Electron Diffraction: The Davisson–Germer Experiment

The peak Davisson and Germer stumbled onto grew into an entire toolkit. Because a slow electron's wavelength is atom-sized and electrons are easy to focus with magnetic lenses, electron diffraction became a routine way to read out how atoms are arranged. Low-energy electron diffraction (LEED) and reflection high-energy electron diffraction (RHEED) are standard instruments in every surface-science and semiconductor-growth lab, mapping the atomic order of a surface layer by layer — the direct descendants of a beam bouncing off nickel. The same wave nature is what gives the electron microscope its power: because electrons can carry wavelengths thousands of times shorter than visible light, an electron microscope resolves detail far finer than any optical one, letting us image individual atoms and the machinery inside cells. Electron diffraction crystallography now solves the structures of proteins and tiny crystals too small for X-rays. And the deepest legacy is conceptual: the double-slit experiment has since been run with single electrons, then neutrons, then whole molecules, each building up an interference pattern one particle at a time. Davisson and Germer's crystal was the first of those experiments, and the wave-particle duality it nailed down is the bedrock of all of quantum mechanics.

Start with the one idea this whole experiment is built to test. A moving electron of momentum $p$ is supposed to have a de Broglie wavelength $\lambda = h/p$, where $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant. For anything you can see or hold this wavelength is fantastically tiny — a thrown cricket ball has $\lambda$ around $10^{-34}\ \mathrm{m}$, unmeasurably small — which is exactly why matter looks like particles. But an electron is light, and if you slow it to a modest speed its wavelength swells to a fraction of a nanometre: roughly the size of an atom, roughly the spacing between atoms in a crystal. That is the crucial coincidence. To diffract a wave you need a grating whose slits are spaced about one wavelength apart, and a crystal hands you exactly that grating for free — its atoms sit in neat rows a couple of ångström apart. So the plan writes itself. Take electrons, give them a known, tunable energy by accelerating them through a voltage $V$, and fire them at a crystal. If de Broglie is right, the electrons carry a wavelength you can dial with the voltage, and the crystal's regular rows of atoms will diffract that wave into a pattern of preferred directions — sharp peaks of scattered intensity at particular angles, with near-darkness in between. If de Broglie is wrong and electrons are only tiny bullets, they should ricochet off the surface into a smooth, featureless spray, brightest straight back and fading gently to the sides, with no special angle singled out. The two pictures make flatly different predictions, and a single detector swung around the crystal can tell them apart.

When Davisson and Germer actually ran this — after the lucky accident that turned their nickel into large crystals — the result was unmistakably the wave picture. Sweeping the detector around, they found the scattered electron count was low at most angles but rose to a distinct hump, and at an accelerating voltage of 54 V that hump sharpened into a clear peak centred on a scattering angle of 50°. Change the voltage and the peak moved and changed height, exactly as a diffraction maximum should when you change the wavelength. Electrons — indivisible lumps of charge with a definite mass — were interfering with themselves off the rows of a crystal. The only way a stream of particles produces a diffraction peak is if each particle travels as a wave, and that wave is what de Broglie had named.

To turn this into a quantitative test we need the electron's wavelength expressed through the one knob we control, the accelerating voltage $V$. An electron of charge $e = 1.60\times10^{-19}\ \mathrm{C}$ pushed through a potential difference $V$ gains kinetic energy $K = eV$; at these low voltages it stays comfortably non-relativistic, so its momentum follows from $K = p^2/2m_e$, giving $p = \sqrt{2m_eK} = \sqrt{2m_e e V}$. Feed that into de Broglie's $\lambda = h/p$:

where: $h$ is Planck's constant, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$ the electron mass, $e$ the electron charge, and $V$ the accelerating voltage in volts. The physical reading is that a bigger voltage means a faster, higher-momentum electron and therefore a shorter wavelength — turn up the voltage and the matter wave gets tighter. Everything on the right except $V$ is a constant of nature, so we can pack them into a single number. Working the constants out (done in the derivation below) collapses equation (1) into the shortcut every exam expects:

where: $\lambda$ comes out in ångström ($1\ \text{Å} = 10^{-10}\ \mathrm{m} = 0.1\ \mathrm{nm}$) when $V$ is in volts. This is the workhorse of the whole chapter: it says a $100\ \mathrm{V}$ electron has $\lambda = 12.27/\sqrt{100} = 1.23\ \text{Å}$, comfortably atom-sized. For the Davisson–Germer voltage,

That $1.67\ \text{Å}$ is a prediction. It uses nothing but the accelerating voltage and three fundamental constants; it never once looks at the crystal or the diffraction angle. If matter waves are real, the wavelength the crystal reports back must match it.

Now the independent measurement. The regularly spaced surface atoms of the nickel crystal act as a reflection diffraction grating for the incoming matter wave, in exactly the way a ruled grating acts on light. Label the spacing between adjacent rows of surface atoms $a$; for the nickel face used, crystallography gives $a = 2.15\ \text{Å}$ (a quoted crystal constant, measured beforehand by X-rays). The electron beam comes in along the normal, and each row of atoms re-radiates the wave. Two neighbouring rows send their scattered wavelets toward the detector with a path difference of $a\sin\phi$, where $\phi$ is the scattering angle measured from the incident direction. Constructive interference — a bright diffraction maximum — occurs when that path difference is a whole number of wavelengths:

where: $a$ is the grating spacing, $\phi$ the angle of the maximum, $n = 1, 2, 3, \dots$ the order, and $\lambda$ the wavelength doing the diffracting. This is the ordinary grating equation; the only novelty is that the wave passing through it is made of electrons. Put in the first-order peak Davisson and Germer measured, $n = 1$ and $\phi = 50^\circ$, with $\sin 50^\circ = 0.766$:

This $1.65\ \text{Å}$ is a measurement. It comes entirely from the crystal geometry and the observed angle; it never uses the voltage or de Broglie's formula. Set the two numbers side by side — the prediction $1.67\ \text{Å}$ from the voltage, the measurement $1.65\ \text{Å}$ from the diffraction peak — and they agree to within about 1%. Two completely separate roads to the same wavelength, meeting. That is the confirmation of matter waves, and it is why this single experiment is worth a whole article.

Board level — confirming the match (CBSE and ICSE pattern)

In the Davisson–Germer experiment, electrons accelerated through $54\ \mathrm{V}$ strike a nickel crystal whose surface rows are spaced $a = 2.15\ \text{Å}$, and a first-order intensity maximum is found at a scattering angle of $50^\circ$. (a) Find the wavelength from the diffraction data. (b) Find the de Broglie wavelength from the voltage. (c) Comment on the comparison. Take $h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$, $e = 1.60\times10^{-19}\ \mathrm{C}$.

Step 1. Diffraction wavelength from the grating condition $a\sin\phi = n\lambda$ with $n = 1$: $\lambda = (2.15\ \text{Å})(\sin 50^\circ) = (2.15)(0.766) = \boxed{1.65\ \text{Å}}$.

Step 2. de Broglie wavelength from the voltage using the shortcut $\lambda = 12.27/\sqrt{V}$ Å: $\lambda = 12.27/\sqrt{54} = 12.27/7.35 = \boxed{1.67\ \text{Å}}$.

Step 3. Compare: $1.65\ \text{Å}$ from the crystal versus $1.67\ \text{Å}$ from the voltage — a difference of about $1\%$, well inside experimental error. The two independent routes agree, which is precisely the evidence that the electron travelled as a wave of wavelength $h/p$.

Competitive level — reading the voltage back out (JEE and NEET pattern)

A Davisson–Germer-type setup gives a first-order electron-diffraction peak at $\phi = 50^\circ$ off a crystal with row spacing $a = 2.15\ \text{Å}$. Treating the measured wavelength as exact, (a) find $\lambda$, and (b) deduce the accelerating voltage that produced these electrons. (c) How does it compare with the nominal $54\ \mathrm{V}$?

Step 1. Wavelength from the peak: $\lambda = a\sin\phi = (2.15)(\sin 50^\circ) = 1.65\ \text{Å}$, as before.

Step 2. Invert the de Broglie shortcut $\lambda = 12.27/\sqrt{V}$ to solve for the voltage: $\sqrt{V} = 12.27/\lambda = 12.27/1.65 = 7.44$, so $V = 7.44^2 = \boxed{55\ \mathrm{V}}$.

Step 3. The reconstructed voltage, $55\ \mathrm{V}$, sits within about $2\%$ of the applied $54\ \mathrm{V}$. Squaring turns a $1\%$ wavelength difference into a $2\%$ voltage difference — the diffraction data and the electrical measurement of the same beam are mutually consistent, closing the loop the other way round. Timing note: about 90 seconds. The trap is stopping at $\lambda$; the marks are in inverting $\lambda = 12.27/\sqrt{V}$ to $V = (12.27/\lambda)^2$. Keep $12.27$ memorised so you never re-derive it under time pressure.

Competitive level — the voltage-scaling of the matter wave (JEE Advanced flavour)

Without recomputing constants, find the de Broglie wavelength of an electron accelerated through $100\ \mathrm{V}$, and hence the ratio of the $54\ \mathrm{V}$ wavelength to the $100\ \mathrm{V}$ wavelength. Verify the ratio two ways.

Step 1. At $100\ \mathrm{V}$: $\lambda = 12.27/\sqrt{100} = 12.27/10 = \boxed{1.23\ \text{Å}} = 0.123\ \mathrm{nm}$. Higher voltage, shorter wave, as $\lambda \propto 1/\sqrt{V}$ demands.

Step 2. Ratio by dividing the two wavelengths: $\lambda_{54}/\lambda_{100} = 1.67/1.23 = 1.36$.

Step 3. Ratio the other way, straight from $\lambda \propto 1/\sqrt{V}$: $\lambda_{54}/\lambda_{100} = \sqrt{V_{100}/V_{54}} = \sqrt{100/54} = \sqrt{1.85} = 1.36$. The two roads agree, confirming the inverse-square-root scaling. A lower voltage always gives the longer matter wave. Timing note: about 60 seconds. Never re-evaluate $12.27$ per part — the whole family of answers rides on $\lambda \propto 1/\sqrt{V}$, so ratios are pure $\sqrt{V}$ arithmetic.

φ is the scattering angle, not the Bragg angle. The $50^\circ$ is measured from the incident beam. If a problem then asks for a Bragg treatment, convert to the glancing angle $\theta = 90^\circ - \phi/2 = 65^\circ$ before using $2d\sin\theta = n\lambda$. Plugging $50^\circ$ into the Bragg formula is the classic mark-loser here.

Do not confuse the two spacings. The surface-grating spacing $a = 2.15\ \text{Å}$ goes with $a\sin\phi = n\lambda$; the interplanar (Bragg) spacing $d = 0.91\ \text{Å}$ goes with $2d\sin\theta = n\lambda$. They are different lengths for the same crystal and are not interchangeable. Match each spacing to its own formula.

Keep the wavelengths independent in your explanation. When asked why the experiment confirms de Broglie, stress that $1.67\ \text{Å}$ came from the voltage alone and $1.65\ \text{Å}$ from the crystal alone. Saying "we used $\lambda = h/p$ to get the diffraction wavelength" describes a circular non-experiment and loses the conceptual marks.

Use the right mass in $\lambda = 12.27/\sqrt{V}$. That coefficient is baked for an electron. For a proton the constant becomes $0.286$, not $12.27$ — a factor of about $43$, from $\sqrt{m_p/m_e}$. Blindly reusing $12.27$ for a proton is a favourite distractor.

Board scope. The Davisson–Germer experiment and electron diffraction are examinable on CBSE, ICSE, JEE (Main and Advanced), NEET, and A-Level (CIE 9702). They are not required by AP Physics 2 or the IB syllabus, which cover matter waves ($\lambda = h/p$) but not this specific apparatus — do not spend AP or IB time memorising the $54\ \mathrm{V}$/$50^\circ$ numbers as an exam objective. IGCSE carries no dual-nature content at all.

Texts. NCERT Physics Class 12, Chapter 11 ("Dual Nature of Radiation and Matter"), the section on the Davisson–Germer experiment · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave-Particle Duality" · D. J. Griffiths, Introduction to Quantum Mechanics (for the university-level treatment of electron diffraction).

Provenance. Standard results with locked anchors. Recompute: $\lambda_{\text{deB}}(54\ \mathrm{V}) = 12.27/\sqrt{54} = 1.67\ \text{Å}$; the constant $h/\sqrt{2m_ee} = 12.27\ \text{Å}\cdot\sqrt{\mathrm{V}}$; $K = eV = 8.64\times10^{-18}\ \mathrm{J}$, $p = 3.97\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$, $\lambda = h/p = 1.67\ \text{Å}$; diffraction $\lambda = 2.15\sin 50^\circ = 1.65\ \text{Å}$; Bragg cross-check $2(0.91)\sin 65^\circ = 1.65\ \text{Å}$ with $\theta = 90^\circ - \phi/2 = 65^\circ$; reconstructed $V = (12.27/1.65)^2 = 55\ \mathrm{V}$; $\lambda(100\ \mathrm{V}) = 1.23\ \text{Å}$; ratio $\lambda_{54}/\lambda_{100} = \sqrt{100/54} = 1.36$; proton constant $0.286\ \text{Å}\cdot\sqrt{\mathrm{V}}$.

Prerequisites. de Broglie Matter Waves & the Dual Nature of Matter — where $\lambda = h/p$ and the $12.27/\sqrt{V}$ shortcut are established; this article is its experimental confirmation. Leads to. The Double-Slit Experiment with Single Particles — the university seam where electron interference is built up one particle at a time and duality is pushed to its strangest limit (go-deeper enrichment, not an exam dependency here).