
Optional unit · institution-dependent · Interactive lesson
Matter waves and the de Broglie wavelength
Electrons are not tiny bullets that occasionally misbehave. Fire them at a crystal and they come back in the same sharp pattern X-rays make. This lesson turns that experiment into one number, λ = h/p, and then shows why a cricket ball obeys the same rule and still flies dead straight.
UPII-15.05 · Lesson 5 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Give a particle a momentum and you have given it a wavelength.
The key idea
The de Broglie wavelength of a particle is Planck's constant divided by its momentum, λ = h/p — the wavelength that governs how that particle interferes and diffracts.
Example. An electron accelerated from rest through 150 V has p = √(2mₑeV) = 6.62 × 10⁻²⁴ kg m s⁻¹, so λ = 6.626 × 10⁻³⁴ ÷ 6.62 × 10⁻²⁴ = 1.00 × 10⁻¹⁰ m = 0.100 nm, about one atomic spacing.
The complete physical model and assumptions
How far matter-wave behaviour reaches is decided by one number, h = 6.626 × 10⁻³⁴ J s. De Broglie's 1924 step was to read the photon relation p = h/λ not as a property of light but as a property of momentum, so that any particle of momentum p arrives with a wavelength λ = h/p and a frequency f = E/h. The hypothesis is silent about what is waving — that is the next topic — but it is sharp about where the wave will show itself, because a wave betrays itself only through interference, and interference needs structure comparable to λ. An electron accelerated through a hundred volts has λ ≈ 0.1 nm, the spacing of atoms in a crystal, which is why Davisson and Germer saw electrons come off nickel in sharp Bragg peaks in 1927. Since h is minute, λ collapses the moment p is of everyday size: a 0.160 kg ball at 30 m s⁻¹ has λ = 1.4 × 10⁻³⁴ m, twenty orders of magnitude below a nucleus, and no aperture can diffract it. That is where the model hands back to mechanics — not by failing, but because λ/D → 0 makes every diffraction angle unmeasurable, exactly as geometrical optics emerges from wave optics. Only the arithmetic has a limit: λ = h/p always holds, while p = √(2mK) fails once K approaches mc².
1Where λ = h/p comes from
Light had already forced the issue. A photon's energy and momentum obey E = pc, and E = hf with f = c/λ, so p = hf/c = h/λ — a momentum expressed entirely by a wavelength. In 1924 de Broglie asked which half of that statement is a fact about light. His hypothesis: neither. The relation belongs to momentum itself, so any particle of momentum p is accompanied by a wave of wavelength λ = h/p, whose frequency is f = E/h. Notice what the hypothesis does not claim. It does not say what is waving, and it does not say the wave travels at the particle's speed — the phase speed fλ = E/p is not v, and it is the group speed of a packet of such waves that keeps pace with the particle. What it does claim is testable: fix p and the diffraction angles are fixed with it.
2Putting numbers on it: λ falls as 1/p
Work in momentum, not speed. For mass m and kinetic energy K well below mc², p = √(2mK), so λ = h/√(2mK): the wavelength falls as 1/√K and as 1/√m. An electron accelerated from rest through a p.d. V gains K = eV, which gives the shortcut λ = 1.226 nm/√V with V in volts — 0.137 nm at 80.0 V, 0.100 nm at 150 V, 5.48 pm at 50.0 kV. A neutron is 1839 times heavier, so at the same kinetic energy its wavelength is √1839 = 42.9 times shorter: at 25 meV an electron has 7.76 nm against the neutron's 0.181 nm. A 0.160 kg ball at 30 m s⁻¹ carries p = 4.80 kg m s⁻¹, so λ = 6.626 × 10⁻³⁴ ÷ 4.80 = 1.4 × 10⁻³⁴ m. The scale is set by h, and h is tiny: matter wavelengths reach atomic size only for light particles moving slowly.
3The evidence: diffraction from a crystal
A wave announces itself only through interference, and that needs structure comparable to λ. At 0.1 nm no manufactured slit will serve, but a crystal is a ready-made grating with atoms about 0.2 nm apart. Davisson and Germer fired 54 eV electrons at a nickel crystal in 1927 and found a sharp peak in the scattered intensity at 50° from the incident beam, where classical scattering predicts a smooth falloff. The surface rows are d = 0.215 nm apart and constructive interference needs d sinφ = λ, giving λ = 0.215 × sin 50° = 0.165 nm. De Broglie's formula predicts 1.226/√54 = 0.167 nm. Two independent numbers agreeing to about 1% is what turned a hypothesis into physics. G. P. Thomson got there the other way, firing electrons through thin metal foils and photographing concentric rings identical in form to X-ray powder patterns. Neutrons, helium atoms and whole C₆₀ molecules (λ = 2.8 pm at 200 m s⁻¹) have been diffracted since.
4Fitting a wave into a space makes energy discrete
A wave confined to a region can only take the wavelengths that fit, and λ = h/p turns that geometry into a statement about momentum. Trap an electron between walls a distance L apart: the standing waves need λ = 2L/n for n = 1, 2, 3 …, so p = nh/2L and K = n²h²/(8mL²). For L = 0.50 nm the lowest level is (6.626 × 10⁻³⁴)² ÷ (8 × 9.109 × 10⁻³¹ × (0.50 × 10⁻⁹)²) = 2.41 × 10⁻¹⁹ J = 1.50 eV — an electronvolt-sized gap conjured from nothing but a length and a fitting condition. Wrap the same argument round a circle and a Bohr orbit must hold a whole number of wavelengths, 2πr = nλ, which is exactly the angular-momentum rule mvr = nh/2π that Bohr had to assume. At the first Bohr radius 2πr = 0.3325 nm, and h/mv for that electron is also 0.3325 nm. Quantisation stops looking like an extra postulate and starts looking like a boundary condition.
5Where the wave model hands back to mechanics
An opening of width D spreads a beam by roughly θ ≈ λ/D radians, and everything classical follows from that ratio going to zero. The 0.160 kg ball has λ = 1.4 × 10⁻³⁴ m, so a 1 m doorway spreads it by 10⁻³⁴ rad, which is not merely small but unmeasurable — more than twenty orders of magnitude below any angle ever resolved. Even a 1 μg grain of dust drifting at 1 mm s⁻¹ has λ = 6.6 × 10⁻²² m. The classical limit is therefore not a different physics: it is λ/D → 0, in the same way geometrical optics is the λ → 0 limit of wave optics and rays replace wavefronts. Read the other way it is a design rule. A 50 kV electron has λ = 5.36 pm, about 10⁵ times shorter than green light, which is why an electron microscope resolves atomic columns — and why, inside its column, those same electrons can be traced as rays.
6When p = mv is not good enough
λ = h/p is exact; p = mv and p = √(2mK) are not. Once K is a noticeable fraction of the rest energy mc² — 511 keV for an electron — use E² = (pc)² + (mc²)² with E = K + mc², which gives pc = √(K² + 2Kmc²) and so λ = hc/√(K² + 2Kmc²), with hc = 1239.8 eV nm a convenient constant. The correction is invisible on a bench and decisive in an instrument: at 1.0 kV the non-relativistic wavelength is 0.05% too large, at 10 kV 0.5%, at 50 kV 2.4% (5.485 pm against the true 5.355 pm) and at 200 kV 9.3%. A 50 kV electron is already travelling at 0.413c. Nothing about the wave hypothesis changes here — only the arithmetic that converts an accelerating voltage into a momentum.
Key equations
Read the units, assumptions and limits before substituting.
De Broglie relation
λ = h/p
Converts a momentum straight into the wavelength that sets its diffraction angles.
λ in m, p in kg m s⁻¹, h = 6.626 × 10⁻³⁴ J sWavelength from kinetic energy
λ = h/√(2mK)
The working form, because beams are specified by an energy rather than a momentum.
K in J, m in kg; valid while K ≪ mc², since it uses p = √(2mK)Electron through a potential difference
λ = h/√(2mₑeV)
A bench voltage tunes λ: 150 V gives 0.100 nm, right on atomic spacings.
V in volts, mₑ = 9.109 × 10⁻³¹ kg; numerically λ = 1.226 nm/√VBragg condition for the diffracted beam
2d sinθ = nλ
Turns a measured peak angle into a wavelength — the test de Broglie's hypothesis had to pass.
d = plane spacing in m, θ measured from the planes, n = 1, 2, 3 …Relativistic wavelength
λ = hc/√(K² + 2Kmc²)
Needed above a few kV: at 50 kV the non-relativistic value is 2.4% too large.
K and mc² in the same energy unit; mₑc² = 511 keV, hc = 1239.8 eV nmClassical limit
θ ≈ λ/D
Wave behaviour survives only while λ is within a few orders of D; below that, rays.
D = width of the aperture or obstacle in m, θ the spreading angle in radInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Slide the voltage from 20 V up to 500 V: the drawn wave shortens against a fixed plane spacing and the reflected beam swings down towards the incident line, which it now leaves by {2*deg(asin(0.613/(d*√(v))))|1}° — the classical limit arriving. Then widen d and watch θ shrink again.
Live results
- WAVELENGTH λ = h/p
- 0.100 nm
- RATIO λ / 2d
- 0.250
- BRAGG ANGLE θ
- 14.5 °
- BEAM DEFLECTION 2θ
- 29.0 °
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: 150 V; Crystal plane spacing d: 0.20 nm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest wreck here is borrowing the photon formula. For light E = pc, so λ = hc/E, and the temptation is to apply that to an electron with E set to the kinetic energy: a 150 eV electron then gets λ = 1239.8 ÷ 150 = 8.27 nm instead of 0.100 nm, wrong by a factor of 83. The fault is not the constant hc but the relation E = pc, which holds only for a massless particle. An electron obeys E² = (pc)² + (mc²)², so pc = √(K² + 2Kmₑc²) = √(150² + 2 × 150 × 511000) eV = 1.238 × 10⁴ eV — eighty-three times the kinetic energy, because nearly all of the momentum is bought against the rest energy. Feed that back in and λ = hc/pc = 1239.8 ÷ 12382 = 0.100 nm. λ = h/p never changes; what changes is how you get p.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An electron-diffraction demonstration tube accelerates electrons from rest through 80.0 V and fires them at a graphite film whose atomic planes are 0.213 nm apart. Find the electron's momentum and de Broglie wavelength, and decide whether that film can diffract the beam.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The electron gains K = eV = 80.0 × 1.602 × 10⁻¹⁹ = 1.282 × 10⁻¹⁷ J, far below mₑc² = 8.19 × 10⁻¹⁴ J, so the non-relativistic p = √(2mₑK) is safe.
- p = √(2 × 9.109 × 10⁻³¹ × 1.282 × 10⁻¹⁷) = √(2.335 × 10⁻⁴⁷) = 4.83 × 10⁻²⁴ kg m s⁻¹.
- λ = h/p = 6.626 × 10⁻³⁴ ÷ 4.83 × 10⁻²⁴ = 1.37 × 10⁻¹⁰ m = 0.137 nm. The shortcut 1.226/√80.0 nm returns the same 0.137 nm.
- Compare wavelength with structure: λ/2d = 0.137 ÷ 0.426 = 0.322, comfortably below 1, so a first-order Bragg beam exists at θ = arcsin 0.322 = 18.8° to the planes. A wavelength within a factor of two of the spacing is exactly the condition for visible diffraction.
Answer. p = 4.83 × 10⁻²⁴ kg m s⁻¹ and λ = 0.137 nm; the 0.213 nm planes do diffract it, the first order leaving at 18.8° to the planes.
MediumExample 2
A reactor beam line selects neutrons of speed 2.20 × 10³ m s⁻¹ (mₙ = 1.675 × 10⁻²⁷ kg). Find their de Broglie wavelength and the first-order Bragg angle from crystal planes 0.282 nm apart, then find their kinetic energy in meV and compare it with kT at 300 K, where k = 1.381 × 10⁻²³ J K⁻¹.
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.
- At 2.2 km s⁻¹ a neutron is nowhere near relativistic, so p = mₙv = 1.675 × 10⁻²⁷ × 2.20 × 10³ = 3.685 × 10⁻²⁴ kg m s⁻¹.
- λ = h/p = 6.626 × 10⁻³⁴ ÷ 3.685 × 10⁻²⁴ = 1.798 × 10⁻¹⁰ m = 0.1798 nm.
- First-order Bragg: sinθ = λ/2d = 0.1798 ÷ (2 × 0.282) = 0.3188, so θ = 18.6° to the planes.
- K = ½mₙv² = 0.5 × 1.675 × 10⁻²⁷ × (2.20 × 10³)² = 4.053 × 10⁻²¹ J; dividing by 1.602 × 10⁻¹⁹ gives 2.53 × 10⁻² eV = 25.3 meV.
- kT at 300 K is 1.381 × 10⁻²³ × 300 = 4.143 × 10⁻²¹ J = 25.9 meV. The neutron carries almost exactly kT — it has been slowed to room temperature in a moderator, and that is what puts its wavelength at atomic spacings.
Answer. λ = 0.1798 nm, first-order Bragg angle θ = 18.6°, and K = 25.3 meV, within 3% of kT = 25.9 meV at 300 K.
HardExample 3
A transmission electron microscope accelerates electrons from rest through 50.0 kV. (a) Find the de Broglie wavelength non-relativistically. (b) Correct it, given mₑc² = 511 keV and hc = 1239.8 eV nm. (c) The beam then passes a 50.0 μm aperture: estimate the spreading θ ≈ 1.22 λ/D and say what it means for the instrument.
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.
- Non-relativistic: λ = 1.226 nm/√V = 1.226 ÷ √50000 = 1.226 ÷ 223.6 = 5.48 × 10⁻³ nm = 5.48 pm.
- Test that assumption: K/mₑc² = 50.0 ÷ 511 = 0.098, so γ = 1.098 and v = 0.413c. Ten per cent of the rest energy should cost a few per cent in λ.
- Do it properly: pc = √(K² + 2Kmₑc²) = √(50.0² + 2 × 50.0 × 511) = √(2500 + 51100) = √53600 = 231.5 keV.
- λ = hc/pc = 1239.8 eV nm ÷ 2.315 × 10⁵ eV = 5.355 × 10⁻³ nm = 5.36 pm, so the non-relativistic value was 2.4% too large.
- Spreading at the aperture: θ ≈ 1.22 × 5.355 × 10⁻¹² ÷ (50.0 × 10⁻⁶) = 1.31 × 10⁻⁷ rad, about 7.5 × 10⁻⁶ degrees.
- That angle is far below anything the lenses control, so between aperture and specimen the electrons can be traced as straight rays — the classical limit. At the specimen, where the spacing is 0.2 nm, λ/d = 0.027 and the same wave diffracts strongly. Against 550 nm light, 5.36 pm is 1.0 × 10⁵ times shorter, which is where the resolving power comes from.
Answer. (a) 5.48 pm (b) 5.36 pm, the non-relativistic value being 2.4% high (c) θ ≈ 1.3 × 10⁻⁷ rad, so the beam is ray-like in the column yet still diffracts at the specimen.
Quick check
Select an answer and explain it before checking.
Review & continue
The de Broglie wavelength of a particle is Planck's constant divided by its momentum, λ = h/p — the wavelength that governs how that particle interferes and diffracts.
- 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 II completion tasks.