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How can light and matter behave as both waves and particles, and what is the de Broglie wavelength?

Wave-particle duality: the dual nature of light, the de Broglie wavelength of moving particles, electron diffraction as evidence, and the conditions under which wave or particle behaviour dominates.

A focused answer to the Eduqas A-Level Physics Component 3 wave-particle duality content, covering the dual wave and particle nature of light, the de Broglie wavelength of moving particles, electron diffraction as experimental evidence, and when wave or particle behaviour dominates.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to explain that light shows both wave and particle behaviour, state and use the de Broglie relation λ=hp\lambda = \frac{h}{p} for moving particles, describe electron diffraction as evidence that particles have a wave nature, and explain when wave or particle behaviour is observed.

The answer

The dual nature of light

The de Broglie wavelength

Electron diffraction

When wave or particle behaviour dominates

Examples in context

Wave-particle duality is a cornerstone of quantum mechanics. The wave nature of electrons makes the electron microscope possible: because the de Broglie wavelength is far shorter than visible light, it resolves structures down to the atomic scale, revealing viruses, surfaces and crystal defects invisible to optical microscopes. Neutron and electron diffraction are standard tools for determining the structure of crystals and biological molecules.

Try this

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

  • Cue. λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}.

Q2. An electron has momentum 3.0×10−24 kg m s−13.0 \times 10^{-24}\ \text{kg m s}^{-1}. Find its de Broglie wavelength (h=6.63×10−34 J sh = 6.63 \times 10^{-34}\ \text{J s}). [2 marks]

  • Cue. λ=hp=6.63×10−343.0×10−24=2.2×10−10 m\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.0 \times 10^{-24}} = 2.2 \times 10^{-10}\ \text{m}.

Q3. State what electron diffraction demonstrates about electrons. [1 mark]

  • Cue. That electrons (normally treated as particles) also have a wave nature.

Exam-style practice questions

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

Eduqas 20194 marksAn electron is accelerated through a potential difference of 2.5 kV2.5\ \text{kV}. Calculate its de Broglie wavelength. Take h=6.63×10−34 J sh = 6.63 \times 10^{-34}\ \text{J s}, me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\ \text{kg} and e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}.
Show worked answer →

Kinetic energy gained equals the work done by the field: Ek=eV=1.6×10−19×2500=4.0×10−16 JE_k = eV = 1.6 \times 10^{-19} \times 2500 = 4.0 \times 10^{-16}\ \text{J}.

Momentum from Ek=p22mE_k = \dfrac{p^2}{2m}: p=2mEk=2×9.11×10−31×4.0×10−16=7.29×10−46=2.70×10−23 kg m s−1p = \sqrt{2mE_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 4.0 \times 10^{-16}} = \sqrt{7.29 \times 10^{-46}} = 2.70 \times 10^{-23}\ \text{kg m s}^{-1}.

de Broglie wavelength: λ=hp=6.63×10−342.70×10−23=2.5×10−11 m\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{2.70 \times 10^{-23}} = 2.5 \times 10^{-11}\ \text{m}.

Markers reward Ek=eVE_k = eV, finding the momentum from p=2mEkp = \sqrt{2mE_k}, and λ=hp≈2.5×10−11 m\lambda = \frac{h}{p} \approx 2.5 \times 10^{-11}\ \text{m}.

Eduqas 20213 marksElectrons passing through a thin graphite film produce a series of concentric rings on a fluorescent screen. Explain what this observation demonstrates about the nature of electrons.
Show worked answer →

The rings are a diffraction pattern, identical in form to the pattern produced when X-rays of similar wavelength pass through the same crystalline material.

Diffraction is a wave property, occurring when waves pass through gaps comparable to their wavelength (here the regular spacing of atoms in the graphite). The fact that electrons, which carry mass and charge and are normally treated as particles, produce a diffraction pattern shows that they also exhibit wave behaviour: this is wave-particle duality.

Markers reward identifying the rings as a diffraction pattern, stating diffraction is a wave property, and concluding that electrons (particles) show wave behaviour.

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