Drift of electrons and origin of resistivity

A copper wire glows to life the instant you flip a switch — the bulb lights almost immediately, as if the electrons inside had been given a starting gun. Yet if you could actually track a single free electron in that wire, you would find it crawling forward at a pace slower than a snail: about $0.02\ \text{cm}$ every second. So what exactly is racing down the wire at nearly the speed of light, and what is really doing the slow crawling? The answer is one of the cleanest microscopic pictures in all of school physics. Inside every metal there is a sea of free electrons already moving — wildly, randomly, at enormous thermal speeds of hundreds of kilometres per second, colliding constantly with the vibrating positive ions of the crystal lattice. Switch on a battery and nothing about that chaos changes very much; what changes is that the electric field itself appears everywhere in the wire almost instantly, and every electron now feels a tiny, steady nudge on top of its frantic zig-zagging. The bulb lights immediately because the field arrives immediately. The current, though, is carried by the excruciatingly slow drift of the whole electron sea — a net creep superimposed on the chaos. This single idea — random thermal motion plus a small field-driven drift, interrupted by frequent collisions with the lattice — does two jobs at once. It explains, from first principles, why a current flows at all when you connect a battery, and it explains why a wire resists that current: the same collisions that limit the drift velocity are exactly what we experience macroscopically as resistivity. Ohm's law stops being a rule you memorise and becomes a rule you can derive, in five short algebraic steps, from the motion of electrons and the timing of their collisions. That is the story this article tells, ending in the same numbers computed for copper that you may already know: a drift velocity of about $2.2\times10^{-4}\ \text{m/s}$ and a resistivity of about $1.7\times10^{-8}\ \Omega\text{m}$ .

Key takeaways

Free electrons in a metal move randomly and fast, but with zero average velocity until an electric field is applied — random thermal speeds are huge (~ $10^5\ \text{m/s}$ ), but they point every which way and cancel out.
An applied field superimposes a tiny net drift, $v_d = eE\tau/m$ , on top of the random motion, where $\tau$ is the average time between collisions (relaxation time).
Current follows directly from drift velocity: $I = nAev_d$ , linking the microscopic electron picture to the macroscopic ammeter reading.
Substituting $v_d$ into $I = nAev_d$ derives Ohm's law from first principles and identifies resistivity as $\rho = \dfrac{m}{ne^2\tau}$ .
Resistivity is literally a measure of how often electrons collide with the lattice — shorter $\tau$ (more frequent collisions) means higher $\rho$ .
Rising temperature shortens $\tau$ (more vigorous lattice vibrations, more collisions), which is the microscopic reason metallic resistance grows with temperature.
Key idea: current is the slow drift $v_d = eE\tau/m$ of a randomly-moving electron sea between collisions with the lattice, and substituting this drift into $I = nAev_d$ derives both Ohm's law and the microscopic origin of resistivity, $\rho = m/(ne^2\tau)$ .

The core insight

Electrons already move randomly and fast inside a metal; an applied field adds only a tiny drift $v_d = eE\tau/m$ between collisions, and turning that drift into current via $I = nAev_d$ shows resistivity is nothing but $\rho = m/(ne^2\tau)$ — a bookkeeping of how often electrons collide with the lattice.

Where it fits in the syllabus

This is a core conceptual topic inside the Current Electricity chapter of CBSE Class 12 Physics (2025-26 syllabus), sitting directly after the definition of current and Ohm's law and directly before mobility. The prescribed scope is exactly this: describe the free-electron model of conduction, define drift velocity $v_d$ and relaxation time $\tau$ , derive the microscopic current relation $I = nAev_d$ , and use it to derive the resistivity expression $\rho = m/(ne^2\tau)$ — including the qualitative explanation of why metallic resistivity rises with temperature. Nothing here strays outside the board syllabus: no mobility numerics (that is the next node), no superconductivity, no quantum treatment of the electron gas.

The physical picture

Picture the inside of a metal wire as a fixed lattice of positive ions, vibrating in place, immersed in a dense gas of free electrons — typically about one free electron per atom. Even with no battery connected, these electrons are not still. They dash about with high thermal speeds (of order $10^5\ \text{m/s}$ at room temperature), colliding with lattice ions roughly every $10^{-14}$ seconds and bouncing off in a new, essentially random direction each time. Because the directions are random and equally likely in every direction, if you add up the velocity vectors of all the free electrons in a small volume, the sum is zero — no net motion, no current, however furious the individual electrons are.

\vec{v}_\text{average} = 0 \quad \text{(no field applied)}

where: $\vec{v}_\text{average}$ is the vector average of the random thermal velocities of all free electrons in a small volume. This is a statement about directions cancelling, not about the electrons standing still — each one is moving very fast, just not in any preferred direction. Now close the switch. A battery sets up an electric field $E$ inside the wire, and — crucially — this field appears throughout the wire almost instantaneously (the field propagates near the speed of light along the conductor, which is why the bulb lights "immediately"). Every free electron, of charge magnitude $e$ and mass $m$ , now feels a steady force $F = eE$ and hence a steady acceleration:

a = \dfrac{eE}{m}

where: $E$ is the applied electric field inside the wire (V/m), $e$ is the elementary charge ( $1.6\times10^{-19}\ \text{C}$ ), and $m$ is the electron mass ( $9.1\times10^{-31}\ \text{kg}$ ). But this acceleration does not build up indefinitely — a collision with a lattice ion randomises the electron's velocity again, wiping out whatever extra speed it had picked up, before the process restarts. The electron accelerates, collides, restarts; accelerates, collides, restarts. The net effect, averaged over the whole electron sea, is a small constant creep in the direction opposite the field (electrons carry negative charge, so they drift against $E$ ), called the drift velocity $v_d$ .

Figure

Two panels comparing electron paths with no field and with an applied field, the first showing a jagged random zig zag path returning near its start, the second showing the same jagged zig zag but drifting steadily to the right under the field

If $\tau$ is the relaxation time — the average time between successive collisions — then the average extra velocity an electron carries just before its next collision, averaged over the whole population, is:

v_d = \dfrac{eE\tau}{m}

where: $v_d$ is the drift velocity, $E$ the field, $\tau$ the relaxation time (of order $10^{-14}\ \text{s}$ in a metal), $e$ the electron charge, and $m$ the electron mass. Physically this says the drift velocity is what you get if the electron accelerates freely for one average collision-interval and then has that velocity erased. It is proportional to $E$ (double the field, double the drift) and inversely proportional to $\tau$ 's reciprocal — more frequent collisions (smaller $\tau$ ) mean less time to accelerate between hits, hence smaller drift.

The physical picture (continued): why the bulb lights instantly

It is worth being explicit about the apparent paradox in the overview. The drift velocity computed above turns out to be of order $10^{-4}\ \text{m/s}$ — at that speed an electron would take hours to crawl the length of a household wire. But the light comes on in a fraction of a second because it is the electric field, not any individual electron, that has to reach the filament, and the field is established throughout the circuit at a speed close to the speed of light. Every free electron in the wire — not just those near the switch — starts drifting together, almost simultaneously, the moment the field arrives. The current is not "electrons travelling from the switch to the bulb"; it is the whole electron sea shifting together, everywhere at once, at a very slow individual pace.

Derivation ADeriving Ohm's law and resistivity from the drift-velocity model

The goal of this derivation is to start purely from the microscopic behaviour of a single free electron between collisions — its acceleration under the field $E$ and the average time $\tau$ before that motion is randomised again — and build up, step by step, to two macroscopic results that a voltmeter and ammeter would actually measure: the linear relationship between current and voltage that we call Ohm's law, and an explicit formula for resistivity $\rho$ purely in terms of microscopic quantities ( $n$ , $e$ , $m$ , $\tau$ ). This is the derivation that upgrades Ohm's law from an empirical rule to a consequence of electron dynamics, and it is the one result in this chapter you are expected to reproduce from scratch.

Step 1. Write the drift velocity of a single electron. Under a field $E$ , an electron of charge magnitude $e$ and mass $m$ accelerates at $a = eE/m$ . If it starts each free flight with (on average) zero velocity component along the field — because the previous collision randomised its direction — then just before its next collision, after an average time $\tau$ , it has picked up an average extra velocity $v_d = a\tau = \dfrac{eE\tau}{m}$ . This $v_d$ , averaged over the entire electron population, is the drift velocity of the electron sea as a whole.

Step 2. Relate drift velocity to current. Consider a conductor of cross-sectional area $A$ carrying $n$ free electrons per unit volume. In a small time interval $\Delta t$ , each electron advances a distance $v_d\Delta t$ along the wire, so every electron inside a cylindrical volume $A\,v_d\,\Delta t$ crosses a given cross-section during that interval. The number of such electrons is $n(Av_d\Delta t)$ , and each carries charge $e$ , so the charge crossing the section in time $\Delta t$ is $\Delta Q = neAv_d\Delta t$ . Dividing by $\Delta t$ gives the current: $I = \dfrac{\Delta Q}{\Delta t} = nAev_d$ .

Step 3. Substitute the microscopic drift velocity from Step 1 into the current expression from Step 2. $I = nAe\left(\dfrac{eE\tau}{m}\right) = \dfrac{nAe^2\tau}{m}\,E$ . This already has the shape of Ohm's law — current directly proportional to field — but it needs to be rewritten in terms of $V$ and $R$ to match the circuit-level statement of the law.

Step 4. Convert field to voltage and read off resistance. For a conductor of length $L$ with a uniform field, $E = V/L$ . Substituting, $I = \dfrac{nAe^2\tau}{mL}\,V$ , which rearranges to $V = \left(\dfrac{mL}{nAe^2\tau}\right)I$ . Comparing with $V = IR$ identifies the resistance as $R = \dfrac{mL}{nAe^2\tau}$ — a constant (for fixed $n$ , $\tau$ at a given temperature), which is exactly the statement of Ohm's law: current is proportional to voltage with a temperature-dependent but voltage-independent proportionality constant $1/R$ .

Step 5. Extract resistivity and state the physical origin. Since $R = \rho L/A$ for a uniform conductor, comparing with $R = \dfrac{mL}{nAe^2\tau}$ from Step 4 gives directly $\rho = \dfrac{m}{ne^2\tau}$ . This is the key result of the whole model: resistivity is not an independent mystery property of a material, it is completely determined by the electron density $n$ , the electron's charge and mass ( $e$ , $m$ , both universal constants), and the relaxation time $\tau$ — the average time between an electron's collisions with the lattice. A metal resists current because its electrons keep colliding with the vibrating lattice ions; the more often they collide (smaller $\tau$ ), the larger the resistivity. Since heating a metal makes the lattice ions vibrate more violently, collisions become more frequent, $\tau$ shrinks, and $\rho = m/(ne^2\tau)$ predicts — correctly — that a metal's resistivity rises with temperature.

Worked examples

Board level - drift velocity in a copper wire

A copper wire has free-electron density $n = 8.5\times10^{28}\ \text{m}^{-3}$ and cross-sectional area $A = 1\times10^{-6}\ \text{m}^2$ . Find the drift velocity of the electrons when a current of $3\ \text{A}$ flows through the wire.

Step 1. Write down the microscopic current relation and solve for $v_d$ : from $I = nAev_d$ , $v_d = \dfrac{I}{nAe}$ .

Step 2. Substitute the numbers. $v_d = \dfrac{3}{(8.5\times10^{28})(1\times10^{-6})(1.6\times10^{-19})}$ .

Step 3. Simplify the denominator in stages: $8.5\times10^{28}\times1\times10^{-6} = 8.5\times10^{22}$ ; then $8.5\times10^{22}\times1.6\times10^{-19} = 1.36\times10^{4}$ .

Step 4. Divide. $v_d = \dfrac{3}{1.36\times10^{4}} \approx 2.2\times10^{-4}\ \text{m/s}$ . Answer: the drift velocity is about $2.2\times10^{-4}\ \text{m/s}$ , roughly $0.02\ \text{cm}$ per second — confirming that drift is astonishingly slow next to how quickly the bulb responds, because it is the field, not the electrons, that arrives quickly.

Competitive level - resistivity of copper from the free-electron model

For the same copper wire ( $n = 8.5\times10^{28}\ \text{m}^{-3}$ ), the relaxation time is $\tau = 2.5\times10^{-14}\ \text{s}$ and the electron mass is $m = 9.1\times10^{-31}\ \text{kg}$ . Estimate the resistivity of copper using the free-electron model, and then find the electric field needed to drive the same $3\ \text{A}$ current found above through this wire if its length is $2\ \text{m}$ .

Step 1. Compute $e^2$ first, since it appears squared in the resistivity formula: $e^2 = (1.6\times10^{-19})^2 = 2.56\times10^{-38}\ \text{C}^2$ .

Step 2. Build up the denominator $ne^2\tau$ in stages to avoid an arithmetic slip: $ne^2 = (8.5\times10^{28})(2.56\times10^{-38}) = 2.176\times10^{-9}$ ; then $ne^2\tau = (2.176\times10^{-9})(2.5\times10^{-14}) = 5.44\times10^{-23}$ .

Step 3. Apply $\rho = \dfrac{m}{ne^2\tau} = \dfrac{9.1\times10^{-31}}{5.44\times10^{-23}} \approx 1.67\times10^{-8}\ \Omega\text{m}$ , matching the textbook resistivity of copper ( $\approx1.7\times10^{-8}\ \Omega\text{m}$ ) closely — a strong internal check that the model and the relaxation time quoted are consistent.

Step 4. Use $\rho$ to find $R$ for the given wire, then $V$ , then $E = V/L$ . $R = \rho L/A = (1.67\times10^{-8})(2)/(1\times10^{-6}) = 3.34\times10^{-2}\ \Omega$ . With $I = 3\ \text{A}$ , $V = IR = 3\times3.34\times10^{-2} \approx 0.10\ \text{V}$ , so $E = V/L = 0.10/2 = 0.05\ \text{V/m}$ . Answer: $\rho \approx 1.67\times10^{-8}\ \Omega\text{m}$ , and a field of only about $0.05\ \text{V/m}$ is needed — consistent with how small a battery voltage drives an easily-detectable current in a good conductor. Timing note: in a numerical-value or assertion-reason question under time pressure, do not recompute $\rho$ from scratch if the value has already been given or asked for in an earlier part — the whole point of a multi-part competitive question is to chain Step 3's result straight into Step 4 without repeating powers-of-ten arithmetic; budget under two minutes for the full chain by pre-simplifying $ne^2\tau$ mentally in one pass rather than three separate multiplications.

Common traps

Confusing drift velocity with the electrons' actual (thermal) speed — drift velocity ( $\sim10^{-4}\ \text{m/s}$ ) is a tiny net creep riding on top of a much larger random thermal speed ( $\sim10^5\ \text{m/s}$ ); the two are different quantities with different orders of magnitude, and a question asking for one should never be answered with the other.
Believing electrons themselves travel the length of the wire in the time it takes a bulb to light — it is the electric field that establishes itself almost instantly through the wire; the electrons are still creeping at drift-velocity pace throughout.
Getting the resistivity formula upside down — it is $\rho = \dfrac{m}{ne^2\tau}$ , not $\rho = \dfrac{ne^2\tau}{m}$ . Sanity-check with the physics: a larger $\tau$ means fewer collisions, which should give lower resistance, and only $\rho = m/(ne^2\tau)$ (where larger $\tau$ sits in the denominator) gives that direction correctly.
Forgetting the negative sign is a direction-only subtlety — because electron charge is negative, electrons physically drift opposite to $E$ , but conventional current $I = nAev_d$ uses the magnitude $e$ and is taken along $E$ ; do not introduce a stray minus sign into the current formula itself.
Treating $\tau$ as if it were the electron's total travel time between two ends of the wire — $\tau$ is the average time between two successive collisions, a microscopic quantity of order $10^{-14}\ \text{s}$ , unrelated to how long the electron takes to traverse the whole conductor.

Common trapDrift velocity is not the speed of the current signal — the electric field, not the electron, is what reaches the bulb quickly

It is tempting to picture current as electrons physically racing from the battery to the bulb the instant the switch closes. They do not — an individual electron drifts at only about $10^{-4}\ \text{m/s}$ and would take a very long time to physically cross a household wire. What actually happens almost instantly (near the speed of light) is that the electric field is established throughout the entire circuit, so every free electron everywhere in the wire begins drifting together at essentially the same moment. The bulb lights quickly because the whole electron sea starts moving together quickly, not because any single electron sprints from the switch to the filament.

You should be able to…

Describe the free-electron model of conduction in a metal, including why random thermal motion alone carries no net current.
Define drift velocity $v_d$ and relaxation time $\tau$ , and state (and derive) $v_d = eE\tau/m$ .
Derive and use the microscopic current relation $I = nAev_d$ to connect the electron picture to the ammeter reading.
Reproduce the full derivation from $v_d$ and $I = nAev_d$ through to $\rho = m/(ne^2\tau)$ , without skipping the $E = V/L$ and $R = \rho L/A$ substitution steps.
Explain, using $\tau$ , why the resistivity (and hence resistance) of a metal increases with temperature.
Correctly distinguish drift velocity, thermal velocity, and the speed at which the electric field itself is established in the wire.

Ask the tutor: Want to work through more drift-velocity or resistivity numbers, or see exactly how this same model leads into mobility next? Ask the AI tutor for guided practice.

If electrons move so slowly, how does electricity travel down a wire so fast?

Two different things are happening at two different speeds, and mixing them up is the single most common confusion in this topic. The individual electrons drift at only about $10^{-4}\ \text{m/s}$ — genuinely slower than a snail. But the electric field that pushes them is established throughout the entire wire almost instantly, at a speed close to the speed of light, because it is an electromagnetic disturbance propagating through the circuit, not a particle physically travelling along it. So every free electron in the wire, not just those near the switch, starts drifting together the moment the field arrives. What travels fast is the signal to start moving; what moves slowly is the electron itself.

Why does the collision with the lattice "reset" the electron's velocity instead of just slowing it down a bit?

On average, a collision with a lattice ion sends the electron off in a direction that is essentially random with respect to where it was heading — not always backwards, just uncorrelated with the field direction. When you average over a huge number of such random post-collision directions, the field-direction component of that "leftover" velocity averages to zero. That is exactly why we say the electron effectively "restarts" its acceleration from zero along the field each time: not because the collision perfectly stops it, but because whatever component it keeps is, on average across all electrons, unrelated to the applied field and so contributes nothing to the net drift.

Is tau the same for every electron, or an average?

It is a statistical average, not a fixed number for any one electron. In reality, individual electrons collide at different, essentially random intervals — some sooner, some later than average — because the exact instant of each collision depends on the electron's chaotic thermal path through a lattice of vibrating ions. The relaxation time $\tau$ used in $v_d = eE\tau/m$ is the average time between collisions across the whole population of free electrons in the metal, and it is this average that correctly predicts the average drift velocity, current, and resistivity — the formulas describe the electron sea collectively, not any single electron's trajectory.

How is this different from just measuring resistivity with a multimeter — why bother with the electron picture at all?

A multimeter gives you the macroscopic number $\rho$ for a given material, but it cannot tell you why different metals have different resistivities, or why resistivity depends on temperature the way it does. The drift model answers both: $\rho = m/(ne^2\tau)$ shows resistivity depends on how many free electrons a metal has per unit volume ( $n$ ) and how often those electrons collide with the lattice ( $1/\tau$ ). Two metals with different $n$ or different collision rates will have different $\rho$ , and heating any metal shortens $\tau$ (more lattice vibration, more collisions), which is exactly why resistance rises with temperature — a fact you can only explain, not just tabulate, once you have the microscopic picture.