# What is the wave particle duality of electrons?

Oct 13, 2014

All particles of matter, including electrons, exhibit characteristics of waves. The most obvious of these is the ability of particles to interfere with each other with a characteristic wavelength given by the de Broglie relation

$\lambda = \frac{h}{p}$

where $h$ is Planck's constant, $h = 6.626 \times {10}^{- 34} J - s$, and $p$ is the momentum of the particle. If $p$ is expressed in units of $\frac{k g - m}{s}$ then the wavelength $\lambda$ is obtained in units of meters.

Example:

If electrons initially at rest are accelerated through a potential of $10 V$, then the kinetic energy of each electron would be $10 e V = 1.602 \times {10}^{- 18} J$. Using the classical mechanics relation between kinetic energy and momentum, we obtain $p = {\left(2 m E\right)}^{\frac{1}{2}} = 1.708 \times {10}^{- 24} \frac{k g - m}{s}$.

Using the deBroglie relation, we can calculate the characteristic wavelength of the electrons as $\lambda = \frac{h}{p} = 3.878 \times {10}^{- 10} m = 0.3878 n m$

If such a beam of electrons were diffracted from an ordered crystal, they would form a pattern similar to that for X-rays of the same wavelength ($0.3878 n m$).