Waves, probability and quantum tunnelling
Why can the Sun shine, when its protons are far too slow to reach each other?
Prior knowledge
- Light consists of photons, which are detected one by one, $E = hf$chapter Light as particles: the photoelectric effect
- Two positive nuclei repel each other, and a slower particle turns back further awaychapter From Rutherford's nucleus to Bohr's atom
- Kinetic energy $E_\text{k} = \tfrac{1}{2}mv^2$ and momentum $p = mv$ (earlier physics courses)
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
The centre of the Sun is about $15.7$ million kelvins. That sounds extremely hot. But fusion needs two positive nuclei to come almost into contact, and at this temperature the protons are far too slow to push so close against their electric repulsion. By the rules of classical physics, fusion in the Sun should practically never happen; physicists met this puzzle in the 1920s. Yet the Sun has been shining for about $4.6$ billion years.
How can protons do something that classical physics says they cannot?
Review: how close can a nucleus come?
Talk in pairs (2 minutes)
What does impossible mean in physics? Think of a ball rolling towards a hill that is too high for it. Is it impossible that it gets over, or only very unlikely? Tell your partner what you think.
Sentence starters
- I think it is impossible, because …
- I think it is only unlikely, because …
- It depends on …
Predict
Make your predictions before investigating 2 questions
Prediction 1: electrons one at a time
Prediction 2: a wall that is too high
Investigate
Electrons one at a time
Light passing through two narrow slits makes a pattern of bright and dark bands on a screen. Thomas Young showed such interference of light around 1801–1803, and it became the classic proof that light is a wave. In 1909 Geoffrey Taylor repeated a two-slit experiment with light so dim that only a few photons were on their way at any moment, and the bands still appeared.
In 1924 the French physicist Louis de Broglie suggested that electrons might be waves too. In 1927 Clinton Davisson and Lester Germer, and independently George Paget Thomson and Alexander Reid, saw electrons diffract from crystals, just like waves. Later, Merli, Missiroli and Pozzi (1974) sent electrons one at a time through an electron biprism, which acts like two slits. The figure shows what such an experiment looks like.
Each electron arrives at one single point, like a particle. But where it arrives is not decided like a bullet's path: after thousands of electrons, the dots form the bands of a wave pattern. At the Fotoni lab visit you can study a double-slit experiment repeated with single photons (unit Fotoni lab visit: modern physics experiments).
Particle or wave?
Discuss with a partner, then write three sentences about the figure with $30$, $300$ and $3000$ electrons. What in the result looks like a particle? What looks like a wave? Can you predict where the next single electron will land?
Word bank
a dot · a band · interference · a pattern builds up · to land · likely / unlikely · probability · at random
Sentence starters
- Each electron lands …, which looks like a particle.
- After thousands of electrons, …, which looks like a wave.
- We cannot predict …, but we can predict …
Model answer
Each electron lands as one small dot at one place, which looks like a particle. After thousands of electrons, the dots build up bright and dark bands, an interference pattern, which looks like a wave. We cannot predict where the next single electron will land, but we can predict where it is likely and unlikely to land: in the bright bands many electrons arrive, in the dark bands almost none.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary.
Through the wall
Roll a ball towards a hill. If its kinetic energy is less than the energy needed to reach the top, it always rolls back. In classical physics the region beyond the hill is simply forbidden. A small particle such as an electron behaves differently. Its wave does not stop at the edge of an energy barrier: inside the barrier the wave shrinks quickly, and if the barrier is thin, a small wave comes out on the other side. So the particle has a chance of appearing beyond the barrier. This is quantum tunnelling.
How large is the chance? The table gives approximate model values for an electron whose energy is $1\ \text{eV}$ below the top of a barrier, for different barrier widths.
| Barrier width (nm) | Chance of passing |
|---|---|
| 0.25 | 7.7 % |
| 0.50 | 0.60 % |
| 0.75 | 0.046 % |
| 1.00 | 0.0035 % |
Tunnelling is not just a theory. In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, explained alpha decay with it: an alpha particle escapes from a nucleus by tunnelling through a barrier that it could never climb. The scanning tunnelling microscope of Gerd Binnig and Heinrich Rohrer (1981, Nobel Prize 1986) measures the tunnelling current between a sharp tip and a surface and can show single atoms. The flash memory in your phone stores data by moving electrons through thin barriers by tunnelling.
Investigate: how fast does the chance fall?
Values calculated by a model
Use the table in Through the wall. Look at how the chance of passing changes each time the barrier becomes $0.25\ \text{nm}$ wider.
By what factor does the chance of passing become smaller when the barrier width grows from $0.25\ \text{nm}$ to $0.50\ \text{nm}$? (For example, type $2$ if the chance is halved.)
Hint 1: a guiding question
A factor is found by dividing, not by subtracting.
Hint 2: what is the connection
Divide the larger chance by the smaller one: $7.7\ \% : 0.60\ \%$.
Hint 3: an intermediate step
Check the next step too: $0.60\ \% : 0.046\ \%$. Is it about the same factor?
Show the solution
Given
The chance is $7.7\ \%$ at $0.25\ \text{nm}$ and $0.60\ \%$ at $0.50\ \text{nm}$ (model values). Find the factor.
Principle
A factor of change is the ratio of the two values.
Formula
$\text{factor} = 7.7\ \% : 0.60\ \%$.
Unit conversions
None: both values are percentages.
Substitution
$7.7 : 0.60 = 12.8$.
Answer
The chance becomes about $13$ times smaller (two significant figures, like the table). The next steps give the same factor: $0.60 : 0.046 \approx 13$ and $0.046 : 0.0035 \approx 13$.
Reasonableness
The same factor at every step is the sign of exponential decrease: every extra $0.25\ \text{nm}$ makes passing about $13$ times rarer. Doubling the width does not halve the chance; here it makes it $13$ times smaller.
Theory summary
Key idea
Light and matter are neither ordinary waves nor ordinary particles. This is called wave–particle duality. An electron, a proton or a photon is detected at one point, like a particle, but how it travels is described by a wave. In 1926 Max Born proposed what the wave means: the square of its amplitude at a place gives the probability of finding the particle there. Quantum physics cannot say where one electron will land, but it predicts the pattern of many very accurately.
Every moving particle has a de Broglie wavelength. When the wavelength is similar to the size of an object in its way, such as the distance between atoms or the width of a barrier, the wave nature becomes important: diffraction, interference and tunnelling. For a ball or a car the wavelength is so tiny that we never notice it.
The de Broglie wavelength
A particle with the momentum $p = mv$ has the wavelength
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$
where $h = 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}$. If the kinetic energy $E_\text{k} = \tfrac{1}{2}mv^2$ is known, the momentum is $p = \sqrt{2mE_\text{k}}$. The chance of tunnelling falls very fast, roughly exponentially, when the barrier becomes wider or higher, or when the particle is heavier.
Words of the lesson
Tunnelling and the Sun
At the centre of the Sun the temperature is about $15.7$ million kelvins. A typical kinetic energy of a particle there is of the order of $k_\text{B}T \approx 1.35\ \text{keV}$. Two protons repel each other, because both are positive. To fuse, they must come within a few femtometres, where the strong nuclear force acts. At $2\ \text{fm}$ the electric potential energy of two protons is about $720\ \text{keV}$, more than $500$ times the typical energy. In classical physics a proton with $1.35\ \text{keV}$ turns back at about $1000\ \text{fm}$ and never gets close.
Gamow's tunnelling (1928) solved the puzzle. A proton's de Broglie wavelength at these energies is hundreds of femtometres, so its wave reaches far into the barrier, and there is a tiny chance that two protons tunnel close enough to fuse. In 1929 Robert Atkinson and Fritz Houtermans used this idea to make the first calculation of fusion in stars. Each meeting has only a tiny chance, but the Sun's core contains an enormous number of protons, so fusion goes on all the time. But tunnelling only lets two protons get close. Two protons alone cannot stay together: fusion succeeds only if one of them turns into a neutron at that moment, $\text{p} + \text{p} \to \text{d} + \text{e}^+ + \nu$, and this change is a very slow process of the weak interaction. So an average proton in the Sun's core waits about $9$ billion years before it fuses, and this is why the Sun shines steadily for so long. The chapter The Energy of the Stars continues this story.
Back to your predictions
Compare your predictions with the evidence. Then answer the opening question in one or two sentences: why can the Sun shine, when its protons are far too slow to reach each other?
Would you change your prediction?
Optional Louis de Broglie
People in this story
Louis de Broglie (1892–1987)
Louis de Broglie was born in Dieppe, in Normandy, into the noble de Broglie family. He first studied history and took a degree in it in 1910, but then turned to physics. In his doctoral thesis of 1924, Recherches sur la théorie des quanta, he proposed that every particle, such as an electron, also behaves like a wave. Three years later experiments with crystals confirmed it. He received the Nobel Prize in Physics in 1929 "for his discovery of the wave nature of electrons". His idea is the basis of tunnelling.
Apply
Worked example
Problem
In an electron microscope, electrons are accelerated from rest through a voltage of $100\ \text{V}$. Calculate their de Broglie wavelength and compare it with the distance between atoms in a crystal, about $0.1\text{–}0.3\ \text{nm}$.
Step 1/7
Given: ${U = 100\ \text{V}}$, electron mass ${m = 9.109 \cdot 10^{-31}\ \text{kg}}$, charge ${e = 1.602 \cdot 10^{-19}\ \text{C}}$, ${h = 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$. Find the wavelength $\lambda$.
Step 2/7
Principle: The electric field does the work $eU$ on the electron, which becomes its kinetic energy (conservation of energy). The de Broglie wavelength follows from the momentum.
Step 3/7
Formula: Solved in symbols:
$$\begin{aligned} E_\text{k} &= eU \\ p &= \sqrt{2mE_\text{k}} = \sqrt{2meU} \\ \lambda &= \frac{h}{p} = \frac{h}{\sqrt{2meU}} \end{aligned}$$
Step 4/7
Unit conversions: None are needed: all quantities are in SI units. The answer in metres is changed to nanometres with $1\ \text{nm} = 10^{-9}\ \text{m}$.
Step 5/7
Substitution: $$\begin{aligned} E_\text{k} &= 1.602 \cdot 10^{-19}\ \text{C} \\ &\quad \cdot 100\ \text{V} \\ &= 1.602 \cdot 10^{-17}\ \text{J} \\[6pt] 2mE_\text{k} &= 2 \cdot 9.109 \cdot 10^{-31}\ \text{kg} \\ &\quad \cdot 1.602 \cdot 10^{-17}\ \text{J} \\ &= 2.919 \cdot 10^{-47}\ \text{kg}^2 \cdot \text{m}^2/\text{s}^2 \\[6pt] p &= \sqrt{2.919 \cdot 10^{-47}} \\ &\quad \cdot \sqrt{\text{kg}^2 \cdot \text{m}^2/\text{s}^2} \\ &= 5.402 \cdot 10^{-24}\ \text{kg} \cdot \text{m/s} \\[6pt] \lambda &= \frac{6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}{5.402 \cdot 10^{-24}\ \text{kg} \cdot \text{m/s}} \\ &= 1.227 \cdot 10^{-10}\ \text{m} \\ &= 0.123\ \text{nm} \end{aligned}$$
Step 6/7
Answer: The electrons have the de Broglie wavelength $\lambda \approx 0.123\ \text{nm}$, three significant figures like the constants. It is of the same size as the distance between atoms in a crystal, so the electrons diffract from crystals, as Davisson and Germer saw.
Step 7/7
Reasonableness: Check the units:
$$\text{C} \cdot \text{V} = \text{J} = \text{kg} \cdot \text{m}^2/\text{s}^2$$
So the square root gives $\text{kg} \cdot \text{m/s}$, a momentum, and $\text{J} \cdot \text{s}$ divided by $\text{kg} \cdot \text{m/s}$ gives $\text{m}$. The speed is $p/m \approx 5.9 \cdot 10^{6}\ \text{m/s}$, about $2\ \%$ of the speed of light, so classical momentum is good enough. The wavelength is much shorter than that of visible light ($400\text{–}750\ \text{nm}$), which is why electron microscopes can see much smaller details than light microscopes.
Apply: the wave of a proton in the Sun
A proton (mass $1.673 \cdot 10^{-27}\ \text{kg}$) at the centre of the Sun has the kinetic energy $1.35\ \text{keV}$. Calculate its de Broglie wavelength in femtometres ($1\ \text{fm} = 10^{-15}\ \text{m}$). Compare it with the classical turning distance, about $1000\ \text{fm}$.
Hint 1: a guiding question
Change the energy to joules: $1\ \text{keV} = 1000\ \text{eV}$ and $1\ \text{eV} = 1.602 \cdot 10^{-19}\ \text{J}$.
Hint 2: what is the connection
$p = \sqrt{2mE_\text{k}}$ and $\lambda = h/p$.
Hint 3: an intermediate step
$E_\text{k} = 2.163 \cdot 10^{-16}\ \text{J}$ and $p = 8.507 \cdot 10^{-22}\ \text{kg} \cdot \text{m/s}$.
Show the solution
Given
$m = 1.673 \cdot 10^{-27}\ \text{kg}$, $E_\text{k} = 1.35\ \text{keV}$, $h = 6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}$. Find the de Broglie wavelength $\lambda$.
Principle
Every particle has the de Broglie wavelength $\lambda = h/p$. The momentum follows from the kinetic energy, $E_\text{k} = p^2/(2m)$; at this speed classical momentum is accurate.
Formula
$p = \sqrt{2mE_\text{k}}$ and $\lambda = h/p = h/\sqrt{2mE_\text{k}}$.
Unit conversions
and at the end $1\ \text{fm} = 10^{-15}\ \text{m}$.
Substitution
$$\begin{aligned} 2mE_\text{k} &= 2 \cdot 1.673 \cdot 10^{-27}\ \text{kg} \\ &\quad \cdot 2.163 \cdot 10^{-16}\ \text{J} \\ &= 7.237 \cdot 10^{-43}\ \text{kg}^2 \cdot \text{m}^2/\text{s}^2 \\ p &= \sqrt{7.237 \cdot 10^{-43}\ \text{kg}^2 \cdot \text{m}^2/\text{s}^2} \\ &= 8.507 \cdot 10^{-22}\ \text{kg} \cdot \text{m/s} \\ \lambda &= \frac{6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}{8.507 \cdot 10^{-22}\ \text{kg} \cdot \text{m/s}} \\ &= 7.79 \cdot 10^{-13}\ \text{m} \end{aligned}$$
Answer
$\lambda \approx 7.79 \cdot 10^{-13}\ \text{m} = 779\ \text{fm}$
Three significant figures like the energy. It is almost as long as the classical turning distance of about $1000\ \text{fm}$.
Reasonableness
The proton is much heavier than the electron of the worked example, but also has more energy; its wavelength, $779\ \text{fm}$, is shorter than the electron's $0.123\ \text{nm} = 123\,000\ \text{fm}$, as expected. Compare: $779\ \text{fm}$ is about the same size as the classical turning distance of about $1000\ \text{fm}$, so the proton's wave reaches far into the barrier, and tunnelling becomes possible.
Optional Think further: a tennis ball
Common misconception
"In tunnelling the particle borrows energy to climb over the barrier" or "it finds a hole". Neither is true. The particle's wave reaches through the barrier, and there is a probability of finding the particle on the other side, with the same energy it had before. Most particles still bounce back.
A second mistake is to think that tunnelling happens only in theory, or to big objects too. It is measured every day in microscopes, electronics and radioactive decay, but its chance falls so fast with mass and width that a ball never tunnels through a wall. A third: wave–particle duality does not mean that an electron is sometimes a wave and sometimes a particle. It is always the same quantum object; the wave tells us the probabilities.
Check your understanding
1. Concept: what the dots tell us
2. Tunnelling: a wider barrier
3. The Sun
4. Explain: why the Sun shines
Explain to a younger student, in four or five sentences, why fusion in the Sun is possible. Use the words barrier, classical physics, tunnelling and probability, and at least one number from this unit.
Sentence starters
- Two protons repel each other, so …
- In classical physics, a proton with … would …
- In quantum physics, however, …
- The probability is very small, but …
- So the Sun can …
Model answer
Two protons are both positive, so they repel each other, and there is an energy barrier of about $720\ \text{keV}$ before they are close enough to fuse. In classical physics, a proton with a typical energy of $1.35\ \text{keV}$ would turn back roughly $1000\ \text{fm}$ away and never fuse. In quantum physics, however, a proton also behaves like a wave, and its wave reaches into the barrier. So there is a small probability that two protons tunnel through the barrier and come close enough to fuse. The probability is very small, but the Sun's core contains so many protons that fusion goes on all the time, and the Sun can shine for billions of years.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary.
Summary
- Light and matter show wave–particle duality: they are detected as particles, but they travel as waves. In the double-slit experiment single electrons build up an interference pattern.
- The wave gives the probability of finding the particle (Born, 1926).
- Every particle has a de Broglie wavelength $\lambda = h/p$.
- Quantum tunnelling: a particle has a small chance to pass a barrier that it could not climb in classical physics. The chance falls very fast with the width and height of the barrier and the mass of the particle.
- In the Sun, protons with about $1.35\ \text{keV}$ face a barrier of hundreds of $\text{keV}$. Tunnelling, and an enormous number of protons, let fusion happen anyway.
Le coin français
Louis de Broglie wrote his famous thesis in French: Recherches sur la théorie des quanta. In French, tunnelling is l'effet tunnel. Listen, repeat, and then say the last sentence to your partner. 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.