The Wave Nature of Matter
De Broglie's relationship applies to absolutely everything with momentum — a thrown ball, a car, a person walking. So why has nobody ever seen a car diffract around a lamp post? The answer is entirely a matter of scale, and it explains why electrons, of all things, are the natural choice for demonstrating matter waves.
A relationship that applies to everything — but shows up almost nowhere
The de Broglie relationship, $\lambda = h/p$, doesn't come with any exceptions or restrictions on what kind of particle it applies to. It holds for electrons, for protons, for whole atoms, and — in principle — for a thrown ball or a moving car, since all of these have momentum. And yet electron diffraction rings are a standard laboratory demonstration, while nobody has ever observed a car, or even a dust grain, diffracting around an obstacle. The relationship is universal; the observable consequences are not. The reason lies entirely in just how small the Planck constant $h$ is.
The condition for wave effects to show up
Wave behaviour — diffraction, interference, anything that only a wave can produce — only becomes noticeable when a particle's de Broglie wavelength is comparable to a relevant length scale that the particle is interacting with. For electron diffraction, that relevant length scale is the spacing between atoms in the graphite crystal lattice, of order $0.1$–$0.3\,\text{nm}$ ($10^{-10}\,\text{m}$). The worked example in the de Broglie lesson found an electron accelerated through 2500 V has a wavelength of $2.46\times10^{-11}\,\text{m}$ — genuinely comparable to, if a little smaller than, that atomic spacing. That closeness in scale is exactly why electron diffraction produces a clear, visible pattern, while nothing remotely similar ever shows up for a ball, a car, or a person.
It's worth being precise about what's changing and what isn't here. The de Broglie relationship $\lambda=h/p$ applies identically to the ball and to the electron — there's no separate "quantum version" of the formula for small particles. What changes is only whether the resulting wavelength happens to be comparable to anything the particle interacts with. For the ball, the answer is emphatically no; for the electron passing through a crystal lattice, the answer is yes.
Why the electron specifically
Given that the comparable-scale condition is what matters, the obvious question is why electrons, rather than any other particle, are the standard choice for demonstrating matter waves in the laboratory. Two practical reasons make electrons especially convenient.
First, mass: the closed form $\lambda = h/\sqrt{2meV}$ shows that, for a fixed accelerating voltage $V$, a lighter mass $m$ gives a longer wavelength. That's the useful direction here — it means a modest, easily achievable voltage already gives an electron a wavelength down at atomic scale, without needing enormous voltages. An electron is roughly 1800 times lighter than a proton, so for the same accelerating voltage, an electron's de Broglie wavelength comes out far longer than a proton's would — comfortably into the range that atomic spacings can diffract, whereas a proton accelerated through the same voltage would need a much higher voltage still to reach a comparably useful wavelength.
Second, practicality: electrons are charged, so they are easily produced (thermionic emission from a heated filament, exactly as in an electron gun), easily accelerated to a precise, controllable speed using a known voltage, and easily focused and detected. No other particle combines "light enough for an easily-achievable wavelength" with "simple to produce, accelerate and control in an ordinary laboratory" as conveniently as the electron does — which is exactly why electron diffraction, rather than (say) diffraction of protons or whole atoms, became the standard classroom and historical demonstration of matter waves.
Wave-particle duality is not a special property of light or of electrons specifically — it's a general feature of all matter and radiation. Every particle has both a wave character (a de Broglie wavelength, set by its momentum) and a particle character (a definite energy, momentum, and — for matter — mass). Which character dominates in a given situation depends entirely on scale: the wave nature only becomes significant, and only produces observable effects like diffraction, when the de Broglie wavelength is comparable to the physical scale of whatever the particle is interacting with. At everyday, macroscopic scales, the particle description is all that's ever needed; at atomic scales, both descriptions matter.
Key takeaways
- The de Broglie relationship $\lambda=h/p$ applies to every particle with momentum, without exception — including everyday macroscopic objects.
- Because $h$ is so small, everyday objects have de Broglie wavelengths many orders of magnitude smaller than themselves or anything they interact with, so no wave effect is ever observable for them.
- Wave behaviour only becomes noticeable when the de Broglie wavelength is comparable to a relevant length scale the particle interacts with — atomic spacing, for electron diffraction.
- Electrons are especially convenient for demonstrating matter waves: being roughly 1800 times lighter than a proton, a modest accelerating voltage already gives them a wavelength down at atomic scale.
- Electrons are also easy to produce, accelerate to a controllable speed, and focus in an ordinary laboratory, unlike heavier or uncharged particles.
- The underlying relationship never changes between the macroscopic and atomic cases — only whether the resulting wavelength happens to be comparable to something the particle interacts with.
- Wave-particle duality is a universal principle applying to all matter and radiation; which character (wave or particle) dominates depends entirely on scale.