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Wave–Particle Nature and Matter Waves

Key idea: Explain electron-diffraction evidence and apply de Broglie wavelength using momentum, kinetic energy, accelerating potential, and relativistic corrections.

  • Advanced Physics
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Learning objectives

  • Extend wave-particle reasoning through de Broglie wavelength, wave packets, and matter-wave evidence.

Matter waves are not ripples of material in the classical sense. A quantum state propagates and produces diffraction or interference, while each particle is detected in one localised event. The de Broglie relation connects the wavelength governing that wave-like behaviour to the particle’s momentum.

Optional extension

This lesson extends the quantum content inherited from H2 Physics. It is useful preparation for university quantum mechanics, but it is not an additional content topic in the 2027 H3 Physics 9814 syllabus.

1. Definitions and model

  • Wave–particle behaviour: quantum objects can produce wave phenomena such as diffraction and interference, yet transfer energy and momentum in localised detection events.

  • de Broglie wavelength:

    λ = h/p

    where λ is in metres, h = 6.63 × 10⁻³⁴ J s, and p is the magnitude of momentum in kg m s⁻¹.

  • Non-relativistic regime: v≪ c, or equivalently K≪ mc², so p = mv and K = p²/(2m) are suitable approximations.

  • Relativistic regime: use the total-energy relation

    (pc)² = E²-(mc²)²

    rather than p = mv.

2. What the evidence shows

A collimated electron beam passes through thin polycrystalline graphite. Randomly oriented crystallites diffract electrons into cones that meet the fluorescent screen as concentric rings around the undeflected central spot.
Each electron produces one localised screen hit. After many electrons, randomly oriented graphite crystallites produce stable concentric diffraction rings.

When a beam of electrons passes through thin polycrystalline graphite, the regularly spaced atomic planes act as diffraction structures. If the electron wavelength is comparable with the plane spacing, constructive interference occurs at particular angles. Because the small crystallites have many orientations, the diffracted directions form cones; those cones intersect the fluorescent screen as rings.

The two observations must be stated together:

  • each electron arrives as a localised detection event;
  • many detections build a reproducible diffraction pattern.

Diffraction is evidence for wave-like propagation. Localised arrival is particle-like. The pattern does not require electrons to collide with one another: it can build when only one electron is inside the apparatus at a time.

A lower-momentum particle paired with a longer wave and a higher-momentum particle paired with a shorter wave
The de Broglie wavelength is inversely proportional to momentum. Greater momentum means more closely spaced phase variation and a shorter wavelength.

3. Choosing the momentum model

A. Given speed at low energy

For v≪ c,

p = mv

and therefore

λ = h/mv

B. Given kinetic energy at low energy

Starting from K = p²/(2m),

p = square root of 2mK

so

λ = h/(square root of 2mK)

C. Accelerated from rest through a potential difference

A particle of charge magnitude |q| gains kinetic energy

K = |q|V

if other energy transfers are negligible. In the non-relativistic regime,

λ = h/(square root of 2m|q|V)

For an electron, use |q| = e; the kinetic-energy gain is positive even though the electron’s charge is negative.

D. Relativistic correction

Write total energy as E = K + mc². Substitution into the energy–momentum relation gives

pc = square root of (K(K + 2mc²))

and hence

λ = hc/(square root of (K(K + 2mc²)))

This expression approaches the non-relativistic result when K≪ mc².

4. Common mistakes

  • Saying the particle follows a visible sinusoidal path. The drawn wave represents phase or probability-amplitude behaviour, not a classical trajectory.
  • Using p = mv without checking whether K is small compared with mc².
  • Substituting electronvolts into an SI formula without converting units, unless every energy term is kept consistently in electronvolts.
  • Concluding that large momentum produces strong diffraction. Large momentum produces a shorter wavelength, so a smaller lattice spacing is needed for comparable diffraction.
  • Calling the ring brightness a path taken by one electron. It represents the distribution of many localised detections.

5. Exam and data skills

  • For evidence questions, name the observation and inference: “electrons produce diffraction, a wave phenomenon; therefore their propagation has wave-like behaviour.”
  • For calculations, state the regime before selecting momentum: “Since K≪ mc², use the non-relativistic expression.”
  • Compare λ with the relevant spacing d. Diffraction becomes prominent when they are of similar order.
  • Use limits as checks: increasing V increases momentum and decreases wavelength; as K/(mc²) → 0, the relativistic expression must recover the classical one.

6. Worked examples

Modelled example 1

Electron wavelength from speed

Core

Problem

An electron moves at 2.0 × 10⁶ m s⁻¹. Calculate its de Broglie wavelength. Take mₑ = 9.11 × 10⁻³¹ kg.
Study the worked solution

  1. Choose the momentum model

    Method

    Use p = mₑv.

    Reason

    The speed is much smaller than c, so the non-relativistic model is suitable.

    Working

    v/c ≈ 0.0067≪1

  2. Calculate momentum

    Method

    p = 1.82 × 10⁻²⁴ kg m s⁻¹.

    Reason

    Momentum is mass multiplied by speed in this regime.

    Working

    p = (9.11 × 10⁻³¹)(2.0 × 10⁶) = 1.82 × 10⁻²⁴

  3. Find wavelength

    Method

    λ = 3.6 × 10⁻¹⁰ m.

    Reason

    The de Broglie wavelength is h/p.

    Working

    λ = (6.63 × 10⁻³⁴)/(1.82 × 10⁻²⁴) = 3.6 × 10⁻¹⁰ m

Guided practice 2

Electron accelerated through 150 V

About 6 min

Problem

An electron is accelerated from rest through 150 V. Find its wavelength using the non-relativistic model. Take e = 1.60 × 10⁻¹⁹ C.

Try this before viewing the solution

Hints

Hint 1: convert voltage to energy
The positive kinetic-energy gain is K = eV using the charge magnitude.
Hint 2: link energy to momentum
Use p = square root of 2mₑK before applying λ = h/p.
View solution step by step

  1. Find kinetic energy

    Method

    K = 2.40 × 10⁻¹⁷ J.

    Reason

    An electron accelerated from rest gains energy eV.

    Working

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

  2. Find momentum

    Method

    p = 6.61 × 10⁻²⁴ kg m s⁻¹.

    Reason

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

    Working

    p = square root of 2mₑK = 6.61 × 10⁻²⁴

  3. Find wavelength and check regime

    Method

    λ = 1.00 × 10⁻¹⁰ m.

    Reason

    K/(mₑc²) ≈ 2.9 × 10⁻⁴, so the approximation is well justified.

    Working

    λ = h/p = 1.00 × 10⁻¹⁰ m

Common misconception 3

Relativistic correction at 100 kV

Find and correct the mistake

Learner claim

An electron is accelerated from rest through 100 kV. A learner assumes p = square root of 2mₑK without checking the regime. Use mₑc² = 511 keV and hc = 1.240 keV nm to calculate the relativistic wavelength and assess the approximation.

Try this before viewing the solution

Appropriate momentum model

View solution step by step

  1. Set the kinetic energy

    Method

    K = 100 keV.

    Reason

    An electron crossing 100 kV gains 100 keV.

    Working

    K = eV = 100 keV

  2. Use relativistic momentum

    Method

    pc = 335 keV.

    Reason

    K/(mₑc²) ≈ 0.196 is not negligible.

    Working

    pc = square root of (K(K + 2mₑc²)) = square root of (100(100 + 2 × 511)) = 335 keV

  3. Calculate wavelength

    Method

    λ = 3.70 pm.

    Reason

    λ = hc/(pc) with consistent keV units.

    Working

    λ = 1.240/335 nm = 3.70 × 10⁻³ nm = 3.70 pm

  4. Assess the shortcut

    Method

    The non-relativistic 3.88 pm result is about 4.9% too large.

    Reason

    The classical model underestimates momentum at this energy.

    Working

    percentage error ≈ 4.9%

Examiner practice 4

Electron and proton at the same kinetic energy

3 marks

Examination question

An electron and a proton have the same non-relativistic kinetic energy. State which has the longer de Broglie wavelength and find λₑ/λₚ. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Express mass dependence

    1 mark

    Method

    λ ∝ 1/square root of m at fixed K.

    Reason

    p = square root of 2mK and λ = h/p.

    Working

    λ = h/(square root of 2mK)

  2. Compare particles

    1 mark

    Method

    The electron has the longer wavelength.

    Reason

    Its mass is smaller.

    Working

    mₑ < mₚ ⇒ λₑ > λₚ

  3. Find the ratio

    1 mark

    Method

    λₑ/λₚ ≈ 42.8.

    Reason

    The common kinetic energy cancels.

    Working

    λₑ/λₚ = square root of (mₚ/mₑ) ≈ 42.8

Challenge 5

Deciding whether diffraction is plausible

Minimal support

Independent transfer

Electrons have wavelength 2.0 × 10⁻¹⁰ m and encounter crystal-plane spacing 2.5 × 10⁻¹⁰ m. Decide whether significant diffraction is plausible and justify the inference from the two scales.

Try this before viewing the solution

Hints

Hint 1: compare orders of magnitude
Diffraction is prominent when wavelength and structure spacing are comparable.
View solution step by step

  1. Compare scales

    Method

    The wavelength and plane spacing are both of order 10⁻¹⁰ m.

    Reason

    Their ratio is 2.0/2.5 = 0.80, so neither scale is negligible relative to the other.

    Working

    λ/d = 0.80

  2. Infer the wave effect

    Method

    Significant diffraction is plausible.

    Reason

    Path differences from neighbouring planes can be comparable with a wavelength, allowing constructive and destructive interference.

    Working

    λ∼ d ⇒ measurable diffraction structure

7. Mind stretchers

Mind stretcher 1: One particle at a timeExtension

Why does reducing the beam intensity until only one electron is present not eliminate the eventual diffraction pattern?

Answer

The pattern is not created by forces between different electrons. Each electron’s quantum state evolves through the apparatus and gives a probability distribution for its detection position. One trial produces one point; many independent trials sample the same distribution and reveal the diffraction pattern.

Mind stretcher 2: Why a moving ball does not visibly diffractExtension

A 0.10 kg ball moves at 10 m s⁻¹. Estimate its wavelength and explain the classical limit.

Answer

Its momentum is 1.0 kg m s⁻¹, so

λ = h/p = 6.6 × 10⁻³⁴ m

No practical aperture or environmental isolation can resolve interference on this scale. For macroscopic objects, the de Broglie wavelength is therefore far too small for wave effects to be observable in ordinary conditions.

8. Optional extension: crystal planes

For scattering from parallel crystal planes separated by d, constructive interference can be described by Bragg’s condition,

nλ = 2d sin θ

where θ is the glancing angle to the planes. This connects a measured diffraction angle with the particle wavelength, but always check how an experimental diagram defines its angle.

Next: return to the Matter Waves Hub, or continue to Quantum Theory of Light.

Continue with the next resource in this course.

Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics