Wave-Particle Duality

By the early 20th century, physics faced a remarkable paradox. Light, long established as a wave (demonstrated by interference and diffraction), also behaved as a stream of particles (demonstrated by the photoelectric effect). Meanwhile, electrons, long established as particles, turned out to exhibit wave behaviour. This dual nature of matter and radiation — wave-particle duality — is one of the cornerstones of quantum physics.

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Light as a Wave

The wave nature of light is demonstrated by phenomena that can only be explained by waves:

• Interference: Young's double-slit experiment produces alternating bright and dark fringes — a hallmark of wave superposition
• Diffraction: Light spreads when passing through narrow slits or around obstacles
• Polarisation: Light can be polarised, which is only possible for transverse waves

These experiments conclusively show that light propagates as a wave, with properties such as wavelength, frequency, amplitude, and phase.

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Light as a Particle

The photoelectric effect (and other experiments such as the Compton effect) revealed that light also behaves as a stream of discrete particles — photons. Each photon carries energy:

$E = hf = \frac{hc}{\lambda}$E=hf=λhc​

and momentum:

$p = \frac{h}{\lambda} = \frac{E}{c}$p=λh​=cE​

The photon model explains:

• The existence of a threshold frequency in the photoelectric effect
• Instantaneous emission of photoelectrons
• The independence of maximum kinetic energy on light intensity

Light is not simply a wave or simply a particle. It exhibits both behaviours, depending on the experiment.

graph TD
    A["Observation / Experiment"] --> B{"What behaviour\nis observed?"}
    B -->|"Interference, diffraction,\npolarisation"| C["WAVE model\napplies"]
    B -->|"Photoelectric effect,\nCompton scattering,\nblackbody spectrum"| D["PARTICLE (photon)\nmodel applies"]
    C --> E["Described by:\nλ, f, amplitude, phase\nv = fλ"]
    D --> F["Described by:\nE = hf, p = h/λ\ninteracts with single electron"]
    E --> G["Wave-particle duality:\nboth descriptions are needed"]
    F --> G

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de Broglie's Hypothesis

In 1924, Louis de Broglie proposed a bold and elegant idea: if waves (light) can behave as particles, perhaps particles can behave as waves. He suggested that every moving particle has an associated wavelength — the de Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{h}{mv}$λ=ph​=mvh​

where:

• h = 6.63 × 10⁻³⁴ J s (Planck constant)
• p = momentum of the particle (kg m s⁻¹)
• m = mass of the particle (kg)
• v = speed of the particle (m s⁻¹)

This wavelength is associated with all moving objects, not just subatomic particles. However, for macroscopic objects the wavelength is unimaginably small — far too tiny to observe.

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Worked de Broglie Wavelength Calculations

Worked example 1: Calculate the de Broglie wavelength of an electron travelling at 2.0 × 10⁶ m s⁻¹. (Mass of electron = 9.11 × 10⁻³¹ kg)

λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.0 × 10⁶)

λ = 6.63 × 10⁻³⁴ / (1.822 × 10⁻²⁴) = 3.64 × 10⁻¹⁰ m ≈ 0.36 nm

This is comparable to atomic spacing in a crystal, which is why electrons can be diffracted by crystal lattices.

Worked example 2: An electron is accelerated through a potential difference of 150 V. Calculate its de Broglie wavelength.

First find the kinetic energy: KE = eV = 1.60 × 10⁻¹⁹ × 150 = 2.40 × 10⁻¹⁷ J

Find the speed: ½mv² = KE → v = √(2KE/m) = √(2 × 2.40 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = √(5.27 × 10¹³) = 7.26 × 10⁶ m s⁻¹

λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 7.26 × 10⁶) = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ = 1.00 × 10⁻¹⁰ m = 0.10 nm

This is the same order of magnitude as X-ray wavelengths and atomic spacings.

Alternative approach using momentum directly: Since KE = p²/(2m), we get p = √(2mKE), so:

λ = h/p = h/√(2mKE) = h/√(2meV)

λ = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 2.40 × 10⁻¹⁷) = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ = 1.00 × 10⁻¹⁰ m ✓

Worked example 3: Calculate the de Broglie wavelength of a 70 kg person walking at 1.5 m s⁻¹.

λ = h/(mv) = 6.63 × 10⁻³⁴ / (70 × 1.5) = 6.63 × 10⁻³⁴ / 105 = 6.3 × 10⁻³⁶ m

This is about 10²¹ times smaller than the diameter of a proton (~10⁻¹⁵ m). It is utterly undetectable — no experiment could ever observe the wave nature of a walking person. Quantum wave behaviour is only significant for very small, very light particles.

Worked example 4: A neutron (mass = 1.675 × 10⁻²⁷ kg) has a de Broglie wavelength of 0.15 nm. Calculate its speed.

v = h/(mλ) = 6.63 × 10⁻³⁴ / (1.675 × 10⁻²⁷ × 0.15 × 10⁻⁹)

v = 6.63 × 10⁻³⁴ / 2.51 × 10⁻³⁷ = 2640 m s⁻¹ ≈ 2.6 km s⁻¹

These are "thermal neutrons" — neutrons at roughly room temperature, widely used in neutron diffraction experiments to study crystal structures.

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Why Wavelength Depends on Momentum

The de Broglie relation λ = h/p has a beautiful implication: the faster a particle moves (greater momentum), the shorter its de Broglie wavelength. This mirrors the relationship between photon energy and wavelength — higher energy means shorter wavelength.

Particle	Typical speed	de Broglie wavelength	Observable wave effects?
Electron (thermal)	~10⁵ m s⁻¹	~7 nm	Yes — electron diffraction
Electron (accelerated, 100 V)	~6 × 10⁶ m s⁻¹	~0.12 nm	Yes — comparable to atomic spacing
Proton (thermal)	~500 m s⁻¹	~0.8 nm	Yes — neutron/proton diffraction
Cricket ball (30 m s⁻¹)	30 m s⁻¹	~10⁻³⁴ m	No — completely undetectable
Human walking	1.5 m s⁻¹	~10⁻³⁶ m	No — unimaginably small

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Electron Diffraction

The wave nature of electrons was confirmed experimentally by Davisson and Germer (1927), who directed a beam of electrons at a nickel crystal and observed a diffraction pattern — just like X-ray diffraction. The pattern of maxima and minima matched the prediction from the de Broglie wavelength.

In a modern electron diffraction experiment:

1. Electrons are accelerated through a known potential difference V
2. The beam strikes a thin polycrystalline graphite target
3. A circular diffraction pattern appears on a fluorescent screen behind the target
4. The ring pattern matches the prediction for waves of wavelength λ = h/√(2meV)

Key observations:

• Increasing the accelerating voltage increases the electron speed, decreases the de Broglie wavelength, and the diffraction rings become smaller (less spreading — same as for any wave when λ decreases relative to the slit/gap size)
• Decreasing the accelerating voltage makes the rings larger (more spreading)

This is direct, compelling evidence that electrons — particles with mass and charge — exhibit wave behaviour.

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Wave-Particle Duality: The Full Picture

Wave-particle duality applies to everything:

• Photons (light) exhibit wave behaviour (interference, diffraction, polarisation) and particle behaviour (photoelectric effect, Compton scattering)
• Electrons exhibit particle behaviour (deflection in electric and magnetic fields, tracks in cloud chambers) and wave behaviour (electron diffraction)
• All matter has an associated de Broglie wavelength, but wave behaviour is only detectable when the wavelength is comparable to the scale of the interaction (e.g., atomic spacing)

The question "is it a wave or a particle?" is the wrong question. Nature does not fit neatly into either classical category. The correct description depends on the experiment:

• When we observe propagation and interference, the wave model applies
• When we observe detection and interaction with matter, the particle model applies

Real-world application: Electron microscopes. Since electrons can have wavelengths much shorter than visible light (~0.1 nm vs ~500 nm), electron microscopes achieve much higher resolution than optical microscopes. A transmission electron microscope (TEM) can image individual atoms. This is only possible because of the wave nature of electrons — the image is formed by electron wave diffraction and interference.

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Comparing Photons and Electrons

Property	Photon	Electron
Rest mass	0	9.11 × 10⁻³¹ kg
Speed	Always c in vacuum	Variable, always < c
Energy	E = hf	KE = ½mv² (or eV)
Momentum	p = h/λ = E/c	p = mv
de Broglie wavelength	λ = h/p = hc/E	λ = h/(mv) = h/√(2mKE)
Wave behaviour shown by	Interference, diffraction	Electron diffraction
Particle behaviour shown by	Photoelectric effect	Deflection in fields, discrete charge

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The Uncertainty Principle (Conceptual Extension)

Wave-particle duality leads to Heisenberg's uncertainty principle: there is a fundamental limit to how precisely we can simultaneously know both the position and momentum of a particle.

$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$Δx⋅Δp≥4πh​

The more precisely we know the position (small Δx), the less precisely we can know the momentum (large Δp), and vice versa. This is not a limitation of our instruments — it is a fundamental property of nature arising from the wave nature of matter.

If a particle has a very well-defined wavelength (and therefore momentum), its wave function extends over a large region of space — the position is very uncertain. If the position is well-defined (particle localised in a small region), the wave function must be composed of many wavelengths — the momentum is very uncertain.

Common exam mistake: Students sometimes say "light is a wave when it travels and a particle when it is detected." This is a useful heuristic but is an oversimplification. Wave-particle duality means that the complete description of light (or any quantum object) requires both models. The choice of model depends on the experiment, not on what the light is "doing" at a given moment.

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A-Level Deep Dive: Wave-Particle Duality

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in its closing sub-strands on the photon model and matter waves, requires students to use the de Broglie equation $\lambda = h/p$λ=h/p; explain how electron diffraction provides evidence for the wave nature of particles; and combine wave-particle ideas with photon energy and the photoelectric equation (refer to the official Pearson Edexcel specification document for exact wording). Although wave-particle duality appears at the end of Topic 5, the underlying ideas are load-bearing throughout: the photon model anchors the photoelectric effect ($KE_{\max} = hf - \phi$KEmax​=hf−ϕ), the energy-level emission spectra ($\Delta E = hf$ΔE=hf), and the de Broglie wavelength of accelerated electrons ($\lambda = h/\sqrt{2meV}$λ=h/2meV​). Wave-particle fluency is also load-bearing in Topic 11 (nuclear and particle physics) when discussing matter probes in colliders, and in Topic 12 (turning points) for the historical-conceptual narrative of quantum theory. The Edexcel data-and-formulae booklet lists $\lambda = h/p$λ=h/p, $E = hf$E=hf and $hf = \phi + \tfrac{1}{2}mv_{\max}^2$hf=ϕ+21​mvmax2​, but does not list the combined form $\lambda = h/\sqrt{2meV}$λ=h/2meV​ — that one must be derived in working from $eV = \tfrac{1}{2}mv^2$eV=21​mv2 and $p = mv$p=mv.

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Worked example with full mark scheme

Question (8 marks):

An electron is accelerated from rest through a potential difference of $250\ \text{V}$250 V in an evacuated tube. The electron beam is then directed at a thin polycrystalline graphite target, where the carbon-atom lattice spacing is approximately $2.1 \times 10^{-10}\ \text{m}$2.1×10−10 m. Take the electron mass as $9.11 \times 10^{-31}\ \text{kg}$9.11×10−31 kg, the electron charge as $1.60 \times 10^{-19}\ \text{C}$1.60×10−19 C and the Planck constant as $6.63 \times 10^{-34}\ \text{J s}$6.63×10−34 J s.

(a) Show that the de Broglie wavelength of the accelerated electrons is approximately $7.8 \times 10^{-11}\ \text{m}$7.8×10−11 m. (4)

(b) State and justify whether a clear diffraction pattern would be observed when the beam strikes the graphite target. (2)

(c) Calculate the wavelength of a photon whose energy equals the kinetic energy gained by the electron, and comment briefly on the part of the electromagnetic spectrum to which this photon belongs. (2)

Solution with mark scheme:

(a) Step 1 — equate the work done by the field with the kinetic energy gained.

$eV = \tfrac{1}{2}mv^2 \implies v = \sqrt{\frac{2eV}{m}}$eV=21​mv2⟹v=m2eV​​

M1 — correct energy-balance equation written down. The most common slip is to use $eV = mv$eV=mv (mixing momentum with energy), which produces a wavelength two orders of magnitude wrong.

Step 2 — convert directly to momentum without explicitly evaluating $v$v.

$p = mv = \sqrt{2meV}$p=mv=2meV​

M1 — recognising the useful shortcut. Candidates who first compute $v$v and then $p$p score the same marks but lose time and risk arithmetic slips.

Step 3 — substitute into the de Broglie equation.

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 250}}$λ=ph​=2meV​h​=2×9.11×10−31×1.60×10−19×250​6.63×10−34​

M1 — algebraically correct substitution.

Step 4 — evaluate.

The denominator: $2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 250 = 7.29 \times 10^{-47}$2×9.11×10−31×1.60×10−19×250=7.29×10−47, so $\sqrt{7.29 \times 10^{-47}} = 8.54 \times 10^{-24}\ \text{kg m s}^{-1}$7.29×10−47​=8.54×10−24 kg m s−1.

$\lambda = \frac{6.63 \times 10^{-34}}{8.54 \times 10^{-24}} = 7.76 \times 10^{-11}\ \text{m}$λ=8.54×10−246.63×10−34​=7.76×10−11 m

A1 — final answer agrees to 2 sf with the printed value $7.8 \times 10^{-11}\ \text{m}$7.8×10−11 m.

(b) Step 1 — compare the wavelength to the lattice spacing.

The de Broglie wavelength $\lambda \approx 7.8 \times 10^{-11}\ \text{m}$λ≈7.8×10−11 m is comparable to the lattice spacing $d \approx 2.1 \times 10^{-10}\ \text{m}$d≈2.1×10−10 m (same order of magnitude, with $\lambda/d \approx 0.37$λ/d≈0.37). M1 — explicit comparison made.

Step 2 — state the diffraction criterion.

A clear diffraction pattern is observed when the wavelength is comparable to (within roughly a factor of 10 of) the spacing of the diffracting structure. Since this condition is met, a clear ring pattern is expected on the fluorescent screen. A1 — correct conclusion linked to the criterion.

(c) Step 1 — compute the kinetic energy in joules.

$KE = eV = 1.60 \times 10^{-19} \times 250 = 4.00 \times 10^{-17}\ \text{J}$KE=eV=1.60×10−19×250=4.00×10−17 J

Step 2 — apply $E = hc/\lambda_{\text{photon}}$E=hc/λphoton​.

$\lambda_{\text{photon}} = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{4.00 \times 10^{-17}} = 4.97 \times 10^{-9}\ \text{m} \approx 5\ \text{nm}$λphoton​=Ehc​=4.00×10−176.63×10−34×3.00×108​=4.97×10−9 m≈5 nm

M1 — correct substitution. A1 — value $\approx 5\ \text{nm}$≈5 nm, identified as far ultraviolet / soft X-ray boundary. The pedagogical point: at the same energy, the photon wavelength ($\sim 10^{-9}\ \text{m}$∼10−9 m) is roughly two orders of magnitude larger than the electron's de Broglie wavelength ($\sim 10^{-10}\ \text{m}$∼10−10 m), because the photon carries all its energy as $hf$hf while the electron carries it kinetically with a much larger momentum.

Total: 8 marks (M3 A1 M1 A1 M1 A1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): In an electron-diffraction tube, electrons are accelerated through a variable potential difference $V$V and strike a thin graphite film. A circular ring pattern appears on a fluorescent screen at a fixed distance $L$L behind the film. The radius $r$r of the first-order ring is measured for several values of $V$V.

(a) Show that the radius $r$r is expected to be inversely proportional to $\sqrt{V}$V​. (3)

(b) State the most appropriate quantities to plot to obtain a straight-line graph that confirms the de Broglie relation, and describe how the gradient could be used. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1) — writing $\lambda = h/\sqrt{2meV}$λ=h/2meV​, so $\lambda \propto 1/\sqrt{V}$λ∝1/V​.
• M1 (AO2) — for small diffraction angles, $r \approx L\sin\theta$r≈Lsinθ, and the diffraction condition $d\sin\theta = n\lambda$dsinθ=nλ gives $\sin\theta \propto \lambda$sinθ∝λ at fixed $d$d and $n$n.
• A1 (AO2) — combining: $r \propto \lambda \propto 1/\sqrt{V}$r∝λ∝1/V​, so $r \propto V^{-1/2}$r∝V−1/2, as required.

(b)

• M1 (AO3) — plotting $r$r on the vertical axis against $1/\sqrt{V}$1/V​ on the horizontal axis is expected to give a straight line through the origin.
• M1 (AO3) — the gradient equals $L \cdot h / (d \sqrt{2me})$L⋅h/(d2me​) (from combining $r = L\lambda/d$r=Lλ/d and $\lambda = h/\sqrt{2meV}$λ=h/2meV​, with $n = 1$n=1).
• A1 (AO3) — knowing $L$L and $d$d, the gradient can be rearranged to give an experimental value for the Planck constant $h$h, providing a quantitative test of the de Broglie relation.