Electric current > Conductivity. Resistance. Sources of electromotive force > Problem 8.2.23

Statement

$8.2.23.$

Why is the kinetic energy of current carriers associated with their ordered
motion not taken into account when considering the electric current in a sub-
stance?

Estimate the kinetic energy of one electron (in electron volts) at a
current of I = 100 A in a sodium wire with a cross-section of S = 1 mm2The
number of conduction electrons per unit volume of the wire ne= 2.5·10^22cm^−3.

Solution

Let's start with the first part: Why is the kinetic energy of ordered motion not considered in electric current?

When a potential difference is established, the free electrons in a conductor acquire a small net drift velocity v_d superimposed on their random thermal motion. The thermal velocity of electrons at room temperature is on the order of 10^5\ \text{m/s}, corresponding to a thermal kinetic energy on the order of kT \approx 0.025\ \text{eV}. In contrast, the drift velocity is several orders of magnitude smaller.

The kinetic energy associated with ordered motion is proportional to v_d^2, so it is extremely small compared with thermal energy and, above all, with the Fermi energy in metals (several eV). Consequently, when studying the energy balance of a circuit, the contribution of this ordered kinetic energy is absolutely negligible compared with the interaction energies and the energy dissipated by the Joule effect. For this reason, models ignore this small net kinetic energy and focus on the charge flow and collision losses.

Second part: Estimation of the kinetic energy of one electron

Data:

Current:$ I = 100\ \text{A}$
Cross‑section of the sodium wire:$ S = 1\ \text{mm}^2 = 1 \times 10^{-6}\ \text{m}^2$
Density of conduction electrons: $n_e = 2.5 \times 10^{22}\ \text{cm}^{-3} = 2.5 \times 10^{28}\ \text{m}^{-3}$
Elementary charge: $e = 1.6 \times 10^{-19}\ \text{C}$
Electron mass: $m_e = 9.11 \times 10^{-31}\ \text{kg}$

Drift velocity
The current is related to the drift velocity by:

$I = e\, n_e\, S\, v_d$

Solving for $v_d$

$v_d = \frac{I}{e, n_e, S}
= \frac{100}{(1.6\times10^{-19}) \times (2.5\times10^{28}) \times (10^{-6})}
= \frac{100}{4.0 \times 10^3}
\approx 0.025\ \text{m/s} = 2.5\ \text{cm/s}$

Kinetic energy of one electron

The kinetic energy is:

$K = \frac{1}{2} m_e v_d^2
= \frac{1}{2} \times 9.11\times10^{-31} \times (0.025)^2
\approx \frac{1}{2} \times 9.11\times10^{-31} \times 6.25\times10^{-4}
\approx 2.85 \times 10^{-34}\ \text{J}$

Conversion to electron‑volts

$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$

Then:

$K = \frac{2.85 \times 10^{-34}}{1.6 \times 10^{-19}} \approx 1.8 \times 10^{-15}\ \text{eV}$

Rounding, we get:

$\boxed{K \approx 2 \times 10^{-15}\ \text{eV}}$

Answer

$\boxed{K \approx 2 \times 10^{-15}\ \text{eV}}$