Black-body radiation: Stefan-Boltzmann and Wien

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A perfect **absorber and emitter** of radiation — it absorbs every wavelength that hits it and, when hot, radiates over all wavelengths. Stars are a good model.

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The total power (luminosity) radiated by a black body is $L = \sigma A T^4$ — surface area × temperature to the fourth power × the Stefan-Boltzmann constant. **Given** in the data booklet.

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The **Stefan-Boltzmann constant**, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given).

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As **T⁴** — doubling the kelvin temperature multiplies the power by 2⁴ = **16**.

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The peak wavelength and absolute temperature multiply to a constant: $\lambda_{max} T = 2.9 \times 10^{-3}$ m K. **Given** in the data booklet.

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They are **inversely** related — a **hotter** body has a **shorter** peak wavelength (bluer light).

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**Kelvin (K)** — never °C. Convert with K = °C + 273.

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It gets **taller** (more total power, Stefan-Boltzmann) and its **peak shifts to a shorter wavelength** (Wien).

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$\lambda_{max} = \dfrac{2.9 \times 10^{-3}}{5800} = 5.0 \times 10^{-7}$ m (500 nm).

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Write $L = \sigma A T^4$ for each and **divide** one by the other — σ cancels, leaving a ratio of areas and T⁴.

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Rising T shifts the spectrum's peak to shorter wavelengths (Wien) and adds power across all wavelengths (Stefan-Boltzmann), so the visible colour shifts red → orange → white.

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The sphere's surface area, $A = 4\pi r^2$, so $L \propto r^2 T^4$.