Unit 4: Revolutions in modern physics

QLDPhysicsSyllabus dot point

Topic 2: Quantum theory

Explain wave-particle duality through de Broglie's matter-wave hypothesis $\lambda = h/p$, applying it to electron diffraction and to the quantum nature of matter

A focused answer to the QCE Physics Unit 4 dot point on wave-particle duality. de Broglie's hypothesis $\\lambda = h/p$, Davisson-Germer electron diffraction, the matter-wave interpretation of Bohr orbits, and the electron microscope application.

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What this dot point is asking

QCAA wants you to explain de Broglie's hypothesis that matter has wave-like properties, apply λ=h/p\lambda = h/p to calculate de Broglie wavelengths, and discuss the experimental evidence for matter waves (Davisson-Germer). Cross-link: see the de Broglie wavelength calculator.

de Broglie's hypothesis

In 1924, Louis de Broglie proposed that every particle of momentum pp has an associated wavelength:

λ=hp\lambda = \frac{h}{p}

The motivation was symmetry. Photons (classically waves) had been shown by Einstein to have particle properties with p=h/λp = h / \lambda. de Broglie inverted: if light has particle properties, perhaps matter has wave properties.

Computing the de Broglie wavelength

For a non-relativistic particle of mass mm and speed vv:

p=mv,λ=hmvp = m v, \quad \lambda = \frac{h}{m v}

For an electron accelerated through a potential VV from rest:

Ek=eV,p=2meV,λ=h2meVE_k = e V, \quad p = \sqrt{2 m e V}, \quad \lambda = \frac{h}{\sqrt{2 m e V}}

For an electron: λ1.226/V\lambda \approx 1.226 / \sqrt{V} nm (where VV is in volts). At 100 V, λ0.123\lambda \approx 0.123 nm; at 1000 V, λ0.039\lambda \approx 0.039 nm.

These wavelengths are comparable to atomic spacings, which is why electron beams diffract from crystals.

The Davisson-Germer experiment (1927)

Davisson and Germer directed a beam of low-energy electrons (around 54 eV) at a nickel crystal. They observed a peak in scattered intensity at a specific angle, matching the Bragg diffraction prediction for waves of wavelength equal to the de Broglie value.

The result confirmed:

  • Electrons exhibit wave behaviour (diffraction).
  • The de Broglie wavelength λ=h/p\lambda = h/p is correct.

G. P. Thomson independently obtained similar results passing electrons through a thin metal foil and observing concentric diffraction rings on a photographic plate. (J. J. Thomson, his father, had discovered the electron as a particle in 1897; G. P. Thomson confirmed it as a wave 30 years later.)

Both received Nobel Prizes (Davisson 1937, G. P. Thomson 1937; J. J. Thomson 1906).

Wave-particle duality

Combining the photoelectric effect (light has particle properties) with the de Broglie hypothesis (matter has wave properties) produces the modern picture of wave-particle duality:

  • Light (classically a wave) shows particle properties: photoelectric effect, atomic spectra, photon momentum.
  • Matter (classically a collection of particles) shows wave properties: diffraction, interference, atomic standing waves.

Neither pure wave nor pure particle model is complete for either light or matter. Both are quantum objects whose wave or particle behaviour depends on the experimental context.

Matter waves and Bohr's quantisation

de Broglie's hypothesis explains Bohr's seemingly ad hoc quantisation of atomic angular momentum.

Bohr postulated L=mvr=nL = m v r = n \hbar where =h/(2π)\hbar = h / (2\pi). de Broglie observed that this is equivalent to:

2πrn=nλ2 \pi r_n = n \lambda

That is, the circumference of the orbit equals an integer number of de Broglie wavelengths. Only orbits that support a stable standing wave for the electron are allowed; other orbits would destructively interfere with themselves.

The matter-wave picture transforms the Bohr quantisation from a postulate into a physical consequence of the wave nature of the electron.

Why no diffraction of macroscopic objects

Diffraction is observable when the wavelength is comparable to a slit or obstacle.

For an electron (λ1010\lambda \sim 10^{-10} m), atomic-scale crystals are matched. Diffraction is observable.

For a tennis ball (v=30v = 30 m/s, m=0.05m = 0.05 kg, p=1.5p = 1.5 kg m/s, λ=h/p4×1034\lambda = h/p \approx 4 \times 10^{-34} m), no physical aperture is small enough. The wave nature exists in principle but is utterly unobservable.

The classical-quantum boundary depends on the ratio λ/d\lambda / d where dd is the relevant geometric scale. For macroscopic objects, λd\lambda \ll d and classical physics applies. For atomic-scale particles in atomic-scale environments, λd\lambda \sim d and quantum effects dominate.

The electron microscope

The resolution of a light microscope is limited by diffraction: features smaller than approximately λ\lambda cannot be resolved. With visible light (λ500\lambda \sim 500 nm), resolution is around 200 nm.

An electron microscope uses an electron beam in place of light. Electrons accelerated through tens or hundreds of kilovolts have de Broglie wavelengths in the picometre range, giving atomic-scale resolution.

The transmission electron microscope (TEM) can image individual atoms; the scanning electron microscope (SEM) images surface topology with nanometre resolution. Both rely on the de Broglie wavelength of accelerated electrons.

Worked example: neutron diffraction

Thermal neutrons at room temperature have Ek0.025E_k \approx 0.025 eV. Neutron mass mn=1.67×1027m_n = 1.67 \times 10^{-27} kg.

Ek=0.025×1.6×1019=4×1021E_k = 0.025 \times 1.6 \times 10^{-19} = 4 \times 10^{-21} J.

p=2mEk=2×1.67×1027×4×1021=1.34×10473.66×1024p = \sqrt{2 m E_k} = \sqrt{2 \times 1.67 \times 10^{-27} \times 4 \times 10^{-21}} = \sqrt{1.34 \times 10^{-47}} \approx 3.66 \times 10^{-24} kg m s1^{-1}.

λ=h/p=6.626×1034/3.66×10241.8×1010\lambda = h/p = 6.626 \times 10^{-34} / 3.66 \times 10^{-24} \approx 1.8 \times 10^{-10} m =0.18= 0.18 nm.

Thermal neutrons therefore diffract from crystals. Neutron diffraction is a standard technique in materials science.

Common errors

Confusing E=hfE = hf (photon) with Ek=p2/(2m)E_k = p^2/(2m) (matter). Photons use E=hfE = hf and λ=c/f\lambda = c/f. Matter uses p=2mEkp = \sqrt{2 m E_k} for non-relativistic particles.

Mass in grams. Always use kg in λ=h/p\lambda = h/p.

Treating matter as either particles or waves. Both descriptions are needed. The electron is a quantum object that exhibits particle properties in detection (a click) and wave properties in propagation (diffraction).

Forgetting the macroscopic-scale collapse. The wave nature exists for all matter but is observable only for atomic-scale particles in atomic-scale environments. Macroscopic objects do not show measurable diffraction.

In one sentence

de Broglie's matter-wave hypothesis λ=h/p\lambda = h/p assigns every particle a wave-like wavelength, confirmed for electrons by the Davisson-Germer experiment (1927); together with the photon model of light, it establishes wave-particle duality (both light and matter are quantum objects with both wave and particle properties), explains the quantisation of atomic orbits as standing-wave conditions, and underpins the resolution advantage of electron microscopes over light microscopes.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style4 marksAn electron is accelerated from rest through a potential difference of $200$ V. (a) Calculate its kinetic energy. (b) Calculate its momentum. (c) Calculate its de Broglie wavelength.
Show worked answer →

(a) KE. Ek=eV=1.6×1019×200=3.2×1017E_k = e V = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17} J. (Equivalent to 200 eV.)

(b) Momentum. Ek=p2/(2m)E_k = p^2 / (2m), so p=2mEkp = \sqrt{2 m E_k}.

p=2×9.11×1031×3.2×1017=5.83×10477.64×1024p = \sqrt{2 \times 9.11 \times 10^{-31} \times 3.2 \times 10^{-17}} = \sqrt{5.83 \times 10^{-47}} \approx 7.64 \times 10^{-24} kg m s1^{-1}.

(c) de Broglie wavelength. λ=h/p=6.626×1034/7.64×10248.67×1011\lambda = h/p = 6.626 \times 10^{-34} / 7.64 \times 10^{-24} \approx 8.67 \times 10^{-11} m =0.087= 0.087 nm.

Markers reward the Ek=eVE_k = eV relation, the momentum from 2mEk\sqrt{2 m E_k}, and the de Broglie wavelength.

Useful shortcut: for an electron, λ1.226/V\lambda \approx 1.226 / \sqrt{V} nm with VV in volts. So at 200 V, λ0.087\lambda \approx 0.087 nm.

2023 QCAA-style3 marksExplain how the Davisson-Germer experiment (1927) supports wave-particle duality for matter.
Show worked answer →

Davisson and Germer directed a beam of low-energy electrons (around 54 eV) at a nickel crystal and measured the angular distribution of scattered electrons.

They observed a peak in scattering intensity at a specific angle (about 50 degrees). The peak position matched the prediction from Bragg diffraction of waves with wavelength equal to the de Broglie value (λ=h/p\lambda = h/p).

This was a direct experimental demonstration that electrons exhibit wave behaviour: they diffract from crystal lattices in the same way X-rays of equivalent wavelength do. Since electrons had been established as particles (J. J. Thomson, 1897), the diffraction result confirmed that matter has both particle and wave properties, consistent with de Broglie's hypothesis (1924).

Markers reward the experimental setup, the diffraction result, and the explicit "matter has wave properties" interpretation.

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