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Electron Diffraction & Single-Particle Interference

Key idea: Explain how electron diffraction and single-particle double-slit interference provide evidence for the wave nature of particles, and use λ = h/p to solve problems (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply de Broglie wavelength and wave-particle evidence.

1. Definitions (Must Know)

  • de Broglie wavelength of a particle:
    • λ = h/p
    • where h is Planck’s constant and p is the momentum.
  • Diffraction: spreading/bending of a wave when it passes through a narrow gap or around an obstacle; significant when the gap/spacing is comparable to λ.
  • Interference: pattern formed when waves superpose, producing alternating regions of constructive and destructive interference.
  • Wavefunction, ψ: describes the state of a particle; probability density is:
    • |ψ|²

2. Key Ideas (What Earns Marks)

Evidence for photon and matter-wave behaviourTwo evidence chains connect threshold-frequency photoemission to photons and electron diffraction with localised detections to matter-wave behaviour.Light: particulate evidenceThreshold frequencyno emission when f < f₀, however intensePhoton modelone quantum has energy E = hfphoton momentum p = E/c = h/λElectrons: wave evidenceDiffraction and interferencepatterns build from one detection at a timeMatter-wave modelde Broglie wavelength λ = h/pdetection remains localised
Scroll diagram horizontally to read all labels.
No single classical model explains every observation: threshold-frequency photoemission reveals photon behaviour, while diffraction and single-particle interference reveal wave behaviour.
  • Evidence for wave nature of particles:
    • electrons show diffraction (e.g. thin crystal/graphite) and interference patterns.
  • Single-particle double-slit:
    • even when particles pass through “one at a time”, the detection pattern builds up into an interference pattern.
  • Superposition (core idea for explanation): with two paths, the probability amplitudes add:

ψ = ψ₁ + ψ₂

|ψ|² = |ψ₁ + ψ₂|²; = |ψ₁|² + |ψ₂|²; + 2 Re(ψ₁ψ₂∗).

The last term is the interference term; it can be positive or negative.

3. Detailed Explanations

A. Electron diffraction (why it’s “wave” evidence)

When an electron beam passes through a thin crystalline material, the atoms are arranged in regular spacings (like a 3D “diffraction grating”). If the electron de Broglie wavelength λ is comparable to those spacings, the beam diffracts and you observe a diffraction pattern.

Key exam statement:

  • “Diffraction is a wave phenomenon; observing diffraction of electrons implies electrons have wave properties.”

B. Single-particle double-slit (what it shows)

If electrons are fired so that they reach the screen one at a time, you still get:

  1. Individual detection events (each electron is detected as a localised hit).
  2. After many electrons, the hits build up into an interference fringe pattern.

This is the key “wave + particle” message:

  • propagation is described by a wavefunction (wave-like),
  • detection is localised (particle-like).

C. What changes when you close a slit (and why it matters)

  • With one slit/path available, you get a single-slit distribution (no two-path interference term).
  • With both slits open, you get an interference pattern because the probability amplitudes superpose.

4. Common Mistakes

  • Saying the electron “splits into two halves”. Better wording: “the wavefunction has two contributions (two paths) that superpose”.
  • Mixing up intensity (classical waves) with probability density (|ψ|²) for particles.
  • Forgetting that λ = h/p means larger momentum → smaller wavelength → less diffraction.

5. Exam Tips

  • Use mark-scheme phrasing:
    • “Diffraction/interference are wave phenomena; observing them for electrons implies wave nature of particles.”
    • “Single-particle interference supports the superposition of probability amplitudes.”
  • If you need a quick physics check:
    • increase particle speed → increase p → decrease λ → pattern becomes less spread out.

6. Worked Examples

Modelled example 1

de Broglie wavelength from momentum

Core

Problem

An electron has momentum p = 3.0 × 10⁻²⁴ kg m s⁻¹. Find its de Broglie wavelength.
Study the worked solution

  1. Use the de Broglie relation

    Method

    λ = h/p.

    Reason

    A particle’s matter-wave wavelength is fixed by its momentum.

    Working

    λ = (6.63 × 10⁻³⁴)/(3.0 × 10⁻²⁴)

  2. Evaluate and interpret

    Method

    λ = 2.21 × 10⁻¹⁰ m.

    Reason

    This is comparable to crystal atomic spacings, allowing observable diffraction.

    Working

    λ = 2.21 × 10⁻¹⁰ m

Guided practice 2

How does speeding up electrons affect diffraction?

About 4 min

Problem

Electrons are accelerated to a higher non-relativistic speed before passing through the same thin crystal. Predict the qualitative change in the diffraction pattern and explain the chain.

Try this before viewing the solution

Pattern change

Hints

Hint 1: momentum first
Higher speed gives greater non-relativistic momentum.
Hint 2: then wavelength
Use λ = h/p, then compare that wavelength with the unchanged crystal spacing.
View solution step by step

  1. Update momentum

    Method

    Momentum increases.

    Reason

    For the stated non-relativistic case, p = mv.

    Working

    higher v → higher p

  2. Update wavelength

    Method

    The de Broglie wavelength decreases.

    Reason

    Wavelength is inversely proportional to momentum.

    Working

    λ = h/p

  3. Predict the pattern

    Method

    The diffraction becomes less pronounced, with smaller angular spread or closer fringes.

    Reason

    The shorter wavelength is less comparable with the fixed crystal spacing.

    Working

    higher v → smaller λ → less spread

Common misconception 3

Electron wavelength from accelerating voltage (non-relativistic)

Find and correct the mistake

Learner claim

An electron is accelerated through 150 V. A learner writes p = eV and substitutes it into λ = h/p. Diagnose the error and estimate the wavelength non-relativistically. Take h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg.

Try this before viewing the solution

View solution step by step

  1. Diagnose the units

    Method

    eV is kinetic energy, not momentum.

    Reason

    Charge times potential difference has unit joules.

    Working

    K = eV = (1.60 × 10⁻¹⁹)(150) = 2.40 × 10⁻¹⁷ J

  2. Convert energy to momentum

    Method

    p = 6.61 × 10⁻²⁴ kg m s⁻¹.

    Reason

    For a non-relativistic electron, K = p²/(2mₑ).

    Working

    p = square root of 2mₑK = square root of ((1.822 × 10⁻³⁰)(2.40 × 10⁻¹⁷)) = 6.61 × 10⁻²⁴ kg m s⁻¹

  3. Find wavelength

    Method

    λ = 1.00 × 10⁻¹⁰ m.

    Reason

    Only after finding momentum can the de Broglie relation be applied.

    Working

    λ = (6.63 × 10⁻³⁴)/(6.61 × 10⁻²⁴) = 1.00 × 10⁻¹⁰ m

Examiner practice 4

When does diffraction become significant?

2 marks

Examination question

Electrons have de Broglie wavelength λ = 2.0 × 10⁻¹⁰ m. Would you expect significant diffraction from a crystal with atomic spacing d = 2.5 × 10⁻¹⁰ m? Justify. [2 marks]

Try this before viewing the solution

View solution step by step

  1. Compare the scales

    1 mark

    Method

    d and λ are comparable.

    Reason

    Both are of order 10⁻¹⁰ m and differ by only a factor of 1.25.

    Working

    d/λ = 2.5/2.0 = 1.25

  2. Infer diffraction

    1 mark

    Method

    Significant diffraction is expected.

    Reason

    Wave diffraction is appreciable when wavelength is comparable to the relevant spacing.

    Working

    λ∼ d → significant diffraction

Challenge 5

What happens if you measure “which slit”?

Minimal support

Independent transfer

In a double-slit experiment with electrons, a detector is added that determines which slit each electron passes through. Predict the accumulated detection pattern and explain the change using probability amplitudes.

Try this before viewing the solution

Hints

Hint 1: compare available coherence
Ask whether the two path amplitudes can still contribute a two-path interference term.
View solution step by step

  1. Predict the observation

    Method

    The two-path interference fringes disappear, leaving a non-interference distribution formed from the slit alternatives.

    Reason

    Which-path information distinguishes the alternatives that previously interfered.

    Working

    with path information: no two-path fringes

  2. Connect to amplitudes

    Method

    The coherent two-path cross term no longer contributes to the observed distribution.

    Reason

    Without which-path information, amplitudes superpose before |ψ|² is formed; the path measurement removes that two-path interference.

    Working

    |ψ₁ + ψ₂|² → |ψ₁|² + |ψ₂|²

7. Mind Stretchers

Mind stretcher 1: Why doesn’t “two slits” mean two electrons?Extension

Explain why two paths can still lead to one detected electron.

Show Answer

The wavefunction can have contributions from both paths (superposition), but the particle is detected as a single localised event because measurement yields one outcome. Over many electrons, the distribution of many single detections matches |ψ|², which includes the interference term from the two-path superposition.

Mind stretcher 2: Do you need many electrons at once to get interference?Extension

You reduce the beam so that only one electron is in the apparatus at a time. Would you still expect an interference pattern after a long time? Explain.

Show Answer

Yes. Each detection is a single localised event, but after many events the distribution builds up to match |ψ|² for the two-slit setup, which includes the interference term.

The pattern does not require electrons to interact with each other; it is a property of the probability amplitudes for each electron.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027