General shape of black-body curves, use of Wien’s displacement law to…

The idea

A black body absorbs all incident radiation and emits a continuous spectrum whose shape depends only on absolute temperature. Plotting intensity against wavelength gives a curve that rises steeply to a single peak then falls away at long wavelengths; hotter bodies peak at shorter wavelengths and have greater total intensity, so curves for different temperatures never cross. Wien’s displacement law states λmaxT = 2.9 × 10⁻³ m K, where λmax is the peak wavelength in metres and T is thermodynamic temperature in kelvin. To estimate a source’s temperature, read λmax from the curve or spectrum, convert to metres, then divide the constant by λmax.

For example: A star peaking at 5.0 × 10⁻⁷ m gives T = 2.9 × 10⁻³ ÷ 5.0 × 10⁻⁷ = 5800 K.

What earns the marks

• States that a black body absorbs all incident radiation and emits a continuous spectrum depending only on its absolute temperature.
• Describes the curve shape: intensity rises with wavelength to a single maximum, then decreases at longer wavelengths, with no crossing of curves for different temperatures.
• States Wien’s displacement law as λmaxT = 2.9 × 10⁻³ m K and identifies λmax as the wavelength of peak intensity.
• Converts a peak wavelength into metres and divides the constant by it to estimate T, or rearranges to find λmax from T.
• Explains that a higher temperature gives a shorter peak wavelength and a greater area under the curve, linking this to total emitted power.
• Uses correct units: λmax in m, T in K, and the constant in m K.

Common mistakes and how to put them right

• Using the wavelength in nanometres or micrometres directly in Wien’s law.

  Put it right: The error is failing to convert to metres, so the correction is to multiply by 10⁻⁹ or 10⁻⁶ before dividing.

• Treating the peak wavelength as the maximum intensity value.

  Put it right: The error is confusing the x-coordinate of the peak with the y-coordinate, so the correction is to read λmax from the wavelength axis.

• Assuming curves for different temperatures cross or that a cooler body can have a shorter peak wavelength.

  Put it right: Remember that higher temperature always shifts the peak to shorter wavelength.

Exam tips

• Sketch the axes with wavelength on the x-axis and intensity on the y-axis, label the peak and mark λmax clearly before substituting numbers.
• Show the rearrangement of λmaxT = 2.9 × 10⁻³ m K explicitly, then substitute values with units so the kelvin unit follows through.
• When comparing two sources, quote both peak wavelengths and state the inverse relationship: the shorter λmax corresponds to the higher temperature.