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Davisson and Germer's discovery, Bragg's law for electrons and neutrons, electron double slits built up one arrival at a time, and diffraction as a structure tool.
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By the end of this lesson you will be able to predict diffraction angles and fringe spacings for electrons and neutrons, and explain single-particle interference.
From the last lesson you can compute an electron's de Broglie wavelength. From the waves and optics lessons you know the conditions for constructive interference from gratings, double slits and crystal planes. This lesson applies them unchanged to electrons and neutrons, and tells the story of how matter waves were discovered by accident at Bell Labs.
| Term | What it means |
|---|---|
| Diffraction | The spreading and interference of waves passing openings or scattering from regular structures. |
| Bragg's law | $2d\sin\theta = n\lambda$, the condition for waves reflecting from planes spaced $d$ apart to reinforce. |
| Lattice spacing | The distance between planes or rows of atoms in a crystal. |
| Order | The whole number $n$ of wavelengths in the path difference. |
| Fringe spacing | The distance $\lambda L/d$ between bright bands behind two slits. |
| Electron biprism | A charged wire that splits an electron beam in two, acting like a double slit. |
| Single-particle interference | An interference pattern built up from particles arriving one at a time. |
A matter wave obeys exactly the interference conditions a light wave does. For waves scattered by a row of atoms spaced $d$ apart, or passing a grating, bright directions satisfy
$$d\sin\theta = n\lambda.$$
For waves reflected from successive planes of atoms spaced $d$ apart, Bragg's law gives
$$2d\sin\theta = n\lambda,$$
with $\theta$ measured from the planes. Behind two slits a distance $d$ apart, bright fringes on a screen $L$ away are spaced $\lambda L/d$. With $\lambda = h/p$, these conditions predict where electrons will land. In 1927 Davisson and Germer found electrons of $54$ eV scattered strongly at $50°$ from nickel, exactly where $\lambda = 0.167$ nm predicts. Each electron lands at one spot; the pattern emerges from many.
Another way: picture
Picture ocean waves passing through two gaps in a breakwater. Beyond it, the waves from the two gaps overlap, reinforcing in some directions and cancelling in others, making a fan of calm and rough stripes. Replace the ocean with a beam of electrons and the breakwater with two microscopic slits, and the same stripes appear, drawn one electron at a time.
Another way: steps
Clinton Davisson and Lester Germer at Bell Telephone Laboratories in New York City were studying how electrons bounce off metals when, in 1925, an air leak oxidized their nickel target. Heating it to clean it turned the metal into a few large crystals. Afterward, the scattered electrons came off in sharp beams at particular angles, instead of spreading smoothly.
Davisson learned of de Broglie's idea at a conference in 1926 and realized what the beams meant. In 1927 they showed that electrons of $54$ eV produced a peak at $50°$, matching $\lambda = h/p = 0.167$ nm for nickel's surface spacing of $0.215$ nm. Davisson shared the 1937 Nobel Prize with George Paget Thomson, who had shown electron diffraction through thin metal films.
A crystal is a stack of planes of atoms. A wave striking them at a glancing angle $\theta$ reflects a little from each plane. Waves from adjacent planes travel an extra $2d\sin\theta$; when that equals a whole number of wavelengths, all the reflections reinforce and a strong beam emerges. At other angles they cancel.
William Henry Bragg and his son William Lawrence Bragg worked this out for X-rays in 1913 and used it to find the first crystal structures. The same law applies to electrons and neutrons. Because $\sin\theta$ cannot exceed 1, only wavelengths shorter than $2d$ produce Bragg reflections, which is why the wavelength must be comparable to atomic spacings.
The double-slit experiment with electrons, which Richard Feynman called the heart of quantum mechanics, was a thought experiment for decades because the slits must be extraordinarily fine. Claus Jönsson made it real in 1961 with slits a fraction of a micrometer wide, and in 1989 Akira Tonomura's group sent electrons through an electron biprism one at a time.
Each electron made a single dot on the detector. At first the dots looked random. As thousands accumulated, bright and dark fringes emerged, spaced exactly $\lambda L/d$. No electron interfered with another; each one's wave passed both sides and set the odds of where that electron would land.
Neutrons slowed to thermal speeds have wavelengths near $0.18$ nm and diffract off crystals just as X-rays do, but they scatter from nuclei rather than electrons. That makes them sensitive to light atoms like hydrogen, which X-rays barely see, and to magnetism, since the neutron carries a magnetic moment. Clifford Shull at Oak Ridge pioneered the method in the 1940s and shared the 1994 Nobel Prize.
Whole atoms diffract too. Beams of helium and sodium atoms have been sent through gratings and made to interfere, and atom interferometers now measure gravity and rotation with extraordinary precision, because an atom's wave accumulates a phase that depends on the forces it feels.
Checking an answer. $\sin\theta$ must be at most 1. Higher energy must give smaller angles. The angle in Bragg's law is from the planes, not the normal. And fringe spacings for electrons through micrometer slits must come out in micrometers or less.
The interference conditions come purely from geometry: two waves reinforce when their paths differ by a whole number of wavelengths. Nothing in that argument cares whether the wave is light, sound or a matter wave. What de Broglie's relation adds is the wavelength to use.
The small-angle formula $y = n\lambda L/d$ replaces $\sin\theta$ and $\tan\theta$ by $\theta$, valid when $\theta$ is much less than a radian. For electrons through micrometer slits the angles are microradians, so the approximation is essentially exact.
If an electron is detected at one point, which slit did it pass? Quantum mechanics says the question has no answer unless something measures it. The electron's wave passes both slits; the particle is found at one place on the screen, with odds given by the combined wave.
If a detector at the slits reveals which one each electron used, the interference vanishes, and the pattern becomes two overlapping blobs. This is not a flaw in the detectors. Any measurement that could distinguish the paths destroys the interference, a principle that lesson 12 examines.
Diffraction turns the wavelength of a probe into a ruler for measuring spacings. From the angles of the diffracted beams, crystallographers work backward to the positions of atoms. X-ray diffraction found the double helix of DNA, from Rosalind Franklin's images; electron diffraction finds the structure of surfaces and thin films; neutron diffraction finds hydrogen and magnetic order.
The three probes complement each other. X-rays scatter from electrons, electrons from the electric potential of atoms, neutrons from nuclei. A structure confirmed by all three is known with great confidence.
Every imaging system that uses waves blurs details smaller than about a wavelength, because a wave passing a small feature spreads by an angle of about $\lambda/d$. Telescopes, cameras, the eye, and electron microscopes all face this limit, known as the diffraction limit.
Chip makers print circuit features with ultraviolet light of $13.5$ nm, the shortest wavelength that can be focused well with mirrors, and features of a few nanometers need clever tricks beyond that. Electron-beam lithography, with wavelengths of picometers, can write finer features but far more slowly. The trade-off between wavelength and speed shapes the whole semiconductor industry.
A single slit diffracts too. Waves from different parts of an opening of width $a$ cancel in directions where $a\sin\theta = \lambda$, so the central bright band spreads over an angle of about $\lambda/a$ on each side. A narrower slit gives a wider spread. For electrons of $0.05$ nm through a slit $50$ nm wide, the half-angle is about a milliradian.
This spreading is not a quirk of slits. It is the wave form of a deep rule: confining a particle's position across a beam, by passing it through a narrow opening, gives it a spread of sideways momentum. The sideways momentum needed to reach the first dark band is about $p\lambda/a = h/a$, so the spread in momentum times the width of the slit is about $h$. The next lesson makes this precise as Heisenberg's uncertainty relation, which says the product can never be made smaller than about $\hbar/2$, however cleverly the experiment is arranged. Single-slit diffraction of electrons is one of the cleanest demonstrations of it.
Chip plants such as Intel's in Hillsboro, Oregon, grow crystalline layers atom by atom and check them as they grow. In reflection high-energy electron diffraction, a beam of electrons strikes the surface at a grazing angle and the diffracted spots appear on a phosphor screen. Sharp, bright spots mean a smooth, orderly crystal; streaks and blurring mean roughness.
The spots even flicker as each new layer of atoms fills in, letting engineers count layers one at a time. Low-energy electron diffraction, the direct descendant of Davisson and Germer's experiment, maps the arrangement of atoms in the top layer of a surface, which governs how catalysts and semiconductor interfaces behave.
Atom interferometers split clouds of cold atoms with laser pulses, send the two parts along different paths, and recombine them. The interference fringes shift if gravity or rotation acted differently on the two paths. Researchers at Stanford University built a ten-meter atom fountain that measures gravity to a part in a trillion.
Such devices can sense underground density changes for oil and mineral surveys, and may one day navigate submarines and aircraft without GPS. They are Davisson and Germer's discovery turned into a precision instrument: matter waves as rulers of the forces acting on them.
It is tempting to explain electron interference by picturing each electron splitting into two halves that go through the two slits and recombine. Experiments show otherwise: every electron is detected whole, at a single point, and never as two half-charges. The pattern appears only statistically, after many arrivals.
A related error is to think the electrons interfere with each other. Sent one at a time, hours apart, they still build the same pattern. Each electron's own wave, spread over both slits, sets the probabilities.
Electrons are accelerated through $54$ V. Find their wavelength.
$\lambda = \dfrac{1.226}{\sqrt{54}} = 0.167\ \text{nm}$
de Broglie shortcut.
Nickel's surface rows are $0.215$ nm apart. Write the condition.
$D\sin\phi = \lambda$
First order.
Solve for the sine.
$\sin\phi = \dfrac{0.167}{0.215} = 0.776$
Below 1, so the beam exists.
Find the angle.
$\phi = 50.9°$
They measured $50°$.
Try $30$ V instead.
$\lambda = 0.224\ \text{nm} > 0.215\ \text{nm}: \text{no first-order beam}$
The wave is too long for this spacing.
Thermal neutrons of $0.180$ nm strike a crystal with planes $0.300$ nm apart. Write Bragg's law.
$2d\sin\theta = n\lambda$
Crystal planes.
Find the first-order sine.
$\sin\theta_1 = \dfrac{0.180}{0.600} = 0.300$
$n = 1$.
Find the first-order angle.
$\theta_1 = 17.5°$
From the planes.
Find the second-order sine.
$\sin\theta_2 = 0.600$
$n = 2$.
Find the second-order angle.
$\theta_2 = 36.9°$
Farther out.
Find the highest order.
$n \le \dfrac{2d}{\lambda} = 3.3 \Rightarrow n = 3$
$\sin\theta \le 1$.
Electrons are accelerated through $600$ V. Find their wavelength.
$\lambda = \dfrac{1.226}{\sqrt{600}} = 0.0500\ \text{nm}$
de Broglie.
The slits are $1.0$ μm apart. Find the first fringe's angle.
$\theta_1 = \dfrac{0.0500 \times 10^{-9}}{1.0 \times 10^{-6}} = 5.0 \times 10^{-5}\ \text{rad}$
$\lambda/d$.
Find the spacing on a screen $2.0$ m away.
$\Delta y = 2.0 \times 5.0 \times 10^{-5} = 0.10\ \text{mm}$
$L\theta$.
Compare with light through the same slits.
$\lambda = 500\ \text{nm}: \Delta y = 1.0\ \text{m}$
Ten thousand times wider.
Explain the difficulty.
$\text{electron fringes are tiny}$
Magnifying electron lenses are needed.
Send electrons one at a time and describe the screen.
$\text{single dots, building up to fringes}$
Each electron's wave passes both slits.
Cover one slit and describe the pattern.
$\text{a single broad blob, no fringes}$
Interference needs both paths open.
Write Bragg's law for the first order.
$2d\sin\theta = \lambda$
$n = 1$.
Solve for the sine.
$\sin\theta = \dfrac{0.10}{0.50} = 0.20$
$\lambda/2d$.
Find the angle.
Electrons accelerated through $400$ V diffract off a crystal. The voltage is raised to four times as much. What happens to the diffraction angles?
Complete the worked solution: electrons accelerated through $100$ V pass through two slits $0.500$ μm apart onto a screen $2.00$ m away. Find their wavelength in nm, the angle of the first bright fringe beside the center in μrad, and the fringe spacing on the screen in μm.
Find the electron wavelength.
$\lambda = \dfrac{1.226}{\sqrt{V}} =$ l
In nm.
Divide by the slit separation.
$\theta_1 \approx \dfrac{\lambda}{d} =$ a
In μrad.
Multiply by the screen distance.
$\Delta y = L\theta_1 =$ y
In μm.
Explain why real experiments magnify.
$\text{fringes this fine need electron lenses to spread them}$
As in electron biprism experiments.
Match each diffraction situation to its condition or significance.
| $2d\sin\theta = n\lambda$ | $d\sin\theta = n\lambda$ | $\lambda L/d$ | electrons diffract with $\lambda = h/p$ | |
|---|---|---|---|---|
| Bragg reflection | ||||
| grating or surface row | ||||
| double-slit fringe spacing | ||||
| Davisson–Germer |
Electrons accelerated through $100$ V reflect from crystal planes $0.200$ nm apart. Fill in their wavelength in nm, $\sin\theta$ for the first-order Bragg reflection, and the angle $\theta$ in degrees.
| value | |
|---|---|
| wavelength (nm) | |
| sin θ | |
| Bragg angle (degrees) |
Electrons of wavelength $0.1$ nm pass through two slits $0.5$ μm apart onto a screen $1$ m away. Write the distance of the $n$th bright fringe from the center, in μm, as a formula in $n$.
Answer:
In Davisson and Germer's experiment, electrons accelerated through $40$ V strike a nickel crystal whose surface atoms sit in rows $0.215$ nm apart. At what angle from the incident beam does the first diffracted beam appear, in degrees?
Answer: degrees
An engineer at a chip plant in Hillsboro, Oregon, checks the crystal quality of a silicon wafer with an electron beam accelerated through $100$ V. At what Bragg angle, in degrees, do the electrons reflect from silicon's planes $0.3135$ nm apart in first order?
Answer: degrees
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Electrons of wavelength $0.02$ nm pass through two slits $0.2$ μm apart onto a screen $1.5$ m away. Write the distance of the $n$th bright fringe from the center, in μm, as a formula in $n$.
Answer:
You can apply diffraction to matter waves. Explain to someone how an interference pattern can form from electrons sent one at a time.
21. Your turn: electrons of wavelength $0.10$ nm strike crystal planes $0.25$ nm apart. Find the first-order Bragg angle., step 3
$\theta = 11.5°$
From the planes.