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Learn9702Paper 4 · A Level Structured QuestionsQuantum physicsWave–particle duality
Quantum physics · Subtopic 2 of 3

Wave–Particle Duality

Light behaves as a wave (diffraction, interference) and as particles (the photoelectric effect). Matter does too: de Broglie proposed that any moving particle has a wavelength λ=hp=hmv\lambda=\dfrac hp=\dfrac h{mv}. Electron diffraction through thin graphite proves it: electrons form rings like waves passing through a grating. Faster electrons have shorter wavelengths, so the rings get smaller.

Cambridge 9702 Paper 4Edexcel YPH11 WPH12AQA 7408 Paper 1AQA 7408 Paper 3B≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State evidence for the wave and particle nature of light.
  • Describe electron diffraction as evidence for matter waves.
  • Use λ=hp\lambda=\dfrac hp.
  • Link accelerating p.d. to de Broglie wavelength.
Pause and think

If everything moving has a wavelength, why don't you diffract when you walk through a doorway?

Section 2 of 14

Core idea

Start hereElectrons that diffract

Fire a beam of electrons through a thin slice of graphite and rings appear on the screen, exactly like light through a grating. Particles, it turns out, also travel as waves.

The big idea

de Broglie: λ=hp\lambda=\frac hp. Electron diffraction proves particle waves. Light shows both wave (interference) and particle (photoelectric) behaviour.

1. Electron diffraction

Electron diffraction rings on a screen
Raising the accelerating voltage increases the electrons' momentum, shortens their wavelength, and the rings shrink.
  • Rings like X-ray diffraction patterns.
  • Higher voltage → more momentum → shorter λ\lambda → smaller rings.

2. Calculating λ

Wavelength from accelerating voltage
  1. eV=12mv2eV=\tfrac12mv^2
    Energy gained from the field.
  2. 12mv2=p22m\tfrac12mv^2=\frac{p^2}{2m}
    p=mvp=mv.
  3. p=2meVp=\sqrt{2meV}
    Rearrange.
  4. λ=h2meV\lambda=\frac h{\sqrt{2meV}}
    de Broglie.
Worked exampleElectron through 2000 V

Find λ\lambda.

Goal
λ\lambda.
Strategy
Formula.
  1. p=2(9.11×10−31)(1.6×10−19)(2000)=2.41×10−23p=\sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})(2000)}=2.41\times10^{-23}
    kg m s−1^{-1}.
  2. λ=6.63×10−342.41×10−23=2.7×10−11 m\lambda=\frac{6.63\times10^{-34}}{2.41\times10^{-23}}=2.7\times10^{-11}\text{ m}
    de Broglie.
Answer2.7×10−112.7\times10^{-11} m.
✓ CheckComparable with atomic spacing (~10−1010^{-10} m): diffraction visible ✓.
de Broglie wavelength of an electron vs accelerating voltage
10002000300040005000246810120accelerating p.d. (V)wavelength (×10⁻¹¹ m)λ = h/√(2meV)
At a few thousand volts the wavelength is about 2×10−112\times10^{-11} m, similar to the spacing of atoms, so crystals diffract them.
Edge caseA cricket ball Wavelength ~10−3410^{-34} m: far too small to diffract through anything.

Summary

  • λ=h/p\lambda=h/p.
  • Diffraction shows particle waves.
Key words
Wave–particle duality
Both wave and particle behaviour.
HistoryA PhD thesis that won a Nobel. De Broglie proposed matter waves in his 19241924 PhD thesis. His examiners weren't sure, so they sent it to Einstein, who called it "a first feeble ray of light on this worst of our physics enigmas". Davisson and Germer confirmed electron diffraction in 19271927, and de Broglie got the 19291929 Nobel Prize.
Fun factElectron microscopes. Electrons accelerated through 100100 kV have wavelengths of about 4×10−124\times10^{-12} m, thousands of times shorter than visible light. That's why electron microscopes can image individual viruses and even rows of atoms that an optical microscope could never resolve.
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