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

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.

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 $$\rho=\frac{1}{\sigma}=\frac {m^*}{nq^2 \tau}$$
• @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. Feb 1, 2018 at 5:13