Problem — Black body at different temperatures

Problem

The emission curve of a black body at $T = 6000$ K has a peak at $\lambda \approx 480$ nm. Estimate the peak at $T = 3000$ K using Wien’s law and qualitatively describe the change in apparent colour.

Solution

Step 1 — Wien’s law. The emission peak satisfies $\lambda_\text{max}\,T = \text{constant}$. So the product is the same at the two temperatures:

$\lambda_1 T_1 = \lambda_2 T_2$

Step 2 — Work out the new peak.

$\lambda_2 = \lambda_1\,\frac{T_1}{T_2} = 480\;\text{nm}\cdot \frac{6000}{3000} = 480\cdot 2 = 960\;\text{nm}$

$\ev{\lambda_\text{max}(3000\;\text{K}) \approx 960\;\text{nm (infrared)}}$

Step 3 — Apparent colour. At $6000$ K the peak ($480$ nm) is in the visible range (blue-white): the body appears bright and whitish, like the Sun. Halving the temperature to $3000$ K shifts the peak to $960$ nm, beyond red, into the infrared: most of the energy is no longer visible. In the visible range only the short-wavelength tail remains, dominated by red: the body appears red-orange and much dimmer (the total power, $\propto T^4$, drops by a factor of $16$). This is why a heated metal goes from red to white-blue as the temperature rises.

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Topics: Fisica quantistica Concepts: Corpo nero e ipotesi di Planck Skills: Impostazione simbolica · Lettura dei grafici