F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsBlackbody Radiation and Planck's Quantum Hypothesis
F1.1 F2.1 F3.1About 70 minutes
Learning goals- I can describe blackbody radiation and how its spectrum changes with temperature.
- I can use Wien's law to relate the peak wavelength of a glowing object to its temperature.
- I can explain why classical physics failed to predict the blackbody spectrum (the ultraviolet catastrophe).
- I can state Planck's quantum hypothesis, E = nhf, and explain why it was a revolution in scientific thought.
Words to know
thermal radiation
Electromagnetic radiation given off by every object because of its temperature.
blackbody
An ideal object that absorbs all radiation that falls on it and emits a spectrum that depends only on its temperature.
blackbody spectrum
The graph of emitted intensity against wavelength for a blackbody at one temperature.
Wien's law
λmax T = 2.90 × 10⁻³ m·K: the peak wavelength of a blackbody is inversely proportional to its kelvin temperature.
ultraviolet catastrophe
The failed classical prediction that a blackbody should emit unlimited energy at short (ultraviolet) wavelengths.
quantum (plural quanta)
The smallest amount of energy that can be exchanged at a given frequency: E = hf.
Planck's constant (h)
The constant that sets the size of quanta: h = 6.63 × 10⁻³⁴ J·s.
quantized
Able to take only certain separate (discrete) values, not any value in between.
Why does a stove element turn red?
Turn on an electric stove element and watch. At first you see nothing, but you can feel warmth from a distance. Then it glows dull red, then bright orange. If it could get hotter still, it would look yellow-white, like the filament of an old light bulb or the Sun.
Every object gives off electromagnetic radiation because of its temperature. You do too: at about 37 °C you glow brightly in the infrared, which is how a thermal camera "sees" people in the dark on a February night. This glow is called thermal radiation.
Around 1900, physicists could measure this glow very precisely, but the best theories of the day (Newton's mechanics, Maxwell's electromagnetism and the physics of heat) could not explain it. Fixing that problem started the first of the two great revolutions of modern physics: quantum mechanics. (The other, special relativity, has its own unit in this course: Special Relativity.)
Remember from earlier units
In the unit The Wave Nature of Light you saw that light is an electromagnetic wave with c = fλ, where c = 3.00 × 10⁸ m/s in a vacuum. Visible light runs from about 400 nm (violet) to 700 nm (red). Infrared has longer wavelengths; ultraviolet, X-rays and gamma rays have shorter ones. In Grade 11 (SPH3U, Thermal and Nuclear Energy) you used the kelvin scale: T(K) = T(°C) + 273.
The ideal glowing object: a blackbody
The colour of a real object depends partly on what it is made of. Physicists simplify by imagining a blackbody: an object that absorbs every wavelength that hits it and reflects none. A good model is a small hole in a hollow box with blackened walls. Light that enters bounces around inside and is absorbed before it can escape.
When a blackbody is heated, the radiation coming out of the hole depends only on the temperature, not on the material. The Sun, a glowing stove element and the faint microwave glow left over from the early universe all behave close to blackbodies.
The graph below shows measured blackbody spectra. Notice three things:
- Each curve has a peak: one wavelength where the object emits most strongly.
- As the temperature rises, the peak moves to shorter wavelengths (red → orange → yellow-white → blue-white).
- A hotter object emits much more at every wavelength. The total power grows very quickly with temperature.
Wien's displacement law
The peak wavelength and the kelvin temperature are inversely proportional:
λmax T = 2.90 × 10⁻³ m·K
Double the temperature and the peak wavelength is cut in half. Astronomers use this to find the surface temperatures of stars from their colours: reddish Betelgeuse is cooler than our yellow-white Sun, and blue-white Rigel is hotter.
The total power radiated also climbs steeply. For a blackbody, the power is proportional to the fourth power of the kelvin temperature (the Stefan-Boltzmann law, P = σAT⁴, with σ = 5.67 × 10⁻⁸ W/(m²·K⁴)). Doubling T multiplies the power by 2⁴ = 16.
Worked exampleExample 1: The temperature of the Sun
The Sun's spectrum peaks at about 500 nm. Estimate the temperature of its surface.
- Write Wien's law: λmax T = 2.90 × 10⁻³ m·K.
- Convert the wavelength to metres: 500 nm = 500 × 10⁻⁹ m = 5.00 × 10⁻⁷ m.
- Solve for T: T = (2.90 × 10⁻³ m·K) ÷ (5.00 × 10⁻⁷ m) = 5.80 × 10³ K.
About 5800 K (roughly 5500 °C). This matches other measurements of the Sun's surface temperature.
Worked exampleExample 2: Why you glow in the infrared
Skin is at about 34 °C. At what wavelength does a person's thermal radiation peak? In what part of the spectrum is that?
- Convert to kelvins: T = 34 + 273 = 307 K.
- λmax = (2.90 × 10⁻³ m·K) ÷ 307 K = 9.45 × 10⁻⁶ m.
- That is about 9.4 μm (9400 nm), far longer than red light at 700 nm.
About 9.4 μm, in the infrared. Thermal cameras are built to detect wavelengths near 8 to 14 μm for exactly this reason.
The ultraviolet catastrophe
Classical physicists modelled the walls of the blackbody as tiny charged oscillators (vibrating charges) that send out electromagnetic waves. Using the ideas of heat that worked so well for gases, they assumed each possible wave pattern inside the box gets the same average share of energy, and that an oscillator can have any amount of energy.
The result, called the Rayleigh-Jeans law, matched the measurements at long wavelengths. But at short wavelengths it predicted intensity growing without limit. A warm oven would blast out infinite ultraviolet energy. This absurd prediction was later nicknamed the ultraviolet catastrophe. Real blackbody curves rise, peak and then fall back toward zero at short wavelengths.
Planck's desperate idea
In 1900 the German physicist Max Planck found a formula that fit the measured curves perfectly. To explain it, he had to make an assumption he did not like: an oscillator of frequency f can only have energies that are whole-number multiples of a smallest amount, which he called a quantum:
E = nhf, where n = 0, 1, 2, 3, ...
The constant h = 6.63 × 10⁻³⁴ J·s is now called Planck's constant.
Why does this fix the catastrophe? At high frequency, one quantum hf is a large "lump" of energy. The walls at a given temperature rarely have enough thermal energy to hand an oscillator such a big lump, so high-frequency waves are hardly excited at all. The curve falls back to zero at short wavelengths, just as measured.
Planck thought of his quanta as a mathematical trick about the oscillators in the walls. He later described it as "an act of desperation." It took Einstein, five years later, to suggest that light itself comes in quanta (next lesson).
Worked exampleExample 3: How big is a quantum?
(a) Find the size of one energy quantum for an oscillator vibrating at 5.0 × 10¹⁴ Hz (the frequency of orange light). (b) Find the quantum for a playground swing with frequency 0.50 Hz. (c) Why do we never notice quantized energy in a swing?
- (a) E = hf = (6.63 × 10⁻³⁴ J·s)(5.0 × 10¹⁴ Hz) = 3.3 × 10⁻¹⁹ J.
- (b) E = hf = (6.63 × 10⁻³⁴ J·s)(0.50 Hz) = 3.3 × 10⁻³⁴ J.
- (c) A swinging child has energy of tens of joules. Steps of 3.3 × 10⁻³⁴ J are about 10³⁵ times smaller than that, far too small to detect, so the energy looks continuous.
(a) 3.3 × 10⁻¹⁹ J (b) 3.3 × 10⁻³⁴ J (c) The steps are unimaginably small compared with everyday energies, so quantization only shows up for atoms, electrons and light.
Why this was a revolution (F1.1)
For 200 years physics had assumed nature is continuous: energy, like water from a tap, could be turned up or down by any amount. Planck's hypothesis said that at the smallest scale energy comes in packets. Nobody, including Planck, wanted to believe it. But it was the only idea that matched the data. This is a model case of how science changes: a careful measurement that the best theory cannot explain forces scientists to accept a new conceptual model, even a strange one.
Just for funA blackbody is the ultimate goalie: every bit of radiation that comes at it gets stopped. It never lets a single wavelength bounce back.
Hands-on activityWatch a stove element change colour
You need: an electric stove with coil elements (or a toaster) and a grown-up. Safety: elements get hot enough to cause serious burns. A grown-up should run the stove; never touch the element, and keep paper, towels and sleeves away.
- In a dim kitchen, have the grown-up turn one element to low. Hold the back of your hand about 30 cm above it (not closer). After a minute you feel warmth but see little or no glow. That is infrared radiation.
- Turn it to high. Note the colour every 30 seconds: no glow → dull red → bright red-orange.
- Optional: look at the element through a phone camera. Many phone cameras show a faint glow slightly before your eyes do, because the sensor responds a little way into the near infrared.
- Turn the stove off and watch the colour fade in reverse.
What to look for: as the element gets hotter, it gets brighter and its colour shifts from red toward orange. A coil element on high reaches very roughly 700 °C to 900 °C (about 1000 K to 1200 K). By Wien's law its peak is near 2.4 μm to 2.9 μm, still in the infrared: you only see the short-wavelength tail of its spectrum, which is why it looks red rather than white.
Your turn
1A star's spectrum peaks at 290 nm. What is its surface temperature?
T = (2.90 × 10⁻³ m·K) ÷ (2.90 × 10⁻⁷ m) = 1.00 × 10⁴ K. It would look blue-white.
2Betelgeuse has a surface temperature of about 3500 K. Find its peak wavelength and say which part of the spectrum it is in.
λmax = (2.90 × 10⁻³) ÷ 3500 = 8.29 × 10⁻⁷ m = 829 nm, in the near infrared. Its visible light is weighted to the red end, so it looks reddish.
3Find the energy of one quantum for an oscillator at 6.0 × 10¹⁴ Hz.
E = hf = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴) = 4.0 × 10⁻¹⁹ J.
4An oscillator with f = 2.0 × 10¹³ Hz drops from n = 5 to n = 3. How much energy does it lose?
ΔE = (5 − 3)hf = 2(6.63 × 10⁻³⁴)(2.0 × 10¹³) = 2.7 × 10⁻²⁰ J.
5A blackbody's kelvin temperature is tripled. By what factor does (a) its peak wavelength change, (b) its total radiated power change?
(a) λmax ∝ 1/T, so it becomes 1/3 as long. (b) P ∝ T⁴, so it becomes 3⁴ = 81 times larger.
6In your own words, what was the ultraviolet catastrophe?
Classical physics predicted that a hot object should emit more and more intensity at shorter and shorter wavelengths, adding up to infinite energy. Real objects emit very little at short wavelengths: the curve peaks and falls. The theory was clearly wrong.
7Why did Planck's hypothesis make the intensity fall at short wavelengths?
At short wavelength (high f) one quantum hf is large. The thermal energy available at a given temperature is rarely enough to supply such a big packet, so those high-frequency oscillations are rarely excited and little radiation comes out.
More practice: warm-up
1What is the value of Planck's constant?
2Which is hotter: a star that looks red or one that looks blue-white?
The blue-white star. Its peak is at a shorter wavelength, so its temperature is higher.
3Convert 727 °C to kelvins.
4What does "quantized" mean?
Able to have only certain separate values, not any value in between (like the steps of a staircase rather than a ramp).
5What does an ideal blackbody do with radiation that falls on it?
It absorbs all of it and reflects none.
More practice: on your own
1The faint microwave glow from the early universe (the cosmic microwave background) is a nearly perfect blackbody at 2.725 K. Find its peak wavelength.
λmax = (2.90 × 10⁻³) ÷ 2.725 = 1.06 × 10⁻³ m (about 1.1 mm), in the microwave region. That is why it was first detected with radio antennas.
2A pizza oven at 500 °C. Find its peak wavelength. Would you expect it to glow visibly in a dark room?
T = 773 K; λmax = 2.90 × 10⁻³ ÷ 773 = 3.75 × 10⁻⁶ m (3.75 μm), infrared. Only a tiny fraction of its output reaches 700 nm, so at most a very faint dull-red glow could be seen in a completely dark room; mostly it is not visibly glowing.
3An oscillator has quanta of 2.65 × 10⁻¹⁹ J. What is its frequency, and what is the wavelength of light with that frequency?
f = E/h = 2.65 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 4.00 × 10¹⁴ Hz. λ = c/f = 3.00 × 10⁸ ÷ 4.00 × 10¹⁴ = 7.50 × 10⁻⁷ m (750 nm, just beyond red).
4Two stars are the same size. Star A is 4000 K and star B is 8000 K. Compare their peak wavelengths and their total power.
B's peak wavelength is half of A's (λ ∝ 1/T). B radiates 2⁴ = 16 times the power of A.
5A 0.20 kg mass on a spring oscillates at 2.0 Hz with energy 0.10 J. If its energy is quantized, what is n (the number of quanta)? Comment.
One quantum = hf = 1.33 × 10⁻³³ J. n = 0.10 ÷ 1.33 × 10⁻³³ = 7.5 × 10³¹. The steps are so tiny compared with the total that the energy appears continuous; quantum effects are invisible at everyday scales.
Challenge
1The Sun (T ≈ 5800 K) and a red dwarf star (T ≈ 2900 K) have radii in the ratio 5 : 1. Compare the total power they emit.
P ∝ AT⁴ ∝ r²T⁴. Ratio = (5)² × (5800/2900)⁴ = 25 × 2⁴ = 25 × 16 = 400. The Sun emits about 400 times as much power.
2A science-fiction writer says that "a thermal camera works by detecting visible light from people." Use Wien's law to correct this.
At about 307 K, λmax ≈ 2.90 × 10⁻³ ÷ 307 ≈ 9.4 μm, which is infrared, more than ten times longer than visible light. A person's visible glow is far too faint to see. Thermal cameras detect infrared near 8 to 14 μm. (Ordinary cameras see people only by reflected visible light.)
3Explain how Planck's hypothesis is an example of a "new conceptual model" changing scientific thought. Include what evidence forced the change and why scientists resisted it.
Evidence: precise measurements of blackbody spectra that peaked and fell at short wavelengths, which the classical model (continuous energy, equal sharing among all waves) could not reproduce: it predicted the ultraviolet catastrophe. New model: energy of an oscillator is quantized, E = nhf. Resistance: it contradicted the long-held idea that energy is continuous, and even Planck saw it as a desperate trick. It became accepted because it matched the data exactly and later explained other effects (the photoelectric effect, atomic spectra), showing how a new model can reorganize a whole field.
Quiz
1. According to Wien's law, if the kelvin temperature of a blackbody doubles, its peak wavelength
2. Which statement describes Planck's quantum hypothesis?
3. What was the "ultraviolet catastrophe"?
4. A star's spectrum peaks at 725 nm. Its surface temperature is about
5. Why don't we notice energy quantization in a swinging pendulum?
6. Which object is closest to an ideal blackbody?
Big idea (F1.1, F3.1): hot objects glow with a spectrum that depends on temperature. Classical physics could not explain the shape of that spectrum; Planck's 1900 idea that energy comes in packets, E = nhf, could. This is where quantum mechanics begins.
How to help: ask your teen to explain, without notes, why a stove element goes from invisible to red to orange as it heats up, and what Planck had to assume. The activity needs adult supervision at the stove.
Common difficulties:
- Forgetting to convert °C to kelvins before using Wien's law, or nanometres to metres.
- Thinking the peak wavelength is the only wavelength emitted. Every blackbody emits a whole range; the peak is just the strongest.
- Thinking quantization means energy is small. It means energy comes in separate steps of size hf; the steps are only noticeable for atoms and light.
- Mixing up the direction of Wien's law: hotter means shorter peak wavelength.