Davisson-Germer experiment — electron diffraction: Exercises

Before we start, one shared toolbox — every symbol we will lean on, defined once so nothing is used unearned.

The figure above is your map: the two roads to $λ$ . Every exercise walks one road, or checks that both meet.

L1 — Recognition

(Can you spot which idea applies? No heavy algebra yet.)

A beam of electrons is fired at a nickel crystal and the detector records a sharp bump of current at one special angle. Which single word names this bump, and which model of the electron — particle or wave — is required to explain it?

Recall Solution 1.1

The bump is a diffraction peak. A stream of pure particles would smear out smoothly (like sand thrown at a wall), so only the wave model can produce a sharp intensity peak. This is the whole point of the experiment: it forces the wave picture onto matter. See Wave-particle duality.

Problem 1.2

Two students disagree. One says "Davisson–Germer proves light is a particle." The other says "it proves electrons are waves." Who is right?

Recall Solution 1.2

The second student. Davisson–Germer fires electrons (matter) and finds wave behaviour. It is the mirror image of the Photoelectric effect, which shows light (a wave) acting as particles. Together they are the two halves of duality.

L2 — Application

(Plug into the formulas correctly.)

Problem 2.1

Find the de Broglie wavelength of an electron accelerated through $V = 100$ V.

Recall Solution 2.1

Use the shortcut, because it already bakes in $h$ , $m$ , $e$ : $λ = \frac{12.27}{V} = \frac{12.27}{100} = \frac{12.27}{10} = 1.227 \overset{˚}{A}$ Why this tool? We are given a voltage and want a wavelength — the "theory road" formula is the exact bridge between them.

Problem 2.2

An electron has been accelerated so that its de Broglie wavelength is $λ = 1.00$ Å. What voltage produced it?

Recall Solution 2.2

Invert the shortcut. Start from $λ = 12.27/ V$ , so $V = 12.27/ λ$ , and squaring: $V = (\frac{12.27}{λ})^{2} = (\frac{12.27}{1.00})^{2} = 150.6 V$ Why square? $λ$ depends on $1/ V$ , so undoing it needs the inverse operations in reverse order: reciprocal, then square.

Problem 2.3

In the surface-grating picture the peak obeys $a sin ϕ = nλ$ . With row spacing $a = 2.15$ Å, first order $n = 1$ , and $λ = 1.67$ Å, find the scattering angle $ϕ$ .

Recall Solution 2.3

$sin ϕ = \frac{nλ}{a} = \frac{1 \times 1.67}{2.15} = 0.7767$ $ϕ = arcsin (0.7767) = 51. 0^{\circ}$ What is $arcsin$ doing here? $sin ϕ$ tells us the "height ratio" that this angle makes; $arcsin$ is the reverse question — "which angle has that sine?" — so it hands us $ϕ$ back.

L3 — Analysis

(Now the geometry and the sign/limit reasoning bite.)

Problem 3.1

The observed peak in the original experiment sits at $ϕ = 5 0^{\circ}$ . Using the surface-grating relation with $a = 2.15$ Å and $n = 1$ , find the experimentally-measured $λ$ . Then compare it to the theory-road value at $V = 54$ V and comment.

Recall Solution 3.1

Experiment road: $λ = a sin ϕ = 2.15 \times sin 5 0^{\circ} = 2.15 \times 0.766 = 1.647 \overset{˚}{A} \approx 1.65 \overset{˚}{A}$ Theory road: $λ = \frac{12.27}{54} = \frac{12.27}{7.348} = 1.670 \overset{˚}{A}$ The two independent roads give $1.65$ Å and $1.67$ Å — agreement to about $1%$ . That match is the proof of matter waves. The tiny residual gap is refraction of the electron as it crosses into the crystal's inner potential.

Problem 3.2

Convert the scattering angle $ϕ = 5 0^{\circ}$ into the Bragg glancing angle $θ$ , using $θ = 9 0^{\circ} - ϕ /2$ . Explain in words which angle is measured from what.

Recall Solution 3.2

$θ = 9 0^{\circ} - \frac{5 0 ^{\circ}}{2} = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ The reference matters: $ϕ$ is measured from the incident beam (the straight line the electrons came in on), while the Bragg $θ$ in Bragg's law is measured from the crystal plane itself. The figure below shows why bisecting $ϕ$ and subtracting from $9 0^{\circ}$ lands you on the plane's glancing angle.

Problem 3.3

If the accelerating voltage $V$ is increased, does the surface-grating peak move to a larger or smaller $ϕ$ ? Also state what happens in the extreme limit.

Recall Solution 3.3

Follow the chain of causes:

Larger $V$ → larger momentum $p = 2 m e V$ → smaller $λ$ (because $λ = h / p$ ).
In $a sin ϕ = nλ$ , a smaller $λ$ forces a smaller $sin ϕ$ , hence a smaller $ϕ$ for the same order $n$ .
Limiting case: push $V$ high enough that $λ < a / n$ can no longer be satisfied for that order — $sin ϕ$ would need to exceed... no, it just keeps shrinking toward $ϕ \to 0$ ; but the first-order peak crowds toward the incident beam and eventually the order becomes unresolvable and the peak effectively disappears. Always name the convention: here $ϕ$ is the surface-grating incident-beam angle.

L4 — Synthesis

(Combine several tools in one chain.)

Problem 4.1

Starting from scratch, predict the surface-grating peak angle for electrons accelerated through $V = 54$ V, using $a = 2.15$ Å. Show every link: voltage → $λ$ → $ϕ$ .

Recall Solution 4.1

Step 1 — voltage to wavelength (theory road): $λ = \frac{12.27}{54} = 1.670 \overset{˚}{A}$ Step 2 — wavelength to angle (grating road): $sin ϕ = \frac{λ}{a} = \frac{1.670}{2.15} = 0.7767 \Rightarrow ϕ = arcsin (0.7767) = 51. 0^{\circ}$ Step 3 — compare to reality: observed $ϕ \approx 5 0^{\circ}$ . The $1^{\circ}$ gap is inner-potential refraction. The prediction is essentially confirmed. This single chain is the Davisson–Germer argument end-to-end.

Problem 4.2

A researcher wants electrons whose wavelength is $λ = 0.10$ Å (comparable to hard X-rays, useful for an Electron microscope). What accelerating voltage is required, and what does this tell you about resolving power?

Recall Solution 4.2

Invert the shortcut: $V = (\frac{12.27}{λ})^{2} = (\frac{12.27}{0.10})^{2} = (122.7)^{2} = 15, 055 V \approx 1.5 \times 1 0^{4} V$ Interpretation: a modest $\sim 15$ kV gives a wavelength $\sim 5000 \times$ smaller than visible light. Because resolving power improves as wavelength shrinks, this is why electron microscopes see far finer detail than optical ones. (At this energy we are still non-relativistic: $15$ keV $≪ 511$ keV rest energy.)

L5 — Mastery

(Push to edge cases, cross-checks, and design.)

Problem 5.1

Show that the Bragg cross-check is consistent with the surface-grating answer. Using Ni(111) plane spacing $d = 2.03$ Å and $θ = 6 5^{\circ}$ , find the wavelength Bragg's law returns for $n = 1$ and $n = 2$ , and explain which order corresponds to the $\sim 1.65$ Å result.

Recall Solution 5.1

First order ( $n = 1$ ): $λ = \frac{2 d s i n θ}{n} = 2 (2.03) sin 6 5^{\circ} = 4.06 \times 0.9063 = 3.679 \overset{˚}{A}$ Second order ( $n = 2$ ): $λ = \frac{2 d s i n θ}{2} = \frac{3.679}{2} = 1.840 \overset{˚}{A}$ Neither is exactly $1.65$ Å — because the clean $1.65$ Å comes from the surface-grating relation with row spacing $a = 2.15$ Å, not from Bragg with $d = 2.03$ Å. The Bragg picture confirms diffraction is happening at a sensible order but does not (with these numbers) reproduce the surface value directly. The lesson: state which model and which spacing you are using before quoting a wavelength.

Problem 5.2

An electron ( $V = 54$ V, $λ = 1.67$ Å) is confined to the crystal surface with a positional uncertainty of $Δ x = 1.0$ Å (about one atomic spacing). Estimate the minimum uncertainty in its momentum, and compare it to the electron's own momentum. What does this say about how "sharp" the diffraction peak can ever be?

Recall Solution 5.2

From the Heisenberg uncertainty principle, $Δ x Δ p \geq ℏ/2$ with $ℏ = h /2 π = 1.055 \times 1 0^{- 34}$ J·s: $Δ p \geq \frac{ℏ}{2 Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 \times 1.0 \times 1 0 ^{- 10}} = 5.27 \times 1 0^{- 25} kg\cdotm/s$ The electron's own momentum is $p = h / λ = 6.626 \times 1 0^{- 34} /1.67 \times 1 0^{- 10} = 3.97 \times 1 0^{- 24}$ kg·m/s. Ratio: $Δ p / p = 5.27 \times 1 0^{- 25} /3.97 \times 1 0^{- 24} \approx 0.13$ , i.e. about $13%$ . So localising the electron to one atom smears its momentum (and hence its diffraction angle) by roughly $13%$ — a real, unavoidable width to the peak. Peaks are sharp because the beam samples many atoms at once, not one.

Problem 5.3

Design check: to keep the first-order surface-grating peak at a comfortable $ϕ = 4 5^{\circ}$ using $a = 2.15$ Å, what wavelength and what voltage do you need?

Recall Solution 5.3

Wavelength: $λ = a sin ϕ = 2.15 \times sin 4 5^{\circ} = 2.15 \times 0.7071 = 1.520 \overset{˚}{A}$ . Voltage: $V = (12.27/ λ)^{2} = (12.27/1.520)^{2} = (8.072)^{2} = 65.2 V$ . So a slightly higher voltage than the historic $54$ V pulls the peak in from $5 0^{\circ}$ to $4 5^{\circ}$ — exactly the "higher $V$ ⇒ smaller $ϕ$ " behaviour from Problem 3.3, now used by design.

Exercises — Davisson-Germer experiment — electron diffraction