Explain wave-particle duality and apply the de Broglie wavelength to matter

What this dot point is asking

WACE wants you to explain that wave and particle models are complementary, apply the de Broglie relation, and cite electron diffraction as evidence. This dot point unifies the wave and photon strands of the unit.

Duality of light

The wave model explains interference, diffraction and polarisation, while the particle (photon) model explains the photoelectric effect and the line spectra of atoms. Neither model alone covers all observations, so light is described as having a dual nature: it shows wave properties in some experiments and particle properties in others, but never both in the same measurement.

De Broglie's matter waves

Louis de Broglie reasoned that if waves can act as particles, particles might act as waves. He assigned every moving particle a wavelength

where $p=mv$ is the momentum and $h$ is Planck's constant. Because $h$ is tiny, the wavelength is utterly negligible for everyday objects, which is why a cricket ball shows no wave behaviour. For light, fast particles such as electrons it becomes comparable to atomic spacings and measurable.

Why everyday objects show no wave behaviour

A moving car has a de Broglie wavelength around $10^{-38}\ \text{m}$, far smaller than any aperture it could pass through, so it never diffracts. An electron accelerated through a modest voltage has a wavelength near $10^{-10}\ \text{m}$, similar to the spacing of atoms in a crystal, so a crystal acts as a diffraction grating for it.

Electron diffraction as evidence

When a beam of electrons is fired at a thin crystal or graphite film, it produces concentric diffraction rings on a screen, exactly the pattern expected when waves pass through a regular array of slits. Particles alone could not interfere to form rings, so this confirms that matter has a wave nature, validating de Broglie's idea.

From accelerating voltage to wavelength

A very common multi-step question gives an accelerating voltage and asks for the de Broglie wavelength, chaining several relationships. The reliable route is: find the kinetic energy gained, $E_k=qV$; convert to speed using $E_k=\tfrac{1}{2}mv^2$, so $v=\sqrt{2qV/m}$; find the momentum $p=mv$; then apply $\lambda=h/p$. A useful shortcut is to combine these into $\lambda=\dfrac{h}{\sqrt{2mqV}}$, which goes straight from voltage to wavelength in one substitution. This shows the wavelength falls as the accelerating voltage rises, which is exactly why electron microscopes use high voltages: a shorter de Broglie wavelength gives finer resolving power than visible light.

Complementarity: one model at a time

A subtle but examinable point is that wave and particle behaviours are complementary, never observed simultaneously. In a given experiment light (or matter) reveals either its wave nature or its particle nature, depending on what is being measured, but no single measurement shows both at once. The double-slit experiment shows interference (wave) when you do not detect which slit the light went through; the photoelectric effect shows discrete photons (particle) when light exchanges energy with electrons. The two pictures are not contradictory but together provide a complete description, with each appropriate to different situations. Stating that the models are complementary, and giving one wave experiment and one particle experiment as examples, is exactly the structure high-band duality answers use.

Using the relation correctly

Use $\lambda=h/mv$ with momentum in $\text{kg m s}^{-1}$, and remember that a larger momentum gives a shorter wavelength. If a voltage is given instead of a speed, first find the speed from $qV=\tfrac{1}{2}mv^2$, then the momentum.