---
title: De Broglie Waves and Electron Diffraction
module: The Wave Nature of Matter
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  In 1924 de Broglie proposed that every particle carries a wave of wavelength
  h/p. The hypothesis explains Bohr's quantized orbits as standing waves, and
  Davisson and Germer, then G. P. Thomson, confirmed it by diffracting electrons
  from crystals exactly as X-rays diffract. We derive the electron wavelength,
  work the Bragg analysis of the data, and give the relativistic form.
topics: [The Wave Nature of Matter]
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 5 — The Wavelike Properties of Particles; §5-1 The de Broglie Hypothesis"
  - book: Tipler & Llewellyn
    ref: "§5-2 Measurements of Particle Wavelengths; The Davisson-Germer Experiment"
draft: false
---

By 1924 radiation had a settled double life. Interference and diffraction made
light a wave; the [photoelectric effect](/quantum-mechanics/old-quantum-theory/the-photoelectric-effect-and-the-photon)
and the [Compton effect](/quantum-mechanics/old-quantum-theory/x-rays-and-the-compton-effect)
made it a stream of photons carrying energy $E = hf$ and momentum $p = h/\lambda$.
Louis de Broglie, in his doctoral thesis, proposed that the symmetry runs both
ways: if a wave can behave like a particle, a particle should behave like a wave.
Matter, he argued, carries a wave whose wavelength is fixed by the particle's
momentum through the same relation that holds for light.[^db-hyp]

## The de Broglie relations

De Broglie assigned to a particle of total energy $E$ and momentum $p$ a wave of
frequency and wavelength

$$
f = \frac{E}{h}, \qquad \lambda = \frac{h}{p}.
$$

> **Definition (de Broglie wavelength).** The wavelength $\lambda = h/p$
> associated with any particle of momentum $p$. Its numerical smallness for
> ordinary objects follows from the smallness of $h = 6.63 \times 10^{-34}\,
> \text{J·s}$, which is why matter waves are invisible at everyday scales.

For a photon the two relations are not new. A photon has zero rest energy, so its
energy and momentum satisfy $E = pc$, and Einstein's quantization $E = hf$ gives

$$
E = pc = hf = \frac{hc}{\lambda}
\quad\Longrightarrow\quad
\lambda = \frac{h}{p},
$$

which is the second de Broglie relation. The content of the hypothesis is that the
identical relation holds for an electron, a proton, a baseball — anything with
momentum. De Broglie confirmed by a relativistic argument that $\lambda = h/p$
transfers unchanged from massless to massive particles.

The frequency relation $f = E/h$ resists direct experimental test, because a
free particle's total energy $E$ includes its rest energy and the zero of energy is
a matter of convention. Wavelength, tied to momentum, is the measurable quantity,
and every experiment below measures $\lambda$.

## Matter waves and the Bohr condition

De Broglie's first success was to make the [Bohr model's](/quantum-mechanics/old-quantum-theory/the-old-quantum-theory-bohr-and-sommerfeld)
quantization rule look inevitable. Bohr had postulated, without deeper reason, that
the electron's orbital angular momentum is quantized in units of $\hbar$:

$$
m v r = n \hbar = \frac{n h}{2\pi}, \qquad n = 1, 2, 3, \dots
$$

Rewrite this with $p = mv$ and $\hbar = h/2\pi$:

$$
n \frac{h}{2\pi} = p r
\quad\Longrightarrow\quad
n \frac{h}{p} = 2\pi r
\quad\Longrightarrow\quad
n \lambda = 2\pi r.
$$

The quantization condition says that an integer number of de Broglie wavelengths
fits around the circumference of the orbit. A stable orbit is a standing wave: the
electron wave joins onto itself in phase after one loop. Any non-integer number of
wavelengths interferes destructively with itself on successive loops and cancels.
Bohr's arbitrary rule becomes the resonance condition for a wave confined to a
ring.[^db-hyp]

$$
% caption: A stable Bohr orbit holds an integer number of de Broglie wavelengths
% (here n = 3) around its circumference; the wave closes on itself in phase.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% reference orbit
\draw[black, dashed] (0,0) circle (2);
% standing wave: r = 2 + 0.32 sin(3 theta)
\draw[acc, very thick] plot[domain=0:360, samples=220]
  ({(2 + 0.34*sin(3*\x))*cos(\x)}, {(2 + 0.34*sin(3*\x))*sin(\x)});
% nucleus
\fill[acc] (0,0) circle (2.4pt);
\node[anchor=north, font=\footnotesize] at (0,-0.15) {nucleus};
\node[anchor=west, font=\footnotesize, text=acc] at (2.5,1.6) {electron wave};
\node[anchor=west, font=\footnotesize, text=black] at (2.05,-1.7) {mean orbit};
\end{tikzpicture}
$$

## The scale of matter waves

The wave properties of light went unnoticed until slits comparable to the
wavelength were available; the same holds for matter. Diffraction spreads a wave
of wavelength $\lambda$ through a slit of width $a$ into angles $\theta$ with
$\sin\theta \sim \lambda/a$. When $\lambda \ll a$ the spreading is unobservable and
the wave travels in straight rays. Because $h$ is so small, $\lambda = h/p$ is
minuscule for any macroscopic object, so its diffraction is undetectable.

> **Example (de Broglie wavelength of a Ping-Pong ball).** A $2.0\,\text{g}$ ball
> struck at $5\,\text{m/s}$ has
> $$
> \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}\,\text{J·s}}{(2.0 \times 10^{-3}\,\text{kg})(5\,\text{m/s})} = 6.6 \times 10^{-32}\,\text{m}.
> $$
> This is seventeen orders of magnitude below nuclear dimensions — smaller than
> any aperture that exists — so the ball never diffracts.

An electron is different. Give it a modest kinetic energy and its wavelength lands
near atomic dimensions, where crystals supply a natural diffraction grating.

## Wavelength of an accelerated electron

Accelerate an electron from rest through a potential difference $V_0$. It gains
kinetic energy $E_k = eV_0$, and for $eV_0 \ll mc^2$ the motion is nonrelativistic,
so

$$
E_k = \frac{p^2}{2m} = eV_0
\quad\Longrightarrow\quad
p = \sqrt{2 m e V_0}.
$$

Substituting into $\lambda = h/p$ and writing it through the electron's rest energy
$mc^2 = 0.511 \times 10^6\,\text{eV}$ and $hc = 1.24 \times 10^3\,\text{eV·nm}$,

$$
\lambda = \frac{h}{p} = \frac{hc}{\sqrt{2 (mc^2) e V_0}} = \frac{1.226}{\sqrt{V_0}}\,\text{nm}
\qquad (V_0 \text{ in volts}, \; eV_0 \ll mc^2).
$$

> **Example (wavelength of a $10\,\text{eV}$ electron).** With $V_0 = 10\,\text{V}$,
> $$
> \lambda = \frac{1.226}{\sqrt{10}}\,\text{nm} = 0.39\,\text{nm}.
> $$
> The momentum route gives the same value:
> $p = \sqrt{2mE_k} = 1.71 \times 10^{-24}\,\text{kg·m/s}$ and
> $\lambda = h/p = 0.39\,\text{nm}$. This is the order of an atomic diameter and of
> the spacing between planes of atoms in a crystal.

The coincidence of the electron wavelength with the crystal-plane spacing
$d \approx 0.1$–$0.2\,\text{nm}$ is the whole reason the hypothesis could be tested
in the 1920s. A crystal is a three-dimensional diffraction grating already used to
diffract X-rays of the same wavelength.

## The Davisson-Germer experiment

C. J. Davisson and L. H. Germer confirmed de Broglie's relation directly in 1927.
They fired low-energy electrons from an electron gun at a single crystal of nickel
and measured the scattered intensity as a function of the angle $\phi$ between the
incident beam and the detector.

$$
% caption: The Davisson-Germer apparatus. Electrons from a gun strike a nickel
% single crystal; a movable detector records scattered intensity versus angle.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% electron gun
\draw[black] (-4.4,1.4) rectangle (-3.0,2.2);
\node[align=center, font=\scriptsize] at (-3.7,1.8) {electron\\gun};
% incident beam
\draw[->, acc, very thick] (-3.0,1.8) -- (-0.35,0.35);
\node[acc, font=\scriptsize, anchor=south] at (-1.8,1.15) {incident beam};
% crystal
\fill[acc!12] (-0.7,-0.7) rectangle (0.7,0.2);
\draw[black] (-0.7,-0.7) rectangle (0.7,0.2);
\node[font=\scriptsize, anchor=north] at (0,-0.75) {Ni crystal};
% surface normal
\draw[black, dashed] (0,0.2) -- (0,2.4);
% scattered beam to detector
\draw[->, acc, very thick] (0.35,0.35) -- (2.9,1.7);
% detector
\draw[black] (2.9,1.4) rectangle (4.2,2.2);
\node[align=center, font=\scriptsize] at (3.55,1.8) {ion\\chamber};
% angle arc
\draw[black] (0,1.5) arc (90:32:1.5);
\node[font=\scriptsize] at (0.55,1.85) {angle};
\end{tikzpicture}
$$

For $54\,\text{eV}$ electrons the intensity peaked sharply at a scattering angle
$\phi = 50^\circ$. To explain the peak, treat the regularly spaced atomic planes as
Bragg reflectors. Constructive interference between waves reflected from successive
planes of spacing $d$ requires the Bragg condition, with $\theta$ the grazing angle
to the planes:

$$
n \lambda = 2 d \sin\theta.
$$

$$
% caption: Bragg reflection. Waves scattered from successive atomic planes of
% spacing d reinforce when their path difference is a whole number of wavelengths.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% three planes of atoms
\foreach \y in {0,0.9,1.8}{
  \foreach \x in {0,0.7,1.4,2.1,2.8,3.5,4.2}{
    \fill[black] (\x,\y) circle (1.4pt);
  }
}
\node[font=\scriptsize, anchor=west] at (4.4,0.9) {spacing d};
\draw[black, <->] (4.35,0) -- (4.35,0.9);
% incident and reflected rays on the top two planes
\draw[->, acc, very thick] (0.4,3.0) -- (2.1,1.8);
\draw[->, acc, very thick] (2.1,1.8) -- (3.8,3.0);
\draw[->, acc, thick] (0.9,3.0) -- (2.45,0.9);
\draw[->, acc, thick] (2.45,0.9) -- (4.0,3.0);
\node[acc, font=\scriptsize] at (0.55,2.75) {in};
\node[acc, font=\scriptsize] at (3.9,2.75) {out};
% grazing angle marker
\draw[black] (1.35,1.8) arc (180:213:0.75);
\node[font=\scriptsize] at (1.05,1.55) {angle};
\end{tikzpicture}
$$

The geometry of the crystal ties the plane spacing $d$ to the surface atomic
spacing $D$ by $d = D\sin\alpha$, where $\alpha$ is the angle the planes make with
the surface. Working the trigonometry through, the Bragg condition becomes a
relation in the measured scattering angle $\phi = 2\alpha$:

$$
n\lambda = 2 d \sin\theta = 2 D \sin\alpha \cos\alpha = D \sin 2\alpha = D \sin\phi.
$$

For nickel, X-ray diffraction gives $D = 0.215\,\text{nm}$. The $n = 1$ peak at
$\phi = 50^\circ$ then implies

$$
\lambda = D \sin\phi = (0.215\,\text{nm})\sin 50^\circ = 0.165\,\text{nm}.
$$

The de Broglie prediction for $54\,\text{eV}$ electrons is
$\lambda = 1.226/\sqrt{54}\,\text{nm} = 0.167\,\text{nm}$. The two agree to about one
percent, the small deficit coming from refraction of the electron wave as it enters
the metal, which shortens $\lambda$ slightly inside the crystal.

$$
% caption: Scattered intensity versus detector angle for 54 eV electrons. The
% peak near 50 degrees matches the Bragg condition for the de Broglie wavelength.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,0) -- (6.4,0) node[right, font=\scriptsize] {detector angle};
\draw[->, black] (0,0) -- (0,3.4) node[above, font=\scriptsize] {intensity};
% intensity curve peaking near x=3.3 (=50 deg)
\draw[acc, very thick] plot[domain=0:6, samples=140]
  (\x, {0.35 + 2.7*exp(-2.6*(\x-3.3)^2)});
% mark peak
\draw[black, dashed] (3.3,0) -- (3.3,3.05);
\node[font=\scriptsize, anchor=north] at (3.3,-0.05) {50};
\node[font=\scriptsize, anchor=north] at (0.2,-0.05) {0};
\node[font=\scriptsize, anchor=north] at (6.0,-0.05) {90};
\end{tikzpicture}
$$

Davisson and Germer then held the detector fixed and varied the accelerating
voltage. Because $\lambda \propto V_0^{-1/2}$, plotting intensity against
$1/\lambda \propto V_0^{1/2}$ produced a series of equally spaced peaks at
successive integers $n$, and the measured wavelengths tracked $\lambda = 1.226/
\sqrt{V_0}\,\text{nm}$ across the whole range up to $400\,\text{eV}$.

$$
% caption: Measured electron wavelength versus one over the square root of the
% accelerating voltage. The data fall on the de Broglie line of slope 1.226 nm.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (5.6,0) node[right, font=\scriptsize] {one over root V};
\draw[->, black] (0,0) -- (0,3.4) node[above, font=\scriptsize] {wavelength};
% de Broglie line
\draw[acc, very thick] (0,0.15) -- (5.2,3.15);
% data points
\foreach \x/\y in {0.7/0.55, 1.5/1.05, 2.3/1.55, 3.1/1.98, 3.9/2.42, 4.6/2.78}{
  \fill[black!70] (\x,\y) circle (1.7pt);
}
\node[acc, font=\scriptsize, anchor=west] at (3.6,2.6) {de Broglie};
\end{tikzpicture}
$$

## Thomson diffraction and other particles

In the same year G. P. Thomson (son of J. J. Thomson, who had shown the electron
to be a particle) demonstrated the wave nature of $10$–$40\,\text{keV}$ electrons in
transmission. Passing the beam through a thin polycrystalline metal foil produced
concentric diffraction rings, exactly like the Laue rings from X-rays through the
same foil. The many randomly oriented crystallites each satisfy the Bragg condition
at some grazing angle $\theta$, and each scatters into a cone of half-angle $2\theta$
about the beam, so the pattern is a set of rings.

$$
% caption: Electron transmission through a polycrystalline foil gives concentric
% rings, each ring a Bragg cone from crystallites at one orientation.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% incident beam
\draw[->, acc, very thick] (-4.0,0) -- (-2.2,0);
\node[acc, font=\scriptsize, anchor=south] at (-3.1,0.05) {electron beam};
% foil
\draw[black, very thick] (-2.2,-0.9) -- (-2.2,0.9);
\node[font=\scriptsize, anchor=north] at (-2.2,-0.95) {foil};
% screen with rings
\draw[black] (2.0,-1.7) rectangle (2.0,1.7);
\fill[acc!6] (2.0,0) circle (1.7);
\draw[black] (2.0,0) circle (1.7);
\foreach \r in {0.5,0.95,1.35}{ \draw[acc] (2.0,0) circle (\r); }
\fill[acc] (2.0,0) circle (2pt);
\node[font=\scriptsize, anchor=north] at (2.0,-1.85) {screen};
% cones
\draw[black] (-2.2,0) -- (2.0,0.95);
\draw[black] (-2.2,0) -- (2.0,-0.95);
\draw[black] (-2.2,0) -- (2.0,0);
\end{tikzpicture}
$$

The hypothesis is not confined to charged particles. Because neutral atoms and
molecules cannot be accelerated electrostatically, their wavelengths were harder to
reach, but Stern and Estermann diffracted thermal beams of helium atoms and
hydrogen molecules from a lithium fluoride crystal in 1930, finding $\lambda \approx
0.1\,\text{nm}$ as predicted from their thermal energy of about $0.03\,\text{eV}$.
Diffraction of protons and neutrons followed. In every case the measured wavelength
matched $\lambda = h/p$.

| Particle | Typical energy | de Broglie $\lambda$ | Diffractor |
| --- | --- | --- | --- |
| Electron (Davisson-Germer) | $54\,\text{eV}$ | $0.167\,\text{nm}$ | Ni single crystal |
| Electron (Thomson) | $10$–$40\,\text{keV}$ | $\sim 0.05\,\text{nm}$ | metal foil |
| He atom (Stern-Estermann) | $0.03\,\text{eV}$ | $\sim 0.10\,\text{nm}$ | LiF surface |
| Thermal neutron | $0.06\,\text{eV}$ | $0.12\,\text{nm}$ | polycrystalline Cu |

## Relativistic wavelengths

At high energy the nonrelativistic $\lambda = 1.226/\sqrt{V_0}\,\text{nm}$ fails, and
one must start from the exact energy-momentum relation of
[special relativity](/relativity/foundations/relativistic-momentum-energy),

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

Writing $E = E_0 + E_k$ with rest energy $E_0 = mc^2$ and solving for $p$,

$$
p = \frac{1}{c}\sqrt{2 E_0 E_k + E_k^2}
\quad\Longrightarrow\quad
\lambda = \frac{h}{p} = \frac{hc}{\sqrt{2 E_0 E_k + E_k^2}}.
$$

Dividing through by $E_0$ expresses the result in units of the particle's Compton
wavelength $\lambda_c = h/mc$, giving a single curve valid at any energy:

$$
\frac{\lambda}{\lambda_c} = \frac{1}{\sqrt{2 (E_k/E_0) + (E_k/E_0)^2}}.
$$

For $E_k \ll E_0$ this reduces to the nonrelativistic $\lambda = h/\sqrt{2mE_k}$; for
$E_k \gg E_0$ it approaches $\lambda \approx hc/E_k$, the photon-like limit. A
$150\,\text{GeV}$ cosmic-ray proton, with $E_0 = 0.938\,\text{GeV}$ and $\lambda_c =
1.32\,\text{fm}$, has $E_k/E_0 = 160$, giving $\lambda/\lambda_c \approx 6\times10^{-3}$
and $\lambda \approx 8 \times 10^{-3}\,\text{fm}$ — short enough to probe the interior
of a nucleus.

The de Broglie hypothesis leaves one question open. A single wave of definite
wavelength extends over all space, yet a particle is found at one place. The
[next lesson](/quantum-mechanics/matter-waves/wave-packets-and-the-probability-interpretation) resolves this
by superposing waves into a localized packet and asking what, physically, the wave
amplitude represents.

[^db-hyp]: Tipler & Llewellyn, §5-1 — the de Broglie relations $f = E/h$, $\lambda = h/p$, their derivation from the photon relations, and the reading of Bohr's angular-momentum quantization as the standing-wave condition $n\lambda = 2\pi r$.