de Broglie wavelength

λ=hp\lambda=\dfrac{h}{p}

Result unit: m (metre)

What it describes

Every particle with momentum p corresponds to a matter wave of wavelength λ.

Symbols

λ\lambda
matter wavelength
[m]
hh
Planck's constant
[J·s]
pp
particle momentum
[kg·m/s]

Where it comes from

Not only light, but also particles (electrons) have a wave nature. The greater the momentum, the shorter the wavelength — which is why for macroscopic bodies the wave is immeasurably small.

When to use it

Quantum mechanics, electron microscope, electron diffraction.

Example

An electron with momentum p=2⋅10−24p=2\cdot10^{-24} kg·m/s. Calculate the wavelength (h=6.63⋅10−34h=6.63\cdot10^{-34}).
Substitutionh=6.63⋅10−34h=6.63\cdot10^{-34}, p=2⋅10−24p=2\cdot10^{-24}.
Resultλ≈3.3⋅10−10\lambda\approx3.3\cdot10^{-10} m.

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