Wave-particle duality and matter waves: QCE Physics Unit 4

What this dot point is asking

QCAA wants you to explain de Broglie's hypothesis that matter has wave-like properties, apply $\lambda = h/p$ 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 $p$ has an associated wavelength:

The motivation was symmetry. Photons (classically waves) had been shown by Einstein to have particle properties with $p = h / \lambda$. 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 $m$ and speed $v$:

For an electron accelerated through a potential $V$ from rest:

For an electron: $\lambda \approx 1.226 / \sqrt{V}$ nm (where $V$ is in volts). At 100 V, $\lambda \approx 0.123$ nm; at 1000 V, $\lambda \approx 0.039$ 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 $\lambda = h/p$ 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 $L = m v r = n \hbar$ where $\hbar = h / (2\pi)$. de Broglie observed that this is equivalent to:

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

For a tennis ball ($v = 30$ m/s, $m = 0.05$ kg, $p = 1.5$ kg m/s, $\lambda = h/p \approx 4 \times 10^{-34}$ 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 $\lambda / d$ where $d$ is the relevant geometric scale. For macroscopic objects, $\lambda \ll d$ and classical physics applies. For atomic-scale particles in atomic-scale environments, $\lambda \sim d$ and quantum effects dominate.

The electron microscope

The resolution of a light microscope is limited by diffraction: features smaller than approximately $\lambda$ cannot be resolved. With visible light ($\lambda \sim 500$ 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 $E_k \approx 0.025$ eV. Neutron mass $m_n = 1.67 \times 10^{-27}$ kg.

$E_k = 0.025 \times 1.6 \times 10^{-19} = 4 \times 10^{-21}$ J.

$p = \sqrt{2 m E_k} = \sqrt{2 \times 1.67 \times 10^{-27} \times 4 \times 10^{-21}} = \sqrt{1.34 \times 10^{-47}} \approx 3.66 \times 10^{-24}$ kg m s$^{-1}$.

$\lambda = h/p = 6.626 \times 10^{-34} / 3.66 \times 10^{-24} \approx 1.8 \times 10^{-10}$ m $= 0.18$ nm.

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

Examples in context

Example 1. A Queensland university electron microscope accelerates electrons through $V = 50 \text{ kV}$. Electron momentum is $p = \sqrt{2 m_e e V} = 1.21 \times 10^{-22} \text{ kg m s}^{-1}$; de Broglie wavelength $\lambda = h/p = 5.49 \text{ pm}$, four orders of magnitude shorter than visible light, enabling atomic-scale imaging. The QCAA Unit 4 dot-point relation $\lambda = h/p$ is exactly this calculation.

Example 2. Cosmic-ray neutron diffraction at ANSTO Mt Cotton uses thermal neutrons ($v \approx 2200 \text{ m s}^{-1}$, $m = 1.67 \times 10^{-27} \text{ kg}$) giving $\lambda = h / (mv) = 0.18 \text{ nm}$ - comparable to crystal lattice spacing, so Bragg diffraction occurs. QCAA EA Unit 4 thematic items pair such an experimental wavelength with a structural inference.

Try this

Q1. State the de Broglie hypothesis. [2 marks]

• Cue. $\lambda = h/p$; matter exhibits wave-like properties with wavelength inversely proportional to momentum.

Q2. Calculate the de Broglie wavelength of an electron at $v = 2.0 \times 10^6 \text{ m s}^{-1}$, $m_e = 9.11 \times 10^{-31} \text{ kg}$. [3 marks]

• Cue. $p = m v = 1.82 \times 10^{-24}$; $\lambda = h/p = 3.64 \times 10^{-10} \text{ m}$.

Q3. A $50 \text{ kV}$ electron microscope. (a) Calculate the electron momentum and de Broglie wavelength. (b) Compare with visible-light wavelength ($550 \text{ nm}$) and explain the resolution advantage. (c) Discuss one limitation of the wave-particle duality concept in IA3 reasoning. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $p = 1.21 \times 10^{-22}$, $\lambda = 5.49 \text{ pm}$; (b) $\lambda_{light}/\lambda_e \approx 10^5$; (c) complementarity - wave and particle behaviour mutually exclusive in single experiment.