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De Broglie's $\lambda = h/p$ for electrons, neutrons and everyday objects, the electron shortcut $1.226/\sqrt{V}$ nm, and relativistic matter waves in electron microscopes.
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By the end of this lesson you will be able to compute de Broglie wavelengths of particles, including relativistic ones, and judge when wave effects matter.
From the last unit you know photons carry momentum $p = h/\lambda$, and from the relativity unit the relation $(pc)^2 = K^2 + 2Kmc^2$. From waves you know diffraction and interference. This lesson turns the photon relation around: if waves carry momentum like particles, particles with momentum should behave like waves, with a wavelength set by the same constant $h$.
| Term | What it means |
|---|---|
| de Broglie wavelength | $\lambda = h/p$, the wavelength associated with a particle of momentum $p$. |
| Matter wave | The wave that governs where a particle is likely to be found. |
| Wave–particle duality | The fact that both light and matter show wave and particle behavior. |
| Accelerating voltage | The potential difference that gives a charged particle its kinetic energy. |
| Thermal neutron | A neutron with kinetic energy near $kT$ at room temperature, about $0.025$ eV. |
| Resolution | The smallest detail an instrument can distinguish, limited by its wavelength. |
| Electron microscope | An instrument that images with electron waves instead of light. |
In 1924 Louis de Broglie proposed in his doctoral thesis that the photon relation applies to all matter: a particle of momentum $p$ has a wavelength
$$\lambda = \frac{h}{p}.$$
For slow particles, $p = mv = \sqrt{2mK}$. An electron accelerated from rest through $V$ volts has $K = eV$, and combining the constants gives a handy form:
$$\lambda = \frac{1.226}{\sqrt{V}}\ \text{nm}.$$
An electron through $100$ V has $\lambda = 0.12$ nm, about the spacing of atoms in a crystal. A baseball at $40$ m/s has $\lambda \approx 10^{-34}$ m, far smaller than any nucleus, which is why baseballs never diffract. Wave behavior matters when the wavelength is comparable to the sizes of things the particle encounters.
Another way: picture
Picture a particle as carrying a small wave train along with it, a wave that tells it where it is likely to go. The faster and heavier the particle, the more tightly packed the wave. For a speck of dust the wave is so fine that the particle simply follows a path. For an electron, the wave is as coarse as the gaps between atoms, and it spreads and interferes as it passes through them.
Another way: steps
De Broglie reasoned from symmetry. Einstein had shown that light, a wave, carries momentum in packets of $h/\lambda$. Nature, he argued, should not be so lopsided that only light has two faces. So he assigned every particle a wave, with the same relation between wavelength and momentum.
He also showed that the idea explained Bohr's puzzling rule for the hydrogen atom: if an electron's wave must fit around its orbit a whole number of times, only certain orbits, and so only certain energies, are allowed. Einstein read the thesis and praised it; Erwin Schrödinger built on it to write his wave equation in 1926.
The wavelength depends on momentum, so it is shortest for heavy, fast objects. A $0.15$ kg baseball at $40$ m/s: $\lambda = 6.63 \times 10^{-34}/(0.15 \times 40) = 1.1 \times 10^{-34}$ m. A dust grain of a microgram drifting at $1$ mm/s: $7 \times 10^{-22}$ m. Neither will ever show a measurable wave effect.
An electron in a television tube at $20$ kV: about $9$ pm. A thermal neutron from a reactor: about $0.18$ nm. A nitrogen molecule in room air: about $0.03$ nm. These are the scales of atoms and crystals, and in that world, the wave nature of matter is everyday physics.
For an electron from rest through $V$ volts, $K = eV$ and $p = \sqrt{2m_eeV}$. Putting in the constants,
$$\lambda = \frac{h}{\sqrt{2m_eeV}} = \frac{1.226\ \text{nm}}{\sqrt{V}}.$$
Through $150$ V, $\lambda = 0.100$ nm. Four times the voltage halves the wavelength. The formula assumes the electron is nonrelativistic, which holds to about a percent below $10$ kV. Above that, and certainly in electron microscopes at hundreds of kilovolts, the relativistic momentum must be used.
At high energies, use the invariant from lesson 4: $(pc)^2 = K^2 + 2Kmc^2$. Then $\lambda = hc/pc$, with $hc = 1240$ keV·pm. An electron through $300$ kV has $pc = \sqrt{300^2 + 2 \times 300 \times 511} = 630$ keV and $\lambda = 1.97$ pm; the nonrelativistic formula would give $2.24$ pm, off by $14$ percent.
For very high energies, $K \gg mc^2$, $pc \approx K$, and $\lambda \approx hc/K$, just as for a photon of the same energy. At the Large Hadron Collider, protons of $6.8$ TeV have wavelengths of about $0.2$ attometers, a thousandth of a proton's size, fine enough to probe the quarks inside.
Checking an answer. Heavier or faster particles must have shorter wavelengths. Electrons through tens to hundreds of volts must give about $0.1$ nm. The relativistic wavelength must be shorter than the nonrelativistic estimate. And everyday objects must give absurdly small wavelengths.
De Broglie's relation was a hypothesis until it was tested, and it has passed every test: electrons, neutrons, atoms and even molecules of hundreds of atoms show interference with exactly the predicted wavelengths. It is now a foundation of quantum mechanics, not an extra assumption.
Using $p = \sqrt{2mK}$ requires $K \ll mc^2$: $511$ keV for an electron, $939$ MeV for a neutron. The shortcut $1.226/\sqrt{V}$ is valid only for electrons, since the mass and charge are built into it; a proton through the same voltage has a wavelength $\sqrt{1836} = 43$ times shorter.
The wave nature of electrons sets the size of atoms. Squeezing an electron into a smaller space means shortening its wavelength, which means raising its momentum and its kinetic energy. The electrical attraction of the nucleus pulls the electron in; its wave nature pushes back.
The balance comes at about $0.05$ nm for hydrogen, where the electron's wavelength is about the circumference of its orbit. If electrons were not waves, nothing would stop them spiraling into the nucleus, and matter would collapse. The stability and size of every atom, and so of everything built from atoms, are consequences of $\lambda = h/p$.
The wave nature of matter is not limited to electrons. In 1999 Anton Zeilinger's group in Vienna sent buckyballs, molecules of sixty carbon atoms, through a grating and saw an interference pattern. Later experiments have shown interference with molecules of more than two thousand atoms.
These experiments probe how large an object can be and still show quantum behavior. The obstacle is not size itself but interaction with the environment: a large molecule that bumps into air or emits heat radiation reveals its path and loses its interference. Physicists in the United States and Europe are pushing the limit toward viruses and small particles.
Any wave can resolve details only down to about its own wavelength. Visible light, at $400$ to $700$ nm, cannot image single atoms, which are about $0.1$ nm across. Electrons at $100$ kV have wavelengths near $4$ pm, a hundred thousand times shorter, so electron microscopes can image atoms, limited in practice by lens imperfections rather than wavelength.
The same principle drives particle accelerators. To see inside a proton, about $1$ fm across, requires probes with wavelengths of that size or less, which means momenta of hundreds of MeV/$c$ and more. Larger accelerators are, in effect, finer microscopes.
The shortcut $1.226/\sqrt{V}$ nm belongs to electrons alone, because it has the electron's mass and charge built in. For any other particle, start from $\lambda = h/\sqrt{2mK}$. A proton accelerated through the same voltage as an electron has the same kinetic energy but $1836$ times the mass, so its momentum is $\sqrt{1836} = 43$ times larger and its wavelength $43$ times shorter.
That is why protons and heavier ions make poor waves for imaging crystals at ordinary energies but excellent probes of nuclei at high ones. Neutrons are slowed to low speeds on purpose, in blocks of water or liquid hydrogen called moderators, so that their wavelengths grow to match atomic spacings. Alpha particles from a radioactive source, at about $5$ MeV, have wavelengths near $6$ fm, comparable to a nucleus, which is why Rutherford's alpha particles could reveal the nucleus in 1911 without anyone yet knowing they were waves. The rule in every case is the same: choose a particle and an energy whose wavelength matches the scale of the thing you want to see.
The TEAM microscopes at Lawrence Berkeley National Laboratory's Molecular Foundry accelerate electrons through up to $300$ kV, giving wavelengths near $2$ pm, and with corrected lenses resolve details below $0.05$ nm, less than half the width of a hydrogen atom. Materials scientists use them to see individual atoms in new alloys and the defects in semiconductor chips.
Cryo-electron microscopes, which freeze proteins in thin ice and image them with electrons, have mapped the structures of thousands of biological molecules, including the spike protein of the coronavirus in 2020. The resolution, set by the electrons' tiny wavelength, is now good enough to see where each atom of a drug binds.
The Spallation Neutron Source at Oak Ridge National Laboratory in Tennessee makes intense pulses of neutrons and slows them to wavelengths of a few tenths of a nanometer, matched to the spacings between atoms. Scattered off samples, they reveal crystal structures, magnetic order and the motion of hydrogen atoms that X-rays barely see.
Because neutrons are uncharged and penetrate metal, researchers use them to watch lithium move inside working batteries and to measure stresses deep inside jet engine parts. Each measurement rests on de Broglie's relation: timing a neutron's flight gives its speed, hence its momentum, hence its wavelength.
It is tempting to picture a matter wave as the particle itself snaking along, or to read the wavelength as how big the particle is. Neither is right. The wavelength belongs to a wave of probability: where the wave is large, the particle is likely to be found. The particle is always detected whole, at one place.
A related error is to think only small particles have wavelengths. Every object has one; for large objects it is simply far too small to produce any measurable effect.
An electron is accelerated from rest through $54$ V. Find its kinetic energy in joules.
$K = 54 \times 1.60 \times 10^{-19} = 8.64 \times 10^{-18}\ \text{J}$
$K = eV$.
Find its momentum.
$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 8.64 \times 10^{-18}} = 3.97 \times 10^{-24}\ \text{kg·m/s}$
$\sqrt{2mK}$.
Find its wavelength.
$\lambda = \dfrac{6.63 \times 10^{-34}}{3.97 \times 10^{-24}} = 0.167\ \text{nm}$
$h/p$.
Check with the shortcut.
$\dfrac{1.226}{\sqrt{54}} = 0.167\ \text{nm}$
Agrees.
Note the history.
$\text{Davisson and Germer's electrons, 1927}$
The voltage used to discover electron diffraction.
Find the wavelength of a $0.145$ kg baseball at $40$ m/s.
$\lambda = \dfrac{6.63 \times 10^{-34}}{0.145 \times 40} = 1.1 \times 10^{-34}\ \text{m}$
$h/mv$.
Compare with a proton's size.
$\dfrac{1.1 \times 10^{-34}}{10^{-15}} = 10^{-19}$
Hopelessly small.
Find the speed of a thermal neutron, $K = 0.0253$ eV.
$v = \sqrt{\dfrac{2 \times 0.0253 \times 1.60 \times 10^{-19}}{1.675 \times 10^{-27}}} = 2200\ \text{m/s}$
Nonrelativistic.
Find its wavelength.
$\lambda = \dfrac{6.63 \times 10^{-34}}{1.675 \times 10^{-27} \times 2200} = 0.18\ \text{nm}$
$h/mv$.
Compare with crystal spacings.
$\text{about } 0.1\text{–}0.5\ \text{nm}$
Neutrons diffract off crystals.
State why the neutron shows waves and the ball does not.
$\lambda \text{ comparable to the obstacles}$
Only then are wave effects visible.
Electrons are accelerated through $200$ kV. Compare $K$ with $mc^2$.
$K = 200\ \text{keV} = 0.39\,mc^2$
Relativistic.
Find the momentum $pc$.
$pc = \sqrt{200^2 + 2 \times 200 \times 511} = 494.5\ \text{keV}$
The invariant.
Find the wavelength.
$\lambda = \dfrac{1240}{494.5} = 2.51\ \text{pm}$
$hc/pc$ with $hc = 1240$ keV·pm.
Find the nonrelativistic estimate.
$\dfrac{1240}{\sqrt{2 \times 200 \times 511}} = 2.74\ \text{pm}$
Nine percent too long.
Compare with an atom.
$\dfrac{100\ \text{pm}}{2.51\ \text{pm}} = 40$
Plenty short enough to resolve atoms.
Find the speed.
$\beta = \dfrac{pc}{E} = \dfrac{494.5}{711} = 0.70$
Seventy percent of light speed.
Explain why higher voltage helps.
$\text{shorter wavelength and better penetration}$
Thicker samples and finer detail.
Use the electron shortcut.
$\lambda = \dfrac{1.226}{\sqrt{V}}\ \text{nm}$
Nonrelativistic electrons.
Substitute the voltage.
$\lambda = \dfrac{1.226}{\sqrt{400}}$
$\sqrt{400} = 20$.
Evaluate the wavelength.
An electron has a de Broglie wavelength of $1.8$ nm. Another electron moves twice as fast, still far below light speed. What is its wavelength?
Complete the worked solution: an electron is accelerated from rest through $400$ V. With $m_e = 9.11 \times 10^{-31}$ kg, $e = 1.60 \times 10^{-19}$ C and $h = 6.63 \times 10^{-34}$ J·s, find its kinetic energy in units of $10^{-17}$ J, its momentum in units of $10^{-24}$ kg·m/s, and its wavelength in nm.
Multiply the charge by the voltage.
$K = eV =$ k
In units of $10^{-17}$ J.
Find the momentum from the energy.
$p = \sqrt{2m_eK} =$ p
In units of $10^{-24}$ kg·m/s.
Divide Planck's constant by the momentum.
$\lambda = \dfrac{h}{p} =$ l
In nm.
Compare with an atom's size.
$\text{atoms are about } 0.1\text{–}0.3\ \text{nm across}$
So such electrons diffract off crystals.
Match each statement about matter waves to its formula or fact.
| $h/p$ | $\sqrt{2mK}$ | $1.226/\sqrt{V}$ nm | a wavelength far too small to observe | |
|---|---|---|---|---|
| de Broglie wavelength | ||||
| momentum from energy | ||||
| electron through V volts | ||||
| a baseball |
An electron starts from rest and is accelerated through $2500$ V. With $m_e = 9.11 \times 10^{-31}$ kg, $e = 1.60 \times 10^{-19}$ C and $h = 6.63 \times 10^{-34}$ J·s, fill in its kinetic energy in eV, its momentum in units of $10^{-24}$ kg·m/s, and its de Broglie wavelength in nm.
| value | |
|---|---|
| kinetic energy (eV) | |
| momentum (10⁻²⁴ kg·m/s) | |
| wavelength (nm) |
Write the de Broglie wavelength of a neutron, in nm, as a formula in its speed $v$ measured in km/s, for speeds far below light's. For this particle, $h/m$ works out to $0.3956$ in these units.
Answer:
A neutron has kinetic energy $1$ eV. With $m_n = 1.675 \times 10^{-27}$ kg, $1$ eV $= 1.602 \times 10^{-19}$ J and $h = 6.626 \times 10^{-34}$ J·s, what is its de Broglie wavelength, in nm?
Answer: nm
A transmission electron microscope at Lawrence Berkeley National Laboratory accelerates electrons through $60$ kV. With the electron's rest energy $511$ keV and $hc = 1240$ keV·pm, what is their wavelength, including the relativistic correction, in pm?
Answer: pm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write the de Broglie wavelength of a proton, in nm, as a formula in its speed $v$ measured in km/s, for speeds far below light's. For this particle, $h/m$ works out to $0.3961$ in these units.
Answer:
You can compute matter wavelengths. Explain to someone why electrons diffract off crystals but baseballs never diffract.
21. Your turn: find the wavelength of an electron accelerated through $400$ V., step 3
$\lambda = 0.0613\ \text{nm}$
About half an atomic spacing.