de Broglie Wavelength and Matter Waves

Every moving particle has a wavelength

If light (normally a wave) can behave like a stream of particles (photons), de Broglie proposed the reverse should hold too: every moving particle has an associated wavelength, set by its momentum $p=mv$ :

\lambda = \dfrac{h}{p} = \dfrac{h}{mv}

For everyday objects this wavelength is absurdly small and unobservable. It only becomes significant for very light, fast particles like electrons, which is exactly why electron microscopes work.

Worked example

Find the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V.

Solution: The accelerating field does work $eV$ on the electron, which becomes its kinetic energy:

eV = \dfrac{1}{2}mv^2 \;\Rightarrow\; mv = \sqrt{2meV}

Substituting $p=mv=\sqrt{2meV}$ into the de Broglie relation:

\lambda = \dfrac{h}{\sqrt{2meV}}

Plugging in $h=6.626\times10^{-34}$ J·s, $m=9.11\times10^{-31}$ kg, and $e=1.6\times10^{-19}$ C gives a clean, commonly-used shortcut for electrons specifically:

\lambda \approx \dfrac{1.226}{\sqrt{V}}\text{ nm} \qquad (V \text{ in volts})

For $V=100$ V:

\lambda \approx \dfrac{1.226}{\sqrt{100}} = \dfrac{1.226}{10} \approx 0.1226\text{ nm}

That's comparable to the spacing between atoms in a crystal, which is exactly why fast electrons diffract off crystal lattices the same way X-rays do.