Real stars are not ideal blackbodies, but their spectra are often close enough that Wien’s law gives a useful temperature estimate.

In practice, astronomers:

  • measure the spectrum

  • identify the approximate wavelength where emission is strongest

  • use that value to estimate surface temperature

The result is often an approximation, but it is still very useful for comparing stars.

Several factors can shift or obscure the measured peak:

  • absorption by Earth’s atmosphere

  • detector sensitivity limits

  • emission or absorption lines superimposed on the spectrum

  • the object not behaving like a perfect blackbody

Because of these effects, scientists often correct observational data before applying Wien’s law to estimate temperature.

Practice Questions

A blackbody has its peak wavelength at 800 nm800\ nm. A second blackbody is at twice the absolute temperature of the first. What is the peak wavelength of the second blackbody?

  • 1 mark for stating that peak wavelength is inversely proportional to temperature.

  • 1 mark for obtaining 400 nm400\ nm.

Wien’s constant is b=2.90×103 mKb=2.90\times10^{-3}\ m\cdot K.

A certain hot object has a peak wavelength of 580 nm580\ nm.

(a) Calculate the temperature of the object.

(b) A second object has a peak wavelength of 290 nm290\ nm. Compare its temperature to the temperature of the first object.

(c) Explain why the answer to part (b) does not mean the second object emits radiation only at 290 nm290\ nm.

  • (a) 1 mark for converting 580 nm580\ nm to 5.80×107 m5.80\times10^{-7}\ m

  • (a) 1 mark for using λmaxT=b \lambda_{max}T=b

  • (a) 1 mark for calculating T5.0×103 KT\approx5.0\times10^{3}\ K

  • (b) 1 mark for stating that the second object has twice the temperature of the first

  • (c) 1 mark for explaining that 290 nm290\ nm is the wavelength of maximum intensity, not the only emitted wavelength

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