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22.2Matter waves
In 1923, Louis de Broglie suggested that to any particle of momentum p there corresponds a wave of wavelength given by the formula (h is Planckâs constant):
Î»=hp
The de Broglie hypothesis, as this is known, thus assigns wave-like properties to something that is normally thought to be a particle. This state of affairs is called the duality of matter. All moving particles (not just electrons) are assigned a wavelength.
What does it mean to say that the electron has wave-like properties? One thing it does not mean is to think that the electron oscillates up and down as it moves along. Having wave-like properties means exhibiting the two basic phenomena of waves: diffraction and interference. A wave of wavelength Î» will diffract around an obstacle of size d if Î» is comparable to or bigger than d. In worked example 22.8 we calculate a typical electron wavelength to be of order 10â10 m. This distance is typical of the separation of atoms in crystals, and it is there that electron diffraction and interference will be seen.
But what does a matter wave actually represent? What kind of wave is it? The answer to this question came long after de Broglie announced his hypothesis and played a key role in the development of quantum mechanics, through the work of E. SchrÃ¶dinger and M. Born: the interpretation given was that the matter wave is a wave of probability. The square of the amplitude of the wave is proportional to the probability of finding the particle near a particular point in space at a particular time.
WORKED EXAMPLE 22.8
Find the de Broglie wavelength of an electron that has been accelerated from rest by a potential difference of 54 V.
Answer
The kinetic energy of the electron is given by EK=p22m.
The work done in accelerating the electron through a potential difference V is qV, and this work goes into kinetic energy:
p22m=qVâp=2mqVââââââ
So:
Î»=h2mqVââââââ=6.63Ã10â342Ã9.1Ã10â31Ã1.60Ã10â19Ã54ââââââââââââââââââââââââââââ=1.7Ã10â10Â m
EXAM TIP
It is preferable to use Ek=p22m for kinetic energy rather than EK=12mv2.
The formula Î»=h2mqVââââââ is very useful in paper 1 questions, where it is often required to know that Î»â1Vâââ.
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Experiments showing the wave nature of the electron were carried out in 1927 by Clinton J. Davisson (1881â1958) and Lester H. Germer (1896â1971), and also by George Thomson (1892â1975), son of J. J. Thomson, the discoverer of the electron. In the DavissonâGermer experiment, electrons of kinetic energy 54 eV were directed at a surface of nickel where a single crystal had been grown and were scattered by it (Figure 22.10).

Figure 22.10: The apparatus of Davisson and Germer. Electrons emitted from the hot filament of the electron gun are accelerated through a known potential difference V and are then allowed to fall on a crystal. The positions of the scattered electrons are recorded by a detector.

Figure 22.11 shows rays of a wave incident on layers of crystal atoms. The wave scatters off atoms in two layers. The distance between the layers is d. The path difference between the two rays is indicated in thick lines and equals d sin Î¸ +d sin Î¸ = 2d sin Î¸.

Figure 22.11: Waves scattering off atoms in two layers of the crystal. The path difference is 2d sin Î¸.

When this path difference is an integer number of the wavelength, 2d sin Î¸ = nÎ», constructive interference takes place i.e. a maximum in the intensity of the scattered wave. In the Davisson-Germer experiment it was known that d = 0.107 nm. The first maximum (n = 1) was observed at an angle Î¸ = 50Â° which means that the number of electrons observed at this angle was a maximum. Hence Î» = 2 Ã 0.107 Ã 10â9 Ã sin 50Â° = 1.64 Ã 10â10 m.
We have already calculated the de Broglie wavelength of the electron that had been accelerated by a potential difference 54 V in worked example 22.8; it was found to be 1.7 Ã 10â10 m. This is in excellent agreement with the experiment, thus verifying the de Broglie hypothesis.
CHECK YOURSELF 6
Electrons, protons and alpha particles pass through a narrow slit with the same speed. Which will produce the widest diffraction pattern?
WORKED EXAMPLE 22.9
Show that the Bohr condition for the quantisation of angular momentum is equivalent to 2Ïr = nÎ», where Î» is the de Broglie wavelength of the electron and r the radius of its orbit.
Answer
The Bohr condition is that:
mvr=nh2Ï
This can be rewritten as:
2Ïr=nhmv
But according to de Broglie:
hmv=Î»
So we have the result.
The result of worked example 22.9 shows that the allowed orbits in the Bohr model of hydrogen are those for which an integral number of electron wavelengths fit on the circumference of the orbit. Figure 22.12 shows the electron wave for n = 6. The circle in blue is the actual orbit. The solid red and the dotted red lines show the extremes of the electron wave. This is reminiscent of standing waves: the electron wave is a standing wave on the circumference. We know that standing waves do not transfer energy. This is a partial way to understand why the electrons do not radiate when in the allowed orbits.
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Figure 22.12: The allowed electron orbits are those for which an integral number of electron wavelengths fits on the circumference of the orbit.

THEORY OF KNOWLEDGE
Davisson and Germer investigated the scattering of low-energy electrons from a nickel surface. Initial results showed that, for fixed electron energy, the intensity of the electron beam decreased sharply as the scattering angle Î¸ increased. A container of liquid air was accidentally dropped, breaking the glass jar housing the apparatus and exposing the nickel surface (which was surrounded by vacuum) to air, oxidising it. To remove the oxide, Davisson and Germer heated the surface in an atmosphere of hydrogen. The scattering of electrons was continued but now the results were very different. The intensity of the scattered electron beam varied strongly with scattering angle. After much thought, Davisson and Germer realised that they were dealing with scattering from a single crystal of nickel (that had grown on the surface as a result of heating it). Using crystals of known interatomic spacing, they eventually concluded they were seeing electron diffraction with a wavelength given by the de Broglie formula.
Links
How can electrons diffract? In SectionÂ 14.2 we learned about the principle of superposition and in SectionÂ 14.3 about diffraction and interference of waves. The key idea there was that depending on the path (or phase) difference between two waves that meet the resultant wave may have a large amplitude or zero amplitude. What is the amplitude involved with electrons? It is not the actual displacement of the electron but an abstract quantity called the electron wavefunction, about which you may want to learn more.
TEST YOUR UNDERSTANDING
26A proton and an alpha particle have the same momentum. What is the ratio Î»pÎ»Î± of de Broglie wavelengths?
27A proton and an alpha particle have the same speed. What is the ratio Î»pÎ»Î± of de Broglie wavelengths?
28A proton and an alpha particle have the same kinetic energy. What is the ratio Î»pÎ»Î± of de Broglie wavelengths?
29Consider a brick of mass 0.250 kg moving at 10 m sâ1.
aEstimate its de Broglie wavelength.
bComment on whether it makes sense to treat the brick as a wave.
30aDescribe an experiment in which the de Broglie wavelength of an electron can be measured directly.
bDetermine the speed of an electron whose de Broglie wavelength is equal to that of red light (680 nm).
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31aShow that the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference V is given by Î»=h2meVââââââ.
bCalculate the ratio of the de Broglie wavelength of a proton to that of an alpha particle when both have been accelerated from rest by the same potential difference.
cCalculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 520 V.
32Find the de Broglie wavelength of a proton (mass 1.67 Ã 10â27 kg) whose kinetic energy is 200.0 MeV.
33What is the de Broglie wavelength of an electron in the n = 2 state of hydrogen? Give your answer in terms of the Bohr radius.
34What is the ratio of the de Broglie wavelength of the electron in the n = 2 state of hydrogen to that in the n = 3 state?
35A beam of electrons moving left to right is incident on an aperture of size 8 Ã 10â10 m, as shown in Figure 22.13. Each electron has kinetic energy 5.0 eV.

Figure 22.13: For question 35.

Most of the electrons will be observed within an angle Î¸.
aExplain why the electrons do not continue to move strictly left to right.
bEstimate the angle Î¸.
36This is a continuation of question 35.
The electrons must have acquired a small vertical component of momentum, Îpy, as a result of going through the aperture, as shown in Figure 22.14.

Figure 22.14: For question 36.

If the aperture size is Îy, show that Îy Ã Îpy â h.
37Show that the quantity hG2Ïc5ââââââ has units of time and the quantity hG2Ïc3ââââââ units of length (h = Planck constant, G = Newton constant of gravitation, c = speed of light). These are called Planck time and Planck length. Evaluate these quantities. Do some research on their significance.
