The photoelectric effect showed that light — the classic “wave” — can act like a particle. A young physicist named Louis de Broglie asked the daring flip-side question: if waves can behave like particles, can particles behave like waves? His answer, tucked into his 1924 PhD thesis, was yes — every moving object has a wavelength, set by its momentum. It sounds absurd for a cricket ball, but for something as light as an electron the effect is real and measurable, and it’s the reason electron microscopes exist.
📘 What you need to know
De Broglie proposed that all moving particles have an associated matter wave
The de Broglie wavelength is the wavelength associated with a moving particle
It’s set by the particle’s momentum: λ = h/p = h/mv
Larger momentum (heavier or faster) → shorter wavelength
Linked to kinetic energy by λ = h / √(2mEk)
For an electron accelerated through a p.d. V: Ek = eV, so λ = h/√(2meV)
Everyday objects have wavelengths far too small to ever notice (~10−34 m)
Wave behaviour (diffraction) only shows up when λ is comparable to the gap size — e.g. electrons through an atomic lattice
Electron diffraction is the direct experimental proof that particles have wave properties
The matter-wave idea
De Broglie’s leap was to take the photon relationships for light and apply them, in reverse, to matter. A photon’s momentum is p = h/λ. Rearrange it and you get a wavelength from a momentum — and de Broglie said that rule works for anything with momentum, not just light.
The de Broglie wavelengthλ = h / p = h / mvλ = de Broglie wavelength (m) • h = Planck’s constant • p = mv = momentum
The relationship is an inverse one: the more momentum a particle has, the shorter its wavelength. A slow electron has a long, easy-to-diffract wavelength; a fast one has a short, harder-to-diffract wavelength.
A low-momentum particle (top) has a long, stretched wavelength; a high-momentum particle (bottom) has a short, tightly-packed one. That’s λ = h/p in a picture.
Linking to kinetic energy
Exam questions often give you an energy rather than a speed. Kinetic energy and momentum are connected by Ek = p2/2m, which rearranges to p = √(2mEk). Slot that into λ = h/p:
de Broglie wavelength from kinetic energyλ = h / √(2mEk)for an electron accelerated through p.d. V: Ek = eV → λ = h / √(2meV)
Accelerate through V
Ek = eV
Gain momentum p
λ = h/p
Matter wavelength
WE 1
An electron is accelerated from rest through a potential difference of 100 V. Calculate its de Broglie wavelength. (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Step 1 — kinetic energy gainedEₖ = eV = (1.60 × 10⁻¹⁹)(100) = 1.60 × 10⁻¹⁷ JStep 2 — momentum from p = √(2mEₖ)p = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁷)p = 5.40 × 10⁻²⁴ kg m s⁻¹Step 3 — de Broglie wavelengthλ = h/p = (6.63 × 10⁻³⁴) / (5.40 × 10⁻²⁴)λ = 1.2 × 10⁻¹⁰ mThat’s about 0.12 nm — roughly the spacing between atoms in a crystal! That match is exactly why electrons diffract through crystals. Work energy → momentum → wavelength, one step at a time.
Why we don’t see everyday objects diffract
Plug a cricket ball into λ = h/mv and you get a wavelength so unimaginably tiny that no gap in the universe could ever diffract it. The wave nature is always there — it’s just hopelessly small for anything bigger than a subatomic particle.
WE 2
(a) A cricket ball of mass 0.16 kg is bowled at 40 m s−1. Calculate its de Broglie wavelength. (b) Explain why the ball never shows wave behaviour. (h = 6.63 × 10−34 J s)
(a) Step 1 — de Broglie wavelengthλ = h/mv = (6.63 × 10⁻³⁴) / (0.16 × 40)λ = 1.0 × 10⁻³⁴ m(b) why no wave behaviour?
The wavelength is ~10⁻³⁴ m, unimaginably smaller than any gap or object.
no aperture is anywhere near this size, so diffraction can never be observedFor comparison, an atom is ~10⁻¹⁰ m — still 10²⁴ times BIGGER than this wavelength. The maths says matter waves exist for everything; the numbers say you’ll only ever detect them for the very lightest particles.
Electron diffraction: the proof
The clincher came from firing electrons at a thin film of graphite. If electrons were pure particles, they’d just make a bright spot. Instead they produce concentric diffraction rings on the screen — the unmistakable signature of a wave passing through the regularly-spaced atomic layers, which act like a diffraction grating.
Electrons fired through graphite spread out as waves and form rings on the screen — something only a wave can do. Speed the electrons up (higher voltage) and their wavelength shrinks, so the rings shrink too.
Watch the cause-and-effect chain examiners love: turn up the voltage → electrons move faster → more momentum → shorter de Broglie wavelength → less diffraction → smaller rings. Every link follows from λ = h/p. If you can recite that chain both forwards and backwards, you can answer any electron-diffraction question they throw at you.
WE 3
A proton and an electron are accelerated from rest through the same potential difference. Determine the ratio of their de Broglie wavelengths, λproton : λelectron. (mp = 1.67 × 10−27 kg, me = 9.11 × 10−31 kg)
Step 1 — same p.d. means same kinetic energy
Both gain Eₖ = eV, the same for each (same charge, same V).
Step 2 — wavelength in terms of massλ = h/√(2mEₖ), so λ ∝ 1/√mStep 3 — take the ratioλₚ/λₑ = √(mₑ/mₚ) = √(9.11 × 10⁻³¹ / 1.67 × 10⁻²⁷)λₚ : λₑ = 0.023 : 1The heavier proton has the SHORTER wavelength — about 43× shorter than the electron’s. The key insight: equal p.d. gives equal KE, so the only thing left that differs is the mass, and λ ∝ 1/√m.
⚛ Working a de Broglie question
Speed given?λ = h/mv.
Kinetic energy given?λ = h/√(2mEk).
Accelerated through a p.d.? First Ek = eV, then use the energy form.
Ratio questions? Cancel the shared quantities; often λ ∝ 1/√m for equal energies.
Will it diffract? Compare λ to the gap size — similar sizes means yes.
Voltage ↑? faster → more p → smaller λ → smaller rings.
💡 Top tips
Pick the right form: λ = h/mv for speed, λ = h/√(2mEk) for energy.
Larger momentum → shorter wavelength (it’s an inverse relationship).
For an electron through a p.d.: Ek = eV first, then find λ.
Diffraction is only observable when λ ≈ the gap size (atomic spacing for electrons).
For equal accelerating voltages, λ ∝ 1/√m — heavier means shorter.
⚠ Common mistakes
Forgetting to square-root: p = √(2mEk), not 2mEk
Using λ = h/mv when only the energy is given — convert first
Thinking bigger momentum means bigger wavelength — it’s the opposite
Saying everyday objects have no wavelength — they do, it’s just absurdly tiny
Mixing up the electron mass and proton mass in ratio questions
Forgetting Ek = eV for a particle accelerated through a p.d.
Quick recap: De Broglie said every moving particle is also a wave, with wavelength λ = h/p = h/mv — an inverse link, so more momentum means a shorter wavelength. Using Ek = p2/2m gives λ = h/√(2mEk), and for a particle accelerated through a p.d., Ek = eV. Everyday objects have wavelengths so tiny (~10−34 m) they never diffract, but electrons have wavelengths near atomic spacing — so firing them through graphite gives diffraction rings, the direct proof that matter has a wave nature.
Now we have both halves of the strange truth: light (a wave) acts like particles, and electrons (particles) act like waves. Neither picture alone is the whole story — everything in the quantum world carries both natures at once, showing whichever one the experiment asks for. That deep idea has its own name, and it’s the theme of the next page: Wave-Particle Duality.
de Broglie calculations not sticking?
Book a free meeting and we’ll drill λ = h/p, the kinetic-energy form, and the electron-diffraction cause-and-effect chain.
