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.

Using the relation correctly

Use $\lambda=h/mv$ with momentum in 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.