
Optional unit · institution-dependent · Interactive lesson
Matter waves and electron diffraction
Treat every electron and neutron as a wave of wavelength h/p, then prove it the only way a wave can be proved: with a grating. You will turn a voltage or a temperature into a wavelength, predict where a crystal sends the beam, and know when relativity or the lens, not λ, decides the answer.
UPIII-09.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Optional unit · institution-dependent. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment. Optional work is not a required completion gate.
Build the model
Momentum sets a wavelength; a crystal is the grating fine enough to show it.
The key idea
A matter wave is the wave associated with any particle of momentum p, with de Broglie wavelength λ = h/p; it diffracts from a periodic structure exactly as light does, and visibly so only when λ is comparable to the structure's spacing.
Example. An electron accelerated from rest through 54 V has p = √(2mₑeV) = 3.97 × 10⁻²⁴ kg m s⁻¹ and λ = h/p = 0.167 nm, close to the 0.215 nm spacing of nickel's surface rows, which send it back at 51° from the incident beam.
The complete physical model and assumptions
De Broglie's hypothesis is that every particle carries a wave whose wavelength is fixed by its momentum, λ = h/p, with the same Planck constant that fixes a photon's. That is a prediction, not an analogy, and a wave can be tested only through interference, which needs a structure whose spacing d is comparable to λ. Electrons accelerated through tens of volts have λ = h/√(2mₑeV) of a few tenths of a nanometre, which is why the atom rows of a nickel surface sent Davisson and Germer's 54 eV electrons into a beam at 50°; thermal neutrons, with λ = h/√(2mₙkT) ≈ 0.18 nm at room temperature, Bragg-reflect from bulk lattice planes; and 100 kV electrons, at a few picometres, make the electron microscope possible. The model has three edges. When λ ≪ d the orders collapse onto the direct beam and the particle behaves as a classical ray. At high voltage the momentum must be relativistic, and the slow-limit formula overstates λ by 4.8% at 100 kV. Inside a solid the electron's wavelength is shortened by the inner potential and its wave is scattered many times, so single scattering predicts where the beams go far better than how bright they are.
1Momentum, not size, sets the wavelength
Planck and Einstein gave light E = hf, and Compton's recoil confirmed p = h/λ. De Broglie read the pair as a statement about every particle: a free particle of energy E and momentum p carries a wave with ω = E/ħ and k = p/ħ, so λ = 2π/k = h/p. The step is more than analogy. A plane wave's phase kx − ωt counts the crests passing a point, and a count cannot depend on the observer, so (ω/c, k) must transform between frames exactly as (E/c, p) does, and one constant has to serve light and matter alike. The wavelength therefore belongs to the motion, not to the object: an electron at 1.00 × 10⁶ m s⁻¹ has λ = h/mₑv = 0.727 nm, several atomic diameters, while a 0.145 kg ball at 40 m s⁻¹ has 1.1 × 10⁻³⁴ m, far below any structure that could diffract it.
2Turn an accelerating voltage into a wavelength
The experimenter sets a voltage, not a momentum. An electron released from rest and accelerated through V gains K = eV, and while K ≪ mₑc² its momentum is p = √(2mₑK), so λ = h/√(2mₑeV). Check the dimensions before trusting it: mₑeV has units kg J = kg² m² s⁻², its root is a momentum in kg m s⁻¹, and J s divided by kg m s⁻¹ is m. Inserting the constants gives the working form λ = 1.226 nm/√V with V in volts: 0.388 nm at 10 V, 0.167 nm at 54 V, 0.0388 nm at 1 kV. Remember the inverse square root. Four times the voltage halves λ, so bracketing the 0.1 to 0.4 nm spacings of real crystals takes only about 10 to 150 V, which is why slow electrons became the natural probe of a crystal surface.
3Davisson and Germer: a crystal surface as a grating
Davisson and Germer fired 54 eV electrons normally onto a nickel crystal and swept a detector through the backscattered directions; the count rate peaked 50° from the incident beam. Electrons this slow penetrate only a few atomic layers, so the scatterers act as a reflection grating made of the surface rows, whose spacing, d = 0.215 nm, was already known from X-ray diffraction. Waves leaving neighbouring rows toward angle φ differ in path by d sin φ and reinforce when d sin φ = nλ. The peak therefore implies λ = 0.215 nm × sin 50° = 0.165 nm, and de Broglie predicts 0.167 nm: agreement to 1.3%, inside the width of their broad peak. The peak also rose and fell as the voltage passed through 54 V, as a wavelength sweeping past a fixed grating condition requires. One crystal, probed with photons and then with electrons, gave the same spacing.
4Diffraction is visible only inside a window of λ/d
The grating condition carries its own window. A first-order beam needs sin φ = λ/d ≤ 1, so a wave longer than the spacing produces no diffracted beam at all, only the specular one; on nickel's rows that happens below 32.5 V. At the other end, when λ ≪ d, the orders sit at φ ≈ nλ/d, pressed against the direct beam. Every real beam has an angular spread from its collimation and energy spread, and once nλ/d falls inside it the orders cannot be separated and the particle behaves as a ray. That is the classical limit, the same step that takes wave optics to geometric optics as λ → 0. The ball's 10⁻³⁴ m through a 1 m gap is hopeless. To push the window to heavier particles, slow them: C₆₀ molecules at 200 m s⁻¹ have λ = 2.8 pm, and a 100 nm grating turns them through 28 μrad, small but measurable. What decides is λ/d, never size alone.
5Neutrons: Bragg planes and a wavelength set by kT
A neutron has no charge, so it passes through centimetres of metal and scatters from nuclei rather than from electron clouds; that makes it the probe for bulk structure, for light atoms such as hydrogen, and, through its magnetic moment, for magnetic order. It samples the whole three-dimensional lattice, so the condition is Bragg's, 2d sin θ = nλ, with d the spacing of a family of planes and θ the glancing angle; the beam turns through 2θ. The wavelength comes from temperature. A moderator at T gives the neutrons a Maxwell distribution whose most probable speed carries kinetic energy kT, so λ = h/√(2mₙkT): at 300 K, kT = 25.9 meV and λ = 0.178 nm, matched to interatomic spacings by thermal equilibrium alone. Since λ = h/mₙv = ht/(mₙL), timing the flight over a path L measures λ directly: a 0.178 nm neutron needs 0.450 ms per metre.
6At 100 kV, relativity and the lens take over
An electron microscope accelerates through 100 to 300 kV, where K is no longer small beside mₑc² = 511 keV. λ is still h/p, but p comes from the invariant (pc)² = K² + 2Kmₑc², so λ = hc/√(K² + 2Kmₑc²). The slow-limit result exceeds this by the factor √(1 + K/2mₑc²): 4.8% at 100 kV (3.88 pm against 3.70 pm) and 13.7% at 300 kV. Both limits check, K ≪ mₑc² returning h/√(2mₑK) and K ≫ mₑc² returning hc/K, the photon's value. Wavelength alone would allow δ ≈ 0.61λ/α = 0.23 nm with α = 10 mrad, but magnetic lenses have spherical aberration, a blur Cₛα³ that grows fast with aperture. With Cₛ = 1.0 mm the two blurs are equal at α = 6.9 mrad, each 0.33 nm, and the best resolution scales as λ³⁄⁴ Cₛ¹⁄⁴: about 90 wavelengths, set by the lens.
Key equations
Read the units, assumptions and limits before substituting.
De Broglie relation
λ = h/p
A proton at 1.00 × 10⁶ m s⁻¹ has λ = 0.396 pm, 1836 times shorter than an electron at the same speed; given equal momenta the two would share one λ. Momentum, not mass or size alone, sets the scale.
λ in m; h = 6.626 × 10⁻³⁴ J s; p the particle's momentum in kg m s⁻¹, relativistic γmv when v nears cElectron accelerated through V, slow limit
λ = h/√(2mₑeV)
54 V gives 0.167 nm and 150 V gives 0.100 nm: tens to hundreds of volts put λ at atomic spacings. Four times the voltage halves λ.
mₑ = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C, V the accelerating voltage in V; numerically 1.226 nm/√V; needs eV ≪ mₑc²Relativistic wavelength for kinetic energy K
λ = hc/√(K² + 2Kmₑc²)
K ≪ mₑc² returns h/√(2mₑK); K ≫ mₑc² returns hc/K, the photon value. At 100 kV it gives 3.70 pm where the slow-limit form gives 3.88 pm.
hc = 1239.8 eV nm, mₑc² = 511.0 keV, K = eV the kinetic energy in eV; exact at any speedSurface-grating condition (Davisson–Germer)
d sin φ = nλ
Nickel rows 0.215 nm apart and 54 V electrons give sin φ = 0.776, φ = 51°, against the 50° measured. No order n exists unless nλ ≤ d.
d the surface-row spacing in m, φ the diffracted beam's angle from the normally incident beam, n = 1, 2, 3, …Bragg condition for lattice planes
2d sin θ = nλ
Bulk diffraction of neutrons and fast electrons: thermal neutrons of 0.178 nm meet 0.25 nm planes at θ = 21°, 100 kV electrons at 0.42°.
d the spacing of a family of planes in m, θ the glancing angle between beam and planes; the beam turns through 2θMicroscope resolution, wavelength limit
δ ≈ 0.61λ/α
At 100 kV and α = 10 mrad, δ = 0.23 nm, a thousand times finer than light allows; spherical aberration of the lens, not λ, forbids a larger α.
δ the smallest resolvable separation in m, λ the electron wavelength in m, α the objective aperture half-angle in radInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at Davisson and Germer's 54 V on nickel's 0.215 nm rows: the first order leaves at 51°. Drop below 33 V and λ outgrows d, so that beam vanishes. Raise V toward 600 V and the orders crowd in toward the normal, the road to the classical limit.
Live results
- DE BROGLIE λ
- 0.1669 nm
- FIRST-ORDER φ₁, 90° AT CUTOFF
- 50.9 °
- CUTOFF VOLTAGE, λ = d
- 32.5 V
- ORDERS EACH SIDE
- 1
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Accelerating voltage V: 54 V; Surface row spacing d: 0.215 nm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest slip is to borrow the photon relation E = hc/λ and write an electron's wavelength as hc/K. For a 54 eV electron that gives 23.0 nm instead of 0.167 nm, wrong by a factor of 138. The relation that holds for every particle is λ = h/p; E = pc belongs only to a massless one. A slow electron has K = p²/2mₑ, so p = √(2mₑK), larger than K/c by the factor √(2mₑc²/K), and that factor is exactly the error. The photon formula is not wrong everywhere: it is the K ≫ mₑc² limit of hc/√(K² + 2Kmₑc²), accurate only for electrons of many MeV. A related slip puts the total energy K + mₑc² into hc/E, giving 1.74 pm at 200 kV instead of 2.51 pm.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Electrons are accelerated from rest through 150 V and strike a crystal surface at normal incidence. Its surface atom rows are 0.250 nm apart. Find the de Broglie wavelength and every angle at which a diffracted beam leaves the surface.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Kinetic energy: K = eV = 150 × 1.602 × 10⁻¹⁹ J = 2.403 × 10⁻¹⁷ J. That is 150 eV against mₑc² = 511 keV, so the slow-limit momentum is safe; the relativistic correction is below 0.01%.
- Momentum: p = √(2mₑK) = √(2 × 9.109 × 10⁻³¹ kg × 2.403 × 10⁻¹⁷ J) = √(4.378 × 10⁻⁴⁷ kg² m² s⁻²) = 6.617 × 10⁻²⁴ kg m s⁻¹.
- Wavelength: λ = h/p = 6.626 × 10⁻³⁴ J s ÷ (6.617 × 10⁻²⁴ kg m s⁻¹) = 1.001 × 10⁻¹⁰ m = 0.100 nm; the shortcut 1.226 nm/√150 agrees.
- Grating condition d sin φ = nλ. For n = 1: sin φ₁ = 0.1001/0.250 = 0.4004, so φ₁ = 23.6°. For n = 2: sin φ₂ = 0.801, so φ₂ = 53.2°.
- For n = 3, sin φ₃ = 1.20 > 1, which no angle satisfies, so there is no third order. Each order that exists appears on both sides of the incident beam.
Answer. λ = 0.100 nm; diffracted beams at 23.6° and 53.2° on either side of the incident beam, and no third order.
MediumExample 2
A reactor neutron beam is reflected from a single crystal whose planes are 0.1808 nm apart, set at a glancing angle of 30.0°. Find the first-order wavelength, the neutrons' speed and kinetic energy, and their flight time over a 12.0 m path. What else does the same crystal setting pass, and how does timing the flight separate it?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Bragg condition, n = 1: λ = 2d sin θ = 2 × 0.1808 nm × sin 30.0° = 0.1808 nm.
- Speed from λ = h/(mₙv): v = h/(mₙλ) = 6.626 × 10⁻³⁴ J s ÷ (1.675 × 10⁻²⁷ kg × 1.808 × 10⁻¹⁰ m) = 2188 m s⁻¹, non-relativistic by a wide margin.
- Kinetic energy: K = ½mₙv² = 4.009 × 10⁻²¹ J = 25.0 meV. Setting K = kT gives T = 290 K: these are thermal neutrons, near the most probable speed of a room-temperature moderator.
- Flight time: t = L/v = 12.0 m ÷ 2188 m s⁻¹ = 5.48 ms. Equivalently λ = ht/(mₙL), so the clock reads the wavelength.
- The same angle satisfies 2d sin θ = 2λ′ for λ′ = 0.0904 nm: second-order neutrons with twice the speed and four times the energy, 100 meV. They arrive at 2.74 ms, half the time, so a detector gated on arrival time separates the orders.
Answer. λ = 0.181 nm; v = 2.19 × 10³ m s⁻¹; K = 25.0 meV (thermal, T ≈ 290 K); t = 5.48 ms. Half-wavelength neutrons (0.0904 nm, 100 meV) also pass and arrive at 2.74 ms.
HardExample 3
A transmission electron microscope runs at 200 kV. Find the electron wavelength and the slow-limit formula's error. A polycrystalline film gives a first-order ring of radius 5.33 mm at an effective camera length of 0.500 m: find the plane spacing, and the spacing the slow-limit λ would report. With spherical-aberration coefficient Cₛ = 1.0 mm, estimate the best resolution using the aperture α at which 0.61λ/α equals Cₛα³.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Momentum from the invariant, with K = 200 keV and mₑc² = 511.0 keV: pc = √(K² + 2Kmₑc²) = √(40 000 + 204 400) keV = 494.4 keV. Then λ = hc/pc = 1239.8 eV nm ÷ 494 400 eV = 2.508 × 10⁻³ nm = 2.508 pm.
- Slow limit: λ = 1.2264 nm/√200 000 = 2.742 pm. The ratio 2.742/2.508 = 1.093 = √(1 + K/2mₑc²) = √1.1957, so the slow-limit formula is 9.3% too long.
- Ring geometry: the beam turns through 2θ with tan 2θ = R/L = 5.33 mm ÷ 500 mm = 0.01066, so θ = 5.330 mrad. Bragg, n = 1: d = λ/(2 sin θ) = 2.508 pm ÷ 0.01066 = 235.3 pm = 0.235 nm, consistent with gold's (111) planes.
- With the slow-limit λ the same ring gives d = 2.742 pm ÷ 0.01066 = 257.2 pm, 9.3% too large, enough to misidentify the material. At 200 kV relativity is not optional.
- Aberration balance: α⁴ = 0.61λ/Cₛ = 0.61 × 2.508 × 10⁻¹² m ÷ (1.0 × 10⁻³ m) = 1.530 × 10⁻⁹, so α = 6.254 mrad. Each blur is then 0.61λ/α = 1.530 × 10⁻¹² m ÷ (6.254 × 10⁻³) = 2.446 × 10⁻¹⁰ m.
- So δ ≈ 0.24 nm, about 100 wavelengths. Halving λ would improve it only by 2³⁄⁴ = 1.68, because the balanced blur scales as λ³⁄⁴ Cₛ¹⁄⁴: the lens, not the wavelength, is the limit.
Answer. λ = 2.51 pm (slow limit 2.74 pm, 9.3% high); d = 0.235 nm, where the slow-limit λ would report 0.257 nm; α ≈ 6.3 mrad and δ ≈ 0.24 nm, about 100 wavelengths.
Quick check
Select an answer and explain it before checking.
Review & continue
A matter wave is the wave associated with any particle of momentum p, with de Broglie wavelength λ = h/p; it diffracts from a periodic structure exactly as light does, and visibly so only when λ is comparable to the structure's spacing.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.