Black-Body Radiation and Planck's Quantum Theory

1Reading a temperature from a colour

A blacksmith in a workshop at Gakinjiro, Kigali, pulls a bar of steel from the fire and judges whether it is hot enough to bend just by looking at it. Below about 800 K the bar gives off no light you can see, although you can feel its heat on your face. As it gets hotter it glows dull red, then orange, then yellow-white. The colour is a thermometer. At the end of the nineteenth century the best physicists in Europe could measure this glow very precisely but could not explain its shape. The fix, proposed by Max Planck in December 1900, was that energy comes in lumps. That idea began quantum physics, and this unit follows where it leads: to photons, to the photoelectric effect and to electrons that behave as waves.

2Thermal radiation and the ideal black body

Every object above absolute zero emits thermal radiation: electromagnetic waves produced by the jostling of the charged particles inside it. A room-temperature wall, your skin and the Sun all radiate. What differs is how much they radiate and at which wavelengths.

Surfaces also absorb. A dull black surface absorbs nearly all the radiation that falls on it; a polished silver one reflects most of it. Gustav Kirchhoff showed in 1859 that, at a given temperature and wavelength, a good absorber is an equally good emitter. The perfect absorber is therefore the perfect emitter, and physicists call it a black body.

No real surface is perfectly black, but a hollow box with a tiny hole comes very close. Radiation that enters the hole bounces around inside, losing part of its energy to the wall at each reflection, and almost none finds its way back out. The hole absorbs essentially everything, so the radiation that comes out of it when the walls are heated is black-body radiation, and its spectrum depends only on the temperature of the walls, not on what they are made of.

3The black-body spectrum

When the radiation from a cavity is spread out into a spectrum and its power measured at each wavelength, the curves below are obtained. They are drawn to scale from Planck's law (which you meet later in this lesson).

Three features stand out:

• Each curve has a single peak at a wavelength ${\lambda }_{\max }\backslash lambda\_\{\backslash max\}$, and falls to zero at very short and very long wavelengths.
• As the temperature rises, the peak moves to shorter wavelengths. This is why a heated bar goes from red to yellow-white.
• A hotter body radiates more at every wavelength, so the total area under the curve grows very rapidly with temperature.

Wien's displacement law

The peak wavelength is inversely proportional to the absolute temperature:

The Stefan–Boltzmann law

The total power radiated by a black body of surface area $AA$ at absolute temperature $TT$ is

For a real surface, multiply by its emissivity: $P=\epsilon \sigma A{T}^{4}P\; =\; \backslash varepsilon\; \backslash sigma\; A\; T^4$. An object at temperature $TT$ in surroundings at $T_{0}T\_0$ also absorbs $\epsilon \sigma A{T}_0^4\backslash varepsilon\; \backslash sigma\; A\; T\_0^4$, so the net power it loses is

4The ultraviolet catastrophe

Classical physics treats the radiation in a cavity as standing electromagnetic waves (modes). The shorter the wavelength, the more modes fit in the box: their number per unit wavelength rises as $\frac{1}{{\lambda }^4}\backslash frac\{1\}\{\backslash lambda^4\}$. Classical statistical physics gives every mode the same average energy $kTkT$, where $k=1.38\times {10}^{-23}\mathrm{J}{\mathrm{K}}^{-1}k\; =\; 1.38\; \backslash times\; 10^\{-23\}\backslash ,\backslash mathrm\{J\backslash ,K^\{-1\}\}$ is the Boltzmann constant. The result, the Rayleigh–Jeans law, is

It matches experiment at long wavelengths (the dashed line in the graph), but it grows without limit as $\lambda \to 0\backslash lambda\; \backslash to\; 0$. Added over all wavelengths, it predicts that any warm body radiates an infinite power, mostly in the ultraviolet and beyond. This impossible prediction was later called the ultraviolet catastrophe.

5Planck's quantum hypothesis

Planck found a formula that fitted the data at every wavelength, and then looked for a reason. He found one only by assuming that the oscillating charges in the walls cannot have any energy at all, but only whole-number multiples of a smallest amount, a quantum, proportional to their frequency $ff$:

where $h=6.63\times {10}^{-34}\mathrm{J}\mathrm{s}h\; =\; 6.63\; \backslash times\; 10^\{-34\}\backslash ,\backslash mathrm\{J\backslash ,s\}$ is Planck's constant. An oscillator can therefore change its energy only in steps of $hfhf$.

This removes the catastrophe. A low-frequency mode has a small quantum, $hf\ll kThf\; \backslash ll\; kT$, and is excited easily, as in classical physics. A high-frequency mode needs a quantum $hf\gg kThf\; \backslash gg\; kT$. The chance that the thermal jostling supplies that much energy is about $e^{-hf\mathrm{/}kT}e^\{-hf/kT\}$, which is tiny, so high-frequency modes are almost never excited and the spectrum falls to zero at short wavelengths. Planck's law is

Check the long-wavelength limit. When $\lambda \backslash lambda$ is large, $x=\frac{hc}{\lambda kT}x\; =\; \backslash frac\{hc\}\{\backslash lambda\; kT\}$ is small and $e^x-1\approx xe^\{x\}\; -\; 1\; \backslash approx\; x$. Then

which is the Rayleigh–Jeans law, as it must be. Setting $\frac{d{B}_{\lambda }}{d\lambda }=0\backslash frac\{dB\_\backslash lambda\}\{d\backslash lambda\}\; =\; 0$ gives $\frac{hc}{{\lambda }_{\max }kT}=4.965\backslash frac\{hc\}\{\backslash lambda\_\{\backslash max\}\; kT\}\; =\; 4.965$, so ${\lambda }_{\max }T=\frac{hc}{4.965k}=2.897\times {10}^{-3}\mathrm{m}\mathrm{K}\backslash lambda\_\{\backslash max\}\; T\; =\; \backslash frac\{hc\}\{4.965k\}\; =\; 2.897\; \backslash times\; 10^\{-3\}\backslash ,\backslash mathrm\{m\backslash ,K\}$: Wien's law comes out of Planck's law, with the right constant.