Unit 4: How have new ideas and ways of thinking developed our understanding of the physical world?

VICPhysicsSyllabus dot point

How has understanding of the physical world changed?

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

A focused answer to the VCE Physics Unit 4 dot point on matter waves. Defines the de Broglie wavelength $\\lambda = h / p$, computes electron and other-particle wavelengths, explains the Davisson-Germer experiment as evidence for matter-wave diffraction, and treats the connection to electron microscopy.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answer

Have a quick question? Jump to the Q&A page

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 λ=h/p\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 pp has an associated matter wave of wavelength:

λ=hp\lambda = \frac{h}{p}

where h=6.626×1034h = 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 λ\lambda, was found by Einstein (1905) to behave like a particle (photon) with momentum p=h/λp = h / \lambda (this also follows from E=pcE = pc for massless particles and E=hfE = h f).

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

Computing the de Broglie wavelength

For a non-relativistic particle of mass mm and speed vv:

p=mv,λ=hmvp = m v, \quad \lambda = \frac{h}{m v}

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

λ=h2meV\lambda = \frac{h}{\sqrt{2 m e V}}

For an electron, m=9.11×1031m = 9.11 \times 10^{-31} kg and e=1.6×1019e = 1.6 \times 10^{-19} C, giving the convenient shorthand:

λ1.226 nmV\lambda \approx \frac{1.226 \text{ nm}}{\sqrt{V}}

where VV is in volts. So:

  • 100 V electron: λ0.123\lambda \approx 0.123 nm.
  • 1000 V electron: λ0.039\lambda \approx 0.039 nm.
  • 10000 V electron: λ0.012\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 λ/d\lambda / d where dd is the slit width.

For an electron with λ0.1\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 λ1034\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 102510^{-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 λ/d\lambda / d. For macroscopic objects λd\lambda \ll d and classical mechanics applies. For atomic-scale particles in atomic-scale environments, λ\lambda and dd 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 λ\lambda cannot be resolved. With visible light (λ500\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 λ=h/p\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:

2πrn=nλ2 \pi r_n = n \lambda

Substituting λ=h/(mv)\lambda = h / (m v) gives 2πrnmv=nh2 \pi r_n m v = n h, so mvrn=nh/(2π)=nm 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.

Worked examples

Example 1. Electron wavelength

An electron is accelerated through 200 V. Find its de Broglie wavelength.

Use the shortcut: λ=1.226/200=1.226/14.140.0867\lambda = 1.226 / \sqrt{200} = 1.226 / 14.14 \approx 0.0867 nm = 8.67×10118.67 \times 10^{-11} m.

Example 2. Neutron wavelength

A thermal neutron at room temperature has Ek0.025E_k \approx 0.025 eV (thermal energy). Mass mn=1.67×1027m_n = 1.67 \times 10^{-27} kg.

Convert. Ek=0.025×1.6×1019=4.0×1021E_k = 0.025 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-21} J.

Momentum. p=2mEk=2×1.67×1027×4.0×1021=1.34×10473.66×1024p = \sqrt{2 m E_k} = \sqrt{2 \times 1.67 \times 10^{-27} \times 4.0 \times 10^{-21}} = \sqrt{1.34 \times 10^{-47}} \approx 3.66 \times 10^{-24} kg m s1^{-1}.

Wavelength. λ=h/p=6.626×1034/3.66×10241.81×1010\lambda = h / p = 6.626 \times 10^{-34} / 3.66 \times 10^{-24} \approx 1.81 \times 10^{-10} m =0.181= 0.181 nm.

Thermal neutrons therefore diffract from crystals in the same way X-rays and electrons do. Neutron diffraction is a standard technique in materials science.

Example 3. Tennis ball

50 g tennis ball at 30 m/s. p=1.5p = 1.5 kg m/s, λ=6.6×1034/1.5=4.4×1034\lambda = 6.6 \times 10^{-34} / 1.5 = 4.4 \times 10^{-34} m. Utterly unobservable.

Common errors

Mass in grams. Always use kg in λ=h/(mv)\lambda = h / (m v).

Energy as momentum. EkE_k in eV is not the momentum. Convert EkE_k to joules, then use p=2mEkp = \sqrt{2 m E_k}.

Photon formula for matter. Photons have E=pcE = pc (relativistic). Non-relativistic matter has E=p2/(2m)E = p^2 / (2 m). The two relations give different λ\lambda.

Mixing up the formula. λ=h/p\lambda = h / p, not λ=h×p\lambda = h \times p or λ=p/h\lambda = p / h.

Forgetting relativistic correction at very high energies. For electrons accelerated to MeV energies, the non-relativistic formula breaks down. VCE Physics stays in the non-relativistic regime.

Confusing matter wave with electromagnetic wave. The matter wave is not an electromagnetic wave; it is the quantum probability amplitude. It does not carry energy in the EM sense.

In one sentence

de Broglie's matter-wave hypothesis assigns every particle of momentum pp a wave-like wavelength λ=h/p\lambda = h / p, predicting diffraction of electrons from crystals (confirmed by Davisson-Germer in 1927) and unifying wave and particle pictures through wave-particle duality; the de Broglie wavelength is observable for atomic-scale particles in atomic-scale environments but utterly negligible for macroscopic objects, which is why classical mechanics suffices for everyday motion.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

2024 VCAA4 marksAn electron is accelerated from rest through a potential difference of $100$ V. (a) Calculate the kinetic energy of the electron in eV. (b) Calculate its momentum (in kg m s$^{-1}$). (c) Calculate its de Broglie wavelength.
Show worked answer →

(a) Kinetic energy. Work-energy theorem: Ek=eV=1×100=100E_k = e V = 1 \times 100 = 100 eV.

Convert: 100100 eV ×1.6×1019\times 1.6 \times 10^{-19} J/eV =1.6×1017= 1.6 \times 10^{-17} J.

(b) Momentum. Use Ek=p2/(2m)E_k = p^2 / (2m), so p=2mEkp = \sqrt{2 m E_k}.

Electron mass: m=9.11×1031m = 9.11 \times 10^{-31} kg.

p=2×9.11×1031×1.6×1017=2.92×10475.40×1024p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-17}} = \sqrt{2.92 \times 10^{-47}} \approx 5.40 \times 10^{-24} kg m s1^{-1}.

(c) de Broglie wavelength. λ=h/p\lambda = h / p.

λ=6.626×1034/5.40×10241.23×1010\lambda = 6.626 \times 10^{-34} / 5.40 \times 10^{-24} \approx 1.23 \times 10^{-10} m =0.123= 0.123 nm.

This is comparable to atomic spacings in a crystal lattice, so the electron will diffract from a crystal grating. The 100 V electron diffraction is the basis of the Davisson-Germer experiment.

Markers reward the correct Ek=eVE_k = eV relation, the momentum from 2mEk\sqrt{2 m E_k}, and the de Broglie wavelength via h/ph / p with appropriate scientific notation.

2023 VCAA3 marks(a) State de Broglie's hypothesis. (b) Calculate the de Broglie wavelength of a $0.05$ kg tennis ball moving at $30$ m s$^{-1}$. (c) Explain why we do not observe diffraction effects for the tennis ball.
Show worked answer →

(a) de Broglie's hypothesis. Every particle of momentum pp has an associated matter wave with wavelength λ=h/p\lambda = h / p. Matter therefore has wave-like properties, analogous to the particle-like properties of light established by the photon model.

(b) Tennis ball. p=mv=0.05×30=1.5p = m v = 0.05 \times 30 = 1.5 kg m s1^{-1}.

λ=h/p=6.626×1034/1.54.4×1034\lambda = h / p = 6.626 \times 10^{-34} / 1.5 \approx 4.4 \times 10^{-34} m.

(c) Why no diffraction. Diffraction effects are observable only when the wavelength is comparable to the size of a slit or obstacle. The tennis ball's de Broglie wavelength (1034\sim 10^{-34} m) is more than 102010^{20} times smaller than any atomic-scale structure (1010\sim 10^{-10} m), let alone any macroscopic slit. No physical aperture is small enough to produce observable diffraction of a tennis ball. The wave nature of matter exists in principle for all objects but is utterly negligible for macroscopic ones.

Markers reward the precise de Broglie hypothesis statement, the correct wavelength calculation, and the explicit comparison between λ\lambda and the smallest available diffracting structure.

Related dot points