Wave-particle Duality
Specification: OCR A H556 | Sections: 4.5.1 & 4.5.3 | Teaching time: ~3 hours
- Explain what photons are and use E = hf and E = hc/λ to calculate photon energy
- Optional enrichment: describe the ultraviolet catastrophe and Planck's quantisation
- Describe the electron diffraction experiment through a thin polycrystalline graphite film
- Use the de Broglie equation λ = h/p and relate it to diffraction observations
- Explain what is meant by wave-particle duality for both light and matter
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:
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:
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.
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).
Optional historical extension · 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:
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.
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:
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 charged particle accelerated from rest through an accelerating potential-drop magnitude V, its kinetic-energy gain is |q|V. For electrons and protons |q| = e; for alpha particles |q| = 2e. Neutral particles cannot be accelerated by this electric potential difference. Combining KE = ½mv² with p = mv and λ = h/p:
This form is extremely useful for electron diffraction calculations, where electrons are accelerated through a known potential difference.
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
- Electron gun: A heated filament emits electrons, which are accelerated through a high potential difference (typically 2000–5000 V)
- 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)
- 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. In this optional crystal-plane model, the Bragg condition is:
Here d is the spacing of the reflecting crystal planes, θ is the Bragg angle between the beam and those planes, and n is the diffraction order. The beam deflection angle measured from the original direction is α = 2θ.
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)
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.
Electron Diffraction Simulator
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 behaviour | Particle-like behaviour | |
|---|---|---|
| Light (EM radiation) | Diffraction, interference, polarisation | Photoelectric effect (Compton scattering is a later medical-imaging topic in 6.5.1(c)) |
| Electrons (matter) | Electron diffraction, interference | Cathode 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
For OCR 4.5.3, explain electron diffraction through polycrystalline graphite as wave evidence and connect momentum to wavelength with λ = h/p. The photoelectric effect (4.5.2) supplies particle evidence for light. Compton scattering and neutron diffraction are optional here; do not assume a named exam question without checking its context.
Why Don't We Notice Wave Behaviour of Everyday Objects?
For macroscopic objects, the de Broglie wavelength is far too small to detect:
Wave-particle Duality Explorer
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
- A photon is a quantum of electromagnetic radiation / a discrete packet of electromagnetic energy ✓
- 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 ✓
- 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 ✓
- 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
(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]
(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 ✓
(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. For this illustrative calculation, assume a crystalline target with plane 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]
(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 wavelength and crystal spacing determine a nonzero diffraction angle; use the supplied geometry to predict whether its ring is measurable ✓
- When the wavelength is similar to the size of the gaps between atoms, diffraction occurs ✓
(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]
(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 ✓
(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]
(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 ✓
- λe/λp = √(mp/me) = √(1.67 × 10⁻²⁷ / 9.11 × 10⁻³¹) ✓
- λe/λp = √(1833) = 42.8 ✓ (the electron's wavelength is about 43 times larger)
(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]
(a) [3]:
- λ = h/√(2meV), so λ ∝ 1/√V, therefore λ² ∝ 1/V ✓
- For small-angle diffraction: D = 2L tan α ≈ 2Lα, where α = 2θ, where L is the distance to the screen ✓
- Since 2d sin θ ≈ 2dθ = λ; hence dα ≈ 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 the geometrical factor 4L²/d²; explicitly gradient = 2L²h²/(me d²) ✓
- A straight line through the origin confirms the relationship ✓
• the behaviour of light
• the behaviour of electrons
• the significance of the de Broglie equation
[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 ✓
| Voltage (kV) | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 |
|---|---|---|---|---|---|
| Ring diameter D (mm) | 48 | 39 | 34 | 30 | 28 |
(a) Explain why D decreases as V increases. [3 marks]
(b) The student plots a graph of D² against 1/V. Describe the expected approximate shape in the small-angle limit and justify it. [3 marks]
(c) Use the data to calculate a value for the plane spacing d in this illustrative crystal target. [4 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]:
- Approximately a straight line through the origin in the small-angle limit ✓
- λ = h/√(2meV), so λ ∝ 1/√V, giving λ² ∝ 1/V ✓
- In the small-angle limit D ≈ 2Lλ/d, so D ∝ λ; D² ∝ λ² ∝ 1/V, producing a straight line ✓
(c) [4]: Use the 4.0 kV datum with the declared geometry.
- 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 ✓
- At V = 4.0 kV, use D = 34 mm and L = 0.132 m: α = arctan(D/(2L)) = 0.1281 rad and θ = α/2 = 0.06404 rad ✓
- d = λ/(2 sin θ) = (1.9415 × 10⁻¹¹)/(2 sin 0.06404) = 1.52 × 10⁻¹⁰ m. In the small-angle limit, d ≈ 2λL/D; the factor of two in α = 2θ must be retained ✓
(b) An alpha particle and a proton are both accelerated from rest through the same potential difference of 1000 V. Calculate the ratio λproton/λalpha. [3 marks]
(c) Suggest why diffraction of alpha particles is not commonly used to investigate crystal structures. [2 marks]
(a) [3]:
- KE gained = |q|V for a fixed charged species, so KE ∝ V ✓
- Momentum p = √(2mKE), so p ∝ √(KE) ∝ √V ✓
- Since λ = h/p, and p ∝ √V, therefore λ ∝ 1/√V ✓
(b) [3]:
- Proton gains KEₚ = eV; alpha gains KEα = 2eV. Therefore λ ∝ 1/√(m|q|) at the same voltage ✓
- mα = 4mp (alpha has 2 protons + 2 neutrons ≈ 4 × proton mass) ✓
- λp/λα = √((mα × 2e)/(mp × e)) = √8 = 2.83 ✓ (the proton’s wavelength is approximately 2.83 times the alpha particle’s)
(c) [2]:
- The de Broglie wavelength of alpha particles is much shorter (≈ 0.354 times 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.