Wave-particle Duality — OCR A Level Physics Revision
Module 4 · Electrons, Waves and Photons

Wave-particle Duality

Specification: OCR A H556  |  Sections: 4.5.1 & 4.5.3  |  Teaching time: ~3 hours

By the end of this topic you should be able to…

What Are Photons?

At the start of the 20th century, physicists assumed that electromagnetic radiation was exclusively a wave phenomenon. This model worked well for many phenomena — diffraction, interference, refraction — but it failed catastrophically when applied to the interaction between light and matter.

The solution came from a radical idea: electromagnetic radiation can also behave as discrete packets of energy. These packets are called photons. Each photon carries a fixed amount of energy that depends only on the frequency of the radiation:

Energy of a Photon E = hf   or   E = hc/λ

Where:

  • h = 6.63 × 10⁻³⁴ J s — Planck's constant
  • f — frequency of the radiation (Hz)
  • c = 3.00 × 10⁸ m s⁻¹ — speed of light
  • λ — wavelength (m)

A photon is a quantum of electromagnetic radiation — the smallest possible unit of energy at a given frequency. You cannot have "half a photon" at a particular frequency; the energy comes in indivisible packets.

The Electronvolt

When working with photon energies, joules are often inconveniently small. The electronvolt (eV) is a more practical unit:

Unit Conversion 1 eV = 1.60 × 10⁻¹⁹ J

It is defined as the kinetic energy gained by an electron accelerated through a potential difference of 1 volt: E = eV = (1.60 × 10⁻¹⁹ C)(1 V) = 1.60 × 10⁻¹⁹ J.

📋 Exam Tip

Be fluent converting between eV and joules. A common error is forgetting to multiply by 1.60 × 10⁻¹⁹. If you're given energy in eV and need to use E = hf, you must convert to joules first (since h is in J s).

Worked Example 1
Calculate the energy (in eV) of a photon of violet light with wavelength 400 nm.
E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (400 × 10⁻⁹) ✓
E = 4.97 × 10⁻¹⁹ J ✓
Convert: E = (4.97 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹) = 3.11 eV

The Ultraviolet Catastrophe

Before the photon model, physicists assumed light was purely a wave. This worked for most phenomena, but when they tried to predict how much energy a hot object radiates (black-body radiation), classical physics gave a nonsensical answer: it predicted infinite energy at short wavelengths. This failure was called the ultraviolet catastrophe.

Planck's Solution: Energy Comes in Packets

In 1900, Max Planck fixed the problem with a radical idea. He proposed that the energy of the oscillators in a hot object is quantised — it can only take certain allowed values, not any value at all. These allowed energies are integer multiples of a fundamental unit:

Planck's Energy Quantisation E = nhf   (n = 0, 1, 2, 3, …)

Where h is Planck's constant (6.63 × 10⁻³⁴ J s) and n is an integer. This means energy is delivered in discrete packets of size hf — the same photons we discussed in Part 1.

The key insight: at high frequencies, the energy gap hf becomes so large that oscillators cannot easily jump to higher energy levels. This naturally prevents the energy from diverging to infinity — exactly what classical physics failed to predict.

Black body radiation curves showing Planck's law at different temperatures compared with the classical prediction
Fig 1. Black-body radiation spectra at different temperatures (coloured curves) compared with the classical prediction (dashed). The classical theory wrongly predicts infinite energy at short wavelengths.
💡 Why This Matters

Planck introduced the idea that energy is quantised — it comes in packets, not a continuous flow. This was the birth of quantum physics. The photon model of light (covered in the photoelectric effect resource) builds directly on this idea.

The de Broglie Hypothesis

In 1924, a French physics student named Louis de Broglie proposed a bold idea that extended wave-particle duality beyond light. If waves (photons) can behave as particles, he argued, then perhaps particles can also behave as waves.

De Broglie suggested that any particle with momentum has an associated wavelength, given by:

de Broglie Equation λ = h / p = h / (mv)

Where:

  • λ — de Broglie wavelength (m)
  • h — Planck's constant (6.63 × 10⁻³⁴ J s)
  • p — momentum of the particle (kg m s⁻¹)
  • m — mass of the particle (kg)
  • v — velocity of the particle (m s⁻¹)

What Does This Mean?

The de Broglie wavelength tells us the "size" of the wave associated with a particle. For everyday objects, λ is unimaginably small — far too small to detect. But for subatomic particles like electrons, λ can be comparable to the spacing between atoms in a crystal, making diffraction observable.

Relating λ to Kinetic Energy

For a particle accelerated from rest through a potential difference V, its kinetic energy is eV. Combining KE = ½mv² with p = mv and λ = h/p:

de Broglie Wavelength (using KE) λ = h / √(2mKE) = h / √(2meV)

This form is extremely useful for electron diffraction calculations, where electrons are accelerated through a known potential difference.

Worked Example 2
An electron is accelerated from rest through a potential difference of 5000 V. Calculate its de Broglie wavelength. (me = 9.11 × 10⁻³¹ kg)
KE = eV = (1.60 × 10⁻¹⁹)(5000) = 8.00 × 10⁻¹⁶ J ✓
λ = h / √(2mKE) = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 8.00 × 10⁻¹⁶) ✓
λ = 6.63 × 10⁻³⁴ / √(1.46 × 10⁻⁴⁵) = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³ ✓
λ = 1.74 × 10⁻¹¹ m
📋 Exam Tip

Notice the result: λ ≈ 1.7 × 10⁻¹¹ m. This is on the same order as the spacing between carbon atoms in graphite (about 10⁻¹⁰ m). This is exactly why electron diffraction through graphite works — the de Broglie wavelength is comparable to the atomic spacing.

Electron Diffraction Through Graphite

De Broglie's hypothesis needed experimental confirmation. In 1927, Davisson and Germer demonstrated diffraction of electrons at Bell Labs. A simpler and more common school/college demonstration involves firing electrons through a thin film of polycrystalline graphite.

The Apparatus

  1. Electron gun: A heated filament emits electrons, which are accelerated through a high potential difference (typically 2000–5000 V)
  2. Graphite target: A very thin film of polycrystalline graphite. The carbon atoms are arranged in a hexagonal lattice with spacing of approximately d ≈ 10⁻¹⁰ m (about 0.1 nm)
  3. Fluorescent screen: The diffracted electrons strike a phosphor screen, producing visible rings

What Is Observed

When the electron beam passes through the graphite film, a pattern of concentric rings is observed on the fluorescent screen. This is a diffraction pattern — clear evidence that electrons behave as waves.

Diffraction is a wave property. Particles, in the classical sense, cannot diffract. The fact that electrons produce a diffraction pattern proves they have an associated wavelength — the de Broglie wavelength.

Why Polycrystalline Graphite?

The graphite film consists of many tiny crystals oriented at random angles. Because the crystal orientations are random, the diffraction condition is satisfied for all angles simultaneously, producing concentric rings rather than discrete spots (which you would get with a single crystal).

How the Rings Relate to Atomic Spacing

The diameter of the diffraction rings on the screen depends on the de Broglie wavelength of the electrons and the spacing between the carbon atoms. For the first-order diffraction maximum, the condition is approximately:

Diffraction Condition d sin θ ≈ nλ   (for small angles: dθ ≈ nλ)

Where d is the atomic spacing, θ is the diffraction angle, and n is the order of diffraction.

Effect of Changing the Accelerating Voltage

If the accelerating voltage is increased:

  • Electrons gain more kinetic energy → higher momentum → shorter de Broglie wavelength
  • The diffraction rings become smaller (closer together) because the wavelength is smaller relative to the atomic spacing

If the accelerating voltage is decreased:

  • Electrons have less kinetic energy → lower momentum → longer de Broglie wavelength
  • The diffraction rings become larger (further apart)
💡 Key Point

Electron diffraction provides evidence for the wave-like behaviour of particles. The photoelectric effect provides evidence for the particle-like behaviour of electromagnetic waves. Together, they demonstrate wave-particle duality.

Worked Example 3
Electrons are accelerated through a potential difference of 3000 V and directed at a thin graphite film. The spacing between carbon atoms is approximately 1.5 × 10⁻¹⁰ m. Calculate the de Broglie wavelength of the electrons and determine whether diffraction is likely to be observed.
KE = eV = (1.60 × 10⁻¹⁹)(3000) = 4.80 × 10⁻¹⁶ J ✓
λ = h / √(2mKE) = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 4.80 × 10⁻¹⁶) ✓
λ = 6.63 × 10⁻³⁴ / √(8.75 × 10⁻⁴⁶) = 6.63 × 10⁻³⁴ / 2.96 × 10⁻²³ ✓
λ = 2.24 × 10⁻¹¹ m
Comparison: λ/d = 2.24 × 10⁻¹¹ / 1.5 × 10⁻¹⁰ = 0.15 ✓
Since λ is comparable to d (within an order of magnitude), diffraction will be observed

Electron Diffraction Simulator

Electron Diffraction — Graphite Film
4000 V
Accelerating voltage: 4000 V
de Broglie λ:
Ring diameter (1st):
d-spacing: 1.5 × 10⁻¹⁰ m

Adjust the accelerating voltage to see how the de Broglie wavelength changes and how this affects the diffraction pattern. Higher voltage → shorter wavelength → smaller rings. Lower voltage → longer wavelength → larger rings.

The Concept of Wave-particle Duality

Wave-particle duality is the idea that all entities that carry energy and momentum (both matter and radiation) exhibit both wave-like and particle-like properties, depending on how they are observed.

Wave-like behaviourParticle-like behaviour
Light (EM radiation)Diffraction, interference, polarisationPhotoelectric effect, Compton scattering
Electrons (matter)Electron diffraction, interferenceCathode rays, deflection in E/B fields

The key insight is that both models are needed. Neither the wave model nor the particle model alone can explain all observations. The model you use depends on the experiment you are describing.

When Do We Use Each Model?

  • Wave model for light: when studying propagation (diffraction, interference, refraction, polarisation)
  • Particle model for light: when studying interactions with matter (photoelectric effect, emission/absorption spectra)
  • Particle model for electrons: when studying motion in fields, electric current, collisions
  • Wave model for electrons: when studying diffraction, interference, atomic orbitals, electron microscopy
📋 Exam Tip

A common 2-mark question: "State two pieces of evidence for the wave nature of electrons." Answer: electron diffraction through graphite, and electron interference (e.g., double-slit experiment). For "State two pieces of evidence for the particle nature of electromagnetic radiation": the photoelectric effect, and Compton scattering. Know which is which.

Why Don't We Notice Wave Behaviour of Everyday Objects?

For macroscopic objects, the de Broglie wavelength is far too small to detect:

Worked Example 4
A person of mass 70 kg walks at 1.5 m s⁻¹. Calculate their de Broglie wavelength and comment on why diffraction is not observed.
p = mv = 70 × 1.5 = 105 kg m s⁻¹ ✓
λ = h/p = 6.63 × 10⁻³⁴ / 105 = 6.3 × 10⁻³⁶ m
This wavelength is ~10²⁶ times smaller than an atom — far too small to interact with any physical aperture or grating, so no diffraction can be observed ✓

Wave-particle Duality Explorer

de Broglie Wavelength Calculator

Enter a particle mass and velocity (or an accelerating voltage) to calculate its de Broglie wavelength. Compare different particles using the presets. Notice how much larger the wavelength is for lighter particles — this is why electron diffraction is observable but diffraction of a person is not.

Knowledge Check

1
State what is meant by a photon.
1 mark
  • A photon is a quantum of electromagnetic radiation / a discrete packet of electromagnetic energy ✓
2
Explain what is meant by the ultraviolet catastrophe.
2 marks
  • The classical (Rayleigh-Jeans) theory predicted that the intensity of black-body radiation would increase without limit at short wavelengths (high frequencies) ✓
  • This predicted infinite total energy, which contradicted experimental observations where intensity peaks then falls ✓
3
Explain how the electron diffraction experiment through a thin graphite film provides evidence for wave-particle duality.
3 marks
  • Electrons are particles (they have mass and charge) and can be deflected by electric and magnetic fields ✓
  • When passed through graphite, a diffraction pattern (concentric rings) is observed on the screen ✓
  • Diffraction is a wave property; therefore electrons also behave as waves — demonstrating wave-particle duality ✓
4
In an electron diffraction experiment, the accelerating voltage is increased from 2000 V to 4000 V. State and explain what happens to the diameter of the diffraction rings.
3 marks
  • The rings become smaller / the ring diameter decreases ✓
  • Higher voltage → greater kinetic energy → greater momentum (p = √(2mKE)) ✓
  • Greater momentum → shorter de Broglie wavelength (λ = h/p) → smaller diffraction angle ✓

Exam-Style Questions

1
(a) Calculate the energy in eV of a photon of wavelength 590 nm (yellow light). [2 marks]

(b) A student suggests that a photon of wavelength 590 nm could be split into two photons each of wavelength 1180 nm. Use a calculation to justify whether this is possible. [3 marks]
5 marks

(a) [2]:

  • E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (590 × 10⁻⁹) = 3.37 × 10⁻¹⁹ J ✓
  • E = 3.37 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.11 eV

(b) [3]:

  • Energy of one 1180 nm photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (1180 × 10⁻⁹) = 1.69 × 10⁻¹⁹ J ✓
  • Two photons: 2 × 1.69 × 10⁻¹⁹ = 3.37 × 10⁻¹⁹ J ✓
  • This equals the energy of the original photon (3.37 × 10⁻¹⁹ J), so energy is conserved — yes, this is possible
2
(a) State the de Broglie equation. [1 mark]

(b) Explain why electron diffraction provides evidence for the wave nature of particles. [2 marks]

(c) Electrons are accelerated through a potential difference of 4000 V in a diffraction tube. The graphite target has atomic spacing d = 1.5 × 10⁻¹⁰ m.

    (i) Show that the de Broglie wavelength of the electrons is about 1.9 × 10⁻¹¹ m. [2 marks]

    (ii) Explain why a diffraction pattern is observed in this experiment. [2 marks]
7 marks

(a) [1]:

  • λ = h/p or λ = h/(mv) ✓

(b) [2]:

  • Diffraction is a wave phenomenon ✓
  • Electrons (which are particles with mass and charge) produce a diffraction pattern when passing through graphite, demonstrating they also behave as waves ✓

(c)(i) [2]:

  • KE = eV = (1.60 × 10⁻¹⁹)(4000) = 6.4 × 10⁻¹⁶ J ✓
  • λ = h/√(2mKE) = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 6.4 × 10⁻¹⁶) = 6.63 × 10⁻³⁴ / 3.42 × 10⁻²³ ≈ 1.94 × 10⁻¹¹ m ✓ (≈ 1.9 × 10⁻¹¹ m)

(c)(ii) [2]:

  • The de Broglie wavelength (≈ 2 × 10⁻¹¹ m) is comparable to the atomic spacing (1.5 × 10⁻¹⁰ m) ✓
  • When the wavelength is similar to the size of the gaps between atoms, diffraction occurs ✓
3
A neutron beam is produced in a nuclear reactor. The neutrons have a kinetic energy of 0.025 eV.

(a) Show that the speed of the neutrons is approximately 2200 m s⁻¹. [2 marks]

(b) Calculate the de Broglie wavelength of these neutrons. (mn = 1.67 × 10⁻²⁷ kg) [2 marks]

(c) Neutron diffraction is used to study crystal structures. Suggest why neutrons are more useful than X-rays for studying materials containing light atoms such as hydrogen. [2 marks]
6 marks

(a) [2]:

  • KE = 0.025 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻²¹ J ✓
  • ½mv² = KE → v = √(2KE/m) = √(2 × 4.0 × 10⁻²¹ / 1.67 × 10⁻²⁷) = √(4.79 × 10⁶) ≈ 2190 m s⁻¹

(b) [2]:

  • λ = h/(mv) = 6.63 × 10⁻³⁴ / (1.67 × 10⁻²⁷ × 2190) ✓
  • λ = 6.63 × 10⁻³⁴ / 3.66 × 10⁻²⁴ = 1.81 × 10⁻¹⁰ m

(c) [2]: Accept either point:

  • X-rays interact with electrons; hydrogen has very few electrons, so X-rays are not strongly diffracted by hydrogen atoms ✓
  • Neutrons interact with nuclei (via the strong nuclear force) and are scattered by all nuclei including hydrogen ✓
4
(a) Explain what is meant by wave-particle duality. [2 marks]

(b) A proton and an electron are each accelerated from rest through the same potential difference. Determine the ratio of their de Broglie wavelengths: λelectron / λproton. [4 marks]
6 marks

(a) [2]:

  • All entities that transfer energy (both EM radiation and matter) can exhibit both wave-like and particle-like behaviour ✓
  • The behaviour observed depends on the type of experiment being performed ✓

(b) [4]:

  • Both gain the same KE = eV (same charge, same voltage) ✓
  • λ = h/√(2mKE), so λ ∝ 1/√m ✓
  • λep = √(mp/me) = √(1.67 × 10⁻²⁷ / 9.11 × 10⁻³¹) ✓
  • λep = √(1833) = 42.8 ✓ (the electron's wavelength is about 43 times larger)
5
In an electron diffraction experiment, the accelerating voltage V is varied and the diameter D of the first diffraction ring is measured. A graph of D² against 1/V is plotted.

(a) Show that D² is proportional to 1/V. [3 marks]

(b) State what the gradient of this graph represents, in terms of fundamental constants. [2 marks]
5 marks

(a) [3]:

  • λ = h/√(2meV), so λ ∝ 1/√V, therefore λ² ∝ 1/V ✓
  • For small-angle diffraction: D = 2L tan θ ≈ 2Lθ, where L is the distance to the screen ✓
  • Since d sin θ ≈ dθ = λ (for first order), θ ∝ λ ∝ 1/√V, so D ∝ 1/√V, giving D² ∝ 1/V ✓

(b) [2]:

  • The gradient is proportional to h²/(2me) multiplied by geometrical factors (L²/d²) ✓
  • A straight line through the origin confirms the relationship ✓
6
Discuss the evidence for wave-particle duality. In your answer, you should refer to:

• the behaviour of light
• the behaviour of electrons
• the significance of the de Broglie equation

[6 marks]
6 marks
  • Light exhibits wave properties: diffraction and interference are observed when light passes through slits or gratings ✓
  • Light also exhibits particle properties: the photoelectric effect shows light transfers energy in discrete packets (photons), with one-to-one photon-electron interactions ✓
  • Electrons exhibit particle properties: they have mass and charge, and can be deflected by electric and magnetic fields ✓
  • Electrons also exhibit wave properties: electron diffraction through a thin graphite film produces concentric ring patterns ✓
  • The de Broglie equation λ = h/p links the wave property (wavelength) with the particle property (momentum) through Planck's constant ✓
  • Both models are needed — wave-particle duality means the behaviour observed depends on the type of experiment performed ✓
7
A student investigates how the diameter of the first diffraction ring varies with accelerating voltage in an electron diffraction experiment. The results are shown in the table.

Voltage (kV)2.03.04.05.06.0
Ring diameter D (mm)4839343028

(a) Explain why D decreases as V increases. [3 marks]

(b) The student plots a graph of D² against 1/V. Describe the expected shape of this graph and explain why it is a straight line. [3 marks]

(c) Use the data to calculate a value for the atomic spacing d in the graphite. [4 marks]
10 marks

(a) [3]:

  • Higher voltage → electrons gain more kinetic energy → greater momentum ✓
  • Greater momentum → shorter de Broglie wavelength (λ = h/p) ✓
  • Shorter wavelength → smaller diffraction angle → smaller ring diameter ✓

(b) [3]:

  • The graph is a straight line passing through the origin ✓
  • λ = h/√(2meV), so λ ∝ 1/√V, giving λ² ∝ 1/V ✓
  • Since D ∝ λ (from diffraction geometry), D² ∝ λ² ∝ 1/V, producing a straight line ✓

(c) [4]: (accept a range around 1.0–1.8 × 10⁻¹⁰ m)

  • Using V = 4.0 kV: λ = h/√(2meV) = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 4000) ✓
  • λ = 6.63 × 10⁻³⁴ / 3.42 × 10⁻²³ = 1.94 × 10⁻¹¹ m ✓
  • For first-order diffraction: d sin θ = λ. Using small-angle geometry with D = 34 mm ✓
  • d ≈ λL/(D/2), which gives d ≈ 1.5 × 10⁻¹⁰ m (method and order of magnitude correct) ✓
8
(a) Explain why the de Broglie wavelength of a particle accelerated through a potential difference V is inversely proportional to the square root of V. [3 marks]

(b) An alpha particle and a proton are both accelerated from rest through the same potential difference of 1000 V. Calculate the ratio of their de Broglie wavelengths. [3 marks]

(c) Suggest why diffraction of alpha particles is not commonly used to investigate crystal structures. [2 marks]
8 marks

(a) [3]:

  • KE gained = eV, so KE ∝ V ✓
  • Momentum p = √(2mKE), so p ∝ √(KE) ∝ √V ✓
  • Since λ = h/p, and p ∝ √V, therefore λ ∝ 1/√V ✓

(b) [3]:

  • Same KE (same charge, same voltage) → λ ∝ 1/√m ✓
  • mα = 4mp (alpha has 2 protons + 2 neutrons ≈ 4 × proton mass) ✓
  • λpα = √(mα/mp) = √4 = 2 ✓ (the proton's wavelength is twice the alpha particle's)

(c) [2]:

  • The de Broglie wavelength of alpha particles is much shorter (≈ half that of a proton for the same voltage) ✓
  • This makes diffraction angles very small and difficult to measure; also alpha particles are easily absorbed by matter ✓

Topic Summary

Photons

Discrete packets of EM energy. Energy E = hf. The photon model explains interactions between light and matter that wave theory cannot.

Ultraviolet Catastrophe

Classical theory predicted infinite energy at short wavelengths. Planck resolved this by proposing energy is quantised in units of hf, introducing Planck's constant h.

de Broglie Wavelength

λ = h/p. Every particle with momentum has an associated wavelength. Observable for subatomic particles; negligible for macroscopic objects.

Electron Diffraction

Electrons through polycrystalline graphite produce concentric rings — evidence for wave behaviour of particles. Ring size depends on accelerating voltage.

Equations to Know

E = hf = hc/λ
E = nhf (n = 0, 1, 2, …)
λ = h/p = h/(mv)
λ = h/√(2mKE) = h/√(2meV)
1 eV = 1.60 × 10⁻¹⁹ J
h = 6.63 × 10⁻³⁴ J s