In short

Every moving particle has an associated wavelength — its de Broglie wavelength — given by

\boxed{\;\lambda = \frac{h}{p} = \frac{h}{mv}\;}

where h = 6.626 \times 10^{-34} J·s is Planck's constant and p = mv is the particle's momentum. For an electron accelerated from rest through a potential difference V,

\lambda = \frac{h}{\sqrt{2m_e eV}}.

At V = 150 V this gives \lambda \approx 1 Å — the same scale as atomic spacing, which is why electrons diffract off crystals (Davisson-Germer, 1927). A cricket ball at 140 km/h has \lambda \approx 10^{-34} m, about 10^{24} times smaller than a proton — undetectable by anything. That is the answer to "why don't we see quantum effects in everyday life": the wavelength is there, but it is absurdly small.

The de Broglie relation is also the seed of the Heisenberg uncertainty principle \Delta x \, \Delta p \geq \hbar/2: if a particle has a definite momentum, its wave has a definite wavelength, so the wave is spread over all space and the particle has no definite position.

In 1924, a French graduate student named Louis de Broglie defended a thesis that made his examiners distinctly uncomfortable. The examiners' panel — good physicists all — could not agree on whether the thesis was brilliant or absurd, so they mailed it to Einstein. Einstein replied with one line: "He has lifted a corner of the great veil." De Broglie passed. Five years later he was given a Nobel Prize.

What had he done? He had taken Einstein's photon — a light wave that behaves like a particle with momentum p = h/\lambda — and asked the symmetric question. If a wave can be a particle, can a particle be a wave? If a beam of light of wavelength \lambda carries momentum h/\lambda per photon, then a beam of particles each with momentum p should carry, somehow, a wavelength h/p.

De Broglie could not prove this experimentally in 1924. He had only a formal analogy and the conviction that nature is symmetric enough to allow it. But within three years, Clinton Davisson and Lester Germer in New York — trying to do something else entirely — accidentally diffracted a beam of electrons off a nickel crystal and found the wavelength de Broglie had predicted, to better than one percent. Matter waves were real. The electron, which everyone had spent three decades teaching themselves to think of as a tiny charged ball, was also a wave.

This article is the case for why that idea is not insane, how to compute the wavelength, why you do not see your own wavelength while walking to class, and what all of this has to do with Heisenberg's uncertainty principle.

The symmetry argument — why de Broglie even had the idea

Go back to the photoelectric effect and the Compton effect. Both confirmed that a light wave of frequency f and wavelength \lambda carries, per photon,

E = hf \qquad\text{and}\qquad p = \frac{h}{\lambda}.

Why: the photoelectric effect gave E = hf (photon energy goes into ejecting electrons); the momentum relation follows from relativistic energy-momentum (E = pc for a massless particle, combined with E = hf and f\lambda = c, yields p = h/\lambda). Both relations are tied to the fact that a photon is simultaneously a wave of wavelength \lambda and a particle of momentum p.

These are two sides of a single object. One side is wave-like (the \lambda), the other is particle-like (the p), and Planck's constant h is the exchange rate between them.

Now de Broglie's move. A photon is not special. A photon has zero rest mass, but any object — an electron, a neutron, a tennis ball — also has both energy and momentum. If the wave-particle relation p = h/\lambda holds for a photon, and nature is symmetric, then perhaps the relation runs the other way too: a particle of momentum p has an associated wavelength

\boxed{\;\lambda = \frac{h}{p}\;}

De Broglie had no experiment to justify this. He had only the elegance of the inverse relation and the observation that Planck's constant already links energy to frequency for light; maybe it links momentum to wavelength for matter too. It was a bet on symmetry.

The de Broglie wavelength of an electron

The cleanest calculation — and the one that matches the experiments of Davisson and Germer, and of G. P. Thomson independently — is the wavelength of an electron accelerated through a known potential difference.

Setup: an electron, initially at rest, is accelerated through a potential difference V. Assume the speed remains non-relativistic, i.e. eV \ll m_e c^2 \approx 511 keV. Derive \lambda in terms of V.

Step 1. Energy conservation gives the kinetic energy after acceleration.

An electron of charge e falling through potential difference V gains kinetic energy eV:

K = \tfrac{1}{2} m_e v^2 = eV

Why: the work done by the electric field on the electron as it crosses the gap is eV. The electron starts at rest, so all of that work becomes kinetic energy.

Step 2. Solve for momentum in terms of K.

K = \frac{p^2}{2 m_e} \quad\Longrightarrow\quad p = \sqrt{2 m_e K} = \sqrt{2 m_e e V}

Why: the identity K = p^2/(2m) comes from K = \frac{1}{2}mv^2 = \frac{(mv)^2}{2m} = p^2/(2m). Inverting gives p as a function of K. Substituting K = eV gives p in terms of the experimental knob V.

Step 3. Apply the de Broglie relation.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e e V}}

Step 4. Plug in constants to get a handy numerical formula.

h = 6.626 \times 10^{-34} J·s, m_e = 9.109 \times 10^{-31} kg, e = 1.602 \times 10^{-19} C. Then

\sqrt{2 m_e e} = \sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19}} = \sqrt{2.919 \times 10^{-49}} \approx 5.403 \times 10^{-25} \text{ kg}^{1/2} \text{C}^{1/2}
\lambda = \frac{6.626 \times 10^{-34}}{5.403 \times 10^{-25} \sqrt{V}} \text{ m} = \frac{1.226 \times 10^{-9}}{\sqrt{V}} \text{ m}

Express in ångström (1 \text{ Å} = 10^{-10} m) with V in volts:

\boxed{\;\lambda = \frac{12.26}{\sqrt{V}}\,\text{Å}\;}

Why: the factors in \sqrt{2 m_e e} combine into a single dimensional prefactor. The practical form with \lambda in ångström and V in volts is the one Indian students use on JEE — memorising it saves the constants substitution on every problem.

This is a remarkable formula. Accelerate an electron through V = 150 volts — a battery, really — and its wavelength is exactly 1 Å, the size of a hydrogen atom. That is why electrons diffract off crystals: their wavelengths are already at the scale of atomic spacing, so a crystal is a natural diffraction grating for them. Accelerate through 10 kV (a television tube) and the wavelength drops to 0.12 Å, below atomic size — which is why electron microscopes can image individual atoms, while an ordinary light microscope (wavelength 500 nm) cannot possibly resolve anything smaller than about 10^3 atomic diameters.

Why you don't see your own de Broglie wavelength

Here is the single fact that keeps de Broglie's hypothesis from being obviously wrong: if every particle has a wavelength, why doesn't a cricket ball diffract around a doorway?

A cricket ball of mass m = 0.16 kg moving at v = 140 km/h = 38.9 m/s has momentum

p = mv = 0.16 \times 38.9 = 6.22 \text{ kg·m/s}

and de Broglie wavelength

\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{6.22} \approx 1.07 \times 10^{-34} \text{ m}

Compare that to the size of a proton (\sim 10^{-15} m) and then to the Planck length (\sim 10^{-35} m). The cricket ball's wavelength is somewhere between these — about 10^{-24} times smaller than a proton. For diffraction to be visible, the wavelength must be comparable to the size of the obstacle or aperture. There is no physical aperture in the universe that is 10^{-34} m wide, not even in principle. Hence: no diffraction, no interference, no observable wave behaviour. The cricket ball is a wave, but its wavelength is so small that it behaves exactly like a classical particle to within any conceivable experimental precision.

Now run the same calculation for an electron moving at a much more modest speed — say, the thermal speed at room temperature. The electron has mass m_e = 9.11 \times 10^{-31} kg, so its thermal speed is around 10^5 m/s, and its wavelength comes out to a few nanometres. That is vastly larger than an atom. An electron does not behave classically in the atom — its wavelength is bigger than the nucleus it is supposedly orbiting. This is the first hint that the classical orbit picture of Rutherford's atom is doomed, and that Bohr's quantisation rule is actually a standing-wave condition in disguise.

The interactive below lets you slide the momentum from a single proton to a cricket ball and watch the wavelength shrink across 24 orders of magnitude. Pay attention to the vertical scale — it is logarithmic.

Interactive: de Broglie wavelength vs momentum across 28 decades Log-log plot of wavelength versus momentum. A line of slope minus one shows lambda equals h over p. Markers indicate typical momenta of an electron in an atom, a neutron at thermal speed, a dust grain, and a cricket ball. log₁₀(momentum p / kg·m/s) log₁₀(λ / m) −35 −25 −15 −5 −25 −20 −15 −10 −5 0 electron in atom p ~ 10⁻²⁴, λ ~ 10⁻¹⁰ m thermal neutron p ~ 10⁻²³, λ ~ 10⁻¹¹ m dust grain p ~ 10⁻¹⁵, λ ~ 10⁻¹⁹ m cricket ball p ~ 6, λ ~ 10⁻³⁴ m drag the red point along the p-axis
Drag the red point along the horizontal axis to change $\log_{10} p$. The red line is the de Broglie relation — on a log-log plot, $\lambda = h/p$ is a straight line of slope $-1$ and intercept $\log_{10} h = -33.82$. Note the grey reference points: an electron in an atom (wavelength comparable to atomic size, so quantum effects dominate), a thermal neutron (used in diffraction at BARC and Oak Ridge), a dust grain (wavelength smaller than a proton, so classical), and a cricket ball (wavelength sub-Planckian — forever undetectable).

The single most important takeaway from that plot: the transition from "quantum" to "classical" is not a sharp line. It is a continuous scaling. An electron is quantum because its wavelength is larger than the structures (atoms) it interacts with. A cricket ball is classical because its wavelength is smaller than any structure anywhere in the universe. Same equation, different regimes.

Worked examples

Example 1: Electron in an electron microscope

An electron microscope at the AIIMS Delhi imaging facility uses electrons accelerated through V = 100 kV. Find (a) the de Broglie wavelength of the electrons, (b) the ratio of this wavelength to the wavelength of visible green light (550 nm), which is the best you can do with an optical microscope.

Comparison of optical microscope wavelength to electron microscope wavelengthTwo horizontal bars. The upper bar represents visible green light at 550 nm. The lower bar, visibly microscopic in comparison, represents a 100 kV electron at 0.00388 nm. Scale shown.Optical microscope (green light)λ = 550 nm = 5500 ÅElectron microscope (100 kV electrons)λ = 0.00388 nm = 0.0388 ÅAbout 140 000 times shorter — the reason electron microscopes can image individual atoms.
The electron's de Broglie wavelength at 100 kV is about 140 000 times smaller than visible light's — which is why an electron microscope reveals structures invisible to any optical instrument.

Step 1. Use the JEE formula.

\lambda = \frac{12.26}{\sqrt{V}}\,\text{Å}

At V = 10^5 V:

\lambda = \frac{12.26}{\sqrt{10^5}} = \frac{12.26}{316.2} \approx 0.0388 \text{ Å}

Why: we derived this formula directly above, assuming non-relativistic motion. The electron's kinetic energy here is 100 keV, which is about 20% of its rest mass energy (511 keV) — not fully non-relativistic. For a rigorous answer at 100 kV, a small relativistic correction applies. But the non-relativistic value is within 5%, which is fine for a first pass. (The relativistic answer is \lambda \approx 0.037 Å.)

Step 2. Ratio to green light.

\frac{\lambda_\text{visible}}{\lambda_\text{electron}} = \frac{550 \text{ nm}}{0.00388 \text{ nm}} \approx 1.42 \times 10^5

Why: both wavelengths in the same units, then divide. The factor of 140 000 sets the theoretical resolution limit: an electron microscope can in principle resolve structures 140 000 times smaller than an optical microscope — which is why you can see individual atoms in a high-resolution transmission electron microscope but never in a light microscope, no matter how expensive the optics.

Result: \lambda \approx 0.039 Å; the electron wavelength is about 1.4 \times 10^5 times shorter than green light.

What this shows: de Broglie's equation is not abstract theory — it is the engineering principle behind every electron microscope, every neutron diffraction instrument at BARC, every low-energy electron diffraction (LEED) setup used to study crystal surfaces at IITs. Higher voltage gives shorter wavelength gives finer resolution. You accelerate the electrons specifically to shrink their wavelength.

Example 2: Thermal neutron wavelength — why BARC diffracts neutrons

A thermal neutron — one that has come to thermal equilibrium with a moderator at room temperature (T = 300 K) — has kinetic energy of order k_B T. India's Dhruva reactor at BARC Trombay produces intense beams of thermal neutrons precisely for diffraction studies. Find the de Broglie wavelength of a thermal neutron (m_n = 1.675 \times 10^{-27} kg, k_B = 1.381 \times 10^{-23} J/K).

Neutron diffraction off a crystalA crystal drawn as a grid of dots with spacing d. A thermal neutron beam enters at an angle theta, reflects off crystal planes, and exits at angle theta, producing Bragg diffraction. Labels show lambda comparable to d.d ≈ λthermal neutronλ ≈ 1.8 Ådiffracted beamθ
A thermal neutron with wavelength $\lambda \approx 1.8$ Å scatters off crystal planes whose spacing is also of order 1–3 Å — the two wavelengths match, so constructive interference produces sharp Bragg peaks. This is the operating principle behind neutron diffraction at BARC Trombay.

Step 1. Kinetic energy at thermal equilibrium.

Equipartition gives K \approx \frac{3}{2} k_B T per translational degree of freedom, but the convention in neutron work is to quote K = k_B T — either is fine for an order-of-magnitude calculation. Take K = \frac{3}{2} k_B T for definiteness.

K = \tfrac{3}{2} \times 1.381 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J}

Why: thermal equilibrium means every translational degree of freedom carries \frac{1}{2} k_B T of energy on average; three such degrees give \frac{3}{2} k_B T. The actual neutron distribution at a reactor is thermal, so different neutrons have slightly different wavelengths; this calculation gives the mean.

Step 2. Momentum.

p = \sqrt{2 m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}
p = \sqrt{2.08 \times 10^{-47}} \approx 4.56 \times 10^{-24} \text{ kg·m/s}

Why: same p = \sqrt{2mK} relation we used for electrons. The numbers are different but the step is identical.

Step 3. de Broglie wavelength.

\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-24}} \approx 1.45 \times 10^{-10} \text{ m} = 1.45 \text{ Å}

Why: direct application of \lambda = h/p. The result — 1.45 Å — is comparable to atomic spacing in most crystals (typical lattice constant is 2–5 Å), which is precisely the condition for Bragg diffraction.

Result: \lambda \approx 1.5 Å — matched to crystal spacings.

What this shows: A moderator slows fast fission neutrons down to thermal energies specifically so their wavelength matches crystal spacing. Neutrons are complementary to X-rays: they scatter off nuclei (X-rays scatter off electrons), so they reveal atomic positions X-rays cannot — especially hydrogen and light elements. India's research programme at BARC Trombay has used thermal neutron diffraction since the 1960s to study the structure of materials for reactors, superconductors, and drug crystals. Without de Broglie, there is no neutron diffraction and no such measurement.

Example 3: The cricket ball again — why you are not a wave

A cricket ball of mass 160 g is bowled at 140 km/h. Find its de Broglie wavelength and compare it to (a) the radius of a proton, (b) the Planck length.

Step 1. Momentum.

p = mv = 0.16 \times 38.9 = 6.22 \text{ kg·m/s}

Why: classical momentum. No relativistic correction — 140 km/h is profoundly non-relativistic.

Step 2. de Broglie wavelength.

\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{6.22} = 1.07 \times 10^{-34} \text{ m}

Why: same equation as for any particle. de Broglie does not care whether the "particle" is an electron or a cricket ball — the wavelength is universal.

Step 3. Compare to physical scales.

  • Proton radius: r_p \approx 0.84 \times 10^{-15} m.

    \frac{\lambda}{r_p} = \frac{1.07 \times 10^{-34}}{0.84 \times 10^{-15}} \approx 1.3 \times 10^{-19}

    The cricket ball's wavelength is 10^{19} times smaller than a proton.

  • Planck length: \ell_P \approx 1.6 \times 10^{-35} m.

    \frac{\lambda}{\ell_P} = \frac{1.07 \times 10^{-34}}{1.6 \times 10^{-35}} \approx 6.7

    The cricket ball's wavelength is only a few times larger than the Planck length — the shortest length scale believed to have any physical meaning.

Why: if diffraction requires the wavelength to be comparable to the size of the obstacle, and the cricket ball's wavelength is already smaller than a proton, there is no obstacle in the universe that could diffract it. Not a slit, not a doorway, not an atom. The wave is there mathematically — but it is unobservable in principle, not just in practice.

Result: \lambda \approx 10^{-34} m — unobservably small.

What this shows: The reason classical mechanics works for everyday objects is not that de Broglie was wrong for them — he was right. It is that their wavelengths are so tiny relative to anything in their environment that the wave behaviour averages out and the particle behaviour dominates completely. Quantum mechanics does not stop applying to cricket balls; its effects are just undetectable. This is the correspondence principle — quantum physics reduces to classical physics in the limit where wavelengths shrink below any relevant size scale.

The Heisenberg uncertainty principle — hiding inside \lambda = h/p

Here is one of the most important consequences of matter waves, and it is nearly forced by the very equation you just derived.

Suppose an electron has an exactly definite momentum p. Then its de Broglie wavelength is exactly \lambda = h/p — a single, sharp, sinusoidal wave that repeats forever in every direction. But a wave that repeats forever has no preferred location. The electron is equally "present" everywhere. It has no definite position at all.

Now suppose instead the electron is confined to a small region — say, a wave packet of width \Delta x. A localised wave packet cannot be a single pure wavelength; it has to be a superposition of many wavelengths, and therefore many different momenta. The sharper the localisation, the broader the spread in wavelengths, and hence in momenta. This is just the mathematics of Fourier analysis, which applies to any wave: you cannot have both a definite wavelength and a definite location.

The precise quantitative statement is

\boxed{\;\Delta x \,\Delta p \;\geq\; \frac{\hbar}{2}\;} \qquad \text{where } \hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34} \text{ J·s}

This is the Heisenberg uncertainty principle. It says: the product of the uncertainty in position and the uncertainty in momentum is at least \hbar/2. You cannot make both small at once.

The proof from de Broglie's relation, stripped to its bare bones:

Why: the Fourier inequality \Delta x \Delta k \geq 1/2 is a mathematical theorem about any wave packet — it has nothing to do with quantum mechanics on its own. What quantum mechanics adds is the identification p = \hbar k, which tells you that the wavenumber is the momentum. Together, the pure mathematics of waves becomes a physical constraint on position and momentum.

This is not a limitation of measurement. It is not that our instruments are too crude. It is a property of what it means to be a wave. A particle that is also a wave cannot have both a definite position and a definite momentum because a wave cannot have both a definite position and a definite wavelength.

Quick numerical check

An electron confined to a region the size of a hydrogen atom (\Delta x \sim 5 \times 10^{-11} m) must have momentum uncertainty at least

\Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1.055 \times 10^{-34}}{2 \times 5 \times 10^{-11}} \approx 1.05 \times 10^{-24} \text{ kg·m/s}

The corresponding speed uncertainty is \Delta v = \Delta p/m_e \approx 1.2 \times 10^6 m/s — already a sizeable fraction of the electron's typical speed in the atom. This is the calculation that tells you a classical "orbit" inside an atom is nonsense: the uncertainty in the electron's momentum is already as large as the momentum itself. The electron in a hydrogen atom is not a particle in orbit; it is a smeared-out wave, and the best you can talk about is a probability distribution.

Connecting quantum mechanics to the classical world

So where is the boundary? Why does the uncertainty principle not stop you from pointing at a cricket ball and saying both "it is here" and "it is moving at 38.9 m/s"?

The answer is in the numbers. For a cricket ball with m = 0.16 kg, if you measure its position to within \Delta x = 1 μm = 10^{-6} m (already insanely precise for a moving ball), the minimum momentum uncertainty is

\Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1.055 \times 10^{-34}}{2 \times 10^{-6}} \approx 5 \times 10^{-29} \text{ kg·m/s}

and the corresponding speed uncertainty is

\Delta v = \frac{\Delta p}{m} \approx \frac{5 \times 10^{-29}}{0.16} \approx 3 \times 10^{-28} \text{ m/s}

That is nine orders of magnitude slower than the distance light travels in a billion years. The uncertainty principle is true for the cricket ball; it is simply so far below any measurement resolution you could dream of that it is operationally absent.

The recipe for crossing from quantum to classical is this: if the object's mass is large enough that \hbar/m is small compared to every velocity scale in the problem, and the object's wavelength h/p is small compared to every length scale in the problem, classical mechanics is an excellent approximation. Quantum effects are always present; they are just hidden by the smallness of h.

This is the content of the correspondence principle, first stated by Niels Bohr: every quantum-mechanical prediction must reduce to the corresponding classical prediction in the limit where h can be treated as small. de Broglie's relation \lambda = h/p is the cleanest embodiment: send h \to 0 and the wavelength vanishes, the wave behaviour disappears, and you are left with a classical particle.

Common confusions

If your goal was to understand what a matter wave is, compute de Broglie wavelengths, and see where the uncertainty principle comes from, you have what you need. What follows is for readers who want the relativistic version of the equation, the connection to Bohr's quantisation, and the experimental confirmation by Davisson and Germer.

Relativistic de Broglie wavelength

For a particle with kinetic energy K comparable to or larger than its rest energy mc^2, you cannot use K = p^2/(2m). The relativistic energy-momentum relation is

E^2 = (pc)^2 + (mc^2)^2

with total energy E = K + mc^2. Solving for p:

p = \frac{1}{c}\sqrt{E^2 - (mc^2)^2} = \frac{1}{c}\sqrt{(K + mc^2)^2 - (mc^2)^2} = \frac{1}{c}\sqrt{K^2 + 2 K mc^2}

Why: the relativistic energy-momentum relation is the Pythagorean-like statement E^2 - (pc)^2 = (mc^2)^2. Expanding (K + mc^2)^2 - (mc^2)^2 = K^2 + 2Kmc^2 gives the clean formula.

For an electron accelerated through V, K = eV, and the relativistic wavelength is

\lambda = \frac{h}{p} = \frac{hc}{\sqrt{eV(eV + 2m_e c^2)}}

At V = 10 kV (electron kinetic energy 10 keV ≪ 511 keV), this reduces to the non-relativistic formula to within 1%. At V = 100 kV, the correction is about 5%. At V = 1 MV, the correction is about 30% — you have to use the relativistic form.

de Broglie's standing-wave picture of the Bohr atom

This is de Broglie's first triumph and the bridge to Bohr's model. Recall Bohr's quantisation condition: the angular momentum of the electron is quantised, L = n\hbar. Why n\hbar? Bohr had no answer in 1913 — it was a postulate. de Broglie gave the reason.

If the electron is a wave, then its orbit around the proton must accommodate a standing wave — an integer number of wavelengths fitting exactly around the circumference, so the wave meets itself in phase:

2\pi r = n \lambda

Substitute \lambda = h/p = h/(m_e v):

2\pi r = \frac{n h}{m_e v}
m_e v r = \frac{n h}{2\pi} = n \hbar

But m_e v r = L, the electron's angular momentum about the proton. So

L = n\hbar

Why: Bohr's quantisation becomes a standing-wave condition the moment you accept that the electron is a wave. The permitted orbits are exactly those where the wavelength fits an integer number of times around the circumference. Orbits that don't fit would interfere destructively with themselves and wash out. This is why the hydrogen atom has discrete energy levels — not because of an arbitrary postulate, but because of wave self-interference.

This is the deepest payoff of the de Broglie hypothesis. Bohr's 1913 atom, which had to invoke a mysterious quantisation rule with no justification, becomes an obvious consequence of matter waves. Every quantised system in nature — atoms, nuclei, molecules — is a standing-wave problem.

Davisson-Germer: the experimental confirmation

In 1927, Clinton Davisson and Lester Germer at Bell Labs were studying how electrons scatter off a nickel surface when an accident cracked their vacuum chamber and let oxygen in. Baking the nickel to drive off the oxide converted it from polycrystalline to a few large crystalline grains. When they resumed the experiment, the scattering pattern had sharp peaks where there had previously been a smooth background. The peaks were at angles exactly matching the Bragg condition for X-ray diffraction, but for a wavelength computed from de Broglie's formula. Electrons were diffracting — behaving as waves — with the wavelength de Broglie had predicted three years earlier. George Paget Thomson independently observed electron diffraction by thin metal foils at about the same time. (Thomson's father, J.J. Thomson, had won the Nobel for proving the electron is a particle; the son won it for proving the electron is a wave. Physics is like that.) The article on the Davisson-Germer experiment develops the crystal-diffraction analysis in full.

Wave-packet dispersion and group velocity

A plane matter wave \psi = e^{i(kx - \omega t)} has one wavelength and one frequency; its group velocity v_g = d\omega/dk equals the particle's classical velocity only for the dispersion relation \omega = \hbar k^2/(2m). This turns out to be the Schrödinger dispersion relation, and it emerges from requiring that the de Broglie-plus-Einstein identifications p = \hbar k, E = \hbar \omega reproduce the non-relativistic E = p^2/(2m). The Schrödinger equation is, in a strong sense, the wave equation that makes de Broglie's hypothesis self-consistent. But that is a story for the article on the Schrödinger equation proper.

Where this leads next