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Blackbody radiation with Wien's and Stefan's laws, the ultraviolet catastrophe and Planck's quanta, Compton scattering, and the correspondence principle.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to apply Wien's and Stefan's laws, compute Compton shifts, and explain why classical physics failed for both.
From the last lessons you have photons with $E = hf$, the photoelectric effect and atomic spectra. From thermodynamics you know temperature and energy transfer by radiation. This lesson presents the two other experiments that classical physics could not explain, and which quanta explained exactly: the light from hot bodies, and X-rays bouncing off electrons.
| Term | What it means |
|---|---|
| Blackbody | An ideal object that absorbs all light falling on it and glows with a spectrum set only by its temperature. |
| Wien's law | $\lambda_{\max}T = 2.898 \times 10^{-3}$ m·K: hotter bodies peak at shorter wavelengths. |
| Stefan–Boltzmann law | $P/A = \sigma T^4$, with $\sigma = 5.67 \times 10^{-8}$ W/m²K⁴. |
| Ultraviolet catastrophe | Classical theory's prediction of infinite short-wavelength output. |
| Compton scattering | A photon bouncing off an electron and losing energy to it. |
| Compton wavelength | $h/m_ec = 2.426$ pm, the scale of the electron's Compton shift. |
| Correspondence | Quantum results reducing to classical ones where quanta are negligible. |
Every hot object glows, and an ideal blackbody glows with a spectrum that depends only on its temperature. The peak shifts to shorter wavelengths as it heats, by Wien's law,
$$\lambda_{\max}T = 2.898 \times 10^{-3}\ \text{m·K},$$
and the total power per unit area grows as $\sigma T^4$. Classical physics, which let each mode of light in a hot cavity hold the same average energy, predicted infinite output at short wavelengths. Planck fixed it in 1900 by letting energy come only in quanta $hf$, which starve the high-frequency modes.
In 1923 Arthur Compton found that X-rays scattered by electrons come out with longer wavelengths, shifted by
$$\Delta\lambda = \frac{h}{m_ec}(1 - \cos\theta),$$
exactly as a photon with momentum $h/\lambda$ colliding elastically with an electron would. A wave picture predicts no shift at all.
Another way: picture
Picture a stove element warming up: first it glows dull red, then orange, then yellow-white, as the peak of its glow slides from the infrared toward the blue. Picture a cue ball striking a stationary ball: it bounces off with less energy, more so the harder it glances back. X-ray photons strike electrons the same way, and lose energy, which means longer wavelength.
Another way: steps
A blackbody absorbs everything, so the light it emits is purely its own thermal glow. A small hole in a hot oven is a good blackbody: light entering bounces around inside until absorbed, and light leaving has come to equilibrium with the walls. Stars, the filament of an old light bulb, and the glowing coals of a fire come close.
The measured spectrum rises from zero at long wavelengths, peaks, and falls back to zero at short wavelengths. The Sun's surface, at about $5800$ K, peaks near $500$ nm, in the middle of the visible band; a person at $310$ K peaks near $9.5$ μm, in the infrared; the universe's leftover radiation, at $2.725$ K, peaks near $1$ mm.
Classical physics treated the light in a hot cavity as waves, and the equipartition theorem gave each possible wave pattern the same average energy, $kT$. The number of possible patterns grows without limit at short wavelengths, so the predicted energy did too: the Rayleigh–Jeans law, which matches measurements at long wavelengths but diverges in the ultraviolet.
Planck's fix was to let the walls exchange energy with each wave only in lumps of $hf$. At high frequency a single lump costs far more than $kT$, so those waves are almost never excited, and the spectrum turns over. The fit with $h = 6.63 \times 10^{-34}$ J·s was perfect, and Planck's constant entered physics.
Wien's law and the Stefan–Boltzmann law both follow from Planck's formula. The peak wavelength falls as $1/T$; the total output, the area under the curve, rises as $T^4$. Double the temperature and the peak moves to half the wavelength while the glow becomes sixteen times brighter per square meter.
Astronomers use both. The color of a star gives its surface temperature through Wien's law; its brightness and temperature together give its radius through Stefan's. Betelgeuse, at $3500$ K, looks orange-red and is hundreds of times the Sun's radius; Rigel, at $12{,}000$ K, looks blue-white.
At Washington University in St. Louis in 1922 and 1923, Arthur Compton aimed X-rays from a molybdenum target, wavelength $71$ pm, at a block of graphite and measured the wavelength of the scattered rays at several angles with a crystal spectrometer. Alongside the original wavelength he found a second peak, longer, shifted more at larger angles.
Treating the X-ray as a photon with energy $hc/\lambda$ and momentum $h/\lambda$, colliding with a free electron, and applying relativistic energy and momentum conservation, gives exactly $\Delta\lambda = (h/m_ec)(1 - \cos\theta)$. Compton's result convinced most physicists that photons were real. He shared the 1927 Nobel Prize.
The Compton wavelength of the electron, $2.426$ pm, sets the size of the shift: at most twice that, for a photon scattered straight back. For X-rays of $71$ pm, that is a shift of several percent, easily measured. For visible light at $500$ nm, it is a change of one part in a hundred thousand, invisible in everyday scattering.
That is why the wave picture of light worked so well for visible optics: the photon's momentum is too small to kick an electron noticeably. Quantum effects in scattering show up only when the photon's energy is comparable to the electron's rest energy scale, which for X-rays and gamma rays it becomes.
Checking an answer. Hotter bodies must peak at shorter wavelengths. The Compton shift must be zero at $0°$ and at most $4.852$ pm at $180°$. Scattered photons must have less energy than incident ones. Temperatures must be in kelvin.
Wien's law applies exactly to ideal blackbodies and approximately to real objects, whose emissivity is less than 1 and may vary with wavelength. Stars are close enough that color temperatures agree with other measurements to a few percent.
Compton's formula assumes the electron is free and at rest. Electrons in graphite are bound by a few eV, tiny next to the X-ray's $17$ keV, so the approximation is excellent. Tightly bound inner electrons of heavy atoms act as if the whole atom recoiled, giving a negligible shift, which is why Compton also saw the unshifted peak.
Quantum theory did not throw classical physics away. At long wavelengths, where $hf$ is much less than $kT$, Planck's formula reduces exactly to the classical Rayleigh–Jeans law. At low photon energies, the Compton shift vanishes and classical scattering returns. Niels Bohr called this the correspondence principle: a new theory must reproduce the old one where the old one was tested.
The same pattern appeared in relativity, which reduces to Newton's mechanics at low speeds. Each time physics has advanced, the old theory has survived as a limit of the new, trustworthy within its range.
In 1964 Arno Penzias and Robert Wilson, testing a radio antenna at Bell Labs in Holmdel, New Jersey, found a faint hiss coming equally from every direction. It was the afterglow of the Big Bang: radiation released when the universe cooled enough for atoms to form, stretched by expansion ever since.
NASA's COBE satellite, in 1990, measured its spectrum and found a perfect blackbody at $2.725$ K, the most precise blackbody curve ever seen in nature. John Mather of NASA Goddard and George Smoot of Berkeley shared the 2006 Nobel Prize for the measurement. Planck's formula, devised to explain ovens, describes the entire universe.
An old incandescent bulb is a blackbody made useful, and its inefficiency is Wien's law at work. Its tungsten filament runs near $2800$ K, so its glow peaks around $1000$ nm, in the infrared. Only about a tenth of the radiated power falls in the visible band; the rest warms the room. Running the filament hotter would move the peak toward the visible, but tungsten melts at $3695$ K and evaporates quickly well before that.
That is the physical reason the United States phased out most incandescent bulbs in favor of LEDs, which emit light from a transition across a semiconductor band gap instead of a thermal glow. An LED can put most of its output into a chosen band of wavelengths, and a modern LED bulb gives the light of a $60$ W incandescent bulb for about $9$ W. The difference between a thermal spectrum and a line spectrum, the subject of the last two lessons, is worth billions of dollars a year in electricity.
FLIR Systems, based in Wilsonville, Oregon, makes the thermal cameras used by firefighters, building inspectors and search-and-rescue teams. Human skin at about $305$ K glows most strongly near $9.5$ μm, and a house wall or road near $10$ μm, in the long-wave infrared. The cameras use sensors tuned to $8$ to $14$ μm, where these objects are brightest and the air is transparent.
Because output grows as $T^4$, a small temperature difference makes a large difference in brightness: a $1$ K difference near room temperature changes the glow by about $1.3$ percent, which good cameras can see. Firefighters use them to find people through smoke, and energy auditors to find where heat leaks from a home.
Gamma rays from radioactive tracers are hard to focus, because no lens bends them. Compton cameras track a gamma ray's scattering off an electron in one detector and its absorption in a second, and use the Compton formula to find the angle it came from. NASA's COSI telescope, a Compton instrument built at the University of California, Berkeley, maps gamma rays from the galaxy this way.
In radiation therapy, Compton scattering is the main way high-energy X-rays deposit energy in tissue, knocking electrons loose to do the damage that kills tumors. Planning a treatment means computing millions of such scatterings, each following Compton's formula.
The failures of classical physics can make it seem entirely wrong. It is not: it predicts blackbody spectra correctly at long wavelengths, visible-light scattering correctly, and nearly everything about motors, bridges and planets. It breaks down where a single quantum's energy, $hf$, is comparable to or larger than the energies available, $kT$ or an electron's rest energy.
A related error is to think hotter objects glow redder, as the word "red-hot" might suggest. Red-hot is the coolest visible glow; white-hot and blue-hot are hotter.
The Sun's spectrum peaks near $500$ nm. Find its surface temperature.
$T = \dfrac{2.898 \times 10^{-3}}{500 \times 10^{-9}} = 5800\ \text{K}$
Wien's law.
Find the power per square meter of its surface.
$\sigma T^4 = 5.67 \times 10^{-8} \times 5800^4 = 6.4 \times 10^7\ \text{W/m}^2$
Stefan–Boltzmann.
Find its total output, with radius $6.96 \times 10^8$ m.
$P = 4\pi R^2\sigma T^4 = 3.9 \times 10^{26}\ \text{W}$
Area of the sphere.
Find the intensity at Earth, $1.50 \times 10^{11}$ m away.
$I = \dfrac{3.9 \times 10^{26}}{4\pi(1.50 \times 10^{11})^2} = 1380\ \text{W/m}^2$
Spread over a sphere.
Compare with the measured value.
$1361\ \text{W/m}^2$
Close: the Sun is nearly a blackbody.
A $50.0$ pm X-ray scatters at $90°$. Find the shift.
$\Delta\lambda = 2.426(1 - 0) = 2.43\ \text{pm}$
$\cos 90° = 0$.
Find the scattered wavelength.
$\lambda' = 52.4\ \text{pm}$
Longer.
Find the incident energy.
$E = \dfrac{1240}{50.0} = 24.8\ \text{keV}$
$hc = 1240$ keV·pm.
Find the scattered energy.
$E' = \dfrac{1240}{52.4} = 23.7\ \text{keV}$
Less.
Find the electron's kinetic energy.
$K = 24.8 - 23.7 = 1.1\ \text{keV}$
Energy conservation.
Find the shift for backscattering.
$\Delta\lambda = 2.426 \times 2 = 4.85\ \text{pm}$
The largest possible.
At $5800$ K, find $kT$ in eV.
$kT = 8.617 \times 10^{-5} \times 5800 = 0.50\ \text{eV}$
Boltzmann's constant in eV/K.
Find the energy of a $250$ nm ultraviolet photon.
$E = \dfrac{1240}{250} = 4.96\ \text{eV}$
Ten times $kT$.
Estimate how rarely that mode is excited.
$e^{-4.96/0.50} = e^{-9.9} \approx 5 \times 10^{-5}$
Boltzmann factor.
Find the energy of a $10\ \mu$m infrared photon.
$E = 0.124\ \text{eV}$
A quarter of $kT$.
Compare with classical equipartition there.
$\text{each mode holds about } kT$
Quanta are small, so classical physics works.
Explain the turnover.
$hf \gg kT \Rightarrow \text{modes frozen out}$
Planck's quanta cut off the ultraviolet.
State the lesson.
$\text{classical where } hf \ll kT; \text{ quantum where } hf \gtrsim kT$
The correspondence principle.
Write Wien's law.
$T = \dfrac{2.898 \times 10^{-3}}{\lambda_{\max}}$
Solved for $T$.
Substitute the peak.
$T = \dfrac{2.898 \times 10^{-3}}{966 \times 10^{-9}}$
In meters.
Evaluate the temperature.
A star's glow peaks at $400$ nm. A second star's surface is twice as hot. At what wavelength does its glow peak?
Complete the worked solution: a blackbody is held at $1000$ K. With $\sigma = 5.67 \times 10^{-8}$ W/m²K⁴ and Wien's constant $2.898 \times 10^6$ nm·K, find the power it radiates per square meter in kW/m², the wavelength of its peak in nm, and the factor by which its output grows if its temperature is doubled.
Apply the Stefan–Boltzmann law.
$\dfrac{P}{A} = \sigma T^4 =$ f
In kW/m².
Apply Wien's law.
$\lambda_{\max} = \dfrac{2.898 \times 10^6}{T} =$ l
In nm.
Raise the doubling to the fourth power.
$2^4 =$ r
Output goes as $T^4$.
Place the peak.
$\text{infrared below about } 3900\ \text{K}$
Why most of a stove's glow is invisible.
Match each law or prediction to its statement.
| $\lambda_{\max}T = 2.898 \times 10^{-3}$ m·K | $P/A = \sigma T^4$ | $(h/m_ec)(1 - \cos\theta)$ | infinite ultraviolet output | |
|---|---|---|---|---|
| Wien's law | ||||
| Stefan–Boltzmann law | ||||
| Compton shift | ||||
| classical cavity prediction |
X-rays of wavelength $71.0$ pm scatter off loosely bound electrons at $60°$, where $\cos\theta = 0.5$. With $h/m_ec = 2.426$ pm and $hc = 1240$ keV·pm, fill in the Compton shift in pm, the scattered wavelength in pm, and the scattered photon's energy in keV.
| value | |
|---|---|
| Compton shift (pm) | |
| scattered wavelength (pm) | |
| scattered energy (keV) |
X-rays of wavelength $79$ pm scatter off electrons. With $h/m_ec = 2.426$ pm, write the scattered wavelength in pm as a formula in $c = \cos\theta$.
Answer:
A star's spectrum peaks at $290$ nm. With Wien's constant $2.898 \times 10^{-3}$ m·K, what is its surface temperature, in kelvin?
Answer: K at the surface
An engineer at a thermal-camera maker in Oregon checks where the glow of a block of ice, at $273$ K, peaks. With Wien's constant $2898$ μm·K, at what wavelength does its thermal radiation peak, in μm?
Answer: μm peak
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
X-rays of wavelength $85$ pm scatter off electrons. With $h/m_ec = 2.426$ pm, write the scattered wavelength in pm as a formula in $c = \cos\theta$.
Answer:
You can explain the failures of classical physics. Explain to someone why X-rays come back from graphite with longer wavelengths.
21. Your turn: a red dwarf's spectrum peaks at $966$ nm. Find its temperature., step 3
$T = 3000\ \text{K}$
A cool, red star.