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Laws of thermal radiation

Short answer

Planck's law: E_bλ = C1 λ^{-5}/(e^{C2/λT}-1). Wien's law: λ_max T = 2898 μm·K (peak shifts to shorter λ with T). Stefan-Boltzmann: E_b = σT^4 (integral of Planck). Kirchhoff: emissivity = absorptivity at equilibrium. Lambert: diffuse surface intensity constant, E = πI. Hotter bodies emit at higher frequency and more total energy.

State and explain Planck's law, Wien's displacement law, Stefan-Boltzmann law, Kirchhoff's law and Lambert's cosine law of thermal radiation. Show how Wien's law and the Stefan-Boltzmann law follow from Planck's distribution and explain the relation between temperature, frequency and wavelength of radiation.

Thermal radiation is electromagnetic radiation emitted by a body because of its temperature, in the range about 0.10.1 to 100 μ100\ \mum (ultraviolet, visible and infrared).

Planck's law

It gives the spectral emissive power of a black body at wavelength λ\lambda and absolute temperature TT:

Ebλ=C1 λ−5exp⁡(C2/λT)−1E_{b\lambda} = \frac{C_1\,\lambda^{-5}}{\exp\left(C_2/\lambda T\right)-1}

C1=3.742×108 W μm4/m2C_1 = 3.742\times10^8\ \text{W}\,\mu\text{m}^4/\text{m}^2, C2=14 388 μm KC_2 = 14\,388\ \mu\text{m K}. At every wavelength EbλE_{b\lambda} rises with TT; each curve has a peak.

Wien's displacement law

Setting dEbλ/dλ=0dE_{b\lambda}/d\lambda = 0 gives the wavelength at the peak. With x=C2/λTx = C_2/\lambda T, the condition becomes

x ex=5 (ex−1)  ⇒  x=4.965x\,e^{x} = 5\,(e^{x}-1)\;\Rightarrow\; x = 4.965

so

λmaxT=C24.965=2898 μm K\lambda_{max}T = \frac{C_2}{4.965} = 2898\ \mu\text{m K}

The peak moves to shorter wavelengths as TT rises (a heated iron goes from dull red to white). The sun (5800 K) peaks near 0.5 μ0.5\ \mum, a room-temperature body (300 K) near 9.7 μ9.7\ \mum.

Stefan-Boltzmann law

Integrating Planck's law over all wavelengths gives the total emissive power of a black body. Put x=C2/λTx = C_2/\lambda T, so λ=C2/xT\lambda = C_2/xT:

Eb=∫0∞Ebλ dλ=C1T4C24∫0∞x3ex−1 dx=C1T4C24⋅π415=σT4E_b = \int_0^\infty E_{b\lambda}\,d\lambda = \frac{C_1T^4}{C_2^4}\int_0^\infty\frac{x^3}{e^x-1}\,dx = \frac{C_1T^4}{C_2^4}\cdot\frac{\pi^4}{15} = \sigma T^4 σ=π4C115 C24=5.67×10−8 W/m2K4\sigma = \frac{\pi^4C_1}{15\,C_2^4} = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4

For a real (grey) surface E=εσT4E = \varepsilon\sigma T^4.

Kirchhoff's law

At thermal equilibrium the emissivity of a surface equals its absorptivity, ελ=αλ\varepsilon_\lambda = \alpha_\lambda. For a grey (diffuse) surface ε=α\varepsilon = \alpha at all conditions. A good absorber is a good emitter.

Lambert's cosine law

The radiation intensity II of a diffuse surface is the same in all directions, so emissive power in direction θ\theta from the normal varies as cos⁡θ\cos\theta, and total emissive power E=πIE = \pi I.

Frequency, wavelength and temperature

Wavelength and frequency are related by λ=c/ν\lambda = c/\nu (c=3×108c = 3\times10^8 m/s in vacuum). A hotter body emits at higher frequency (shorter wavelength) for its peak, according to Wien's law, and also emits more energy at all frequencies according to the T4T^4 law.

✓ Answer checked by Claude Sonnet 5.5

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