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Wave Particle Duality

Key idea: Connect photon and wave models: E = hf, p = E/c = h/λ, and de Broglie wavelength λ = h/p for matter waves (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)

A. Wave–particle duality

Wave–particle duality means that the same physical entity can show:

  • wave behaviour (e.g. interference, diffraction),
  • particle behaviour (e.g. localised detection events).

B. Photon energy and momentum

  • photon energy: E = hf
  • photon momentum: p = E/c = h/λ

C. de Broglie wavelength (matter waves)

For a particle with momentum p:

λ = h/p

2. Key Ideas (What Earns Marks)

  • Photoelectric effect supports the particle nature of light (threshold frequency).
  • Interference/diffraction supports the wave nature of light.
  • Electron diffraction and single-particle interference support the wave nature of particles.
  • In many questions, you use:
    • E = hf for photon energy,
    • p = E/c or p = h/λ for photon momentum,
    • λ = h/p for matter waves.
Syllabus link (9478)

Quantum Physics learning outcomes 19a–19e cover the evidence + the key equations above.

3. Detailed Explanations

A. Photon momentum relationship

Using f = c/λ in E = hf:

E = hf = hc/λ; E/c = h/λ

So photon momentum can be written as: p = E/c = h/λ

B. Why “cannot reveal both aspects at once” is a useful sentence

Some experiments are designed to reveal wave behaviour (interference patterns), while others reveal particle behaviour (localised detection and energy transfer events).

In exam answers, you can say:

  • “The experiment reveals wave behaviour (diffraction/interference).”
  • “Detection happens as discrete events (particle-like).”

4. Common Mistakes

  • Mixing up photon wavelength with de Broglie wavelength (the formula looks the same, but the momentum is different).
  • Using p = h/λ without checking units (use SI: m, kg, s).

5. Exam Tips

  • If asked for evidence, name the phenomenon and the conclusion:
    • “threshold frequency → photons”,
    • “electron diffraction → matter waves”.
  • If asked for a calculation, write the equation first, then substitute with units.

6. Worked Examples

Modelled example 1

Photon momentum

Core

Problem

A photon has wavelength λ = 500 nm. Find its momentum. Take h = 6.63 × 10⁻³⁴ J s.
Study the worked solution

  1. Convert wavelength

    Method

    500 nm = 5.00 × 10⁻⁷ m.

    Reason

    The momentum relation requires SI wavelength.

    Working

    500 × 10⁻⁹ m = 5.00 × 10⁻⁷ m

  2. Apply photon momentum

    Method

    p = 1.33 × 10⁻²⁷ kg m s⁻¹.

    Reason

    A photon has momentum h/λ despite having zero rest mass.

    Working

    p = h/λ = (6.63 × 10⁻³⁴)/(500 × 10⁻⁹) = 1.33 × 10⁻²⁷ kg m s⁻¹

Guided practice 2

de Broglie wavelength of a particle

About 4 min

Problem

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

Try this before viewing the solution

Unit: m

Hints

Hint 1: use inverse momentum
Substitute directly into λ = h/p; the electron mass is not needed because momentum is already given.
View solution step by step

  1. Select the matter-wave relation

    Method

    λ = h/p.

    Reason

    de Broglie wavelength is determined by particle momentum.

    Working

    λ = (6.63 × 10⁻³⁴)/(3.0 × 10⁻²⁴)

  2. Evaluate

    Method

    λ = 2.21 × 10⁻¹⁰ m.

    Reason

    This is comparable to atomic spacings, so electron diffraction is feasible.

    Working

    λ = 2.21 × 10⁻¹⁰ m

Common misconception 3

Can a finite-momentum object have zero wavelength?

Find and correct the mistake

Learner claim

A learner claims that a moving elephant has de Broglie wavelength exactly zero because its wavelength is too small to observe. Diagnose the claim. Contrast this with the p → 0 limit.

Try this before viewing the solution

Correct conclusion

View solution step by step

  1. Test finite momentum

    Method

    Apply λ = h/p with finite p.

    Reason

    Since h is positive and non-zero, finite non-zero momentum gives a positive, non-zero wavelength.

    Working

    An exactly zero wavelength would require p → ∞, not merely a large everyday momentum.

  2. Interpret the stationary limit

    Method

    Let momentum approach zero.

    Reason

    The wavelength grows without bound as p → 0; the formula has no finite value at p = 0.

    Working

    p → 0 ⇒ λ = h/p → ∞

Examiner practice 4

Electron wavelength from accelerating voltage

5 marks

Examination question

An electron starts from rest and is accelerated through a potential difference of 150 V. Calculate its de Broglie wavelength. Use h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg and e = 1.60 × 10⁻¹⁹ C. State why a non-relativistic model is suitable. [5 marks]

Try this before viewing the solution

View solution step by step

  1. Connect electrical and kinetic energy

    1 mark

    Method

    Set transferred electrical energy equal to kinetic energy.

    Reason

    The electron starts from rest and the model neglects other transfers.

    Working

    eV = p²/2mₑ

  2. Express momentum

    1 mark

    Method

    Rearrange for p.

    Reason

    de Broglie wavelength is determined by momentum.

    Working

    p = square root of (2mₑ eV)

  3. Substitute into the de Broglie relation

    1 mark

    Method

    Combine λ = h/p with the momentum expression.

    Reason

    This avoids rounding momentum before finding wavelength.

    Working

    λ = h/(square root of (2mₑ eV))

  4. Calculate the wavelength

    1 mark

    Method

    Obtain an atomic-scale wavelength.

    Reason

    All values are in SI units.

    Working

    λ = (6.63 × 10⁻³⁴)/(square root of (2(9.11 × 10⁻³¹)(1.60 × 10⁻¹⁹)(150))) = 1.00 × 10⁻¹⁰ m

  5. Check the model

    1 mark

    Method

    Compare the kinetic energy with electron rest energy.

    Reason

    150 eV is far below 511 keV, so relativistic corrections are negligible at this precision.

    Working

    150 eV≪511 keV.

Challenge 5

Speed from de Broglie wavelength

Minimal support

Independent transfer

An electron has de Broglie wavelength λ = 1.0 × 10⁻¹⁰ m. Estimate its speed, treating it as non-relativistic. Take mₑ = 9.11 × 10⁻³¹ kg and h = 6.63 × 10⁻³⁴ J s.

Try this before viewing the solution

Hints

Hint 1: reverse the chain
Infer momentum from wavelength first, then use p = mₑv.
View solution step by step

  1. Infer momentum

    Method

    p = 6.63 × 10⁻²⁴ kg m s⁻¹.

    Reason

    Rearrange the de Broglie relation as p = h/λ.

    Working

    p = (6.63 × 10⁻³⁴)/(1.0 × 10⁻¹⁰) = 6.63 × 10⁻²⁴ kg m s⁻¹

  2. Infer speed

    Method

    v = 7.3 × 10⁶ m s⁻¹.

    Reason

    For the stated non-relativistic model, v = p/mₑ.

    Working

    v = (6.63 × 10⁻²⁴)/(9.11 × 10⁻³¹) = 7.28 × 10⁶ m s⁻¹

7. Mind Stretchers

Mind stretcher 1: When is diffraction more obvious?Extension

Explain why slower particles tend to show more obvious diffraction effects.

Show Answer

Slower particles have smaller momentum p.

Since λ = h/p, smaller p gives larger wavelength, and diffraction becomes more significant when the wavelength is comparable to the aperture/spacing.

Mind stretcher 2: Why don’t we see wave behaviour for everyday objects?Extension

Explain why wave effects are not obvious for a 0.10 kg ball moving at 10 m s⁻¹.

Show Answer

Its momentum is p = mv = 0.10 × 10 = 1.0 kg m s⁻¹, so: λ = h/p ≈ (6.63 × 10⁻³⁴)/1.0 = 6.6 × 10⁻³⁴ m

This wavelength is unimaginably smaller than any slit/atomic spacing, so diffraction/interference effects are negligible and unobservable.

8. Optional (Enrichment)

A. “Complementarity” wording

Some texts call the “can’t reveal both aspects in one setup” idea complementarity. You do not need that word for most exam questions; describing the observation and conclusion is enough.

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