Consider a copper wire that is carrying, say, a few amperes of current. Is the drift speed ${\mathit{v}}_{\mathbf{d}}$of the conduction electrons that form that current about equal to, much greater than, or much less than the Fermi speed ${\mathit{v}}_{\mathbf{F}}$for copper (the speed associated with the Fermi energy for copper)?

A copper wire is considered.

A drift speed of conduction electron is the average velocity attained by the charged particles, such as electrons, in a material due to an electric field. In general, an electron propagates randomly at the Fermi velocity, resulting in an average velocity of zero.

Formulae:

The drift speed in a material,$v_d=\frac{i}{nAq}$ (i)

The Fermi speed of the given material,$v_F=\sqrt{\frac{2E_F}{m}}$ (ii)

The equation of Fermi energy is$E={\left[\frac{3}{16\sqrt{2\mathrm{\pi }}}\right]}^{2/3}\frac{h^2}{m}{n}^{2/3}$ (iii)

where n is the number of conduction electrons per unit volume, m is the mass of an electron and h is Planck’s constant.

When a potential difference is applied across a conductor, free electrons gain velocity in the direction opposite to the electric field. During the motion, they collide with other electrons, ions, and the walls of the conductor. Due to these collisions, the energy is lost and the speed of electrons is reduced. Thus, there is a definite small drift velocity of free electrons.

From equation (i), we can see that the drift speed is inversely proportional to the value of charge carrier density, n.

While using equation (iii) in equation (ii), we can see that the Fermi speed is proportional to the value of the conduction electrons density.

Thus, comparing both the cases, it can be stated that the drift speed is much less than the Fermi speed.