Explain de Broglie's hypothesis that matter has wave-like properties with wavelength $\lambda = h / p$, and apply it to predict diffraction of electrons and other particles

What this dot point is asking

VCAA wants you to state de Broglie's matter-wave hypothesis, compute the de Broglie wavelength of electrons (and other particles) from $\lambda = h / p$, explain why electron diffraction is observable while tennis-ball diffraction is not, and connect matter waves to the wave-particle duality picture. Cross-link: see the de Broglie wavelength calculator.

de Broglie's hypothesis

In 1924, Louis de Broglie proposed that matter exhibits wave-like properties in the same way light exhibits particle-like properties through the photon model. Specifically: every particle of momentum $p$ has an associated matter wave of wavelength:

where $h = 6.626 \times 10^{-34}$ J s is Planck's constant.

The hypothesis was motivated by symmetry with light. Light, classically a wave with wavelength $\lambda$, was found by Einstein (1905) to behave like a particle (photon) with momentum $p = h / \lambda$ (this also follows from $E = pc$ for massless particles and $E = h f$).

de Broglie inverted the relationship: if light particles have a wavelength $\lambda = h / p$, then particles of matter (electrons, neutrons, atoms) should also have a wavelength $\lambda = h / p$.

Computing the de Broglie wavelength

For a non-relativistic particle of mass $m$ and speed $v$:

For an electron accelerated from rest through a potential difference $V$, the kinetic energy is $E_k = e V$ and the momentum is $p = \sqrt{2 m E_k}$, so:

For an electron, $m = 9.11 \times 10^{-31}$ kg and $e = 1.6 \times 10^{-19}$ C, giving the convenient shorthand:

where $V$ is in volts. So:

• 100 V electron: $\lambda \approx 0.123$ nm.
• 1000 V electron: $\lambda \approx 0.039$ nm.
• 10000 V electron: $\lambda \approx 0.012$ nm.

These wavelengths are comparable to atomic spacings, which is why electron beams diffract from crystals.

Davisson-Germer experiment

In 1927, Davisson and Germer confirmed de Broglie's hypothesis by directing a beam of low-energy electrons (around 54 eV) at a nickel crystal. They observed:

• A peak in the scattered electron intensity at a specific angle (about 50 degrees).
• The peak angle and intensity matched the predictions for Bragg diffraction of waves with wavelength equal to the de Broglie value.

The result: electrons (matter) diffract from a crystal lattice in the same way X-rays (waves) diffract. The Davisson-Germer experiment is the canonical experimental demonstration of matter waves.

A similar result was obtained independently in 1927 by G. P. Thomson, who passed electrons through a thin metal film and observed concentric diffraction rings on a photographic plate. (Notably, J. J. Thomson, the father, had discovered the electron as a particle in 1897; G. P. Thomson, the son, demonstrated it as a wave 30 years later. Both received Nobel prizes.)

Why diffraction of electrons but not tennis balls

Diffraction is observable when the wavelength is comparable to the size of a slit or obstacle. Specifically, the angular spread of diffraction is approximately $\lambda / d$ where $d$ is the slit width.

For an electron with $\lambda \sim 0.1$ nm, diffraction is observable when the slit (or crystal lattice spacing) is around 0.1 nm, which is the typical atomic spacing in a crystal. The Davisson-Germer experiment exploits this match.

For a tennis ball with $\lambda \sim 10^{-34}$ m, no physical slit is small enough to produce observable diffraction. Even a slit of 1 nm width would produce an angular spread of $10^{-25}$ radians, far below any measurable threshold. The wave nature exists, but the effects are negligible.

The classical-vs-quantum boundary is determined by the ratio $\lambda / d$. For macroscopic objects $\lambda \ll d$ and classical mechanics applies. For atomic-scale particles in atomic-scale environments, $\lambda$ and $d$ are comparable and quantum effects dominate.

Wave-particle duality

The de Broglie hypothesis completes the picture of wave-particle duality:

• Light (classically a wave) shows particle properties: photoelectric effect, atomic spectra, photon momentum in Compton scattering.
• Matter (classically particles) shows wave properties: diffraction, interference, quantised atomic orbits explained as standing waves.

In modern quantum mechanics, both light and matter are quantum objects whose particle or wave aspect depends on the experimental setup. The de Broglie wavelength is the de facto wavelength of the matter "wave function", although VCE Physics treats it at the classical-wave level.

Application: electron microscope

A standard light microscope's resolution is limited by diffraction: features smaller than approximately $\lambda$ cannot be resolved. With visible light ($\lambda \sim 500$ nm), the resolution limit is around 200 nm.

An electron microscope uses an electron beam in place of light. Electrons accelerated through tens or hundreds of kilovolts have de Broglie wavelengths in the picometre range, giving resolution down to atomic scale. The Transmission Electron Microscope (TEM) can image individual atoms; the Scanning Electron Microscope (SEM) images surface topology with nanometre resolution.

The electron microscope's resolution advantage is a direct application of $\lambda = h / p$: more energetic electrons have higher momentum and shorter wavelength, hence finer resolution.

Matter waves and atomic orbits

de Broglie's hypothesis also provides a physical interpretation of Bohr's quantised atomic orbits. In the Bohr model, the electron orbits the nucleus at specific radii. In the matter-wave interpretation, only orbits whose circumference is an integer number of de Broglie wavelengths are allowed:

Substituting $\lambda = h / (m v)$ gives $2 \pi r_n m v = n h$, so $m v r_n = n h / (2 \pi) = n \hbar$, recovering Bohr's quantisation of angular momentum.

The standing-wave picture is intuitive: only certain orbits support a stable electron matter wave around the nucleus, in the same way only certain frequencies produce stable resonances on a circular drum head.

Examples in context

Example 1. Electron diffraction at the University of Melbourne School of Physics. A demonstration electron-diffraction tube at the University of Melbourne accelerates electrons through $5000$ V. Kinetic energy is $5000$ eV $= 8.0 \times 10^{-16}$ J. Electron speed $v = \sqrt{2KE/m} = \sqrt{1.76 \times 10^{15}} = 4.19 \times 10^7$ m s$^{-1}$ (non-relativistic). Momentum $p = mv = 9.11 \times 10^{-31} \times 4.19 \times 10^7 = 3.82 \times 10^{-23}$ kg m s$^{-1}$. de Broglie wavelength $\lambda = h/p = 6.63 \times 10^{-34}/3.82 \times 10^{-23} = 1.74 \times 10^{-11}$ m = $17.4$ pm, comparable to graphite layer spacing ($0.335$ nm), so diffraction rings form on the phosphor screen.

Example 2. Cricket ball at the MCG. A $156$ g cricket ball travelling at $35$ m s$^{-1}$ has momentum $p = mv = 0.156 \times 35 = 5.46$ kg m s$^{-1}$. de Broglie wavelength is $\lambda = h/p = 6.63 \times 10^{-34}/5.46 = 1.21 \times 10^{-34}$ m, about $10^{20}$ times smaller than a proton. This wavelength is far smaller than any conceivable aperture, so wave properties of macroscopic objects are utterly undetectable. The contrast with electrons (whose wavelength is $\sim 10^{-11}$ m at typical lab energies, comparable to atomic spacings) explains why quantum behaviour is only observable at microscopic scales.

Try this

Q1. State the de Broglie hypothesis and its mathematical form. [2 marks]

• Cue. Particles have a wave-like character with wavelength $\lambda = h/p$, where $p$ is momentum.

Q2. Calculate the de Broglie wavelength of (a) an electron with kinetic energy $100$ eV, and (b) a proton with the same kinetic energy. [4 marks]

• Cue. (a) $p = \sqrt{2mE} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-17}} = 5.40 \times 10^{-24}$ kg m s$^{-1}$; $\lambda = 0.123$ nm. (b) Mass $\sim 1836$ times larger, $\lambda = 0.123/\sqrt{1836} = 0.0029$ nm.

Q3. Refer to the Melbourne electron-diffraction tube. (a) Calculate the de Broglie wavelength for a $5000$ eV electron. (b) Compare to graphite layer spacing of $0.335$ nm. (c) Explain why electrons but not cricket balls show diffraction. [2+2+2 marks]

• Cue. (a) $17.4$ pm. (b) Comparable order of magnitude, so diffraction rings appear. (c) Cricket-ball $\lambda$ is $10^{-34}$ m, far smaller than any aperture.