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UQMI-01.05 · Foundations of Quantum Mechanics

de Broglie matter waves
& electron diffraction

Give a particle momentum. A crystal can measure its wavelength.

An electron from a 54 V gun has a wavelength of 0.167 nm. Predict its diffraction, test the geometry, and see how one relation connects electrons, neutrons and molecules.

Electron beamλ = h / pCrystalDiffraction ringsMore momentum.Shorter wavelength.

One relation. A measurable prediction.

Matter diffraction puts λ = h/p to an experimental test.

de Broglie’s 1924 proposal extended the photon’s momentum relation to matter. It was a hypothesis, not a consequence of classical mechanics. With a crystal spacing measured independently, its predicted diffraction angles can be checked.

The wavelength

Momentum sets the scale

λ = h/p

More momentum means a shorter wavelength. At equal momentum, different particles have the same wavelength—not necessarily the same scattering intensity.

For a slow electron: λ = 1.2264 nm / √(V/1 V).

The crystal

Paths must arrive in step

2d sin θ = nλ
R = L tan(2θ)

θ is measured to the planes. A forward-screen diffraction ray is deflected through 2θ; a polycrystal produces rings.

Small angles: R ≈ Lλ/d for first order.

The common trap

An electron is not a photon

pelectron ≈ √(2meK)
pphoton = E/c

At 54 eV, an electron has λ = 0.167 nm. A photon carrying 54 eV has λ = 23.0 nm. Equal energy does not mean equal momentum.

Do not use λ = hc/K for a slow electron.

An electron accelerated from rest through 150 V has p = √(2mₑeV) = 6.62 × 10⁻²⁴ kg m s⁻¹, so λ = h/p = 0.100 nm: the wavelength of a 12.4 keV X-ray, and half the 0.20 nm spacing of planes in a typical metal.

Predict before measuring: quadrupling the accelerating voltage approximately halves every diffraction-ring radius.

Explore the diffraction lab

Change one variable, then explain why the rings move.

Calculated teaching model
50–200000 V · logarithmic slider
50–400 pm
50–1000 mm
Side viewScreen · face on2θ = 6.02°Thin filmL = 135 mmAngles enlarged in side view for clarityScreen: ±50.0 mmRing radii share this scale
Fixed ±50 mm screen window: rings beyond it are outside the view. Green: d; blue: d/√3. First-order positions are calculated; brightness and peak width are not predicted.
Wavelength λ22.359 pm
Glancing angle θ3.009°
Inner ring radius R14.230 mm
R√V24.647mm √kV · approximately constant at low voltage, small angles

Predict before changing a control: quadruple V from 1.5 to 6 kV. The wavelength is nearly halved and the rings shrink. Now hold V fixed and vary d: the rings move, but λ does not.

Flat screen; first-order elastic diffraction. The side-view angles are enlarged.

First hold d fixed and increase V. Then hold V fixed and change d. The wavelength belongs to the beam; the spacing belongs to the crystal. The rings are calculated geometry, not measured intensities.

Derive the prediction, one step at a time

V → kinetic energy → momentum → wavelength → angle → ring radius.

1 · Start with momentum, not energy
λ = h/p

The relation applies to photons and massive particles alike. Use the appropriate momentum law for the particle.

Units: J s ÷ (kg m s⁻¹) = m. A wavelength is a length.

2 · Turn accelerating voltage into an electron wavelength
K = eV = p²/(2me)
p = √(2meeV)
λ = h/√(2meeV)

Here e is the positive elementary charge, V is the accelerating potential difference, and the electron starts approximately from rest.

λ = 1.2264 nm / √(V/1 V)

At 54 V, p = 3.97 × 10⁻²⁴ kg m s⁻¹ and λ = 0.1669 nm. At 3.00 kV, λ ≈ 22.4 pm. These are non-relativistic values.

3 · Compare the paths from adjacent planes
θ2θdConstructive interference: 2d sin θ = nλ
θ is the glancing angle to the planes. The outgoing beam is deflected by 2θ. Schematic, not an intensity calculation.
Extra path = 2d sin θ = nλ

The two extra path segments must add to a whole number of wavelengths. θ is the glancing angle between beam and planes, not the detector’s deflection angle.

For a selected plane family and order n, a solution requires nλ ≤ 2d. A condition on that one reflection is not a claim about all possible scattering.

4 · Turn the Bragg angle into a screen radius
2θ = deflection angle
R = L tan(2θ)
R ≈ Lλ/d   (first order, small angles)

In a polycrystalline sample, grains with different orientations send the same reflection around a cone. A flat screen cuts the cone into a ring.

Only a ray with 2θ < 90° meets a screen downstream. The exact geometry is used in the lab; the convenient R√V ≈ constant test additionally needs small angles and non-relativistic momentum.

5 · Use relativistic momentum when the precision needs it
(K + mec²)² = (pc)² + (mec²)²
λ = hc / √[K(K + 2mec²)]
λ = h / √[2meeV(1 + eV/2mec²)]

The correction is continuous, not an on/off threshold. The non-relativistic wavelength is 0.49% too large at 10 kV and 3.84% too large at 80 kV, using the relativistic wavelength as the denominator.

At 80 kV: λNR = 4.336 pm; λrel = 4.176 pm. The exact relation λ = h/p has not failed—the approximation for p has.

Test the model with data

Infer wavelengths, fit h, and inspect what the residuals leave behind.

The starting readings are deterministic synthetic teaching data with stated radius uncertainties. They are not the original Davisson–Germer or Thomson observations. Edit them to explore fit sensitivity.

Synthetic teaching data · editable
Graphite-like spacing d = 213 pm · camera length L = 135 mm
V (V)Radius (mm)u(R) (mm)Edit

Infer λ = 2d sin[½ arctan(R/L)]. Fit against 1/√[V(1 + eV/2mₑc²)]. A through-origin slope predicts 10¹²h/√(2mₑe) when λ is in pm. A free intercept is a diagnostic, not automatically a better physical model.

0070.006150.012220.017290.023370.0291500 V: 31.697 ± 0.123 pm2000 V: 27.316 ± 0.124 pm2500 V: 24.527 ± 0.124 pm3000 V: 22.458 ± 0.125 pm4000 V: 19.311 ± 0.125 pm5000 V: 17.238 ± 0.125 pm6000 V: 15.821 ± 0.125 pm1 / √[V(1 + eV / 2mₑc²)] (V⁻½)Inferred wavelength λ (pm)
● Teaching readings ±1 standard uncertainty · red line: weighted fit. Voltage and geometry treated as exact; calibration uncertainty is not included.
Estimated h (J s)6.6279 × 10⁻³⁴± 0.0134 × 10⁻³⁴ (1 standard uncertainty)
Fit slope1226.77± 2.48 pm √V
χ² / degrees of freedom1.74 / 6Reduced χ²: 0.290
Intercept0.000 pmFixed at zero

With wavelength in picometres, the predicted slope is 10¹²h/√(2mₑe) pm √V. The conversion back to h includes the factor 10⁻¹² m/pm.

A plausible slope is not proof on its own. Inspect residuals, check geometry and calibration, and state the tested range. These synthetic observations were generated from the model; agreement is expected.

The fit uses the exact inverse screen geometry and propagates radius uncertainty into wavelength uncertainty. Voltage, camera length and spacing are treated as known; a real experiment must budget their uncertainties and systematic errors too. In today’s SI, h is exact: this exercise tests the model and calibration.

Practise with three worked examples

Calculate first. Open the solution when you are ready to compare.

Easy

A 250 V electron and an equal-energy photon

An electron is accelerated from rest through 250 V. Find its momentum and de Broglie wavelength, confirm that h/√(2mₑeV) is a length, and compare λ with the wavelength of a photon carrying the same 250 eV.

Show the worked solution →
  1. Energy: the electron gains K = eV = 250 × 1.602 × 10⁻¹⁹ J = 4.005 × 10⁻¹⁷ J, all kinetic, so p = √(2mₑK) = √(2 × 9.109 × 10⁻³¹ kg × 4.005 × 10⁻¹⁷ J) = 8.54 × 10⁻²⁴ kg m s⁻¹.
  2. Wavelength: λ = h/p = 6.626 × 10⁻³⁴ J s ÷ 8.54 × 10⁻²⁴ kg m s⁻¹ = 7.76 × 10⁻¹¹ m = 77.6 pm. The shortcut 1.2264 nm/√250 = 0.0776 nm agrees.
  3. Dimensions: [mₑeV] = kg × J = kg² m² s⁻², so its root is kg m s⁻¹, and J s ÷ kg m s⁻¹ = (kg m² s⁻¹)/(kg m s⁻¹) = m. A length, as a wavelength must be.
  4. Limiting case: eV/2mₑc² = 250 eV ÷ 1.022 × 10⁶ eV = 2.4 × 10⁻⁴, so the relativistic correction shifts λ by about 0.01%; the non-relativistic result is good to three figures.
  5. Photon comparison: λ = hc/E = 1239.8 eV nm ÷ 250 eV = 4.96 nm, 64 times longer. Equal energies do not mean equal wavelengths, because a massive particle carries far more momentum per unit energy.

p = 8.54 × 10⁻²⁴ kg m s⁻¹ and λ = 77.6 pm (0.0776 nm); a 250 eV photon has λ = 4.96 nm, 64 times longer.

The comparison equates the electron’s kinetic energy with the photon’s total energy. The electron’s total energy also includes its rest energy.

Medium

Read graphite’s spacing from a 27 mm ring

A teaching diffraction tube accelerates electrons through 3.60 kV onto a graphite film 0.140 m from a flat screen. The inner ring measures 27.0 mm across. Find the plane spacing d responsible, then predict the ring's diameter at 5.00 kV two ways: from the Bragg geometry, and from the scaling R√V = constant.

Show the worked solution →
  1. Wavelength at 3.60 kV: λ = 1.2264 nm/√3600 = 0.02044 nm = 20.44 pm. The relativistic correction is eV/4mₑc² = 0.18%, below the precision of a ring measured with a ruler.
  2. Geometry: R = 13.5 mm, so tan 2θ = 13.5/140 = 0.09643, giving 2θ = 5.508° and a glancing angle θ = 2.754°.
  3. Bragg with n = 1: d = λ/(2 sin θ) = 20.44 pm ÷ (2 × 0.04805) = 212.7 pm = 0.213 nm, the plane spacing X-ray diffraction gives for graphite.
  4. Bragg at 5.00 kV: λ = 1.2264 nm/√5000 = 17.34 pm, sin θ = 17.34/(2 × 212.7) = 0.04076, θ = 2.336°, and R = 140 mm × tan 4.672° = 11.44 mm, a diameter of 22.9 mm.
  5. Scaling: R√V = 13.5 mm × √3.60 = 25.61 mm √kV, so at 5.00 kV R = 25.61/√5.00 = 11.45 mm, again a diameter of 22.9 mm. The routes differ by 0.1%, the small-angle error at 2θ ≈ 5°, so the scaling law is the de Broglie prediction in a form a ruler can test.

d = 0.213 nm; at 5.00 kV the inner ring is 22.9 mm across by either route, 15% smaller than at 3.60 kV.

27.0 mm is a diameter: use R = 13.5 mm. The screen in this problem is explicitly flat.

Hard

Relativity in an 80 kV microscope

An electron microscope in diffraction mode runs at 80.0 kV with camera length L = 0.600 m on a polycrystalline aluminium film whose (111) planes are 0.2338 nm apart. Predict the (111) ring radius with the non-relativistic and with the relativistic wavelength. The ring is then measured at 10.72 mm: what plane spacing would an analysis using λ = h/√(2mₑeV) report, and how large is its error?

Show the worked solution →
  1. Non-relativistic: λ = 1.2264 nm/√(8.00 × 10⁴) = 4.336 pm.
  2. Relativistic: 1 + eV/2mₑc² = 1 + 80.0 keV/1022 keV = 1.0783, so λ = 4.336 pm/√1.0783 = 4.176 pm. The non-relativistic value is 3.8% too long, close to the estimate eV/4mₑc² = 3.9%.
  3. Ring radius: sin θ = λ/2d = 4.176/467.6 = 0.008931, so θ = 0.5117°, 2θ = 1.0234° and R = 600 mm × tan 1.0234° = 10.72 mm. With 4.336 pm instead, sin θ = 0.009273, 2θ = 1.0626° and R = 11.13 mm.
  4. Small-angle check: Lλ/d = 600 mm × 4.176 pm ÷ 233.8 pm = 10.72 mm, the same to four figures, so at these angles R ∝ λ and every ring inherits the full 3.8% error.
  5. Invert the measurement: 10.72 mm gives 2θ = 1.0236° and sin θ = 0.008932. Then d = 4.336 pm ÷ (2 × 0.008932) = 242.7 pm, while the relativistic 4.176 pm returns 233.8 pm. The non-relativistic analysis overstates d by 3.8%, nearly four times the 1% precision of a careful ring measurement.
  6. Limiting case: the correction vanishes as V → 0, only 0.003% at 54 V, and grows in proportion to V at first, which is why the low-voltage experiments of the 1920s needed no relativity and a modern microscope cannot do without it.

R = 10.72 mm (relativistic) against 11.13 mm (non-relativistic); the non-relativistic analysis of the measured ring reports d = 0.2427 nm, 3.8% too large.

“3.8% too large” is relative to the corrected result. The correction decreases λ from the non-relativistic estimate by about 3.7%.

Check your understanding

Distinguish the wavelength, the glancing angle and the beam deflection.

Question 1/5 · 0/5 correct
Electrons accelerated from rest through 150 V fall on a crystal whose diffracting planes are 0.200 nm apart. Through what angle is the first-order Bragg beam deflected from the incident direction?
Original Bragg-angle question plus four concept checks. This practice does not overwrite saved course progress.

The historical test uses a different geometry

Davisson–Germer’s 50° is an exit angle from the outward surface normal.

At normal incidence, nickel’s surface rows act as a reflection grating. Keep their row spacing D separate from the plane spacing d used in the Bragg diagram.

D sin φ = nλ
D = 0.215 nm · φ = 50°
λmeasured ≈ 0.165 nm · λde Broglie ≈ 0.167 nm

The approximately 1% agreement is a quantitative test of matter waves. In this backreflection geometry, the change between the incoming and outgoing velocity directions is 180° − φ, not φ.

Do not substitute φ = 50° and D = 0.215 nm into 2d sin θ = λ. They describe a surface-grating geometry, not the θ and d in the bulk Bragg sketch.

The complete original pattern also involves internal propagation and multiple scattering. See the 1927 primary report and the Binghamton surface-diffraction treatment.

The same h, from neutrons to molecules

Mass changes the momentum. It does not switch off the wavelength relation.

A thermal neutron

K ≈ 25 meV
λ ≈ 0.18 nm
v ≈ 2200 m s⁻¹

Atomic-scale wavelength at thermal energy. Nuclear and magnetic scattering make neutrons a different structural probe from X-rays.

A C₆₀ molecule

m = 720 u · v = 220 m s⁻¹
λ ≈ 2.5 pm
θ₁ ≈ λ/a = 25 μrad

A 100 nm grating produces resolvable diffraction. The centre-of-mass wavelength can be far smaller than the molecule’s own size.

An everyday ball

m = 0.145 kg · v = 40 m s⁻¹
λ ≈ 1.1 × 10⁻³⁴ m

For accessible structures, λ/d is too small to resolve diffraction. This is a practical classical limit, not a different momentum law.

In the 1999 Vienna C₆₀ experiment, a 100 nm grating and 1.25 m detection distance resolved a molecular diffraction pattern. At 220 m s⁻¹, the nominal first-order displacement is about 31 μm.

Know what the model can claim

Predict geometry accurately; qualify intensities and experimental precision.

  • Momentum: use the relativistic relation when the required precision makes its correction significant.
  • Scattering: the interactive rings are kinematic geometrical predictions, not calculated dynamical intensities.
  • Apparatus: a curved fluorescent bulb needs a different screen correction from a flat camera plane.
  • Material: finite thickness, crystal size, disorder, internal potential and energy spread can affect real features.
  • Interpretation: λ = h/p predicts interference geometry. It does not by itself say what the wavefunction means.

Next: electrons arrive one at a time, yet their detections build an interference pattern. Wave–particle duality and the double slit ask how amplitudes become measured outcomes.

Full university study notes · all six explanations

From the photon's momentum to the electron's wavelength

Compton scattering fixed p = h/λ for light. de Broglie's step was to assert the same relation for matter with nothing adjusted: any particle of momentum p has λ = h/p. For an electron accelerated from rest through V, energy conservation gives eV = p²/2mₑ, so p = √(2mₑeV) and λ = h/√(2mₑeV). Check the dimensions before trusting it: mₑeV is kg × J = kg² m² s⁻², whose root is kg m s⁻¹, a momentum, and J s divided by kg m s⁻¹ is m. With the constants in, λ = 1.226 nm/√(V/1 V), so 54 V gives 0.167 nm and 10 kV gives 12.3 pm. The same formula explains why nobody had noticed: a 0.145 kg ball at 40 m s⁻¹ has λ = 1.1 × 10⁻³⁴ m, diffraction angles of order λ/d vanish, and the straight-line trajectories of classical mechanics are the λ → 0 limit, just as ray optics is the short-wavelength limit of wave optics.

A crystal is a grating with a known period

A ruled grating has lines about a micrometre apart, ten thousand times the 0.1 nm wavelengths here; the planes of atoms in a crystal sit 0.1 to 0.4 nm apart, exactly the right scale. Treat each plane as a weak partial mirror. Waves reflected from neighbouring planes a distance d apart leave in step when the extra path, 2d sin θ, is a whole number of wavelengths: 2d sin θ = nλ, with θ the glancing angle between the beam and the planes. Three features of this condition do the work. The beam is deflected through 2θ, so that is the angle a detector records. A reflection needs sin θ ≤ 1, so it exists only for λ ≤ 2d/n. And d is measured independently, by X-ray diffraction from the same crystal, so an electron experiment has no free parameter: elastic diffraction at that same wavelength and from the same periodic structure has the same nominal geometrical conditions. Different interactions can make different beams visible and give different intensities.

Davisson and Germer: one strong beam at 50°

Davisson and Germer sent 54 eV electrons normally onto nickel and observed a strong outgoing beam about 50° from the outward surface normal. For surface rows spaced D = 0.215 nm, the normal-incidence grating condition D sin φ = λ gives 0.165 nm, compared with the de Broglie prediction 0.167 nm. D is a surface-row spacing, not the interplanar d of bulk Bragg diffraction. Parallel wavevector conservation fixes the ideal surface beam’s exit direction using the vacuum wavelength; an inner potential is not an explanation of this small residual. Internal refraction and multiple scattering matter when interpreting the complete voltage-dependent pattern and intensities.

Thomson's rings, and a test only the voltage can run

G. P. Thomson sent fast electrons through thin polycrystalline films. Crystallites at many orientations produce diffraction cones, one for each accessible plane spacing, and a flat screen cuts the cones into rings with R = L tan(2θ). At small angles, R ≈ Lλ/d, so ring-radius ratios approach inverse plane-spacing ratios. For non-relativistic momentum, λ ∝ V(−1/2), making R√V approximately constant—not exactly constant with the full screen geometry. At the tens-of-kilovolts energies of Thomson’s experiments, use the relativistic wavelength when the correction is significant compared with measurement precision. At 50 kV the non-relativistic estimate is 2.42% too large. Magnetic deflection of the pattern identifies the charged electron beam rather than neutral X-rays. The test is the wavelength-dependent diffraction geometry; peak intensities need a scattering model as well.

Same relation, heavier particles

Nothing in λ = h/p mentions charge or the electron, so the prediction extends to anything with momentum. At fixed kinetic energy λ = h/√(2mK) falls as 1/√m: a neutron, 1839 times heavier, carries 1/43 of an electron's wavelength. To reach 0.18 nm it needs only K ≈ 25 meV, the thermal energy of a room-temperature moderator, at about 2200 m s⁻¹. Neutrons scatter from nuclei rather than from electron clouds, so they locate the hydrogen atoms X-rays barely see, their own magnetic moment probes magnetic order, and being uncharged they can penetrate deeply into many materials: the same Bragg geometry, a different probe. For C₆₀, 720 u moving at 220 m s⁻¹, λ = h/mv = 2.5 pm, roughly 400 times smaller than the molecule's 1 nm diameter; a grating of 100 nm period diffracts it into a first order only 25 μrad from the beam, and that fringe pattern was measured in 1999. Mass makes the angle small, not zero: the classical limit is λ/d → 0, not a change in the physics.

Where the simple model stops

The wavelength relation λ = h/p survives; the approximations used to calculate p and the diffraction pattern have limits. At 54 V the non-relativistic wavelength is only 0.0026% high, but at 80 kV it is 3.84% high relative to the relativistic value. A kinematic crystal model adds amplitudes from single scattering and predicts nominal Bragg directions; electron intensities often require dynamical multiple-scattering theory. Finite crystal size, disorder and beam spread broaden features, and multiple scattering can change which beams become visible. Do not infer exact peak heights or all selection rules from the simple geometry alone. A crystal’s inner potential changes internal momentum and bulk diffraction conditions. For a periodic surface, the ideal exit directions remain constrained by parallel wavevector conservation and the vacuum wavelength.

Complete formula reference · six original relations

de Broglie relation

λ = h/p

λ in m; p in kg m s⁻¹; h = 6.626 × 10⁻³⁴ J s

At equal momentum, particles have equal wavelength. For elastic scattering from the same periodic structure, this fixes the same nominal geometrical diffraction conditions, not identical intensities or visibility of every beam.

Electron accelerated from rest through V

λ = h/√(2mₑeV)

mₑ = 9.109 × 10⁻³¹ kg; e = 1.602 × 10⁻¹⁹ C; V in volts; from eV = p²/2mₑ

Numerically λ = 1.226 nm/√(V/1 V): 54 V gives 0.167 nm and 3.0 kV gives 22.4 pm. Halving λ costs four times the voltage.

Bragg condition

2d sin θ = nλ

d = plane spacing in m; θ = glancing angle between beam and planes; n = 1, 2, 3, …

θ is measured to the planes; the deflection is 2θ. For this selected plane family and order, a reflection requires nλ ≤ 2d.

Ring radius on a screen at distance L

R = L tan 2θ

R and L in m; θ from the Bragg condition; one cone per set of planes in a polycrystalline film

At small angles, R ≈ Lλ/d for first order. With non-relativistic momentum, R√V is approximately constant at fixed d; exact screen geometry and relativistic momentum both modify that scaling.

Relativistic correction

λ = h/√(2mₑeV(1 + eV/2mₑc²))

mₑc² = 511 keV; the bracket is 1.00005 at 54 V and 1.049 at 50 kV

Relative to the exact wavelength, the non-relativistic estimate is 0.29% high at 6 kV, 2.42% at 50 kV and 4.78% at 100 kV. Compare this correction with the required precision; no universal voltage makes it safe to ignore.

Any particle of kinetic energy K

λ = h/√(2mK)

m in kg; K in J; a neutron with K = 25 meV has λ = 0.18 nm

At equal K, λ ∝ 1/√m: a neutron carries 1/43 of an electron's wavelength, which is why neutron diffraction runs at thermal energies, not kilovolts.

Why the photon shortcut fails for a slow electron

The commonest slip is to borrow the photon's relation and write λ = hc/E for an electron. That relation rests on E = pc, which holds only for a massless particle. A slow electron's momentum is √(2mₑK), far larger than K/c, so its wavelength is far shorter: for a 54 eV electron hc/E gives 23.0 nm, while h/√(2mₑK) gives 0.167 nm. The photon formula is too long by the factor √(2mₑc²/K) = 138, and that factor is itself the diagnosis: it compares the electron's rest energy with its kinetic energy, so the photon formula becomes right only when K ≫ mₑc², where the electron is ultra-relativistic and E ≈ pc. Used at 54 eV it would also predict no Bragg reflection from nickel at all, since 23 nm is far more than twice any of its plane spacings, which are about 0.2 nm.

Sources & study notes

Prerequisites: electric potential and electronvolts, conservation of energy, wave interference, basic trigonometry and uncertainty propagation. The data fit treats the voltage and geometry as known.

Interactive plots and starter readings are synthetic teaching models. Historical observations are identified separately. Print this page to create your own PDF; no downloadable notebook or video is implied.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.