Textbooks describe resistance as involving electrons colliding with other ions in metals, resulting in a heating effect, though how exactly is this achieved? Although I am not required to learn the process in detail (why this happens), I feel like I need an explanation/firm understanding!

This is my attempted understanding so far, using the limited knowledge I have. When an electron nears another electron, the force of electrostatic repulsion could be enough to excite that electron to a higher energy level as a result of the kinetic energy obtained, where it would then subsequently return to a lower energy state or enter another ion, losing the transferred potential energy in the form of a photon emission (IR when referring to heat?). What happens to the initial electron that collided, how is the remaining energy not absorbed by the other electron used - does this electron simply 'deflect'?

The highest energy level electrons are de-localised, however, which contradicts my original explanation, as the electrons are not bound to a particular ion. What would result in the EM emission?

I apologise as this is a really superficial understanding, though simply wanted to know the rough reasoning behind the thermal emission beyond secondary school/(don't know the American equivalent!) knowledge. Thank you!

  • $\begingroup$ While your title refers to electron-ion collisions, the body of your question focuses on electron-electron interactions. Which are you interested in? $\endgroup$
    – Jon Custer
    May 26, 2015 at 15:14
  • $\begingroup$ Hi, thanks for the reply. This is the problem that I am having. I do not know what is meant by an electron-ion collision - I assumed that this term refers to the collisions between electrons that are BOUND to the ion and an incoming electron. I do not understand the difference. I would like to focus on electron-electron 'collisions' - thank you! $\endgroup$
    – ZDust
    May 26, 2015 at 15:24
  • $\begingroup$ Technically the electrons couple to the phonons (quantized lattice vibrations) trough imperfections in the crystal structure and the quasi-particle states made by the phonons and electrons couple to the photons in the vacuum around the metal trough charge fluctuations that can be described by QED, but in a sense this is an overkill for your purposes. It is rarely necessary to go beyond the semiclassical black-body radiation picture to "grock" what happens between a metal and the vacuum. $\endgroup$
    – CuriousOne
    May 26, 2015 at 16:40
  • $\begingroup$ @CuriousOne That's exhausting ! :D $\endgroup$
    – Gaurav
    May 27, 2015 at 3:23

1 Answer 1


Take a look at the Drude model. It gives a fairly intuitive way to look at conductivity in solids. Although later proved to be slightly incorrect due to the ignorance of quantum effects, it does the job for a classical explanation.

One can reason as to how heat is generated in the conducter in a classical manner from the Drude model. As the electrons move through the conducter, a few electrons strike the constituent atoms or molecules. Since the atoms/molecules acquire kinetic energy, the temperature of the conducter as a whole increases. Any object having surface area $A$ with temperature $T$ emits electromagnetic radiation with power $$P=A \epsilon \sigma T^4$$ according to the Stefan-Boltzmann law, where $\epsilon$ and $\sigma$ are constants.

  • $\begingroup$ Thanks for the reply, though my question may not be worded clearly. I wanted to know how electron-electron and electron-ion collisions result in the emission of heat - as in what forces are exerted on electrons for them to emit EM radiation of wavelengths that our body interprets as heat. Maybe this question cannot be answered without delving into more advanced concepts? Thanks once again. $\endgroup$
    – ZDust
    May 26, 2015 at 15:45
  • $\begingroup$ @ZDust Take a look at the edited answer. $\endgroup$
    – Gaurav
    May 26, 2015 at 16:47

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