RuukkilabraUpper secondary physics

When classical physics ran out of answers

Why could the best physics of 1900 not explain the glow of a hot piece of metal?

Prior knowledge

  • Light is an electromagnetic wave, and its speed, frequency and wavelength are linked by $c = f\lambda$ (earlier physics courses)
  • The absolute temperature in kelvins is $T = (t + 273.15)\ \text{K}$, where $t$ is the temperature in degrees Celsius (earlier physics courses)

Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 1.

Think about it

On 27 April 1900 the famous physicist Lord Kelvin gave a lecture at the Royal Institution in London. Physics seemed almost complete: Newton's mechanics described the motion of planets and machines, Maxwell's theory described electricity, magnetism and light, and thermodynamics described heat. Kelvin praised this beautiful theory, but he warned of two "clouds" over it, two problems it could not solve. One cloud was about the rule for how energy is shared among the parts of a body. The same rule, applied to the light inside a hot oven, ran into trouble with something very ordinary: the glow of hot objects. Within thirty years both clouds had turned into new physics.

How can something as ordinary as a glowing piece of hot metal break the physics of 1900?

Talk in pairs (2 minutes)

Name three things around you that glow because they are hot. What colour is each glow? Does a hot thing always glow? Write down your ideas: you will come back to them.

Sentence starters

  • Something that glows because it is hot is …
  • Its glow is … , and I think it is about … degrees.
  • I think a hot thing does not always glow, because …

Light is an electromagnetic wave

In 1864 James Clerk Maxwell published a theory that joined electricity, magnetism and light. Light, he said, is an electromagnetic wave: changing electric and magnetic fields that travel through space at the speed $c = 3.00 \cdot 10^{8}\ \text{m/s}$. Between 1886 and 1889 Heinrich Hertz made such waves with electric sparks and detected them: radio waves. For every wave, the speed, the frequency $f$ and the wavelength $\lambda$ are linked by $c = f\lambda$.

Visible light is only a narrow band of the electromagnetic spectrum. Waves shorter than violet light are ultraviolet, X-rays and gamma rays; waves longer than red light are infrared, microwaves and radio waves.

$\gamma$
$\text{X-rays}$
$\text{UV}$
$\text{IR}$
$\text{micro-}$
$\text{waves}$
$\text{radio}$
$10^{-12}$
$10^{-10}$
$10^{-8}$
$10^{-6}$
$10^{-4}$
$10^{-2}$
$10^{0}$
$10^{2}$
$\lambda\ (\text{m})$
$\text{shorter } \lambda\text{, higher } f$
$380\ \text{nm}$
$750\ \text{nm}$
$\text{visible light}$

Review: the electromagnetic spectrum

Predict

Make your predictions before investigating 2 questions

Prediction 1: a hotter glow

A black body (an ideal glowing object that absorbs all light falling on it) is heated from $4000\ \text{K}$ to $6000\ \text{K}$.What happens to the wavelength $\lambda_\text{max}$ at which it radiates most strongly?

Prediction 2: sharing the energy

Think of the radiation in a hot oven as standing waves, like the vibrations of a guitar string. There is no limit to how short a wave can be. Classical physics gives every wave pattern the same average energy. What does this idea predict for the radiation at very short wavelengths?

Investigate

Reading the curves

Every object sends out thermal radiation: electromagnetic radiation produced by the heat motion of its particles. A cold object sends out mostly invisible infrared. A hot one also glows. To compare objects fairly, physicists study an ideal object called a black body. It absorbs all radiation that falls on it, so all the radiation that leaves it is its own thermal radiation. A small hole in the wall of a hot oven is a very good black body, and the Sun and a lamp filament are fairly close to one.

Around 1900, careful measurements in Berlin showed how the power of this radiation is spread over the wavelengths. The figure shows the result as model values from the law that fits those measurements. Study it carefully: how do the curves change when the temperature rises?

$500$
$1000$
$1500$
$2000$
$0.25$
$0.5$
$0.75$
$1$
$0$
$\lambda\ (\text{nm})$
$\text{relative power}$
$\text{visible}$
$6000\ \text{K}$
$5000\ \text{K}$
$4000\ \text{K}$
$\text{model values}$

Investigate: the peak and the temperature

Values calculated by a model

The table gives the peak wavelength of a black body at five temperatures, computed from Planck's law (the law that fits the measurements). Look for a pattern: what happens to $\lambda_\text{max}$ when $T$ doubles? Try multiplying $\lambda_\text{max}$ by $T$ in each row.

Temperature $T$ (K) Peak wavelength $\lambda_\text{max}$ (nm)
2000 1449
3000 966
4000 724
5000 580
6000 483

A hot blue star has a surface temperature of $10\,000\ \text{K}$. Use the pattern in the table to predict its peak wavelength in nanometres.

nm
Hint 1: a guiding question

Compare the rows $3000\ \text{K}$ and $6000\ \text{K}$: what happens to $\lambda_\text{max}$ when $T$ doubles?

Hint 2: what is the connection

The product $\lambda_\text{max} \cdot T$ is the same in every row. Calculate it for one row.

Hint 3: an intermediate step
$483\ \text{nm} \cdot 6000\ \text{K}$
$\approx 2.90 \cdot 10^{6}\ \text{nm} \cdot \text{K}$

Divide this by $10\,000\ \text{K}$.

Show the solution

Given

$T = 10\,000\ \text{K}$ and the table. Find $\lambda_\text{max}$.

Principle

In every row, $\lambda_\text{max} \cdot T$ has the same value, about $2.90 \cdot 10^{6}\ \text{nm} \cdot \text{K}$: the peak wavelength is inversely proportional to the temperature (Wien's law).

Formula

$\lambda_\text{max} T = b$, so $\lambda_\text{max} = b/T$.

Unit conversions

None: the table is in nanometres and kelvins, so $b$ is in $\text{nm} \cdot \text{K}$.

Substitution

$b$$= 483\ \text{nm} \cdot 6000\ \text{K}$
$= 2.898 \cdot 10^{6}\ \text{nm} \cdot \text{K}$

and

$\lambda_\text{max}$$= 2.898 \cdot 10^{6}\ \text{nm} \cdot \text{K} : 10\,000\ \text{K}$
$= 289.8\ \text{nm}$

Answer

$\lambda_\text{max} \approx 290\ \text{nm}$

Given to three significant figures like the table. This is ultraviolet, just shorter than visible violet light ($380\ \text{nm}$).

Reasonableness

The star is $10\,000 : 6000 \approx 1.7$ times as hot as the $6000\ \text{K}$ body, so its peak should be $1.7$ times shorter: $483\ \text{nm} : 1.7 \approx 290\ \text{nm}$. Such a star looks bluish-white.

Describe the graph

Discuss with a partner, then write three sentences that describe the figure Reading the curves. Say what happens to (1) the height of the curves and (2) the position of the peak when the temperature rises, with numbers. (3) Which of the three temperatures glows in the colour closest to white, and why?

Word bank

the peak · the wavelength · the curve rises / falls · higher / lower · to shift (shifts, shifted) · in the infrared · in the visible band · at every wavelength

Sentence starters

  • When the temperature rises from … to …, the curve …
  • The peak shifts from … nm to … nm.
  • The … K curve has its peak in the middle of the visible band, so …

Theory summary

Key idea

A black body sends out radiation at all wavelengths, but not equally. Its spectrum has one peak. When the temperature rises, the body radiates more power at every wavelength, and the peak moves to shorter wavelengths. This is why a heated iron bar first glows dull red, then orange, and finally white: at first only the long-wave edge of its spectrum reaches the visible band, and later the whole band is covered.

The Sun's surface, at about $5800\ \text{K}$, has its peak near $500\ \text{nm}$, in the middle of the visible band. Your skin, at about $300\ \text{K}$, has its peak near $10\ \mu\text{m}$, in the infrared: thermal cameras see it.

The ultraviolet catastrophe

Classical physics tried to explain the curve. In an oven, the radiation forms standing waves between the walls, like the vibrations of a guitar string. There is no limit to how short a standing wave can be, and the shorter the wavelength, the more different wave patterns fit into the oven. Classical physics says that at a temperature $T$ every wave pattern gets the same average share of the energy.

Lord Rayleigh worked out what this means in 1900, and Rayleigh and James Jeans completed the calculation in 1905. At long wavelengths their result agrees with the measurements. At short wavelengths it does not: the predicted power grows without limit, and the total energy of the radiation would be infinite. Every warm oven would flood the room with ultraviolet and X-rays. In 1911 Paul Ehrenfest named this failure the ultraviolet catastrophe.

$1000$
$2000$
$3000$
$0.5$
$1$
$1.5$
$2$
$0$
$\lambda\ (\text{nm})$
$\text{relative power}$
$\text{classical prediction}$
$\text{measured (Planck's law)}$

The measured curve falls to zero at short wavelengths. Classical physics was not just a little inaccurate here; its basic idea gave an impossible answer.

Wien's law and Planck's energy packets

The peak wavelength $\lambda_\text{max}$ is inversely proportional to the absolute temperature $T$ (Wilhelm Wien, 1893):

$$\begin{aligned} \lambda_\text{max} T &= b \\ b &= 2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K} \end{aligned}$$

so $\lambda_\text{max} = b/T$. The temperature must be in kelvins.

On 19 October 1900 Max Planck presented a formula that fitted the Berlin measurements at every wavelength. To derive it, on 14 December 1900, he assumed that the walls of the oven can give and take radiation energy only in packets, or quanta, of size

$$E = hf$$

where $h = 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}$ is Planck's constant. A short wave has a high frequency, so its packet is large. At a temperature $T$ the typical energy available from the heat motion is small, of the order of $k_\text{B}T$, where $k_\text{B} = 1.381 \cdot 10^{-23}\ \text{J/K}$ is the Boltzmann constant. Very few short-wave patterns ever receive a whole packet, so they carry almost no energy. The catastrophe disappears.

Planck called his assumption "an act of despair" and saw it as a mathematical step. Five years later Einstein took the packets seriously as particles of light.

Back to your predictions

Compare your two predictions with the curves and with the classical prediction. Look at the glowing things you named at the start: which was the hottest? Then answer the opening question in one or two sentences: why could the best physics of 1900 not explain the glow of a hot piece of metal?

    Would you change your prediction?

    Apply

    Worked example

    Problem

    A filament lamp has a tungsten filament at a temperature of about $2700\ \text{K}$. At which wavelength does the filament radiate most strongly? How large is one energy packet $hf$ at that wavelength, in electronvolts? Is the peak in the visible band?

    Step 1/7

    Given: ${T = 2700\ \text{K}}$, ${b = 2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}}$, ${h = 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$, ${c = 2.998 \cdot 10^{8}\ \text{m/s}}$. Find the peak wavelength $\lambda_\text{max}$ and the packet energy $E$.

    Step 2/7

    Principle: The filament is close to a black body, so Wien's law gives the peak. The energy of one packet is $E = hf$, and the frequency follows from $c = f\lambda$.

    Step 3/7

    Formula: Solved in symbols:

    $$\begin{aligned} \lambda_\text{max} &= \frac{b}{T} \\ E &= hf = \frac{hc}{\lambda_\text{max}} \end{aligned}$$

    Step 4/7

    Unit conversions: The temperature is already in kelvins. The energy in joules is changed to electronvolts with ${1\ \text{eV} = 1.602 \cdot 10^{-19}\ \text{J}}$.

    Step 5/7

    Substitution: $$\begin{aligned} \lambda_\text{max} &= \frac{2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}}{2700\ \text{K}} \\ &= 1.073 \cdot 10^{-6}\ \text{m} \\[6pt] hc &= 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s} \\ &\quad \cdot 2.998 \cdot 10^{8}\ \text{m/s} \\ &= 1.9865 \cdot 10^{-25}\ \text{J} \cdot \text{m} \\[6pt] E &= \frac{hc}{\lambda_\text{max}} \\ &= \frac{1.9865 \cdot 10^{-25}\ \text{J} \cdot \text{m}}{1.073 \cdot 10^{-6}\ \text{m}} \\ &= 1.851 \cdot 10^{-19}\ \text{J} \\ &= \frac{1.851 \cdot 10^{-19}\ \text{J}}{1.602 \cdot 10^{-19}\ \text{J/eV}} \\ &= 1.155\ \text{eV} \end{aligned}$$

    Step 6/7

    Answer: The filament radiates most strongly at ${\lambda_\text{max} \approx 1.07\ \mu\text{m}}$, that is, ${1070\ \text{nm}}$, and one packet there has ${E \approx 1.16\ \text{eV}}$. Three significant figures, because the temperature is a rounded value. The peak is in the infrared, not in the visible band ($380\text{–}750\ \text{nm}$).

    Step 7/7

    Reasonableness: The filament is about half as hot as the Sun's surface, so its peak should be at about twice the Sun's $500\ \text{nm}$: $1070\ \text{nm}$ fits. A visible photon has $1.6\text{–}3.3\ \text{eV}$, and $1.16\ \text{eV}$ is less, as it should be for infrared. This is why an old filament lamp turns most of its electrical energy into infrared radiation (and heat), not visible light.

    Apply: a thermal camera

    A thermal camera shows that the radiation from a person's forehead has its peak at $\lambda_\text{max} = 9.4\ \mu\text{m}$. Treat the skin as a black body and calculate its temperature in kelvins.

    K
    Hint 1: a guiding question

    Which law links the peak wavelength and the temperature?

    Hint 2: what is the connection

    $T = b/\lambda_\text{max}$ with $b = 2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}$.

    Hint 3: an intermediate step

    Write the wavelength in metres: $9.4\ \mu\text{m} = 9.4 \cdot 10^{-6}\ \text{m}$.

    Show the solution

    Given

    $\lambda_\text{max} = 9.4\ \mu\text{m}$, $b = 2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}$. Find the temperature $T$.

    Principle

    The skin behaves almost like a black body, so Wien's law links the peak and the temperature.

    Formula

    $\lambda_\text{max} T = b$, so $T = b/\lambda_\text{max}$.

    Unit conversions

    $\lambda_\text{max}$$= 9.4\ \mu\text{m}$
    $= 9.4 \cdot 10^{-6}\ \text{m}$

    Substitution

    $$T = \frac{2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}}{9.4 \cdot 10^{-6}\ \text{m}} = 308.3\ \text{K}$$

    Answer

    $T \approx 308\ \text{K}$

    Which is about $35\ ^\circ\text{C}$. The wavelength has two significant figures, so the answer is good to a few kelvins.

    Reasonableness

    The skin of the forehead is a little cooler than the inside of the body ($37\ ^\circ\text{C} = 310\ \text{K}$), so $308\ \text{K}$ is very reasonable.

    Optional Find the error
    A student calculates the peak wavelength of a pizza oven at $400\ ^\circ\text{C}$:

    $$\begin{aligned} \lambda_\text{max} &= \frac{2.898 \cdot 10^{-3}\ \text{m} \cdot \text{K}}{400} \\ &= 7.2\ \mu\text{m} \end{aligned}$$

    What is wrong?

    Common misconception

    "Red hot is the hottest." In everyday life red means danger, but in thermal radiation red is the colour of the coolest glow. A hotter body has its peak at a shorter wavelength, so its glow moves from red through orange and yellow to white, and very hot stars look bluish.

    A second mistake is to put degrees Celsius into Wien's law. The law needs the absolute temperature: $T = (t + 273.15)\ \text{K}$. A third is to think that Planck discovered photons. Planck quantised the exchange of energy in the oven walls and did not believe that light itself is made of particles; that was Einstein's step in 1905.

    Investigate: a timeline of modern physics

    Planck's packets were only the start. In the next thirty years physicists discovered the nucleus, photons, the quantum atom and the wave nature of matter. This chapter follows that story, and the rest of the course shows why it matters for fusion.

    Explore the timeline with a partner: from Röntgen's X-rays in 1895 to Bethe's theory of the Sun in 1938. Filter by theme or search for a name, and open Who was …? on a card to read a short biography.

    Six steps in order

    Speak and write: one life in physics

    Work in pairs. Each of you opens two biographies in the timeline (Who was …?), for example Max Planck, Marie Curie, Ernest Rutherford, Niels Bohr or George Gamow. Tell your partner about one person in about one minute, without reading aloud. Then write three or four sentences of your own about that person: where they worked, what they discovered, and why it matters for the story of fusion.

    Word bank

    to discover · to propose · to explain · to measure · an experiment · a theory · a model · a prize · to emigrate · important for

    Sentence starters

    • … was a … physicist who worked in …
    • In …, she / he discovered / proposed that …
    • This was important because …
    • For fusion, this matters because …

    Check your understanding

    1. Concept: red hot or white hot?

    A blacksmith heats an iron bar. First it glows "red hot", later "white hot". What does this tell you about its temperature?

    Suggestion: one answer does not show everything you can do, but review the theory summary.

    2. Calculation: two temperatures

    Body A has a temperature of $3000\ \text{K}$ and body B $6000\ \text{K}$. How do their peak wavelengths compare?

    Suggestion: one answer does not show everything you can do, but review the theory summary.

    3. Planck's idea

    How did Planck's assumption $E = hf$ remove the ultraviolet catastrophe?

    Suggestion: one answer does not show everything you can do, but review the theory summary.

    4. Explain: what went wrong?

    A friend says: "Classical physics was simply wrong about everything." Do you agree? Write four or five sentences. Say what classical physics explained well, what it predicted for the radiation of a hot oven, how this compared with the measurements, and what Planck changed.

    Sentence starters

    • I partly agree / disagree, because …
    • Classical physics explained … well.
    • For a hot oven, it predicted that …
    • The measurements showed instead that …
    • Planck assumed that …, so …

    Summary

    • Light is an electromagnetic wave, $c = f\lambda$; visible light is a narrow part of the electromagnetic spectrum.
    • Every object sends out thermal radiation. The spectrum of a black body has one peak, and Wien's law $\lambda_\text{max} T = b$ says that the peak moves to shorter wavelengths when the temperature rises.
    • Classical physics shared the energy equally between all wave patterns and predicted unlimited power at short wavelengths, the ultraviolet catastrophe.
    • Planck (1900) solved the problem by assuming that energy is exchanged in packets $E = hf$. This was the beginning of quantum physics.

    Le coin français

    In French, thermal radiation is le rayonnement thermique and a black body is un corps noir. Listen, repeat, and then say the last sentence to your partner. Can you guess its meaning before you read the English? In the hints, capitals show the stressed syllable, and õ, ã and ẽ are nasal vowels: the air goes through the nose and no n is heard.