Example 1
Use h=6.63×10−34Js and electron momentum 3.00×10−24kgms−1. Find wavelength.
λ=ph=2.21×10−10m.
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Year 12 · SACE Stage 2
de Broglie waves and electron diffraction
Calculate matter wavelengths and identify the evidence requiring wave behaviour.
Method and worked examples
Matter with momentum p has de Broglie wavelength λ=ph. For nonrelativistic motion, p= mv, giving λ=(hmv). Larger momentum means shorter wavelength. A massive everyday object has such a small wavelength that ordinary obstacles cannot reveal an observable diffraction pattern. For an electron accelerated from rest through voltage V,p=√2meV in the low-speed approximation.
Electron double-slit patterns and diffraction from crystal planes demonstrate wave behaviour. Davisson–Germer electron scattering showed intensity maxima linked to the crystal spacing and electron wavelength. Individual electrons still arrive as localised detections; many arrivals form a patterned distribution. Electron microscopes use short electron wavelengths, although practical resolution also depends on lenses and specimen preparation. Use relativistic momentum instead of mv if the electron speed is not small compared with c.
Example 1
Example 2
An electron’s accelerating voltage is increased by four while the nonrelativistic model remains valid. Compare wavelengths.
p=√2meV, so p doubles and λ=ph halves.
Watch out: Localised detections and wave-pattern distributions are both features of quantum matter; a single classical model is insufficient.