University Physics IV · Photons and Matter Waves · 4.5

The de Broglie Hypothesis

One relation, λ = h/p, and the whole of matter is inside it. Learn to run it in the direction that pays — a voltage in, a wavelength out — to know when γ has to be carried, and to see why the standing wave wrapped round a Bohr orbit is the one part of the picture the real theory throws away.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Einstein had already forced a wavelength and a momentum onto the same object: light of wavelength λ carries quanta of momentum h/λ, because the massless branch of E² = (pc)² + (mc²)² makes E = pc, and E = hf with fλ = c then fixes p = h/λ. De Broglie's move in 1924 was to read that sentence in the other direction and apply it to things with mass. His argument was not analogy but bookkeeping about phase.

A particle of rest mass m carries an internal oscillation at f₀ = mc²/h in its own rest frame; seen from a lab where it moves at v, that clock is dilated to f₀/γ, while the wave de Broglie attached to it must have the frame's own energy frequency γf₀. Demanding that the wave's phase at the particle's position track the internal oscillation — his "harmony of phases" — leaves exactly one possible wavelength, λ = h/γmv = h/p. What you buy is enormous: every particle now has a wavelength you can compute from a voltmeter reading, and electron diffraction becomes a prediction rather than a surprise.

What you pay is that the hypothesis names a wavelength and nothing else. It does not say what is oscillating, it supplies no equation of motion, its phase velocity ω/k = c²/v exceeds c, and the standing-wave-on-an-orbit picture that first made it famous is precisely the part that does not survive contact with the three-dimensional theory.

Simple definition: The de Broglie hypothesis assigns every particle a wavelength λ = h/p set by its momentum alone, so the wave relation first established for light applies to matter as well.
Example: 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 — an atomic spacing, which is exactly why a crystal diffracts it.

The de Broglie relationλ = h/p, with p = γmv

Momentum sets the wavelength — not speed, not energy. A heavy slow particle can be shorter-wave than a light fast one.

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

Phase harmony, de Broglie's own argumentf = E/h, vₚₕₐₛₑ = fλ = c²/v, v_group = dω/dk = v

The construction that produces λ = h/p also shows that only the group velocity can be the particle's own speed.

E = γmc² is the total energy, so f ≠ K/h; vₚₕₐₛₑ > c carries no signal

Accelerated through a potential differenceλ = h/√(2mqV) → λ = 1.226 nm ÷ √(V/volts) for electrons

Turns a dial setting into a wavelength: 150 V gives 100 pm, 54 V gives 167 pm, 10 kV gives 12.3 pm.

V in volts, q the charge, m the rest mass; valid while qV ≪ mc²

The relativistic correctionλ = h/√(2mqV⋅(1 + qV/2mc²))

At 100 kV the uncorrected 3.878 pm overstates the true 3.701 pm by 4.8%; the error passes 1% above about 20.5 kV.

For electrons qV/2mₑc² = V ÷ 1.022 × 10⁶, with V in volts

Bohr's condition, recovered2πr = nλ ⇔ L = mvr = nħ

Wrapping a whole number of wavelengths round a Bohr orbit reproduces its quantisation — and only its quantisation.

n = 1, 2, 3 …; ħ = h/2π = 1.055 × 10⁻³⁴ J s

Why matter waves stay hiddenλ = h/mv

A 0.145 kg ball at 40 m s⁻¹ has λ = 1.1 × 10⁻³⁴ m, about 10⁻¹⁹ of a nuclear diameter. No aperture can diffract that.

Non-relativistic form. m in kg, v in m s⁻¹, λ in m.

Photons first: where the relation comes from

The relation begins on the massless branch. Put m = 0 into E² = (pc)² + (mc²)² and you get E = pc exactly; combine that with E = hf and fλ = c, and a photon's momentum is p = E/c = hf/c = h/λ. This is not a definition dressed up as physics — Compton's 1923 measurement of the wavelength shift off nearly free electrons had just tested it against a two-body ledger, and it held. Read the chain in the direction it was derived and λ = h/p converts a known photon wavelength into a momentum. De Broglie's proposal was that the equation is more fundamental than the derivation that produced it: that λ = h/p is a statement about any quantum object, and that for an electron you should run it backwards, feeding in a momentum you already know how to compute and reading out a wavelength nobody had thought to look for. Nothing in the photon argument forces this step. It is a hypothesis, and it needed Davisson and Germer to survive.

The phase-harmony argument

De Broglie did not argue by analogy; he argued about phase. Attach to a particle of rest mass m an internal oscillation at its rest-frame frequency f₀ = mc²/h — for an electron, 1.24 × 10²⁰ Hz. In a lab frame where the particle moves at v, that internal clock is time-dilated and ticks at f₀/γ. Separately, let a wave of frequency f = γf₀ = E/h run through space, since it is the energy that transforms like a frequency. These two rates disagree, and de Broglie demanded that they agree at one place: the phase of the travelling wave evaluated at the particle's position x = vt must equal the phase of the internal clock. Writing that out gives γf₀ − v/λ = f₀/γ, so v/λ = f₀(γ − 1/γ) = f₀γβ², and therefore λ = c/(f₀γβ) = h/(γmv) = h/p. The same construction hands you the phase velocity: vₚₕₐₛₑ = fλ = E/p = c²/v, which is greater than c. That is not a contradiction, but it is a warning — the thing that actually moves at the particle's speed is the group velocity, and separating the two is the next topic's business.

From a dial setting to a wavelength

The momentum you can actually control in a laboratory is the one an accelerating voltage supplies. An electron released from rest through a potential difference V arrives with kinetic energy K = eV, and while eV stays far below mₑc² = 511 keV that means p = √(2mₑeV). So λ = h/√(2mₑeV), and folding the constants together gives the number worth memorising: λ = 1.226 nm ÷ √(V in volts). Three checks. At V = 150 V, √150 = 12.25, so λ = 0.100 nm — one atomic spacing, and the reason a crystal is the natural grating for slow electrons. At the 54 V of the Davisson–Germer peak, λ = 1.226/7.348 = 0.167 nm. At V = 10 kV, λ = 1.226/100 = 12.3 pm, some forty times smaller than a 0.5 nm lattice spacing. Notice the exponent: λ ∝ V^(−1/2), so quadrupling the voltage halves the wavelength while doubling it buys only a factor of 1.41. Wavelength is expensive in volts.

Where γ has to be carried

√(2mₑeV) is the non-relativistic momentum, and it starts to lie as soon as eV is a noticeable fraction of mₑc². The exact statement comes from the invariant: (pc)² = (K + mc²)² − (mc²)² = K² + 2Kmc², so p = √(2mK)⋅√(1 + K/2mc²), and λ = h/√(2mₑeV(1 + eV/2mₑc²)). For electrons the correction factor is √(1 + V ÷ 1.022 × 10⁶), with V in volts. Take the 100 kV of a transmission electron microscope. Non-relativistically λ = 1.2264/√(1.00 × 10⁵) nm = 3.878 pm. The correction factor is √(1 + 0.09785) = 1.0478, so the true wavelength is 3.878/1.0478 = 3.701 pm and the shortcut overstates it by 4.8%. Set that factor equal to 1.01 and you locate the one-percent line: eV = (1.01² − 1) × 1.022 MeV = 20.5 keV. Above roughly 20 kV the non-relativistic formula is no longer good enough for anything you would fit a lattice constant to.

Bohr's orbits, recovered and then discarded

The result that made the hypothesis famous is one line of algebra. Insist that the wave close on itself around a circular orbit — a whole number of wavelengths, 2πr = nλ — and substitute λ = h/mv: 2πr = nh/mv, so mvr = nh/2π = nħ. Bohr's angular-momentum postulate, assumed out of nowhere in 1913, drops out of a standing-wave condition. The numbers land exactly. The first Bohr orbit has r₁ = 52.92 pm and v₁ = 2.188 × 10⁶ m s⁻¹, so λ = h/mₑv₁ = 332.5 pm while 2πr₁ = 332.5 pm — one wavelength, closed, to four figures. Now the bad news. The three-dimensional theory keeps the spectrum and throws the picture away: hydrogen's ground state has l = 0, zero orbital angular momentum, so there is nothing circulating, no orbit, and no loop for a wave to wrap around. The real quantisation condition is that ψ be single-valued and normalisable in three dimensions, of which the ring is a one-dimensional caricature. Right answer, wrong reason.

What the hypothesis does not give you

λ = h/p attaches a single number to a particle, and it is worth being precise about how little that number says. It does not say what is waving — de Broglie's phase wave had no interpretation, and the probability-amplitude reading is Born's, two years later. It supplies no equation of motion: knowing a wavelength does not tell you how the wave propagates through a potential, which is the gap Schrödinger closed in 1926. And it is exact only for a definite momentum, which means a plane wave of infinite extent; a genuinely localised electron is a superposition of many momenta, so it carries a spread of wavelengths rather than one, and "the" de Broglie wavelength is only the centre of that spread. Finally, the relation is unforgiving about scale. A 0.145 kg ball thrown at 40 m s⁻¹ has λ = 6.63 × 10⁻³⁴ ÷ 5.8 = 1.1 × 10⁻³⁴ m. There is no aperture in the universe that small, which is why classical mechanics survives the hypothesis intact.

Change one variable at a time

Make the relationship visible.

Interactive model

Accelerating voltage V100 kV

Window width along the beam24 pm

At 5 kV the two curves sit on top of each other. Raise the voltage to 160 kV and the dashed wave visibly lags — then widen the window to 40 pm and they appear to line up again at the right edge. They are a whole cycle apart, not zero, and that is exactly the trap.

λ NON-RELATIVISTIC3.878 pm

λ EXACT3.701 pm

OVERSTATEMENT4.78 %

DASHED WAVE LAGS BY0.30 cycles

Live interpretationλ NON-RELATIVISTIC: 3.878 pm. λ EXACT: 3.701 pm. OVERSTATEMENT: 4.78 %. DASHED WAVE LAGS BY: 0.30 cycles

Catch the common trap

Explain before calculating.

Misconception check

The wavelength relation carries over from light; the frequency relation does not carry over in the form students expect. For a photon E = pc, so λ = h/p and λ = hc/E say the same thing — and that is the last time they agree. Apply λ = hc/E to a 1.00 keV electron and you get 1.24 nm, when the true answer is h/√(2mₑK) = 38.8 pm, wrong by a factor of 32. The trap underneath is the assumption that λf = v for matter waves. It does not: f = E/h uses the total energy γmc², so λf = E/p = c²/v, which is larger than c rather than equal to v. Nor does switching to the kinetic energy repair it — the frequency you get then depends on where you set the zero of energy, and no measurable quantity is allowed to. Only λ = h/p is safe, because only p has an unambiguous value.

A photon of energy 1.00 keV and an electron of kinetic energy 1.00 keV travel down the same beamline. Using hc = 1240 eV⋅nm and mₑc² = 511 keV, which statement about their de Broglie wavelengths is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is accelerated from rest through a potential difference of 150 V. Find its momentum and its de Broglie wavelength, and say whether a relativistic correction is needed. Take h = 6.626 × 10⁻³⁴ J s, mₑ = 9.109 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C.

Check the regime before choosing a formula. K = eV = 150 eV against mₑc² = 511 keV is a ratio of 2.9 × 10⁻⁴, so the correction factor √(1 + K/2mₑc²) differs from 1 by about 7 × 10⁻⁵. The non-relativistic momentum is good to 0.01% here — use it.
Convert the energy: K = 150 × 1.602 × 10⁻¹⁹ = 2.403 × 10⁻¹⁷ J.
Momentum: p = √(2mₑK) = √(2 × 9.109 × 10⁻³¹ × 2.403 × 10⁻¹⁷) = √(4.378 × 10⁻⁴⁷) = 6.617 × 10⁻²⁴ kg m s⁻¹.
Wavelength: λ = h/p = 6.626 × 10⁻³⁴ ÷ 6.617 × 10⁻²⁴ = 1.001 × 10⁻¹⁰ m = 0.100 nm. The shortcut agrees: 1.226 nm ÷ √150 = 1.226/12.25 = 0.100 nm.

Answerp = 6.62 × 10⁻²⁴ kg m s⁻¹ and λ = 0.100 nm = 100 pm, with the relativistic correction below 0.01%. That wavelength is an atomic spacing, which is why a 150 V beam diffracts off a crystal.

MediumThe n = 1 Bohr orbit of hydrogen has radius r₁ = 5.292 × 10⁻¹¹ m and orbital speed v₁ = 2.188 × 10⁶ m s⁻¹. (a) Show that the closure condition 2πr = nλ with λ = h/mv is equivalent to Bohr's postulate mvr = nħ. (b) Check the n = 1 case numerically. (c) State one thing the picture gets wrong.

(a) Substitute λ = h/mv into 2πr = nλ to get 2πr = nh/mv. Multiply both sides by mv/2π: mvr = nh/2π = nħ. The standing-wave closure condition and Bohr's angular-momentum postulate are the same equation written twice.
(b) Momentum on the first orbit: mₑv₁ = 9.109 × 10⁻³¹ × 2.188 × 10⁶ = 1.993 × 10⁻²⁴ kg m s⁻¹.
Wavelength: λ = h/mₑv₁ = 6.626 × 10⁻³⁴ ÷ 1.993 × 10⁻²⁴ = 3.325 × 10⁻¹⁰ m.
Circumference: 2πr₁ = 2π × 5.292 × 10⁻¹¹ = 3.325 × 10⁻¹⁰ m. The ratio 2πr₁/λ = 1.000, so exactly one wavelength closes the loop, as n = 1 demands.
(c) The derivation assumes the electron runs round a circle carrying angular momentum 1ħ. Solving the three-dimensional problem gives a hydrogen ground state with l = 0 — no orbital angular momentum, nothing circulating, and no loop for a wave to close around. The condition delivers the right energies from a picture the theory later discards.

Answer2πr = nλ is algebraically identical to mvr = nħ, and for n = 1 both λ and 2πr₁ come to 3.325 × 10⁻¹⁰ m. But the true ground state has l = 0, so the closed circulating wave that motivates the condition does not exist.

HardA transmission electron microscope accelerates electrons through 100 kV. (a) Give the non-relativistic de Broglie wavelength. (b) Give the exact one, using mₑc² = 511.0 keV and hc = 1239.8 keV⋅pm. (c) By what percentage does the shortcut overstate λ? (d) Above what accelerating voltage does that overstatement first exceed 1%?

(a) λ = 1.2264 nm ÷ √(V in volts). With √(1.00 × 10⁵) = 316.23, λ = 1.2264/316.23 = 3.878 × 10⁻³ nm = 3.878 pm.
(b) Use the invariant with K = 100.0 keV: (pc)² = (K + mₑc²)² − (mₑc²)² = 611.0² − 511.0² = (611.0 − 511.0)(611.0 + 511.0) = 100.0 × 1122.0 = 1.1220 × 10⁵ keV², so pc = 334.96 keV.
Then λ = hc/(pc) = 1239.8 keV⋅pm ÷ 334.96 keV = 3.701 pm.
(c) The ratio is 3.878/3.701 = 1.0478, an overstatement of 4.8%. It agrees with the closed form √(1 + K/2mₑc²) = √(1 + 100.0/1022.0) = √1.0978 = 1.0478.
(d) Set √(1 + eV/2mₑc²) = 1.01, so eV/2mₑc² = 1.01² − 1 = 0.0201 and eV = 0.0201 × 1022.0 keV = 20.5 keV.
So the shortcut is already 1% wrong at 20.5 kV and 4.8% wrong at 100 kV. Since λ sets the diffraction-limited resolution, a lattice spacing extracted with the uncorrected wavelength would come out systematically 4.8% too large.

Answerλₙₒₙ-rel = 3.878 pm, λ_exact = 3.701 pm, an overstatement of 4.8%. The 1% threshold is crossed at about 20.5 kV.