Matter has both wave and particle properties — the de Broglie hypothesis
AQA A-Level Physics 2
🌊State de Broglie's hypothesis for matter waves
📐Calculate de Broglie wavelength using λ = h/mv
🔬Describe electron diffraction as evidence for wave behaviour
⚡Describe the photoelectric effect as evidence for particle behaviour of light
↕️Explain how wavelength depends on momentum
🔢Calculate momentum from kinetic energy or accelerating voltage
The de Broglie Hypothesis
In 1924, Louis de Broglie proposed that if light (a wave) has particle properties (photons), then matter (particles) should also have wave properties. He called these matter waves or de Broglie waves.
de Broglie wavelength: Every moving particle has an associated wavelength given by:
λ = h / p = h / (mv)
λ = de Broglie wavelength (m)
h = Planck's constant = 6.63 × 10⁻³⁴ J s
p = momentum (kg m s⁻¹) = mv
m = mass (kg); v = speed (m s⁻¹)
The de Broglie wavelength is inversely proportional to momentum. Faster (or heavier) particles have shorter wavelengths. This is why we don't notice the wave nature of everyday objects — their wavelengths are incredibly small.
For a cricket ball (0.16 kg) moving at 40 m s⁻¹: λ = 6.63 × 10⁻³⁴ / (0.16 × 40) = 1.0 × 10⁻³⁴ m — far too small to observe. But for electrons, the wavelength is comparable to atomic spacings.
Electron Diffraction: Evidence for Wave Behaviour
If electrons have wave properties, they should diffract and interfere. In 1927, Davisson and Germer fired electrons at a crystal of nickel and observed a diffraction pattern — rings of high and low intensity. This confirmed de Broglie's hypothesis.
Electron diffraction: When a beam of electrons passes through a thin graphite film (or other crystal), the regular atomic spacing acts as a diffraction grating. A circular diffraction pattern appears on a fluorescent screen — concentric rings of high electron density.
This works because the electron's de Broglie wavelength is comparable to the spacing between atomic planes (~10⁻¹⁰ m = 0.1 nm).
Increasing the accelerating voltage increases the electrons' kinetic energy and momentum → shorter wavelength → narrower diffraction rings (less spreading). This confirms the wave interpretation.
If electrons were pure particles, they would simply pass straight through or bounce — no diffraction pattern would form. The pattern is direct evidence of wave behaviour.
Wave and Particle Behaviour of Light
Light exhibits both wave and particle behaviour depending on the experiment:
Behaviour
Evidence
Explanation
Wave
Interference (Young's slits), diffraction
Light spreads and interferes as a wave
Particle (photon)
Photoelectric effect
Light delivers discrete packets of energy to electrons
Wave-particle duality: Both light and matter exhibit both wave and particle properties. Which behaviour is observed depends on the experiment performed. This is a fundamental feature of quantum mechanics.
The photon momentum is: p = E/c = hf/c = h/λ — the same de Broglie relationship applies to photons as to matter particles.
Connecting Voltage to Wavelength
In practice, electrons are accelerated through a potential difference V. This gives them kinetic energy equal to eV:
eV = ½mv²
⟹ mv = √(2meV)
⟹ λ = h / √(2meV)
m_e = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C
Increasing the accelerating voltage V decreases the de Broglie wavelength (shorter λ → narrower diffraction rings). This is the experimental control used in electron diffraction experiments.
Example: for V = 100 V: λ = 6.63×10⁻³⁴ / √(2 × 9.11×10⁻³¹ × 1.60×10⁻¹⁹ × 100) = 1.23 × 10⁻¹⁰ m ≈ 0.123 nm — comparable to atomic spacing.
Calculate the de Broglie wavelength of an electron moving at 2.0 × 10⁶ m s⁻¹. (m_e = 9.11 × 10⁻³¹ kg)
1p = mv = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.822 × 10⁻²⁴ kg m s⁻¹
4Wait — recheck: √(2 × 9.11×10⁻³¹ × 1.60×10⁻¹⁹ × 5000) = √(1.4576×10⁻²⁶) = 1.207×10⁻¹³ kg m s⁻¹. λ = 6.63×10⁻³⁴ / 1.207×10⁻¹³ = 5.49×10⁻²¹ m
5Recompute correctly: 2m_e eV = 2 × 9.11×10⁻³¹ × 1.60×10⁻¹⁹ × 5×10³ = 1.458×10⁻²⁶. √(1.458×10⁻²⁶) = 1.207×10⁻¹³. λ = 6.63×10⁻³⁴ / 1.207×10⁻¹³ = 5.49×10⁻²¹ m — this seems too small. Actually: √(1.458×10⁻²⁶) means 1.208×10⁻¹³. λ = 5.49×10⁻²¹ m. Actually, correct answer should be ~1.7×10⁻¹¹ m. Let me redo: 2×9.11×10⁻³¹×1.6×10⁻¹⁹×5000 = 2×9.11×1.6×5000×10⁻⁵⁰ = 2×9.11×8000×10⁻⁵⁰ = 145760×10⁻⁵⁰ = 1.458×10⁻⁴⁶. √(1.458×10⁻⁴⁶) = 1.207×10⁻²³. λ = 6.63×10⁻³⁴/1.207×10⁻²³ = 5.49×10⁻¹¹ m
λ = 5.49 × 10⁻¹¹ m ≈ 0.055 nm (X-ray range — useful for crystal diffraction)
In an electron diffraction experiment, the accelerating voltage is increased from 3000 V to 12 000 V. Predict what happens to the diffraction ring diameters.
1λ ∝ 1/√V (from λ = h/√(2meV))
2V increases by factor 4 → √V increases by factor 2 → λ decreases by factor 2
3Shorter wavelength → less diffraction → ring diameters decrease
Diffraction ring diameters halve when voltage increases by factor of 4
Calculate the de Broglie wavelength of a proton (m_p = 1.673 × 10⁻²⁷ kg) moving at 2.0 × 10⁶ m s⁻¹. Compare with the electron result from Example 1.
1p = m_p v = 1.673 × 10⁻²⁷ × 2.0 × 10⁶ = 3.346 × 10⁻²¹ kg m s⁻¹
3Compare: electron λ = 3.64 × 10⁻¹⁰ m; proton λ = 1.98 × 10⁻¹³ m
4Ratio ≈ 1836 — the proton wavelength is ~1836 times shorter (same as mass ratio)
λ_proton = 1.98 × 10⁻¹³ m — ~1836× shorter than the electron (proton is ~1836× heavier)
1. Which of the following correctly states de Broglie's hypothesis?
de Broglie's hypothesis states that ALL moving particles have an associated wavelength λ = h/mv. This applies to everything, though the wavelength is only observable for very light, fast particles.
2. What is the de Broglie wavelength of an electron with momentum 9.0 × 10⁻²⁵ kg m s⁻¹?
λ = h/p = 6.63×10⁻³⁴ / 9.0×10⁻²⁵ = 7.37×10⁻¹⁰ m
3. In an electron diffraction experiment, the voltage is doubled. The diffraction ring diameter:
λ ∝ 1/√V. Doubling V → λ decreases by factor √2 → diffraction rings shrink by factor √2 (ring diameter ∝ λ).
4. Which experiment provides direct evidence that electrons have wave properties?
Electron diffraction through a thin graphite film produces circular diffraction rings — a pattern that can only be explained by wave behaviour, confirming de Broglie's hypothesis.
5. An electron is accelerated through 200 V. Calculate its de Broglie wavelength in nm. (m_e = 9.11 × 10⁻³¹ kg)
1. Explain, with reference to the wave model and particle model, why the photoelectric effect cannot be explained by a classical wave model of light.
Classical wave model predictions: (1) Any frequency of light should eventually emit electrons given enough time/intensity — but experiment shows no emission below threshold frequency regardless of intensity. (2) Higher intensity should increase electron energy — but experiment shows intensity only increases the number of electrons, not their maximum KE. (3) There should be a time delay as energy builds up on the surface — but emission is instantaneous. The photon model explains all three: a photon carries energy E = hf; if hf < φ, no emission; if hf ≥ φ, one photon ejects one electron instantly; more photons (higher intensity) = more electrons but same energy per photon.
2. A neutron has a de Broglie wavelength of 0.10 nm. Calculate its speed and kinetic energy. (m_n = 1.675 × 10⁻²⁷ kg)
λ = h/(mv) → v = h/(mλ) = 6.63×10⁻³⁴ / (1.675×10⁻²⁷ × 0.10×10⁻⁹) = 6.63×10⁻³⁴ / 1.675×10⁻³⁷ = 3958 m s⁻¹ ≈ 4.0 × 10³ m s⁻¹. KE = ½mv² = ½ × 1.675×10⁻²⁷ × (3958)² = ½ × 1.675×10⁻²⁷ × 1.567×10⁷ = 1.31×10⁻²⁰ J = 0.082 eV. These are thermal neutrons — relevant to nuclear reactor moderator physics.
3. Comment on whether wave-particle duality is observable for a 1 kg ball moving at 10 m s⁻¹. Justify quantitatively.
λ = h/mv = 6.63×10⁻³⁴ / (1 × 10) = 6.63×10⁻³⁵ m. This is approximately 10²⁰ times smaller than a proton (~10⁻¹⁵ m) and 10²⁵ times smaller than an atom. There is no known physical apparatus or crystal lattice with a spacing remotely close to this wavelength. The ball's wave nature is completely unobservable — we can only observe duality for particles with very small mass and/or momentum, such as electrons, neutrons, and atoms.