Davisson-Germer experiment — electron diffraction: Foundations

This page assumes you have seen nothing. Before we touch the parent note's equations, we build every symbol, picture, and word it uses — in an order where each new idea leans only on the ones before it.

0. The two mental pictures we keep colliding

The whole experiment is a fight between two ways of imagining an electron.

Why do we need both pictures? Because the experiment shows an electron is fired like a particle (a beam from a hot wire) but lands like a wave (a striped fan). Neither picture alone is enough.

1. Angle — how we point at a direction

Everything we measure is an angle, so we pin down what an angle is first.

Two different angles appear in the parent note and mixing them is the classic trap, so we name them carefully with a picture.

Formula Deriving the link between

$ϕ$ and $θ$ — from the picture Look at figure s02. The beam comes in making a glancing angle $θ$ with the plane, and reflects out making the same glancing angle $θ$ on the other side (mirror-like, "specular" scattering).

Step 1 — what is $ϕ$ ? $ϕ$ is the total kink between the incoming and outgoing beam. Because each beam sits $θ$ above the plane, and they open away from each other, the straight-through direction would be $18 0^{\circ}$ ; each beam is bent up by $θ$ from the surface, so the two beams are separated by $ϕ = 18 0^{\circ} - 2 θ .$ Why? Going from the incoming ray to the outgoing ray, you turn by $θ$ to reach the plane and another $θ$ to leave it — but measured as the deflection from straight ahead, the beams differ from a straight line by $2 θ$ , giving $18 0^{\circ} - 2 θ$ .
Step 2 — solve for $θ$ . Rearranging $ϕ = 18 0^{\circ} - 2 θ$ : $2 θ = 18 0^{\circ} - ϕ \Rightarrow θ = 9 0^{\circ} - \frac{ϕ}{2} .$
Check with numbers: the observed $ϕ = 5 0^{\circ}$ gives $θ = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ , a very grazing beam — exactly what the parent note uses.

2. Sine — turning an angle into a length ratio

To connect angles to path lengths we need one tool: the sine.

Key values to keep in your pocket:

$sin 0^{\circ} = 0$ (no slant → no extra path)
$sin 5 0^{\circ} \approx 0.766$
$sin 6 5^{\circ} \approx 0.906$
$sin 9 0^{\circ} = 1$ (biggest possible)

3. Wavelength $λ$ — the size of one ripple

Why we care: diffraction — the striped fan — only appears when $λ$ is comparable to the spacing between the slits (here, atoms). Too-small $λ$ and the wave sails through unbent; too-large and everything blurs. This is why the experiment needs a crystal, not a wide grating.

4. Atom spacings $a$ and $d$ — the "comb teeth"

A crystal is atoms stacked in a perfectly regular grid. Two spacings matter.

Now we connect each spacing to a wavelength — one relation per picture.

Why we need these: the spacings are the "slit spacing" that must be near $λ$ for diffraction, and each relation turns a measured angle into a wavelength.

5. Momentum $p$ and the de Broglie link

Why introduce $p$ before $λ = h / p$ ? Because you cannot claim a particle has a wavelength without first saying which property sets it — and the answer is its momentum.

6. Charge $e$ , voltage $V$ , and energy $e V$

How do we control the electron's momentum? By pushing it through a voltage.

7. Diffraction order $n$ — counting whole wavelengths

Why it exists: bright spots repeat. Waves reinforce whenever the extra path is $1, 2, 3 \dots$ full wavelengths — every one of those gives its own bright angle. Both bright conditions from section 4 carry this counter: $a sin ϕ = nλ$ and $2 d sin θ = nλ$ .

The prerequisite map

Read it top-to-bottom: the two roads (theory via de Broglie, experiment via the grating/Bragg picture) each produce a wavelength, and their agreement is the proof.

Recall checks

Recall Which angle is measured from the incoming beam?

The scattering angle $ϕ$ ::: yes — $ϕ$ is beam-to-beam; $θ$ is measured from the atomic plane.

Recall Derive

$θ$ from $ϕ$ . $θ = 9 0^{\circ} - ϕ /2$ ::: because $ϕ = 18 0^{\circ} - 2 θ$ for mirror-like scatter; rearrange to isolate $θ$ .

Recall What does

$sin ϕ$ physically give us here? The extra path length ::: it converts the beam's slant into the extra distance a wave travels between neighbouring atoms.

Recall If momentum

$p$ doubles, what happens to $λ$ ? It halves ::: because $λ = h / p$ , wavelength is inversely proportional to momentum.

Equipment checklist

Self-test: can you state each in one breath? Reveal to check.

What an angle is and its unit ::: the amount of turning between two directions, in degrees ( $36 0^{\circ}$ = full turn) or radians ( $2 π$ = full turn).
Why the radian matters ::: SI physics and calculators often expect radians; convert with $rad = deg \times π /180$ before using $sin$ .
The difference between $ϕ$ and $θ$ , and their link ::: $ϕ$ from the incident beam, $θ$ from the atomic plane; $ϕ = 18 0^{\circ} - 2 θ$ so $θ = 9 0^{\circ} - ϕ /2$ .
Definition of $sin ϕ$ ::: opposite side over hypotenuse in a right triangle; here it turns slant into extra path.
What wavelength $λ$ is ::: crest-to-crest distance of a wave; measured in ångström, $1 \overset{˚}{A} = 1 0^{- 10}$ m.
Difference between $a$ and $d$ ::: $a$ = surface row spacing (sideways); $d$ = interplanar spacing (layer depth).
The grating and Bragg conditions ::: $a sin ϕ = nλ$ (surface rows) and $2 d sin θ = nλ$ (stacked planes).
Meaning of momentum $p$ ::: mass times velocity, "how much motion."
The de Broglie relation ::: $λ = h / p$ — momentum sets a particle's wavelength.
What KE and $e V$ represent ::: KE = kinetic energy (energy of motion, $\frac{1}{2} m v^{2}$ ); $e V$ = the energy a charge $e$ gains through voltage $V$ .
The full theory-road chain ::: $e V = p^{2} /2 m \Rightarrow p = 2 m e V \Rightarrow λ = h / 2 m e V$ .
What the order $n$ counts ::: how many whole wavelengths fit the extra path; $n = 1, 2, 3 \dots$