A batter drives the ball through the gap between cover and mid-off. The ball goes straight through the gap and on to the boundary. It never fans out behind the gap the way water waves fan out behind an opening in a harbour wall. Yet a beam of electrons sent at a crystal comes off in a few sharp directions, exactly as a wave would. Both are particles. Why does one of them behave like a wave?
Light has momentum, so matter may have a wave
The chapter on photons showed that light, long treated as a wave, also comes in particles. A photon of wavelength \(\lambda\) carries momentum \(p = h/\lambda\), where \(h\) is Planck's constant. In 1924 a young French physicist, Louis de Broglie, turned this relation round. If a wave can behave as a particle, he argued, a particle should also behave as a wave.
Every moving particle has a wave associated with it. The wavelength of this wave is Planck's constant divided by the momentum of the particle, \(\lambda = h/p\).
This wave is called a matter wave, and its wavelength is the de Broglie wavelength. A larger momentum means a shorter wavelength. So a fast particle has a shorter wave than a slow one of the same mass. A heavy particle has a shorter wave than a light one at the same speed.
Try the slider in the figure above. The electron has a mass of \(9.1 \times 10^{-31}\,\text{kg}\). At \(2 \times 10^6\,\text{m/s}\) its momentum is \(1.82 \times 10^{-24}\,\text{kg m/s}\), and its wavelength is \(0.36\,\text{nm}\). Double the speed and the wavelength halves, to \(0.18\,\text{nm}\). These lengths are about the size of an atom.
A cricket ball of mass \(0.16\,\text{kg}\) moves at \(40\,\text{m/s}\). Find its de Broglie wavelength.
The momentum is \(p = mv = 0.16 \times 40 = 6.4\,\text{kg m/s}\).
The wavelength is \(\lambda = h/p = 6.63 \times 10^{-34} / 6.4 = 1.0 \times 10^{-34}\,\text{m}\).
Values used: \(h = 6.63 \times 10^{-34}\,\text{J s}\), \(e = 1.6 \times 10^{-19}\,\text{C}\), \(m_e = 9.1 \times 10^{-31}\,\text{kg}\). The neutron example also uses \(m_n = 1.675 \times 10^{-27}\,\text{kg}\) and \(k_B = 1.38 \times 10^{-23}\,\text{J/K}\).
Why we never see the wave of a ball
A wave shows itself by diffraction: it spreads round an obstacle or out of an opening. The spreading is strong only when the opening is about as wide as the wavelength. A gap much wider than \(\lambda\) lets the wave through in a straight beam, like a particle. The gap between two fielders is about \(1\,\text{m}\) wide. That is \(10^{34}\) times the wavelength of the ball.
Fig 1.1 settles the question. Nothing in nature is anywhere near \(10^{-34}\,\text{m}\) across. Even a nucleus is about \(10^{-14}\,\text{m}\). So the ball's wave can never meet an opening small enough to spread it, and the ball always behaves as a particle. An electron's wave is about \(10^{-10}\,\text{m}\) long, the same as the spacing of atoms in a solid. The rows of atoms in a crystal are just the right openings for it.
Quantum physics does not stop at some size. The cricket ball has a matter wave too; it is only far too short for any effect to be seen. Mass does not switch the wave off. Momentum makes it short.
A moving cricket ball shows no wave behaviour. What is the reason?
From kinetic energy to wavelength
In a laboratory we rarely know the speed of an electron. We know the voltage that accelerated it. So we need the wavelength in terms of the kinetic energy. A charge \(q\) that falls through a potential difference \(V\) gains energy \(qV\); the chapter on electricity treats potential difference in full. Starting from rest, the particle then has \(K = qV\).
\[ K = \dfrac{p^2}{2m} \]
Why. \(K = \tfrac12 mv^2\) and \(p = mv\), so \(K = (mv)^2/2m\).
\[ p = \sqrt{2mK} \]
Why. solve for the momentum.
\[ \lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}} \]
Why. put the momentum into the de Broglie relation.
\[ \lambda = \dfrac{h}{\sqrt{2mqV}} \]
Read it. for a charge accelerated from rest through \(V\), \(K = qV\).
An electron starts from rest and is accelerated through \(100\,\text{V}\). Find its kinetic energy, its speed and its de Broglie wavelength.
\(K = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}\,\text{J}\), which is \(100\,\text{eV}\).
\(v = \sqrt{2K/m}\) and \(\lambda = h/\sqrt{2mK}\).
\(v = \sqrt{2 \times 1.6 \times 10^{-17} / 9.1 \times 10^{-31}} = 5.93 \times 10^6\,\text{m/s}\). Then \(p = mv = 5.40 \times 10^{-24}\,\text{kg m/s}\) and \(\lambda = 6.63 \times 10^{-34} / 5.40 \times 10^{-24} = 1.23 \times 10^{-10}\,\text{m}\).
The electron has \(100\,\text{eV}\), moves at about \(6 \times 10^6\,\text{m/s}\), and has \(\lambda = 0.123\,\text{nm}\), the size of an atomic spacing.
A shortcut for electrons
For an electron the product \(2m_ee\) is fixed, so the whole constant \(h/\sqrt{2m_ee}\) can be worked out once. It is \(1.23 \times 10^{-9}\,\text{m V}^{1/2}\) with the values above, and \(1.227 \times 10^{-9}\) with more exact ones. So \(\lambda = 1.227/\sqrt{V}\,\text{nm}\), or \(12.27/\sqrt{V}\) in angstrom units, where \(1\,\text{Å} = 0.1\,\text{nm}\). An electron of kinetic energy \(K\) electron volts has the same wavelength as one accelerated through \(K\) volts.
The wavelength falls as \(1/\sqrt{V}\). So four times the voltage gives half the wavelength. At \(100\,\text{V}\) the wavelength is \(0.123\,\text{nm}\); at \(10\,000\,\text{V}\) it is \(0.0123\,\text{nm}\), ten times shorter for a hundred times the voltage.
An electron has a kinetic energy of \(150\,\text{eV}\). Find its de Broglie wavelength, in nm.
An electron is accelerated from rest through a voltage \(V\). How must the voltage change to halve its de Broglie wavelength?
An electron starts from rest and is accelerated through \(400\,\text{V}\). Find its de Broglie wavelength, in nm.
Different particles, one rule
The form \(\lambda = h/\sqrt{2mqV}\) compares particles quickly. Through the same voltage, \(\lambda\) is proportional to \(1/\sqrt{mq}\). A proton has the same size of charge as an electron but \(1836\) times the mass. So through the same voltage its wavelength is \(1/\sqrt{1836} = 0.023\) of the electron's. An alpha particle has four times the mass of a proton and twice its charge. Through the same voltage its wavelength is \(1/\sqrt{8} = 0.35\) of the proton's.
The comparison depends on what is kept equal. For an electron and a proton:
- at the same speed, \(\lambda = h/mv \propto 1/m\), so \(\lambda_e : \lambda_p = 1836 : 1\);
- at the same kinetic energy, \(\lambda = h/\sqrt{2mK} \propto 1/\sqrt{m}\), so \(\lambda_e : \lambda_p = \sqrt{1836} : 1 \approx 43 : 1\);
- at the same momentum, \(\lambda = h/p\) is the same, so \(\lambda_e : \lambda_p = 1 : 1\).
Neutrons at room temperature
Neutrons from a reactor are slowed by collisions until they share the temperature of their surroundings. Such a thermal neutron has the average kinetic energy of a gas molecule at that temperature, \(K = \tfrac32 k_BT\). Here \(k_B = 1.38 \times 10^{-23}\,\text{J/K}\) is Boltzmann's constant, and the kinetic theory of gases explains the formula. The same formula gives the wavelength of any gas molecule.
Find the de Broglie wavelength of a thermal neutron at \(300\,\text{K}\). Take \(m_n = 1.675 \times 10^{-27}\,\text{kg}\) and \(k_B = 1.38 \times 10^{-23}\,\text{J/K}\).
The kinetic energy is \(K = \tfrac32 k_BT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\,\text{J}\), about \(0.039\,\text{eV}\).
The momentum is \(p = \sqrt{2m_nK} = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} = 4.56 \times 10^{-24}\,\text{kg m/s}\).
The wavelength is \(\lambda = h/p = 6.63 \times 10^{-34} / 4.56 \times 10^{-24} = 1.45 \times 10^{-10}\,\text{m} = 0.145\,\text{nm}\).
This is again the size of an atomic spacing. That is why beams of slow neutrons from research reactors are used to study crystals.
An electron, a proton and an alpha particle each start from rest and are accelerated through the same voltage. Put them in order of de Broglie wavelength, the longest first.
An electron and a proton have the same kinetic energy. The proton is \(1836\) times as heavy. What is the ratio \(\lambda_e : \lambda_p\) of their de Broglie wavelengths?
A thermal neutron at \(300\,\text{K}\) has a wavelength of \(0.145\,\text{nm}\). A helium atom has about four times the mass of a neutron. Find the de Broglie wavelength of a helium atom at the same temperature, in nm.
Return to the batter's drive at the start of the lesson. The ball's momentum gives it a wavelength of \(10^{-34}\,\text{m}\), and the gap in the field is \(1\,\text{m}\) wide. The ball goes straight through, as a particle. An electron through \(100\,\text{V}\) has a wavelength of \(0.123\,\text{nm}\), and the rows of atoms in a crystal are about that far apart. So the crystal spreads the electron beam into sharp directions, as the next lesson shows.
Louis de Broglie first studied history, then turned to physics after serving as a radio operator in the First World War. In his doctoral thesis of 1924 he proposed that electrons are guided by waves of wavelength \(h/p\). His examiners in Paris did not know what to make of it. Paul Langevin sent a copy to Einstein, who praised it warmly. Within three years electron diffraction had been seen, and in 1929 de Broglie received the Nobel Prize for the idea.
- Using \(\lambda = h/mv\) with the speed of a different particle, or with the mass of the wrong one.Find the momentum of the particle itself first, \(p = mv\) or \(p = \sqrt{2mK}\).
- Writing \(\lambda \propto 1/V\) for an accelerated electron.The kinetic energy is proportional to \(V\), and the momentum to \(\sqrt{V}\). So \(\lambda \propto 1/\sqrt{V}\).
- Thinking that a big body has no matter wave.Every moving body has one; for a big body it is too short to show.