If a photon of light carries momentum despite having no mass, and its momentum is tied to a wavelength, then why should that connection stop at light? De Broglie's answer was that it doesn't — every moving particle has a wavelength, however solid and particle-like it seems.
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De Broglie's hypothesis
Einstein's photon theory established that light, ordinarily pictured as a wave, carries momentum in discrete lumps when it behaves as a particle. In 1924, Louis de Broglie asked the question the other way round: if a particle of light has a wavelength associated with its momentum, could a particle of matter — an electron, say — have a wavelength associated with its momentum too?
De Broglie's hypothesis is that the answer is yes, for literally any moving particle, not just photons:
$\lambda = \dfrac{h}{p}$
where $h = 6.63\times10^{-34}\,\text{J s}$ is the Planck constant and $p$ is the particle's momentum. This is now called the particle's de Broglie wavelength. For a non-relativistic particle of mass $m$ moving at speed $v$, momentum is simply $p = mv$, giving the more familiar form
$\lambda = \dfrac{h}{mv}$
A genuinely new kind of claim
This was a strange proposal at the time. Nothing about a thrown ball, a car, or an electron in a cathode ray tube looks remotely wave-like — they have a definite position, a definite trajectory, and bounce off things like solid objects should. De Broglie's hypothesis says that all of them, without exception, have a wavelength anyway; it's just that for anything with everyday mass, that wavelength is so absurdly small that no wave effect could ever be observed. The next lesson looks at exactly why that is.
De Broglie's hypothesis: every moving particle, not just a photon, has an associated wavelength $\lambda = h/p$ set by its momentum — a wave character that becomes significant only when the particle's mass and speed make $\lambda$ large enough to detect.
Deriving the relationship from the photon case
De Broglie's relationship is not an arbitrary guess — it falls directly out of combining two things already known about photons.
DerivationFrom photon relationships to de Broglie's equation
1
A photon's energy is given by $E = hf$ (Einstein's photon relationship, from the previous lesson).
2
The wave equation for light gives $c = f\lambda$, so $f = c/\lambda$. Substituting into $E=hf$:
$E = \dfrac{hc}{\lambda}$
3
Separately, for any massless particle of radiation, energy and momentum are related by $E = pc$.
4
These are two different expressions for the same energy $E$, so they can be set equal:
$pc = \dfrac{hc}{\lambda}$
5
Cancelling $c$ from both sides and rearranging for $\lambda$:
$\lambda = \dfrac{h}{p}$
This is de Broglie's relationship, derived here purely for a photon. De Broglie's insight was that nothing in this final relationship — unlike the two equations it was built from — mentions light or the speed of light specifically. He proposed that it holds for any particle with momentum $p$, whether or not it has mass, and whether or not it moves anywhere near the speed of light.
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The closed form for an accelerated electron
The most common practical use of de Broglie's relationship in this module is for an electron accelerated from rest through a potential difference $V$ — exactly the situation inside an electron gun. Rather than finding the electron's speed and momentum separately, it's more direct to combine everything into a single formula in terms of $V$.
DerivationDe Broglie wavelength in terms of accelerating voltage
1
The work done by the field accelerating an electron of charge $e$ through potential difference $V$ equals its gain in kinetic energy: $eV = \tfrac{1}{2}mv^2$.
2
Kinetic energy can be rewritten in terms of momentum $p=mv$: $\tfrac{1}{2}mv^2 = \dfrac{(mv)^2}{2m} = \dfrac{p^2}{2m}$, so
$eV = \dfrac{p^2}{2m}$
3
Rearranging for $p$: $p^2 = 2meV$, so
$p = \sqrt{2meV}$
4
Substituting into $\lambda = h/p$ gives the closed form:
$\lambda = \dfrac{h}{\sqrt{2meV}}$
This single formula gives the electron's de Broglie wavelength directly from its accelerating voltage — no separate calculation of speed or momentum needed. It's the version used directly in electron microscope and electron diffraction calculations later in this module.
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Worked exampleDe Broglie wavelength of an accelerated electron
1
An electron, initially at rest, is accelerated through a potential difference of $V = 2500\,\text{V}$ in an electron gun. Find its de Broglie wavelength once accelerated.
2
The work done by the field accelerating the electron equals its gain in kinetic energy: $eV = \tfrac{1}{2}mv^2$, where $e$ is the elementary charge and $m$ is the electron's mass.
This is the standard link between accelerating voltage and speed for a charged particle in an electron gun, exactly as used elsewhere for electron beams and cathode ray tubes.
3
Rearranging for momentum $p = mv$ is more direct than solving for $v$ first: from $eV = \tfrac{1}{2}mv^2 = \dfrac{p^2}{2m}$, so $p = \sqrt{2meV}$.
This is smaller than typical atomic spacing in a crystal (roughly $0.1$–$0.3\,\text{nm}$), which is why higher-voltage electron beams like this one are used in electron microscopy — the shorter wavelength allows finer detail to be resolved than an optical microscope using visible light ever could.
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Two routes to the same wavelength
For any particle, $\lambda = h/p$ always works once momentum is known by whatever means. The closed form $\lambda = h/\sqrt{2meV}$ is just that same equation specialised to the one situation — an electron accelerated from rest through a known voltage — that comes up repeatedly in this module. Both routes must agree: in the worked example above, computing $p$ first and then $\lambda=h/p$ gives exactly the same answer as substituting directly into $\lambda = h/\sqrt{2meV}$.
Key takeaways
De Broglie's hypothesis: every moving particle has an associated wavelength, $\lambda = h/p$, where $p$ is its momentum — not just photons.
This follows by combining the photon relationships $E=hf$, $c=f\lambda$ and $E=pc$: setting the two expressions for $E$ equal and cancelling $c$ gives $\lambda = h/p$ directly.
De Broglie's insight was that this final relationship doesn't reference light specifically, so it should hold for any particle with momentum, massive or not.
For a non-relativistic particle, $p = mv$, so $\lambda = h/(mv)$.
For an electron accelerated from rest through voltage $V$, using $eV = p^2/2m$ gives $p=\sqrt{2meV}$, and hence the closed form $\lambda = h/\sqrt{2meV}$.
Worked example: an electron accelerated through 2500 V has momentum $p\approx2.70\times10^{-23}\,\text{kg}\,\text{m}\,\text{s}^{-1}$ and de Broglie wavelength $\lambda\approx2.46\times10^{-11}\,\text{m}$, comparable to atomic spacing.
