If light, a wave, comes in particles, could particles act like waves? In 1924 Louis de Broglie said yes, using the same link between momentum and wavelength that photons obey. You know momentum $p = mv$ from Linear Momentum; the Double-Slit Interference is the test.
Key idea
Every particle has a wavelength $$\lambda = \frac{h}{p} = \frac{h}{mv}.$$ An electron accelerated from rest through a voltage $V$ has $p = \sqrt{2meV}$, so $\lambda = \dfrac{h}{\sqrt{2meV}}$.
Why it works
Why don't we see a baseball diffract? Its wavelength is absurdly small. A $0.15\ \text{kg}$ ball at $40\ \text{m/s}$ has $\lambda = \frac{6.6 \times 10^{-34}}{6.0} \approx 10^{-34}\ \text{m}$, far smaller than a proton. Waves show their wave nature only when they meet gaps or obstacles about their own size. An electron's wavelength is about the spacing of atoms in a crystal, so crystals diffract electrons, as Davisson and Germer found in 1927. Electron microscopes use electrons whose wavelengths are far shorter than light's, so they can show details far too small for any light microscope.
Common mistake
Thinking only light can act like a wave. Electrons, neutrons, atoms and even large molecules have all been diffracted. Check the size: compare $\lambda = h/p$ with the gaps the particle passes through; when they are similar, expect wave behavior.
Worked example: An electron
Find the wavelength of an electron moving at $1.0 \times 10^{6}\ \text{m/s}$.
- $$\begin{aligned} p &= mv = (9.11 \times 10^{-31})(1.0 \times 10^{6}) \\ &= 9.11 \times 10^{-25}\ \text{kg·m/s}. \end{aligned}$$
- $\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-25}} = 7.3 \times 10^{-10}\ \text{m}$.
- About $0.73\ \text{nm}$, a few atoms across.
Answer: $7.3 \times 10^{-10}\ \text{m}$
Worked example: Through 100 volts
An electron is accelerated from rest through $100\ \text{V}$. Find its wavelength.
- Kinetic energy: $K = eV = 1.60 \times 10^{-17}\ \text{J}$, and $p = \sqrt{2mK}$.
- $$\begin{aligned} p &= \sqrt{2(9.11 \times 10^{-31})(1.60 \times 10^{-17})} \\ &= 5.40 \times 10^{-24}\ \text{kg·m/s}. \end{aligned}$$
- $\lambda = \dfrac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10}\ \text{m}$, about the size of an atom: right for electron microscopes and crystal diffraction.
Answer: $1.23 \times 10^{-10}\ \text{m}$