Skip to main content
EnglandPhysicsSyllabus dot point

Can particles behave like waves?

Wave-particle duality, electron diffraction as evidence for the wave nature of matter, the de Broglie wavelength, and the complementary wave and particle models of light.

A focused answer to the Edexcel 9PH0 wave-particle duality content, covering duality, electron diffraction as evidence for matter waves, the de Broglie wavelength, and the complementary wave and particle models of light.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Edexcel wants you to explain wave-particle duality, describe electron diffraction as evidence for the wave nature of matter, use the de Broglie wavelength λ=hp\lambda = \frac{h}{p}, and appreciate the complementary wave and particle models of light.

The answer

Wave-particle duality

No single classical picture captures both behaviours; they are two complementary aspects of the same quantum reality. Light shows its wave side in diffraction and interference but its particle side in the photoelectric effect; electrons show their particle side in collisions but their wave side in diffraction.

Electron diffraction

This experiment, predicted by de Broglie and confirmed by Davisson and Germer (and by Thomson), was the decisive evidence for matter waves. Increasing the accelerating voltage gives the electrons more momentum, shortening their wavelength and shrinking the diffraction rings towards the centre.

The de Broglie wavelength

Because hh is tiny, everyday objects have an immeasurably small wavelength and never show wave behaviour. Only very low-mass, fast particles such as electrons have wavelengths comparable to atomic spacings, which is why their wave nature is observable. For an electron accelerated through a potential difference VV, the kinetic energy eV=12mv2eV = \frac{1}{2}mv^2 gives the speed and hence the wavelength.

Complementary models of light

The wave model of light explains refraction, diffraction, interference and polarisation; the photon model explains the photoelectric effect, line spectra and the energy of individual quanta. Neither model alone is complete; they are complementary, each correct within its domain. The same is true for matter: an electron is described by a wave for diffraction but as a localised particle when it strikes a detector.

Examples in context

The electron microscope exploits the very short de Broglie wavelength of fast electrons to resolve detail far smaller than any light microscope, revealing viruses and atomic structures. Electron diffraction is used routinely to study crystal structures. Neutron diffraction probes materials where X-rays struggle. The complementary nature of light underlies the design of both lasers (wave coherence) and photon detectors (particle counting), and duality is the conceptual foundation of all of quantum mechanics.

Try this

Q1. State the de Broglie relationship. [1 mark]

  • Cue. λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}, the wavelength of a particle equals the Planck constant divided by its momentum.

Q2. A proton (mass 1.67×10−271.67 \times 10^{-27} kg) moves at 3.0×1053.0 \times 10^{5} m per second. Find its de Broglie wavelength. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s. [2 marks]

  • Cue. λ=hmv=6.63×10−341.67×10−27×3.0×105=1.3×10−12\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 3.0 \times 10^{5}} = 1.3 \times 10^{-12} m.

Q3. Explain why everyday objects do not show wave behaviour. [2 marks]

  • Cue. Their large momentum makes the de Broglie wavelength vanishingly small, far too short to produce any observable diffraction or interference.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksAn electron travels at 2.0×1062.0 \times 10^{6} m per second. Calculate its de Broglie wavelength. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s and the electron mass as 9.11×10−319.11 \times 10^{-31} kg.
Show worked answer →

de Broglie wavelength: λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}.

λ=6.63×10−349.11×10−31×2.0×106=6.63×10−341.82×10−24=3.6×10−10\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.0 \times 10^{6}} = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.6 \times 10^{-10} m.

Markers reward λ=hmv\lambda = \frac{h}{mv}, the momentum in the denominator, and the value about 3.6×10−103.6 \times 10^{-10} m.

Edexcel 20215 marksDescribe the electron diffraction experiment and explain how it provides evidence for the wave nature of matter. State how the diffraction pattern changes if the electrons are accelerated through a larger potential difference.
Show worked answer →

A beam of electrons accelerated through a potential difference is fired at a thin graphite film. On the screen beyond, concentric bright rings appear. Diffraction rings are a wave phenomenon (the regular atomic spacing in graphite acts like a diffraction grating), so the fact that electrons produce them shows electrons have a wave nature.

Larger accelerating voltage gives the electrons more momentum, so by λ=hp\lambda = \frac{h}{p} their de Broglie wavelength is shorter. A shorter wavelength diffracts less, so the rings move closer to the centre (the pattern shrinks).

Markers reward the graphite-film setup, the rings as evidence of wave behaviour, and the shrinking pattern with higher voltage (shorter wavelength).

Related dot points

Sources & how we know this