Davisson-Germer experiment — electron diffraction: Worked examples

Before anything, the symbols we will lean on constantly. Each is defined in plain words and pinned to the picture below.

Definition Every symbol used on this page

$V$ = accelerating voltage (volts). The "push" that speeds up the electron. Bigger push → faster electron.
$λ$ = wavelength (in ångström, $1 \overset{˚}{A} = 1 0^{- 10}$ m). The "size" of the matter wave, from de Broglie hypothesis.
$ϕ$ = scattering angle (degrees), measured from the straight-ahead incident beam to the detector. This is what the movable Faraday cylinder reads.
$a$ = spacing between neighbouring surface atoms (ångström), the "grating pitch."
$n$ = order number — a whole number $1, 2, 3, \dots$ counting how many full wavelengths fit into the path difference. $n = 1$ is the first (brightest) peak, $n = 2$ the second, and so on.
$d$ = spacing between atomic planes (deeper than the surface), used only in the Bragg cross-check.
$θ$ = glancing angle, measured from the atomic plane up to the beam (defined carefully in Ex 5).
$h$ = Planck's constant $= 6.626 \times 1 0^{- 34} J\cdots$ — the fixed number linking wave and particle.
$m$ = particle mass (kg); for the electron $m_{e} = 9.11 \times 1 0^{- 31}$ kg.
$q$ (or $e$ ) = particle charge (coulombs); for the electron $e = 1.6 \times 1 0^{- 19}$ C.
$c$ = speed of light $= 3.0 \times 1 0^{8} m/s$ , the yardstick for "is it relativistic?"

The one working formula, already derived in the parent note (electron-only shortcut): $λ = \frac{12.27}{V} \overset{˚}{A} (V in volts)$ and its parent, the general de Broglie form for any charged particle: $λ = \frac{h}{2 m q V}$

Figure — Davisson-Germer experiment — electron diffraction

Read the figure carefully — every example below is this one picture with different numbers. The blue arrow (top-left) is the incident beam skimming in toward the surface. The yellow dots on the white line are the surface atoms, spaced $a$ apart. The pink arrow leaving the middle atom is the ray that reaches the detector, tilted by the angle $ϕ$ (the pink arc). The short yellow double-arrow at the bottom marks the spacing $a$ . The key thing to read off the picture: a wave scattering from the next atom along travels an extra distance $a sin ϕ$ before heading to the detector — that extra path is what must equal $nλ$ for a bright peak.

The scenario matrix

Every case class this topic can throw at you:

Cell	Case class	What makes it tricky	Example
A	Standard forward ( $V \to λ \to ϕ$ )	the baseline, both roads must agree	Ex 1
B	Reverse (measure $ϕ \to λ \to V$ )	run the formula backwards	Ex 2
C	Different voltage / limiting large $V$	peak moves; check it stays real	Ex 3
D	Degenerate: peak disappears ( $sin ϕ > 1$ )	no solution — must say so	Ex 4
E	Grating vs Bragg convention clash	which angle reference? relate $ϕ$ and $θ$	Ex 5
F	Relativistic edge case (high kV)	when does $2 m e V$ break?	Ex 6
G	Real-world word problem (electron microscope)	translate words → same formula	Ex 7
H	Exam twist: proton, not electron	mass changes everything	Ex 8

Ex 1 — Cell A: standard forward pass

Worked example

$V = 54$ V, nickel row spacing $a = 2.15$ Å. Find the peak angle $ϕ$ . Forecast: guess before reading — will $ϕ$ come out near $5 0^{\circ}$ , or nowhere close?

Step 1. Theory-road wavelength: $λ = \frac{12.27}{54} = 1.67$ Å. Why this step? We need $λ$ first; the accelerating voltage is the only wavelength knob we control.

Step 2. Grating condition, first order ( $n = 1$ ): $a sin ϕ = nλ = λ$ , so $sin ϕ = \frac{λ}{a} = \frac{1.67}{2.15} = 0.777.$ Why this step? Constructive interference from adjacent surface atoms needs the extra path $a sin ϕ$ (the geometry read off the figure) to equal one whole wavelength.

Step 3. $ϕ = arcsin (0.777) = 51. 0^{\circ}$ . Why this step? $arcsin$ answers "which angle has this sine?" — it undoes the $sin$ we built. There are formally two answers: $arcsin$ gives the principal value $5 1^{\circ}$ , and $18 0^{\circ} - 5 1^{\circ} = 12 9^{\circ}$ also has the same sine. We keep $5 1^{\circ}$ because the detector arm physically sweeps only $0^{\circ}$ – $9 0^{\circ}$ on that side of the beam; a $12 9^{\circ}$ reading would send the ray back into the crystal, which is unphysical. (In higher orders, Ex 4, we check whether any value is allowed at all.)

Verify: observed peak is $\approx 5 0^{\circ}$ ; our $5 1^{\circ}$ agrees to within $1^{\circ}$ (the gap is inner-potential refraction). Both roads to $λ$ land together → matter waves confirmed.

Ex 2 — Cell B: reverse pass (peak → voltage)

Worked example A peak is seen at

$ϕ = 4 4^{\circ}$ with $a = 2.15$ Å. What accelerating voltage produced it? Forecast: a smaller angle than $5 0^{\circ}$ — do you expect a higher or lower $V$ ?

Step 1. Grating gives $λ = a sin ϕ = 2.15 \times sin 4 4^{\circ} = 2.15 \times 0.6947 = 1.494$ Å. Why this step? The detector angle is now the given, so we run the geometry forward to recover $λ$ . (The measured $ϕ$ is already the principal, physical value, so no branch ambiguity here.)

Step 2. Invert the voltage formula: $λ = \frac{12.27}{V} \Rightarrow V = \frac{12.27}{λ}$ , so $V = (\frac{12.27}{1.494})^{2} = (8.213)^{2} = 67.5 V .$ Why this step? We square to undo the square root — the algebra that turns a measured $λ$ into the voltage that made it.

Verify: smaller $λ$ (1.49 < 1.67 Å) needs higher $V$ (67.5 > 54 V) — consistent with $λ \propto 1/ V$ . Forecast confirmed: smaller angle ⇒ higher voltage. ✅

Ex 3 — Cell C: raise the voltage, watch the peak slide

Worked example Keep

$a = 2.15$ Å. Compare peak angles at $V = 54$ V and $V = 100$ V. Forecast: does the peak move toward the beam (smaller $ϕ$ ) or away (larger $ϕ$ ) as $V$ rises?

Step 1. At $100$ V: $λ = \frac{12.27}{100} = 1.227$ Å. Why this step? Higher voltage → shorter wave; recompute the wavelength before the angle.

Step 2. First order, $sin ϕ = \frac{1.227}{2.15} = 0.5707 \Rightarrow ϕ = 34. 8^{\circ}$ (principal value; the $14 5^{\circ}$ root is again unphysical). Why this step? Same grating relation; only $λ$ changed.

Step 3. Compare: $54$ V gave $5 1^{\circ}$ , $100$ V gives $34. 8^{\circ}$ . Why this step? Two data points reveal the trend — the peak slides toward the incident beam.

Verify: the figure's two rays confirm the shorter (blue) wave scatters at the tighter angle. As $V \to \infty$ , $λ \to 0$ , $sin ϕ \to 0$ , so $ϕ \to 0$ : the peak collapses onto the straight-through beam. Limiting behaviour handled. ✅

Ex 4 — Cell D: the degenerate case (no peak at all)

Worked example Very slow electrons:

$V = 20$ V, $a = 2.15$ Å. Where is the second-order ( $n = 2$ ) peak? Forecast: trust the formula blindly, or is there a trap?

Step 1. $λ = \frac{12.27}{20} = 2.744$ Å. Why this step? Low voltage → long wavelength — this is exactly when trouble appears.

Step 2. Second order ( $n = 2$ ): $a sin ϕ = 2 λ \Rightarrow sin ϕ = \frac{2 \times 2.744}{2.15} = \frac{5.488}{2.15} = 2.552$ . Why this step? Plug in honestly and read the result.

Step 3. $sin ϕ = 2.552 > 1$ . No angle has a sine above 1 — not the principal value, and not the $18 0^{\circ} - ϕ$ root either, since neither branch can exceed $1$ . So this peak does not exist. Why this step? Because $- 1 \leq sin ϕ \leq 1$ always; a demand outside that range means the order is physically forbidden, and no branch of arcsine can rescue it.

Verify: the largest possible order needs $nλ \leq a$ . Compute the ceiling explicitly: $\frac{a}{λ} = \frac{2.15}{2.744} = 0.78.$ So we need $n \leq 0.78$ . Since $n$ must be a whole number $\geq 1$ (by definition, the order counts complete wavelengths), even the first order fails — at $20$ V this crystal shows no diffraction peak. Degenerate case fully described: state "no solution," never force an answer. ✅

Ex 5 — Cell E: grating angle vs Bragg angle

Worked example The peak sits at

$ϕ = 5 0^{\circ}$ . Convert to the Bragg glancing angle $θ$ , then check the Bragg reading. Forecast: is $θ$ bigger or smaller than the grating $ϕ$ ?

Step 1 — pin down each angle from the figure. There are two different conventions and confusing them is the classic error.

The grating $ϕ$ is measured from the incident beam to the outgoing ray at the surface.
The Bragg $θ$ is the glancing angle: from the atomic plane up to the beam.

Look at the figure. The incident and diffracted rays are mirror images about the plane. Call the glancing angle (beam-to-plane) $θ$ . The angle between the two beams is then $2 θ$ by simple mirror symmetry — the standard Bragg relation $ϕ_{beam-to-beam} = 2 θ ⟹ θ = \frac{ϕ _{beam-to-beam}}{2} .$ Why this step? This is the correct, textbook link: the total deflection between incoming and outgoing beams is twice the glancing angle.

Step 2 — mind which $ϕ$ you were handed. The detector angle quoted for D–G, $ϕ = 5 0^{\circ}$ , is measured from the incident beam to the surface ray, not the full beam-to-beam deflection. The beam-to-beam deflection is $18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$ , so $θ = \frac{18 0 ^{\circ} - ϕ}{2} = \frac{13 0 ^{\circ}}{2} = 6 5^{\circ},$ which is the same as $θ = 9 0^{\circ} - \frac{ϕ}{2} = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Why this step? Both forms agree — the $9 0^{\circ} - ϕ /2$ shortcut is just $\frac{18 0 ^{\circ} - ϕ}{2}$ rewritten. The lesson: always identify whether your given angle is beam-to-surface or beam-to-beam before halving.

Step 3 — Bragg reading. With the correct Ni(111) plane spacing $d = 2.03$ Å, first order: $λ = 2 d sin θ = 2 \times 2.03 \times sin 6 5^{\circ} = 4.06 \times 0.9063 = 3.68 \overset{˚}{A} .$ This corresponds to a higher-order reflection; the $n = 2$ reading $λ = 3.68/2 = 1.84$ Å is the physically comparable one. The clean $1.65$ Å comes from the surface-grating relation, not from Bragg — a deliberate reminder from the parent's Bragg's law discussion.

Verify: $θ = 6 5^{\circ}$ is larger than the beam-to-surface $ϕ = 5 0^{\circ}$ , because the plane angle is measured from a line nearly perpendicular to the beam. Feeding $ϕ = 5 0^{\circ}$ straight into Bragg would give $2 \times 2.03 \times sin 5 0^{\circ} = 3.11$ Å — a wrong number, proving the conversion matters. ✅

Ex 6 — Cell F: relativistic edge case

Worked example At

$V = 10, 000$ V, does the non-relativistic $λ = 12.27/ V$ still hold? Estimate the error. Forecast: modern physics — must we go relativistic, or is $10$ kV still "slow"?

Step 1. Kinetic energy $= e V = 10, 000$ eV $= 10$ keV. Rest energy of electron $= m_{e} c^{2} = 511$ keV. Why this step? The test is the ratio $KE / (m_{e} c^{2})$ ; if it's small, non-relativistic is fine.

Step 2. Ratio $\frac{KE}{m _{e} c ^{2}} = \frac{10}{511} = 0.0196 \approx 2%$ . Why this step? This tells us we are in the low-speed regime, but it is not the size of the wavelength error — see the next step.

Step 3. The relativistic momentum expands as $p = 2 m KE (1 + \frac{KE}{4 m _{e} c ^{2}} + \dots)$ , so the leading fractional correction to $p$ (and hence to $λ$ , since $λ = h / p$ ) is $\frac{Δ λ}{λ} \approx \frac{KE}{4 m _{e} c ^{2}} = \frac{10}{4 \times 511} = 0.0049 \approx 0.5%.$ Why this step? The correction is $KE / (4 m_{e} c^{2})$ , one quarter of the raw energy ratio — a common trap is to quote the $2%$ ratio as the error; the true wavelength shift here is only $\approx 0.5%$ .

Step 4. Non-relativistic $λ = 12.27/ 10000 = 0.1227$ Å; the true value is $\approx 0.5%$ shorter, $\approx 0.1221$ Å. Why this step? Relativity nudges $p$ up, shrinking $λ$ slightly.

Verify: at the original $54$ V, $KE / (4 m_{e} c^{2}) = 54/ (4 \times 511000) = 0.0026%$ — utterly negligible, so the parent note was right to stay non-relativistic. Only near hundreds of kV does the correction grow past $10%$ . Edge case bounded. ✅

Ex 7 — Cell G: real-world word problem (electron microscope)

Worked example An

Electron microscope uses $V = 60, 000$ V. What is the electron wavelength, and why is that spectacular for imaging? Forecast: guess whether it beats visible light ( $\sim 5000$ Å) by a little or by thousands.

Step 1. $λ = \frac{12.27}{60000} = \frac{12.27}{244.9} = 0.0501$ Å. Why this step? Same de Broglie formula — the microscope is just Davisson–Germer at high voltage.

Step 2. Compare to visible light $λ_{light} \approx 5000$ Å: ratio $= 5000/0.0501 \approx 1 0^{5}$ . Why this step? Resolving power scales with $1/ λ$ ; a $1 0^{5}$ -times-shorter wave resolves $1 0^{5}$ -times-finer detail.

Verify: $0.05$ Å is far smaller than an atom ( $\sim 1$ Å), so an electron microscope can, in principle, resolve individual atoms — impossible for light. The word problem reduced to Cell C arithmetic. ✅ (By Ex 6, ignoring relativity here gives $KE / (4 m_{e} c^{2}) = 60/ (4 \times 511) \approx 3%$ error — acceptable for an estimate.)

Ex 8 — Cell H: exam twist — proton instead of electron

Worked example A proton is accelerated through

$V = 54$ V. Find its wavelength and compare to the electron's $1.67$ Å. Forecast: heavier particle — longer or shorter wave than the electron?

Step 1. The $12.27/ V$ shortcut is electron-only (it hid the electron mass). Go back to the general form $λ = \frac{h}{2 m q V}$ . Why this step? Every constant collapsed into $12.27$ assumed $m = m_{e}$ ; a proton needs the general form with its own mass.

Step 2. Proton mass $m_{p} = 1.673 \times 1 0^{- 27}$ kg $= 1836 m_{e}$ ; charge $q = e$ same. So $\frac{λ _{p}}{λ _{e}} = \frac{m _{e}}{m _{p}} = \frac{1}{1836} = \frac{1}{42.8} .$ Why this step? Only the mass changed; $λ \propto 1/ m$ , so the ratio is a clean square root.

Step 3. $λ_{p} = \frac{1.67}{42.8} = 0.0390$ Å. Why this step? Scale the known electron answer by the ratio — faster and less error-prone than re-plugging every constant.

Verify: direct compute $λ_{p} = \frac{6.626 \times 1 0 ^{- 34}}{2 ( 1.673 \times 1 0 ^{- 27} ) ( 1.6 \times 1 0 ^{- 19} ) ( 54 )} = 3.90 \times 1 0^{- 12}$ m $= 0.0390$ Å. ✅ Heavier → shorter wave, forecast confirmed. A proton at the same voltage would need a $\sim 43 \times$ finer grating to diffract — nickel would show no peak.

Recall Which cell does each trap belong to?

Peak vanishes because $sin ϕ > 1$ ::: Cell D (degenerate) Feeding beam-to-surface $ϕ$ straight into Bragg's law ::: Cell E (convention clash) Using $12.27/ V$ for a proton ::: Cell H (mass-dependent constant) Quoting the raw $KE / m_{e} c^{2}$ ratio as the wavelength error ::: Cell F — the true error is one quarter of it

See also: Wave-particle duality, Photoelectric effect, Heisenberg uncertainty principle, and the parent Davisson–Germer topic note.