Davisson-Germer experiment — electron diffraction: Visual walkthrough

Step 1 — Give the electron a push (energy)

WHAT. We start an electron at rest and let a voltage $V$ pull it. "Voltage" is just a hill of electric energy; the electron rolls down it and speeds up.

WHY. We need to control the electron's speed precisely, because speed sets momentum, and momentum sets the wavelength we want to measure. Voltage is the one knob we can dial exactly.

PICTURE. The electron slides down the voltage hill and arrives with speed $v$ .

Figure — Davisson-Germer experiment — electron diffraction

Here $e = 1.6 \times 1 0^{- 19}$ C is a fixed charge, $m = 9.11 \times 1 0^{- 31}$ kg is a fixed mass — only $V$ and $v$ change.

Step 2 — Turn speed into momentum

WHAT. We rewrite the energy equation so that momentum $p$ (the "quantity of motion", $p = m v$ ) appears alone.

WHY. de Broglie's rule is written in terms of $p$ , not $v$ . So we must translate. Momentum is the natural currency of a wave-particle, because a heavier-but-slow and a light-but-fast particle with the same $p$ behave identically here.

PICTURE. The same electron, now labelled by its momentum arrow $p$ pointing along its motion.

Step 3 — Attach the wavelength (de Broglie)

WHAT. We finally invoke the hypothesis on trial: every momentum $p$ carries a wavelength $λ = h / p$ .

WHY. This is the claim the whole experiment exists to verify. We are not proving it yet — we are computing the number it predicts, so we can later compare it to a measurement. The tool is de Broglie's relation because it is the only formula that links a particle's momentum to a wave's wavelength.

PICTURE. The momentum arrow becomes a little wave whose spacing is $λ$ : fast electron $\to$ short wave, slow electron $\to$ long wave.

Convert-to-Å

The

12.27

already bundles

h

m

e

and the Å conversion, so you only feed it

V

in volts.

Step 4 — Meet the crystal ruler

WHAT. We fire this wave at a nickel single crystal. Its surface atoms sit in evenly spaced rows a distance $a$ apart. That spacing is our ruler.

WHY. To measure a wave you need something the same size as its wavelength — like using a comb, not a fence, to feel the teeth. Nickel's row spacing $a \approx 2.15$ Å $\approx λ$ , so the crystal can act as a natural diffraction grating. Anything much coarser would show nothing.

PICTURE. The incoming wavefronts (flat lines) arrive at a row of atoms spaced $a$ ; each atom re-radiates a little wave.

Step 5 — Count the extra path (where the peak comes from)

WHAT. Look at the wave leaving at angle $ϕ$ (measured from the incident beam). The ripple from one atom travels an extra distance $a sin ϕ$ compared to its neighbour.

WHY. Two ripples add up brightly only when they arrive in step — crest on crest. That happens exactly when the extra path is a whole number of wavelengths. This is the single geometric fact that produces the diffraction peak. We use $sin ϕ$ (not $cos$ or $tan$ ) because the extra path is the side of a right triangle opposite the angle $ϕ$ , with the spacing $a$ as the hypotenuse's base — "opposite over hypotenuse" is precisely $sin$ .

PICTURE. A zoom on two neighbouring atoms; the amber segment is the extra path $a sin ϕ$ .

For nickel with $a = 2.15$ Å, at the observed $ϕ = 5 0^{\circ}$ , $n = 1$ : $λ = a sin ϕ = 2.15 \times sin 5 0^{\circ} = 1.65 \overset{˚}{A} .$

Step 6 — The two roads meet

WHAT. Put the two independent numbers side by side.

WHY. This is the verdict. Road 1 (voltage $\to$ de Broglie) and Road 2 (angle $\to$ grating) never talked to each other, yet:

Road	Input	$λ$
Theory ( $h / 2 m e V$ )	$V = 54$ V	$1.67$ Å
Experiment ( $a sin ϕ$ )	$ϕ = 5 0^{\circ}$	$1.65$ Å

PICTURE. Both arrows landing on essentially the same wavelength mark.

Step 7 — Edge cases: what if we change the knob?

WHAT. We check the degenerate and limiting situations so no reader is ambushed.

WHY. A derivation you can't push to its extremes isn't understood. Watch what $a sin ϕ = nλ$ does at the boundaries.

PICTURE. Three panels — high voltage (peak slides in), low voltage, and the "no solution" wall.

The one-picture summary

This single blueprint threads the whole story: voltage $V$ on the left drives the electron, which acquires momentum $p = 2 m e V$ , which de Broglie converts to wavelength $λ = h / p$ (top road). That same wave strikes the crystal ruler $a$ and the geometry $a sin ϕ = nλ$ reads $λ$ straight off the peak angle (bottom road). The two roads meet on one number — proof of matter waves.

Recall Feynman retelling — tell it back in plain words

We rolled an electron down a voltage hill so we'd know exactly how fast it went. Fast means "lots of momentum". De Broglie whispers that anything with momentum secretly carries a wave, and a tiny wave for a fast electron. We threw that wave at nickel, whose atoms are lined up like teeth on a comb spaced just right. Ripples bouncing off neighbouring atoms only cheer together — make a bright bump — when the extra distance between them is a whole number of wavelengths. We measured the angle of that bump, did the "extra path = $a sin ϕ$ " arithmetic, and out popped a wavelength of $1.65$ Å. Meanwhile the voltage-and-momentum arithmetic predicted $1.67$ Å. Two totally different calculations, same answer. That "same answer" is the entire experiment: electrons are waves.

Recall Self-test

Why $sin ϕ$ and not $tan ϕ$ in the extra-path term? ::: The path difference is the side of a right triangle opposite $ϕ$ with the atom spacing as the reference length — opposite/hypotenuse is $sin$ . What makes the peak disappear entirely? ::: If $λ > a$ , then $sin ϕ = λ / a > 1$ has no solution — no order can constructively interfere. Which two independent quantities must agree for matter waves to be "proven"? ::: The theory-road wavelength $h / 2 m e V$ and the experiment-road wavelength $a sin ϕ$ .

Connections: de Broglie hypothesis · Bragg's law · Wave-particle duality · Photoelectric effect (the complementary half) · Electron microscope (this effect put to work) · Heisenberg uncertainty principle.