Wave-Particle Duality
Light and matter travel like waves but arrive like particles. Fire them through a double slit one at a time and watch both behaviours at once.
Double slit, one particle at a time
Each electron lands as a single dot at a random spot. Where it can land is set by a wave that passes through both slits. Watch the fringes build up from the dots.
Side view: source, double slit, screen
Screen, face on. Bars count hits; the line is the predicted |ψ|².
With only a few electrons the dots look random. Keep going: after a few hundred, stripes appear where the waves from the two slits add up.
de Broglie wavelength calculator
Every moving object has a wavelength λ = h / (m v). Choose an object or type your own mass and speed.
Comparable to the spacing of atoms in a crystal (about 10⁻¹⁰ m), so crystals diffract it. This is how electron microscopes and neutron scattering work.
Waves that arrive as particles
Send electrons through two slits so slowly that only one is in the apparatus at a time. Each electron makes a single sharp dot on the screen, just like a particle would. Yet after thousands of dots the screen shows bright and dark stripes, an interference pattern that only waves make.
Quantum mechanics describes each electron by a wavefunction ψ. The wave passes through both slits and interferes with itself. The probability of detecting the electron at a point is |ψ|², so each individual landing spot is random, but the overall pattern is fixed. Bright fringes are places where the two parts of ψ arrive in step; dark fringes are where they cancel.
Add a detector that records which slit the electron used and the fringes disappear, leaving two plain bands. Knowing the path means only one part of ψ reaches the screen, so there is nothing to interfere with. You can see a wave pattern or which-path information, never both.
This trade-off is tied to the Heisenberg uncertainty principle: position and momentum cannot both be known exactly, Δx · Δp ≥ ħ/2. Pinning down which slit (Δx small) disturbs the sideways momentum by enough to wash out the fringes.
Key formulas
de Broglie wavelength
λ = h / p = h / (m v)
- h
- Planck constant, 6.626 × 10⁻³⁴ J·s
- p = m v
- momentum of the particle
Photon energy and momentum
E = h f, p = h / λ = E / c
- f
- frequency; c = speed of light
Double-slit fringe spacing
Δy = λ L / d
- L
- distance from slits to screen
- d
- slit separation
Uncertainty principle
Δx · Δp ≥ ħ / 2
- ħ
- h / 2π = 1.055 × 10⁻³⁴ J·s
About the model
The simulation uses a simplified screen pattern: each slit alone gives a broad band, and with both slits open the two amplitudes interfere with fringe spacing proportional to λ / d. Real patterns have the same structure.How to use this visualizer
- 1Watch the side view: without a detector each particle spreads out as a wave from the source and then from both slits.
- 2Look at the screen. Early on the dots look random; after a few hundred hits the stripes appear. Raise the rate to speed things up.
- 3Compare the bars (counted hits) with the pink line (predicted |ψ|²). They match better the more particles you collect.
- 4Change the wavelength or slit separation. Longer λ or smaller d spreads the fringes apart.
- 5Turn on the which-path detector. Each particle now takes one definite slit, a lamp flashes, and the pattern becomes two bands.
- 6Use the calculator to compare wavelengths of an electron, a neutron and a baseball.
How big are matter waves?
| Object | Mass | λ = h / (m v) | What it means |
|---|---|---|---|
| Electron at 10⁶ m/s | 9.11 × 10⁻³¹ kg | 7.3 × 10⁻¹⁰ m | Atom-sized, diffracts off crystals |
| Thermal neutron, 2200 m/s | 1.67 × 10⁻²⁷ kg | 1.8 × 10⁻¹⁰ m | Used to probe crystal structure |
| C60 molecule, 200 m/s | 1.20 × 10⁻²⁴ kg | 2.8 × 10⁻¹² m | Interference seen in 1999 |
| Baseball, 40 m/s | 0.145 kg | 1.1 × 10⁻³⁴ m | Far too small to ever detect |
The evidence
Light as a wave
Light as particles
Matter as a wave
One at a time
Practice problems
Problem 1: Electron wavelength
Show solutionHide solution
p = m v = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.82 × 10⁻²⁴ kg·m/s
λ = h / p = 6.626 × 10⁻³⁴ / 1.82 × 10⁻²⁴
λ = 3.6 × 10⁻¹⁰ m = 0.36 nm
λ ≈ 0.36 nm, comparable to 0.3 nm, so yes, it diffracts
Problem 2: Why baseballs do not interfere
Show solutionHide solution
p = 0.145 × 40 = 5.8 kg·m/s
λ = 6.626 × 10⁻³⁴ / 5.8 = 1.14 × 10⁻³⁴ m
sin θ ≈ λ / a = 1.14 × 10⁻³⁴ / 1 m
θ ≈ 10⁻³⁴ rad, utterly undetectable
The wave nature is still there, but for everyday masses the wavelength is about 10¹⁹ times smaller than a proton.
Discussion
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