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Wave-particle dualityAQA A-Level Physics: Revision notes

Section 1

Evidence for wave and particle behaviour

Two experiments show that the wave and particle models both have a role.

  • The photoelectric effect suggests electromagnetic waves have a particulate nature: radiation arrives as photons of energy E=hfE = hf, with a threshold frequency and immediate emission
  • Electron diffraction suggests particles have wave properties: a beam of electrons passing through a thin graphite film or crystal produces a pattern of rings

Light and matter both show wave–particle duality. You do not need to know the details of particular diffraction methods.

Key termswave–particle dualityelectron diffraction

Section 2

The de Broglie wavelength

A particle of momentum p=mvp = mv has a de Broglie wavelength

λ=hmv\lambda = \dfrac{h}{mv}

where h=6.63×10−34h = 6.63 \times 10^{-34} J s. The wavelength is smaller for a greater momentum. Everyday objects have a very large momentum, so their wavelength is too small to notice.

For an electron accelerated from rest through a potential difference VV: 12mv2=eV\tfrac{1}{2}mv^2 = eV, so v=2eV/mv = \sqrt{2eV/m}.

Key termsde Broglie wavelengthmomentum
Exam tip

Always find the speed first, then use λ=h/mv\lambda = h/mv. The electron mass is in the data booklet.

Section 3

How diffraction changes with momentum

Diffraction is significant when the wavelength is comparable to the gap (for electrons, the spacing between atoms, about 10−1010^{-10} m).

If the accelerating potential difference is increased:

  • the electrons gain more kinetic energy, so their speed and momentum increase
  • the wavelength λ=h/mv\lambda = h/mv decreases
  • there is less diffraction, so the rings in a graphite experiment get smaller

A particle with a greater mass moving at the same speed has a greater momentum and a smaller wavelength.

Key termsdiffraction

Section 4

Worked example

An electron is accelerated through 250 V.

  • v=2eV/m=(2×1.60×10−19×250)/(9.11×10−31)=9.4×106v = \sqrt{2eV/m} = \sqrt{(2 \times 1.60 \times 10^{-19} \times 250) / (9.11 \times 10^{-31})} = 9.4 \times 10^6 m s⁻¹
  • λ=h/mv=6.63×10−34/(9.11×10−31×9.4×106)=7.7×10−11\lambda = h/mv = 6.63 \times 10^{-34} / (9.11 \times 10^{-31} \times 9.4 \times 10^6) = 7.7 \times 10^{-11} m

This is similar to the spacing between atoms in a crystal, so strong diffraction is seen.

Section 5

How scientific knowledge develops

Ideas about the nature of matter and light change over time. De Broglie's idea made a testable prediction, and electron diffraction experiments supported it.

New results are published so that other scientists can check them. Peer review is the process in which independent experts examine the methods, data and conclusions. Results that other groups can replicate are more reliable. The scientific community validates a new idea in this way, and the idea can still be revised if new evidence appears.

Key termspeer reviewvalidation

Must Know

  • Photoelectric effect: evidence for the particle nature of light
  • Electron diffraction: evidence for the wave nature of matter
  • λ=h/mv\lambda = h/mv, so greater momentum means smaller wavelength
  • Greater accelerating voltage means smaller wavelength and less diffraction
  • New ideas are tested, published, peer reviewed and replicated

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Wave-particle duality

  1. A lecturer explains that the behaviour of electrons and of light cannot be described by waves alone or by particles alone.
    Explain how the photoelectric effect suggests that electromagnetic waves have a particulate nature.2 marks
  2. In a vacuum tube, electrons are accelerated through a potential difference and pass through a thin sheet of graphite. They produce a pattern of concentric rings on a fluorescent screen.
    Explain why the rings move closer to the centre when the accelerating potential difference is increased.2 marks
  3. In an electron diffraction experiment an electron is accelerated from rest through a potential difference of 250 V in a vacuum.
    Calculate the speed of the electron.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).