Question

What is the wave particle duality of electrons?

Answer 1

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=h/p`

where `h` is Planck's constant, `h=6.626times10^(-34)J-s`, and `p` is the momentum of the particle. If `p` is expressed in units of `(kg-m)/s` then the wavelength `lambda` is obtained in units of meters.

Example:

If electrons initially at rest are accelerated through a potential of `10V`, then the kinetic energy of each electron would be `10eV=1.602times10^(-18) J`. Using the classical mechanics relation between kinetic energy and momentum, we obtain `p=(2mE)^(1/2)=1.708times10^(-24)(kg-m)/s`.

Using the deBroglie relation, we can calculate the characteristic wavelength of the electrons as `lambda=h/p=3.878times10^(-10)m=0.3878 nm`

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 nm`).