De Broglie Wavelength
The de Broglie wavelength describes the wave nature of matter particles, the wave-particle duality.
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Formula
\lambda = \frac{h}{p} = \frac{h}{mv}Variables & units – De Broglie Wavelength
| Symbol | Meaning | Unit |
|---|---|---|
| λ | De Broglie wavelength | m |
| h | Planck constant (6.626×10⁻³⁴ J·s) | J·s |
| p | Momentum of the particle (= mv) | kg·m/s |
Derivation & background – De Broglie Wavelength
In his 1924 dissertation, Louis de Broglie postulated that all matter possesses wave properties (by analogy with photons: p = h/λ → λ = h/p). This was confirmed experimentally by the Davisson-Germer experiment (1927): the diffraction of electrons by a nickel crystal.
Exam blueprint
Validity range
Applies to matter particles with momentum p. The h/(mv) form is non-relativistic.
Derivation steps
De Broglie transfers the photon relation p = h/λ to matter particles.
- 1For photons, p = h/λ.
- 2The matter-wave hypothesis therefore sets λ = h/p.
Rearrangements
Momentum from wavelength
Small wavelength means large momentum.
Task variant
Why do heavy everyday objects not show visible matter waves?
Their momentum is macroscopically large, so λ becomes extremely small.
Common mistakes
Confusing h with ℏ.
This formula uses h; ℏ = h/(2π) appears in other quantum formulas.
Exam context
- Often combined with electron diffraction, accelerating voltage or microscope resolution.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Quantum waves
Connects wavelength, momentum and uncertainty.
Worked example
An electron at v = 10⁶ m/s: λ = 6.626×10⁻³⁴ / (9.11×10⁻³¹ × 10⁶) ≈ 0.73 nm, on the order of atomic spacings, which makes the electron microscope possible.
Applications
Electron microscopy (0.1 Å resolution), neutron scattering, quantum computing
Quanta exam set
Curated exam set for "De Broglie Wavelength":
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Which formula describes De Broglie Wavelength?
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How do you rearrange λ = h/p for Momentum from wavelength?
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Which common mistake happens with De Broglie Wavelength?
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Scientific sources
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Frequently asked questions about De Broglie Wavelength
How do you calculate the de Broglie wavelength of a particle?+
Divide Planck constant h = 6.626×10⁻³⁴ J·s by the momentum p of the particle: λ = h/p. For a non-relativistic particle p = m·v, so λ = h/(m·v). An electron with v = 10⁶ m/s has λ = 6.626×10⁻³⁴/(9.11×10⁻³¹·10⁶) ≈ 0.73 nm, which is on the order of atomic spacings and enables electron diffraction. Insert the mass in kilograms and the speed in metres per second, then the wavelength comes out in metres. Small wavelengths mean large momentum, large wavelengths mean small momentum.
Why do everyday objects show no visible matter wave?+
Because their momentum is macroscopically huge and λ = h/p therefore gives an unimaginably small wavelength. Planck constant h is extremely tiny at 6.626×10⁻³⁴ J·s. A football of 0.4 kg at 10 m/s has a momentum of 4 kg·m/s and thus a wavelength of about 1.7×10⁻³⁴ m, billions of times smaller than an atomic nucleus. Such wavelengths cannot be measured in any experiment; there is no object with a suitably small slit. Wave character appears only for very light particles such as electrons, whose small mass leads to measurable wavelengths in the nanometre range.
What is the difference between h and ℏ in quantum formulas?+
h is Planck constant at 6.626×10⁻³⁴ J·s and appears in the de Broglie relation λ = h/p as well as in E = h·f. The reduced Planck constant ℏ is defined as ℏ = h/(2π) ≈ 1.055×10⁻³⁴ J·s. It shows up wherever angular frequencies or angular momenta are involved, for example in the Heisenberg uncertainty relation Δx·Δp ≥ ℏ/2. A common mistake is to confuse h and ℏ; they differ by the factor 2π ≈ 6.28. Check carefully in every formula whether the normal or the reduced constant is required, otherwise your result is off by this factor.
How is the de Broglie wavelength linked to the accelerating voltage?+
When an electron is accelerated by a voltage U, it gains the kinetic energy E_kin = e·U. Through E_kin = p²/(2m) the momentum can be found: p = √(2·m·e·U). Inserting this into λ = h/p gives λ = h/√(2·m·e·U). The wavelength therefore decreases with increasing voltage, because a faster electron has more momentum. This relationship is the basis of the electron microscope: high voltages produce very small wavelengths and thus a much higher resolution than visible light. For very high voltages you also have to apply a relativistic correction.
Why does the de Broglie wavelength support wave-particle duality?+
In 1924 de Broglie transferred the relation p = h/λ, known for photons, to matter particles and postulated that electrons too have a wavelength. Shortly afterwards Davisson and Germer confirmed this experimentally through diffraction of electrons at a crystal lattice, a typical wave phenomenon. This made it clear that particles such as electrons show both particle and wave properties, depending on the experiment. The wavelength λ = h/p links the particle quantity momentum with the wave quantity wavelength through the natural constant h. This duality is a cornerstone of quantum mechanics and explains phenomena such as interference and diffraction of matter.
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How do you calculate with De Broglie Wavelength?
Here is how to work through a typical De Broglie Wavelength (λ = h/p) task step by step:
- 1
Task
Why do heavy everyday objects not show visible matter waves?
Solution path
Their momentum is macroscopically large, so λ becomes extremely small.