Wave–Particle Duality and the de Broglie Wavelength

Introduction

Light, long regarded as a wave, turned out to arrive in particle-like photons. In 1924 Louis de Broglie asked the reverse question: if waves can behave like particles, can particles behave like waves? His answer — that every moving particle has an associated wavelength — was confirmed within three years and became the foundation on which Schrödinger built wave mechanics. This page shows how to calculate matter wavelengths and why they matter so much for electrons.

Core explanation

The de Broglie relation. For a photon, E = hν and E = pc, so the momentum of a photon is p = h/λ. De Broglie proposed that the same relation applies to matter:

λ = h/p = h/(mv)

where p is the momentum of the particle, m its mass and v its speed. The equation links a particle property (momentum) with a wave property (wavelength), with the Planck constant as the conversion factor.

Why we do not notice matter waves in daily life. Because h is tiny, the wavelength is significant only when the momentum is very small. A 0.057 kg tennis ball moving at 50 m s⁻¹ has λ ≈ 2 × 10⁻³⁴ m, far smaller than an atomic nucleus, so no diffraction can ever be observed. An electron has a mass of only 9.109 × 10⁻³¹ kg, and at typical atomic speeds its wavelength is a few hundred picometres — the same size as atoms and bond lengths.

Electrons accelerated through a potential difference. An electron accelerated from rest through a potential difference V gains kinetic energy eV = p²/2m, so p = √(2meV) and

λ = h/√(2mₑeV)

Numerically, λ ≈ 1.226/√V nm with V in volts. A 100 V beam gives λ ≈ 0.123 nm, comparable to the spacing between atoms in a crystal — which is why crystals act as natural diffraction gratings for electrons.

Experimental evidence. In 1927 Davisson and Germer directed 54 eV electrons onto a nickel crystal and found a strong reflected beam at a particular angle, exactly as expected for waves of wavelength 0.167 nm. G. P. Thomson independently observed diffraction rings when electrons passed through thin metal foils. Later experiments sent electrons through two slits one at a time: each electron arrives at a single point on the screen, yet the build-up of many arrivals forms an interference pattern. Each electron behaves as a wave while travelling and as a particle when detected.

Beyond electrons. Neutrons, helium atoms and even large molecules diffract. In 1999 fullerene molecules (C₆₀, mass 720 u) travelling at about 220 m s⁻¹ produced a clear interference pattern with λ ≈ 2.5 pm. Wave–particle duality is a universal property of matter, not a peculiarity of electrons.

Why this matters for chemistry. An electron confined to an atom must fit its wave into a small region. Only certain wavelengths "fit", which is the physical root of quantised energy levels. The shorter the confining region, the shorter the wavelength required, the larger the momentum and hence the larger the kinetic energy. This idea explains why small atoms hold electrons tightly and why confinement raises energy — themes that recur throughout this unit.

Formulae

λ = h/p = h/(mv). For kinetic energy Eₖ: λ = h/√(2mEₖ). For an electron accelerated through V volts: λ = h/√(2mₑeV) ≈ 1.226/√V nm.

Step-by-step reasoning

To calculate a de Broglie wavelength:

1. Convert mass to kilograms (1 u = 1.661 × 10⁻²⁷ kg). 2. Find the momentum: p = mv, or p = √(2mEₖ) if energy is given. 3. Convert any energy in eV to joules first. 4. Divide h by p to obtain λ in metres. 5. Compare λ with the relevant length scale to judge whether wave effects will be observable.

Visual explanation

Imagine the two-slit screen as the simulation shows it: dots appear one by one at apparently random positions. After a few dozen arrivals nothing is visible, but after thousands the dots have gathered into bright and dark stripes. The stripes are the wave; each dot is the particle.

Real-world analogy

A water wave spreads out and interferes after passing two gaps in a harbour wall, but a boat is too large compared with the wavelength to be affected. Electrons are like the water waves passing through atomic-sized gaps; everyday objects are like the boat.

Real-world example

Electron microscopes exploit the short de Broglie wavelength of fast electrons. Electrons accelerated through 200 kV have wavelengths of a few picometres, allowing transmission electron microscopes to image individual columns of atoms in crystals and to determine protein structures by cryo-electron microscopy.

Why?

Why do heavier particles have shorter wavelengths at the same speed? Wavelength is inversely proportional to momentum, and momentum is proportional to mass. Doubling the mass at the same speed doubles the momentum and halves the wavelength.

Common misconception

"An electron is a tiny ball that moves along a wavy path." The wave is not a path; it describes where the electron is likely to be found. The electron does not follow a trajectory at all while it behaves as a wave.

Worked example

Question: Calculate the de Broglie wavelength of an electron accelerated from rest through 100 V.

Reasoning: Eₖ = eV = 1.602 × 10⁻¹⁹ × 100 = 1.602 × 10⁻¹⁷ J. p = √(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁷) = 5.40 × 10⁻²⁴ kg m s⁻¹. λ = 6.626 × 10⁻³⁴ ÷ 5.40 × 10⁻²⁴ = 1.23 × 10⁻¹⁰ m.

Answer: λ ≈ 0.123 nm, similar to interatomic spacings in crystals.

Quick check

1. If the speed of an electron is doubled, what happens to its de Broglie wavelength? Answer: Momentum doubles, so the wavelength is halved.

Exam focus

Always check units: mass in kg, energy in J, answer in m before converting to nm or pm. Be ready to explain why electron diffraction by crystals proves wave behaviour, and why the wavelength of a macroscopic object is unobservable.

Advanced insight

The de Broglie relation applies to the particle's momentum, and for very fast electrons the relativistic momentum must be used. At 200 kV the relativistic correction shortens the wavelength from about 2.7 pm to 2.5 pm — a difference that electron microscopists must include when calibrating their instruments.

Summary

De Broglie proposed that every moving particle has a wavelength λ = h/p. The wavelength is negligible for everyday objects but comparable with atomic dimensions for electrons, which is why electrons show diffraction and interference. Experiments with electrons, neutrons, atoms and even C₆₀ molecules confirm wave–particle duality. Fitting an electron's wave into a confined region leads directly to quantised energy.

Practice questions

1. Calculate the de Broglie wavelength of an electron moving at 1.0 × 10⁶ m s⁻¹. Answer: λ = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 1.0 × 10⁶) ≈ 7.3 × 10⁻¹⁰ m, or 0.73 nm. 2. A H₂ molecule (mass 3.35 × 10⁻²⁷ kg) moves at 1900 m s⁻¹. Find its wavelength. Answer: p = 6.37 × 10⁻²⁴ kg m s⁻¹, so λ = 6.626 × 10⁻³⁴ ÷ 6.37 × 10⁻²⁴ ≈ 1.0 × 10⁻¹⁰ m. 3. A proton and an electron have the same speed. Which has the longer wavelength, and by roughly what factor? Answer: The electron, by a factor equal to the mass ratio, about 1836. 4. Explain how the two-slit experiment with single electrons shows both wave and particle character. Answer: Each electron is detected at a single point (particle behaviour), but many arrivals build up an interference pattern that requires each electron to pass through both slits as a wave.