Do electrons only bump with other electrons in a circuit and that only causes resistance? Or do they bump with the nucleus as well?

Here's a post I have been studying recently:

The path of a typical electron through a wire could be described as a rather chaotic, zigzag path characterized by collisions with fixed atoms. Each collision results in a change in direction of the electron. Yet because of collisions with atoms in the solid network of the metal conductor, there are two steps backwards for every three steps forward.

enter image description here

Does this mean that electrical resistance originates from the electrons' collisions with nuclei as well as other electrons, or just with other electrons?

  • $\begingroup$ The answers heremight be helpful (even if the question is a bit of a mess). $\endgroup$ – The Photon Mar 3 at 1:18
  • $\begingroup$ Short version: the nuclei have a lot more inertia than the other electrons, so they'll cause a much bigger change in the electron's momentum on a collision. But we might rather say that the electron interacts with the phonon states of the crystal lattice than with the individual nuclei. $\endgroup$ – The Photon Mar 3 at 1:20

A conducting electron, in an eigenstate of the periodic potential of a crystalline lattice, will never scatter, assuming the lattice is perfect. This is because its periodic wavefunction already “includes” interaction with the lattice nuclei.

So scattering comes from distortions of the lattice. Think of a sudden distortion like an interface that the electron wave could reflect/scatter from. The model you cite is outdated (as of the middle 20th century), although it works reasonably well, as a practical matter, for simple systems.

Lattice distortions come primarily in the form of defects and lattice compression/expansion waves (i.e. phonons). This is why metals tend to have a higher conductivity at lower temperature, because the thermally-excited lattice waves are less populous, so there’s less electron-phonon scattering. This trend continues for lower and lower temperatures until scattering from lattice defects becomes dominant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.