Matter waves

Light behaves like a particle in the photoelectric effect — photons of energy hνh\nu knock electrons out of a metal. In 1924 Louis de Broglie asked the reverse question: if a wave can behave like a particle, can a particle behave like a wave? He proposed that every moving particle carries a wavelength, λ=h/p\lambda = h/p, turning momentum (a particle idea) and wavelength (a wave idea) into two faces of the same coin. For everyday objects this wavelength is unimaginably small and undetectable, but for an electron accelerated through a modest voltage it lands in the X-ray range — comparable to the spacing between atoms in a crystal. That coincidence is what let Davisson and Germer catch electrons diffracting off a nickel crystal in 1927, turning de Broglie's guess into confirmed physics and completing the symmetry: radiation and matter both show wave–particle duality.

The core insight

Every moving particle of momentum pp has an associated de Broglie wavelength λ=h/p\lambda = h/p, confirmed for electrons by the Davisson–Germer diffraction experiment.

The core idea

The de Broglie hypothesis. De Broglie reasoned by symmetry: photons of momentum pp have wavelength λ=h/p\lambda = h/p (from the photoelectric-effect topic); if that link runs both ways, a particle of momentum p=mvp = mv should equally have a wave character of wavelength

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

where: λ\lambda is the de Broglie wavelength, hh is Planck's constant, mm is the particle's mass, and vv its speed. This is a single formula covering any moving mass — a cricket ball, a car, or an electron — though, as the worked example below shows, only very light particles give a wavelength large enough to matter.

The accelerated-electron case. The most exam-relevant scenario is an electron started from rest and accelerated through a potential difference VV. The electric field does work eVeV on the electron, all of which becomes kinetic energy, so its momentum and wavelength can be written purely in terms of VV — this is worked out fully in the derivation below.

Experimental confirmation. De Broglie's idea was only a hypothesis until 1927, when Clinton Davisson and Lester Germer fired a beam of electrons at a nickel crystal and observed a diffraction pattern — intensity maxima and minima at specific angles, just as X-rays diffract off the regularly spaced planes of atoms in a crystal. A purely classical particle cannot diffract; only a wave can. The wavelength extracted from the diffraction pattern matched the de Broglie wavelength calculated from the electrons' known momentum, so the experiment confirmed the wave nature of matter. The syllabus needs only this qualitative picture and conclusion — no Bragg-angle diffraction calculation is required. Diffraction has since been shown for neutrons and even large molecules, so wave–particle duality is a universal feature of matter, not a quirk of electrons.

Figure
Two symmetric arrows, one showing a photon wave behaving as a particle in the photoelectric effect and the other showing an electron particle behaving as a wave with de Broglie wavelength lambda equals h over p
Wave-particle duality runs both ways radiation (wave) acts as particle photon electron ejected photoelectric effect: E = hν matter (particle) acts as wave electron de Broglie wave lambda = h / p, confirmed by Davisson-Germer
Derivation AWavelength of an electron accelerated through a potential difference V

The goal is to find a formula for the de Broglie wavelength of an electron accelerated from rest through a potential difference VV, using only the work-energy relation and the momentum-kinetic-energy link, so that λ\lambda can be computed directly from VV without separately finding the speed first.

Step 1. The electric field does work eVeV on the electron as it accelerates from rest, and this work becomes kinetic energy: eV=12mv2eV = \frac12 mv^2. Here ee is the electron's charge and mm its mass, so this is just the work-energy theorem applied to an electron starting at rest.

Step 2. Rewrite the kinetic energy in terms of momentum, using p=mvp = mv so that v=p/mv = p/m and KE=12mv2=p22mKE = \frac12 mv^2 = \frac{p^2}{2m}. Substituting into Step 1 gives eV=p22meV = \dfrac{p^2}{2m}, so p2=2meVp^2 = 2meV and p=2meVp = \sqrt{2meV}.

Step 3. Insert this momentum into the de Broglie formula λ=h/p\lambda = h/p from the core idea above:

λ=h2meV\lambda = \frac{h}{\sqrt{2meV}}

Step 4. Substitute the known constants h=6.63×10−34 J ⋅ sh = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}, me=9.11×10−31 kgm_e = 9.11\times10^{-31}\ \text{kg} and e=1.6×10−19 Ce = 1.6\times10^{-19}\ \text{C} and simplify. This collapses to the numerical shortcut, valid whenever VV is in volts and λ\lambda comes out in nanometres:

λ≈1.227V nm\lambda \approx \frac{1.227}{\sqrt{V}}\ \text{nm}

This is the formula used directly in the worked example below, and it shows immediately that raising the accelerating voltage shrinks the wavelength — a higher-energy electron behaves more like a compact particle and less like a spread-out wave.

Worked example

Find the de Broglie wavelength of an electron accelerated through a potential difference of 100 V100\ \text{V}, and compare it with the de Broglie wavelength of a 0.2 kg0.2\ \text{kg} ball moving at 10 m/s10\ \text{m/s}.

Step 1. Find the kinetic energy gained by the electron: KE=eV=(1.6×10−19)(100)=1.6×10−17 JKE = eV = (1.6\times10^{-19})(100) = 1.6\times10^{-17}\ \text{J}.

Step 2. Find the electron's momentum from KE=p2/2mKE = p^2/2m, so p=2m⋅KEp = \sqrt{2m\cdot KE}: p=2×(9.11×10−31)×(1.6×10−17)=2.915×10−47=5.40×10−24 kg ⋅ m/sp = \sqrt{2 \times (9.11\times10^{-31}) \times (1.6\times10^{-17})} = \sqrt{2.915\times10^{-47}} = 5.40\times10^{-24}\ \text{kg}\,\cdot\,\text{m/s}.

Step 3. Find the wavelength: λ=h/p=(6.63×10−34)/(5.40×10−24)=1.23×10−10 m=0.123 nm\lambda = h/p = (6.63\times10^{-34})/(5.40\times10^{-24}) = 1.23\times10^{-10}\ \text{m} = 0.123\ \text{nm}. This matches the quick-formula check, λ≈1.227/100=0.1227 nm\lambda \approx 1.227/\sqrt{100} = 0.1227\ \text{nm}, and is comparable to X-ray wavelengths and to atomic spacing in a crystal — exactly why such electrons diffract off nickel in the Davisson–Germer experiment.

Step 4. Now the ball, moving classically (not accelerated by a potential): pball=mv=(0.2)(10)=2 kg ⋅ m/sp_{ball} = mv = (0.2)(10) = 2\ \text{kg}\,\cdot\,\text{m/s}, so λball=h/p=(6.63×10−34)/2=3.32×10−34 m\lambda_{ball} = h/p = (6.63\times10^{-34})/2 = 3.32\times10^{-34}\ \text{m}.

Final answer: the 100 V100\ \text{V} electron has λ≈0.123 nm\lambda \approx 0.123\ \text{nm}, detectable via crystal diffraction, while the ball's wavelength is ≈3.32×10−34 m\approx 3.32\times10^{-34}\ \text{m} — about twenty orders of magnitude smaller than a nucleus, and utterly undetectable, which is why we never see everyday objects diffract.

Common traps

  • Thinking λ=h/mv\lambda = h/mv applies only to electrons — it applies to any moving mass, but the wave nature is only observable for very small masses or momenta.
  • Skipping straight from VV to λ\lambda without going through momentum — you must first get p=2meVp = \sqrt{2meV} before applying λ=h/p\lambda = h/p.
  • Believing Davisson–Germer "proves electrons are waves, not particles" — it shows electrons exhibit wave-like diffraction under certain conditions, which is the essence of wave-particle duality, not a replacement of the particle picture.
  • Confusing photon momentum (p=h/λp = h/\lambda, wavelength gives momentum) with de Broglie momentum (p=mvp = mv, then λ=h/p\lambda = h/p, momentum gives wavelength) — the direction of reasoning is reversed between the two topics.

Ask the tutor: Want to practice de Broglie wavelength calculations for different particles, speeds, and accelerating voltages, or dig into why the Davisson–Germer angles match the crystal spacing?

Why don't we see a thrown cricket ball diffracting like light does?

Every moving object has a de Broglie wavelength, λ=h/mv\lambda = h/mv, but Planck's constant hh is extremely tiny, about 6.63×10−34 J ⋅ s6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}. For a macroscopic object like a 0.2 kg0.2\ \text{kg} ball moving at ordinary speeds, the momentum mvmv is large, so λ\lambda works out to around 10−34 m10^{-34}\ \text{m} — far smaller than any gap or slit that could diffract it, and completely undetectable. Only very light particles like electrons, with tiny momentum, get wavelengths large enough (comparable to atomic spacing) to show observable wave effects.

Does the Davisson-Germer result mean an electron is really a wave and not a particle?

No. It means an electron shows wave-like behaviour, such as diffraction, under the right conditions, while in other experiments (like tracks in a detector) it behaves as a localized particle. This is wave-particle duality: the same entity can display either character depending on what you measure, and neither description alone is complete. The syllabus asks you to know that the diffraction pattern matched the de Broglie prediction, confirming that matter genuinely has a wave aspect — not that particles have somehow stopped being particles.

Why does the formula for an accelerated electron need momentum, not just speed?

Because the de Broglie relation is naturally written in terms of momentum, λ=h/p\lambda = h/p, and momentum is what connects cleanly to the kinetic energy gained from the accelerating voltage through eV=p2/2meV = p^2/2m. You could work through speed instead (v=2eV/mv = \sqrt{2eV/m}, then p=mvp = mv), but going via KE=p2/2mKE = p^2/2m is the fastest route from a given voltage VV straight to p=2meVp = \sqrt{2meV} and then to λ\lambda, which is why this route is used in the standard derivation.

Before this

Where this leads