University Physics IV · Photons and Matter Waves · 4.6
Electron, Neutron & Molecular Diffraction
This is the experiment that measured a matter wave. Take a crystal whose spacing you already know, fire particles at it, read the angle where they pile up, and compare the wavelength that implies with h/p. Then find out which of the two ways of reading that same angle survives contact with the metal.
Build the model
Connect the measurement to the mechanism.
Diffraction is the experiment that turned λ = h/p from a hypothesis into a measurement, and the reason it could is that the geometry was already understood. A periodic array of scatterers reinforces an outgoing wave wherever the path difference between neighbours is a whole number of wavelengths: D sin φ = nλ for a surface row at normal incidence, nλ = 2d sin θ for a stack of planes. Both are classical optics.
The only quantum content is what you substitute for λ, so an angle measured on a crystal of known spacing becomes a direct reading of the electron's momentum. Davisson and Germer got that reading almost by accident — 54 eV electrons on nickel, a peak at 50°, 2.15 Å × sin 50° = 1.65 Å against h/√(2mₑeV) = 1.67 Å — and the same geometry now runs on thermal neutrons at 1.80 Å and on whole C₆₀ molecules at 2.8 pm. What it costs is a set of conditions the model quietly assumes. λ is fixed by the local momentum, so an electron entering a metal changes wavelength and the Bragg reading needs a refraction correction the grating reading does not.
Fringes need a wavelength comparable to the spacing and a coherence length longer than the path difference. And because λ falls as 1/√(mE), resolution is bought with energy — the sentence that governs every electron microscope ever built.
- Simple definition
- Matter diffraction is the constructive interference of a particle beam scattered from a periodic array, with maxima at exactly the angles a wave of wavelength λ = h/p would choose.
- Example
- 54 eV electrons on a nickel surface whose atom rows are 2.15 Å apart peak at φ = 50°, so D sin φ = 1.65 Å — within 1.3% of h/√(2mₑeV) = 1.67 Å.
The whole measurement: you know D, you read φ off the detector, and λ comes out. If λ > D there is no diffracted order at all.
D = spacing of the surface atom rows (m); φ = angle from the surface normal; n = order number
54 V gives 1.67 Å, matched to crystal spacings; 200 kV gives 2.51 pm, where the naive formula would claim 2.74 pm.
Non-relativistic. Multiply by 1/√(1 + eV/2mₑc²) above about 10 kV; V in volts, λ in ångström.
Refraction at the surface displaces every Bragg peak. Fit the displacement and you measure V₀; ignore it and the λ you infer comes out low.
θ = glancing angle measured outside; V₀ = mean inner potential (≈ 15 V for Ni); λ = vacuum wavelength
Room temperature already puts a neutron at the atomic scale, which is why a reactor moderator is a ready-made source of 1 Å waves.
E = kBT at the moderator temperature; 293 K gives 0.0253 eV, λ = 1.80 Å, v = 2.20 km s⁻¹
Transverse coherence must span one period, longitudinal coherence the path difference — so the highest visible order is about λ/Δλ.
s = source width, L its distance to the grating, D the period, n the order; all lengths in metres
At 200 kV λ = 2.51 pm, but aberrations cap α near 10 mrad, so dₘᵢₙ ≈ 0.15 nm. The lens is the limit, not the wavelength.
α = half-angle of the collection aperture; μ = refractive index of the medium, 1 in vacuum. Here n stays reserved for the diffraction order.
The grating condition is classical; only λ is new
Nothing in D sin φ = nλ came from quantum mechanics. Send a wave at normal incidence onto a row of scatterers spaced D apart: every scatterer is struck on the same wavefront, so the only path difference is accumulated on the way out, and it is D sin φ between neighbours. Reinforcement demands that this be a whole number of wavelengths. That is the optical reflection grating, unchanged. The single quantum input is the substitution λ = h/p, and it turns the apparatus into a measuring instrument. You take D from X-ray crystallography, read φ off a detector, and the grating condition hands you λ; then you compute h/√(2mₑeV) from the accelerating voltage and see whether the two agree. In 1927, 54 eV electrons on nickel with D = 2.15 Å peaked at φ = 50°: the geometry gives 1.647 Å, de Broglie gives 1.669 Å. Note the ceiling built into the condition — sin φ cannot exceed 1, so the highest order is n = ⌊D/λ⌋, and if λ exceeds D nothing diffracts at all.
Two readings of one peak, and why they cannot disagree
The same maximum is often quoted as Bragg reflection instead. Rays leaving at φ from a beam that came in along the normal are reflected off planes tilted φ/2 to the surface, so the glancing angle is θ = 90° − φ/2 = 65°, and those planes cut the surface rows at spacing d = D sin(φ/2) = 2.15 sin 25° = 0.909 Å. Then 2d sin θ = 2(0.909)(0.906) = 1.647 Å — the same number to every digit. It has to be, because 2⋅D sin(φ/2)⋅cos(φ/2) = D sin φ is a trigonometric identity: the two statements are one external geometry written twice, not two independent confirmations of anything. Treating their agreement as evidence is a mistake worth naming now, because the two readings stop being equivalent the moment the beam actually enters the crystal.
The wavelength inside the crystal is not the one outside
A conduction electron sits in a potential well. Nickel's mean inner potential is V₀ ≈ 15 V, so an electron accelerated through V arrives inside with kinetic energy e(V + V₀). At V = 54 V that is 69 eV, and λ falls from 1.669 Å outside to 12.264/√69 = 1.476 Å inside — the crystal has a refractive index μ = λₒᵤₜ/λᵢₙ = √(1 + V₀/V) = 1.130 for these electrons, and the beam refracts at the surface. Which reading survives now depends on which component of k the condition constrains. The surface-grating condition fixes only the component parallel to the surface, and that component is conserved across the step, so the vacuum wavelength is the correct one there and no correction is needed. The Bragg condition also fixes the perpendicular component, which the step does change; Snell's law in the form cos θ = μ cos θᵢₙ converts it into nλ = 2d√(sin²θ + V₀/V). Every Bragg peak is displaced, and fitting that displacement across a range of voltages is how V₀ is measured.
At a fixed wavelength, the energy you need goes as 1/m
λ = h/√(2mE) inverts to E = h²/(2mλ²), so at the atomic scale the energy a probe needs is inversely proportional to its mass. To reach λ = 1.80 Å: a neutron needs 0.0253 eV, which is simply kBT at 293 K; an electron needs 46.4 V; an X-ray photon, obeying E = hc/λ instead, needs 6.89 keV. That one line explains why neutron scattering exists as a field. Thermal neutrons leaving a moderator are already at the right wavelength and travel at 2.20 km s⁻¹, slow enough for a mechanical chopper to select them. They are uncharged, so they penetrate centimetres of material rather than nanometres and report bulk structure. They scatter from nuclei rather than electron clouds, so hydrogen — nearly invisible to X-rays — shows up strongly and neighbouring elements are distinguishable. And they carry a magnetic moment, so magnetic order diffracts too. Electrons scatter some 10⁴ times more strongly than X-rays, which is the opposite bargain: superb sensitivity, a probing depth of a few atomic layers, and a vacuum requirement.
Molecules diffract, and coherence becomes the constraint
Nothing in λ = h/p stops at the electron. A C₆₀ molecule of mass 720 u moving at 200 m s⁻¹ has p = 2.39 × 10⁻²² kg m s⁻¹ and λ = 2.77 pm — about 250 times smaller than the molecule's own 0.7 nm diameter. Diffracting it needs a 100 nm grating, which gives a first-order angle of only λ/D = 2.8 × 10⁻⁵ rad, so the fringes are 35 μm apart a metre and a quarter downstream. Two conditions then decide whether they appear at all. Transversely, the beam's coherence width at the grating, roughly λL/s for a source of width s at distance L, must span at least one period — which is why the collimating slits throw away almost the entire beam. Longitudinally, the coherence length λ²/Δλ must exceed the path difference nλ, so the highest visible order is about λ/Δλ. A thermal C₆₀ beam has Δv/v ≈ 60%, a coherence length under two wavelengths: the first orders survive and the rest wash out. A 54 eV electron beam with a 0.5 V spread has Δλ/λ = ½(ΔV/V) = 0.46%, a coherence length of 36 nm, and room for a couple of hundred orders.
h/p is the floor an electron microscope cannot go under
The diffraction limit does not care what the wave is made of: in Rayleigh's form the smallest resolvable separation is dₘᵢₙ ≈ 0.61λ/(μ sin α), with α the collection half-angle and μ the refractive index of the medium — 1 in the vacuum a microscope column runs in. Put an electron wavelength into it and the numbers turn absurd. A 200 kV electron has λ = 2.51 pm — and note that the non-relativistic 12.264/√V would return 2.74 pm, 9% too long, because at 200 keV against a rest energy of 511 keV the factor √(1 + eV/2mₑc²) is no longer 1. A wavelength of 2.51 pm promises picometre resolution, and no microscope delivers it. Magnetic lenses have severe spherical aberration, which forces α down to about 10 mrad, and 0.61 × 2.51 pm / 0.010 = 0.15 nm — roughly what an uncorrected instrument achieves. Aberration correctors that open α to 40 mrad reach about 40 pm. So h/p sets the floor and the lens decides how far above it you land, the reverse of light optics, where the wavelength is the binding constraint.
Change one variable at a time
Make the relationship visible.
Set V = 54 V and D = 2.15 Å — Davisson and Germer's numbers — and the ray swings out to about 51°, a degree from the 50° peak they measured. Now raise V: λ falls as 1/√V, the ray closes on the normal, and a second order opens just past 130 V. A third needs a wider row as well — take D to 3.4 Å and it appears by 120 V.
WAVELENGTH h/p1.669 Å
sin φ = λ / D0.776
FIRST-ORDER ANGLE φ50.9 °
HIGHEST ORDER n1
Live interpretationWAVELENGTH h/p: 1.669 Å. sin φ = λ / D: 0.776. FIRST-ORDER ANGLE φ: 50.9 °. HIGHEST ORDER n: 1
Catch the common trap
Explain before calculating.
Davisson and Germer's 54 eV peak at φ = 50° from nickel can be read as a surface grating, D sin φ = λ with D = 2.15 Å, or as Bragg reflection from planes of spacing d = 0.91 Å at glancing angle θ = 65°. Both return 1.65 Å. Which of the two readings needs a correction for nickel's inner potential V₀ ≈ 15 V?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyDavisson and Germer fired 54 eV electrons at normal incidence onto a nickel crystal and found a sharp maximum in the scattered intensity at φ = 50° from the surface normal. The surface atom rows are 2.15 Å apart. Find the wavelength that peak implies, compare it with h/p, and say how well they agree.
- The beam arrives along the normal, so every surface atom is struck on the same wavefront and the only path difference is accumulated on the way out: D sin φ between neighbours. First-order reinforcement therefore needs D sin φ = λ.
- From the peak: λ = (2.15 Å)(sin 50°) = 2.15 × 0.7660 = 1.647 Å.
- From de Broglie: λ = h/√(2mₑeV) = 12.264 Å⋅V^½ / √54 = 12.264/7.348 = 1.669 Å.
- The gap is 0.022 Å, or 1.3% of 1.669 Å — comfortably inside the uncertainty of the 1927 angle measurement and of D itself.
AnswerThe peak implies λ = 1.65 Å; h/p predicts 1.67 Å. Agreement at the one-percent level, against a spacing D that X-ray crystallography had already fixed independently, is what promoted the de Broglie wavelength from a postulate to a measured quantity.
MediumA reactor moderator at 293 K delivers neutrons with kinetic energy E = kBT. (a) Find their de Broglie wavelength and speed. (b) Find the first-order Bragg angle from nickel (111) planes, d = 2.034 Å. (c) Find the accelerating potential an electron would need, and the photon energy an X-ray would need, to reach the same wavelength.
- E = kBT = (1.3806 × 10⁻²³ J K⁻¹)(293 K) = 4.045 × 10⁻²¹ J = 0.02525 eV.
- λ = h/√(2mₙE) = 0.2860 Å/√(E/eV) = 0.2860/0.1589 = 1.800 Å. Speed: v = h/(mₙλ) = 6.626 × 10⁻³⁴ / (1.675 × 10⁻²⁷ × 1.800 × 10⁻¹⁰) = 2.20 × 10³ m s⁻¹.
- Bragg at n = 1: sin θ = λ/2d = 1.800/(2 × 2.034) = 0.4425, so θ = 26.3° and the detector sits at a scattering angle 2θ = 52.5°.
- Electron at the same λ: V = (12.264/1.800)² = (6.813)² = 46.4 V. Photon: E = hc/λ = 1240 eV⋅nm / 0.1800 nm = 6.89 keV.
- The three energies span more than five orders of magnitude at one wavelength, because λ = h/√(2mE) makes E ∝ 1/m for a massive probe while a photon pays E = hc/λ regardless.
Answerλ = 1.80 Å at v = 2.20 km s⁻¹; the first-order Bragg angle is θ = 26.3°, so the detector sits at 2θ = 52.5°. The same wavelength costs an electron 46.4 V and an X-ray photon 6.89 keV.
HardA 150 V electron beam is Bragg-reflected from nickel (111) planes lying parallel to the surface, d = 2.034 Å. Nickel's mean inner potential is V₀ = 15 V. (a) Where does the uncorrected Bragg law put the n = 2 peak? (b) Where is it actually found? (c) If you fitted the observed peak with the uncorrected law, what wavelength would you report?
- Outside the crystal, λ = 12.264 Å⋅V^½/√150 = 12.264/12.247 = 1.0014 Å.
- Uncorrected: sin θ = nλ/2d = 2(1.0014)/(2 × 2.034) = 2.0028/4.068 = 0.4923, so θ = 29.5°.
- Inside, the electron also falls through V₀, so its kinetic energy there is e(V + V₀) = 165 eV and λᵢₙ = 12.264/√165 = 0.9548 Å. The refractive index is μ = λ/λᵢₙ = √(1 + V₀/V) = √1.100 = 1.0488.
- Conserving the component of k parallel to the surface gives cos θ = μ cos θᵢₙ, and substituting that into nλᵢₙ = 2d sin θᵢₙ reduces the condition to nλ = 2d√(sin²θ + V₀/V).
- So sin²θ = (0.4923)² − 0.100 = 0.2424 − 0.1000 = 0.1424, sin θ = 0.3773, θ = 22.2°. The peak sits 7.3° below the naive prediction.
- Feeding 22.2° back into the uncorrected law gives λ = 2d sin θ/n = (4.068)(0.3773)/2 = 0.767 Å, 23% below the true 1.0014 Å. Fitting exactly this displacement across a range of voltages is how V₀ ≈ 15 V for nickel was measured.
AnswerUncorrected, 29.5°; actually 22.2° — a 7.3° shift. Reading the observed peak with the uncorrected law reports λ = 0.767 Å instead of 1.0014 Å, low by 23%.