# How do electrons actually move in a wire? [duplicate]

This question already has an answer here:

Do they jump from atom to atom or are they free-flowing? Where does resistance fit in? Do electrons physically HIT the atoms? If so, how do they hit atoms if the nucleus is small and far away from the electron cloud? What makes something more resistive than something else? Is it simply a greater density of atoms, so more obstacles in the way of the electrons? I am trying to fully understand exactly what is going on.

## marked as duplicate by probably_someone, Chris♦, Kyle Kanos, Jon Custer, stafusaFeb 1 '18 at 20:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

The conduction electrons in a metal can be considered to be free. They are not bound to a particular atom. Electrons in a metal are accelerated by an applied electric field, like in a wire, between scattering events with phonons, the quantized vibrations of the crystal lattice of the metal. By these very frequent scattering events, the electrons give up kinetic energy and momentum to the crystal so that on the average an electron (an thus also an ensemble of electrons) reaches a mean velocity $v$, the so-called drift-velocity, which is proportional to the electric field $E$ $$v=\mu E$$ $\mu$ is the electron mobility which is material specific and depends also on temperature. The differences of resistivity between different metals arise from variations in these phonon scattering processes expressed by a mean scattering time $\tau$, differences in the effective mass $m^*$ of the conduction electrons near the Fermi energy, and differences in conduction electron density $n$.

Phenomenologically one can express the mobility $\mu$ by the electron charge $q$, the effective mean scattering time $\tau$, and the effective electron mass $m*$ $$\mu=\frac{q \tau}{m^*}$$ This gives the specific conductivity $$\sigma=n q \mu=\frac{n q^2 \tau}{m^*}$$ where $n$ is the conduction electron density of the metal. The specific resistivity $\rho$ of the metal is thus $$\rho=\frac{1}{\sigma}=\frac {m^*}{nq^2 \tau}$$

PS: In these scattering processes, the electrons do not directly "hit" the atoms , much less their nuclei. The scattering is caused by the thermally produced deviations (vibrations called phonons) from the perfect alignment of the crystal lattice in which an electron would normally move (quantum mechanically) without resistance.

• So, is it basically like a vibrational wave in the wire that reduces the kinetic energy and momentum of the electrons? So, in a resistor, these vibrational waves are stronger, thus reducing the overall electron velocity even more than in a conductor? – Lew Rod Feb 1 '18 at 0:59
• @LewRod- there are many, many vibrational waves impeding the movement of the electrons. Depending on the type, a resistor can just be made of a long metal wire, or of some other higher resistivity material. – freecharly Feb 1 '18 at 1:13
• Then how and why do resistors and wires heat up when more current flows through them? If the electrons do not hit the atoms, how do the resistor's or wire's atoms obtain more kinetic energy and become hotter? By the way, thank you very much for your response. I will upvote it once I have 15 reputation. – Lew Rod Feb 1 '18 at 2:04
• @LewRod - The scattering of the electrons with the lattice vibrations transfers kinetic energy from the electrons to the crystal lattice, increasing the lattice vibrations. These irregular lattice (atom) vibrations are, in essence, the heat energy of the crystal. Thus the wire heats up due to the flow of electrons through it. Each electron passing the wire loses its potential energy $qV$ after stepwise conversion into kinetic energy to the crystal lattice of the wire as heat energy. The more electrons flow, the larger the electric current and the heating of the wire. – freecharly Feb 1 '18 at 5:13