Wave and Particle Nature of Light

Lessons

Senior 5 · Physics

Wave and Particle Nature of Light

5 lessons

Lesson 5 of 5

The Wave Nature of Matter

25 min

A father, a son and the electron

In 1897 J. J. Thomson showed that cathode rays are streams of tiny charged particles, electrons, and won the 1906 Nobel Prize for it. In 1927 his son, George Paget Thomson, fired electrons through thin metal foils in Aberdeen and photographed rings exactly like those made when X-rays are diffracted by crystals. In the same year Clinton Davisson and Lester Germer, in New York, saw electrons bounce off a nickel crystal in strong beams at particular angles. G. P. Thomson and Davisson shared the 1937 Nobel Prize for showing that the electron is a wave. Both father and son were right. This lesson explains how.

De Broglie's hypothesis

In 1924 Louis de Broglie, a doctoral student in Paris, argued from symmetry. Light, long known as a wave, had turned out to carry momentum $p = \frac{h}{\lambda}$ like a particle. Perhaps particles, in turn, have a wavelength given by the same relation turned round. He proposed that any particle of momentum $p$ has an associated wave, a matter wave, of wavelength

\lambda = \frac{h}{p} = \frac{h}{mv}

For an electron accelerated from rest through a potential difference $V$ (slow enough for classical mechanics), the kinetic energy is $eV = \frac{p^2}{2m}$ , so $p = \sqrt{2meV}$ and

\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.23 \times 10^{-9}}{\sqrt{V}}\,\mathrm{m} \qquad (V \text{ in volts})

At 100 V this is 0.123 nm, about the spacing of atoms in a crystal. That is why crystals diffract electrons, just as they diffract X-rays of similar wavelength.

Definition

The de Broglie wavelength of a particle of momentum $p$ is $\lambda = \frac{h}{p}$ . Wave–particle duality is the fact that both radiation and matter show particle properties in some experiments (photoelectric effect, Compton effect) and wave properties in others (interference, diffraction). Example: an electron moving at $1.0 \times 10^{6}\,\mathrm{m\,s^{-1}}$ has $\lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(1.0 \times 10^{6})} = 0.73$ nm.

Worked example 1: checking Davisson and Germer

Davisson and Germer accelerated electrons through 54 V. The rows of atoms on their nickel surface, 0.215 nm apart, acted like a diffraction grating, and the strongest reflected beam was at 50° to the incident direction, so that $d\sin\theta = \lambda$ . Compare the measured wavelength with de Broglie's prediction.

Solution

Measured: $\lambda = d\sin\theta = 0.215 \times \sin 50^\circ = 0.165$ nm.

Predicted: $\lambda = \frac{h}{\sqrt{2meV}} = \frac{6.63 \times 10^{-34}}{\sqrt{2(9.11 \times 10^{-31})(1.60 \times 10^{-19})(54)}} = 1.67 \times 10^{-10}\,\mathrm{m} = 0.167$ nm.

The two agree to within about 1.5%, well inside the uncertainty of the experiment.

The electron diffraction tube

In a school laboratory, the effect is shown with an electron diffraction tube. Electrons from a heated filament are accelerated through a few kilovolts and pass through a very thin film of graphite. Graphite is polycrystalline: it is made of tiny crystals pointing in every direction. For each set of atomic planes, of spacing $d$ , some crystals are tilted at just the angle $\theta$ that satisfies the Bragg condition $2d\sin\theta = \lambda$ , and they deflect electrons through $2\theta$ . Because the crystals point in all directions around the beam, the deflected electrons form a cone, which strikes the screen as a ring.

For small angles, $\sin\theta \approx \theta$ and the ring radius is $r \approx L \times 2\theta$ , so

r \approx \frac{L\lambda}{d}

Raising the accelerating voltage shortens the wavelength ( $\lambda \propto \frac{1}{\sqrt{V}}$ ), and the rings shrink. Electrons made the rings, but only waves make rings.

Worked example 2: ring sizes in the tube

In the tube above, electrons are accelerated through 4.0 kV, the screen is 135 mm from the film, and the graphite planes responsible for the inner ring are 0.213 nm apart. (a) Find the electron wavelength. (b) Find the radius of the inner ring. (c) Find its radius when the voltage is raised to 6.0 kV.

Solution

(a) $\lambda = \frac{1.23 \times 10^{-9}}{\sqrt{4000}} = 1.94 \times 10^{-11}$ m $= 19.4$ pm.

(b) $r = \frac{L\lambda}{d} = \frac{(0.135)(1.94 \times 10^{-11})}{0.213 \times 10^{-9}} = 1.23 \times 10^{-2}$ m $= 12.3$ mm. (The outer ring, from planes 0.123 nm apart, has $r = 21.3$ mm.)

Worked example 3: why a football does not diffract

Find the de Broglie wavelength of a 0.45 kg football kicked at 20 m s⁻¹, and explain why its wave nature is never noticed.

Solution

$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(0.45)(20)} = 7.4 \times 10^{-35}$ m.

Diffraction is noticeable only when a wave passes an opening or obstacle comparable in size to its wavelength. Nothing is anywhere near $10^{-34}$ m across (a proton is about $10^{-15}$ m), so the football behaves exactly as a classical particle. Wave behaviour matters only for very light particles: electrons, neutrons, atoms.

Matter waves and Bohr's orbits

De Broglie's idea also explained the rule Bohr had simply assumed for the hydrogen atom (Unit 6). If the electron in an orbit is a wave, the orbit can last only if the wave joins smoothly onto itself after one trip round, making a standing wave. Otherwise the wave would interfere with itself and cancel. So a whole number of wavelengths must fit round the circumference:

2\pi r = n\lambda = \frac{nh}{mv} \quad \Rightarrow \quad mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \ldots

This is exactly Bohr's quantisation of angular momentum.

A check. In the ground state of hydrogen, $r = 0.0529$ nm and the electron's kinetic energy is 13.6 eV. The circumference is $2\pi(0.0529) = 0.332$ nm, and the de Broglie wavelength at 13.6 eV is $\frac{1.23}{\sqrt{13.6}} = 0.333$ nm: exactly one wavelength fits, as $n = 1$ requires.

Using electron waves

The finest detail any microscope can show is limited by the wavelength it uses. Visible light (400–700 nm) cannot resolve anything much smaller than about 200 nm. Electrons accelerated through 100 kV have a wavelength of about 0.004 nm, so electron microscopes can show viruses, the structure of cell membranes and even rows of atoms. Electron and neutron diffraction are also used to work out the arrangement of atoms in new materials.

Warning

Use momentum, not energy. $\lambda = \frac{h}{p}$ , never $\frac{h}{E}$ . For an electron given energy $e V$ , first find $p = \sqrt{2meV}$ . And the shortcut $\frac{1.23}{\sqrt{V}}$ nm works only for electrons and only well below about 50 kV; above that the electron's speed is a large fraction of $c$ and relativistic momentum must be used.

Tip

Same $h$ , two directions. For a photon, $p = \frac{h}{\lambda}$ gives a wave a momentum. For a particle, $\lambda = \frac{h}{p}$ gives momentum a wavelength. It is one equation read both ways.

The Electron Diffraction Tube, SSERC TV. Watch the rings on the screen shrink as the accelerating voltage is increased, because a higher voltage gives the electrons more momentum and so a shorter wavelength.

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Practice

State de Broglie's hypothesis, and name one experiment that confirmed it.
Find the de Broglie wavelength of an electron accelerated through 150 V.
Through what potential difference must an electron be accelerated to have a wavelength of 0.10 nm?
A neutron ( $m = 1.67 \times 10^{-27}$ kg) moves at 2700 m s⁻¹. Find its de Broglie wavelength, and say why such neutrons are useful for studying crystals.
An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength, and by what factor? ( $\frac{m_p}{m_e} = 1836$ )
In an electron diffraction tube, a ring has radius 15 mm at 3.0 kV. Predict its radius at 5.0 kV.
Use the standing-wave picture to derive $mvr = \frac{nh}{2\pi}$ , and explain why an orbit with $2\pi r = 3.5\lambda$ cannot exist.
Keza says: "The electron is a particle when it hits the screen and a wave when it goes through the graphite, so it changes from one to the other." Evaluate this statement.

Show answersHide answers

Every moving particle has an associated wave of wavelength $\lambda = \frac{h}{p}$ . Confirmed by Davisson and Germer (electrons reflected from a nickel crystal) or by G. P. Thomson (electrons through thin foils), both in 1927.
$\lambda = \frac{1.23 \times 10^{-9}}{\sqrt{150}} = 1.0 \times 10^{-10}$ m $= 0.10$ nm.
$\sqrt{V} = \frac{1.23}{0.10} = 12.3$ , so $V = 151 \approx 150$ V.
$\lambda = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27})(2700)} = 1.5 \times 10^{-10}$ m $= 0.15$ nm, about the spacing of atoms in a crystal, so the neutrons are diffracted strongly.
$p = \sqrt{2mE_k}$ , so at equal $E_{k}$ , $\lambda \propto \frac{1}{\sqrt{m}}$ . The electron's wavelength is longer, by $\sqrt{1836} = 42.8$ times.
$r \propto \frac{1}{\sqrt{V}}$ : $r = 15 \times \sqrt{\frac{3.0}{5.0}} = 11.6 \approx 12$ mm.
A stable orbit must hold a standing wave, so $2\pi r = n\lambda = \frac{nh}{mv}$ , giving $mvr = \frac{nh}{2\pi}$ . With 3.5 wavelengths, after one trip the wave arrives out of phase with itself, and the successive passes cancel, so no steady state exists.
Not a good description. The electron does not switch between two kinds of object. Its behaviour is described at all times by a wave that tells where it is likely to be found; that wave spreads through the graphite and interferes, and the electron is detected as a single localised event at the screen. Wave and particle are two complementary aspects of one quantum object, and which one shows depends on the experiment.

Key points

De Broglie: every particle has $\lambda = \frac{h}{p}$ ; for an electron through $V$ , $\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.23}{\sqrt{V}}$ nm.
Electron diffraction (Davisson–Germer, G. P. Thomson, the school tube) confirms it; ring radius $r \approx \frac{L\lambda}{d} \propto \frac{1}{\sqrt{V}}$ .
Standing electron waves explain Bohr's rule $mvr = \frac{nh}{2\pi}$ .
Wave–particle duality: light and matter show both behaviours; wave effects matter only when $\lambda$ is comparable with the size of the apparatus.

1A father, a son and the electron

2De Broglie's hypothesis

3The electron diffraction tube

4Matter waves and Bohr's orbits

5Using electron waves

A father, a son and the electron

De Broglie's hypothesis

The electron diffraction tube

Matter waves and Bohr's orbits

Using electron waves