Does the number of electrons colliding in a wire get doubled when length of wire is made twice with its area of cross-section remain constant.

My calculations for this are:-

For wire of length $L$ and area of cross-section:

Let the average distance after travelling which electron will collide $=d$

The number of free electrons in wire $=n $

Time taken between two successive collisions of electron $=τ$

Electric field across it $=E_1$

And Voltage apllied across wire $=V$

Average velocity of electron $= E_1eτ/2m$ The number of collision of electron in time $t = E_1eτtn/2md$

When using a wire of length 2L but with same area of cross-section and same voltage apllied across it:

Number of free electrons $ =2n$

Electric field inside wire $= E_2=V/2L=E_1/2$

Average velocity of electrons $= E_2eτ/2m=E_1eτ/2×2m$

And at last, the number of collisions $= E_2eτt×2n/2md=E_1eτtn/2md$

We can see here that number of collision remains same when length of wire is changed.

*Edit: There are lot of errors above. So I am writing correct version down( correct in my sense)

For wire of length $L$:

The number of free electrons in wire $=n$

Time taken between two successive collisions of electron (it is constant for a material for reason I don't know)$=τ$

Electric field across it $=E_1$

And Voltage apllied across wire $=V$

The number of collision of electron in time $t = tn/ \tau $

When using a wire of length 2L and area of cross section A but with same area of cross-section and same voltage apllied across it:

Number of free electrons $ =2n $

Average velocity of electrons $= t2n/τ$

Finally, why is the $\tau$ uneffected by Electric field, $E = V/L$? ($L$is the length of wire).

  • $\begingroup$ The electric field, more precise the electric potential difference between the ends of the wire has its origin in the source, a battery or an electric generator. The potential difference does not changes with different lengths of the wire. $\endgroup$ Jan 5, 2022 at 5:34
  • $\begingroup$ Yeah you can see that i have used this fact in my derivation of number of collision. Can you please focus on the answer needed for question $\endgroup$ Jan 6, 2022 at 15:35
  • $\begingroup$ Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! $\endgroup$
    – jng224
    Jan 7, 2022 at 9:59
  • $\begingroup$ How to write proportional? $\endgroup$ Jan 8, 2022 at 8:22
  • 1
    $\begingroup$ \propto renders as $\propto$ $\endgroup$
    – Shub
    Jan 8, 2022 at 9:59

1 Answer 1


I think that the main observation is that in a typical metal the velocity of the conduction electrons (the Fermi velocity) is much, much bigger than the drift velocity. The Fermi velocity is on the order of $10^6$ m/s. The drift velocity depends on many details, like the mobility and the actual field, but a reasonable number is $\sim 1$ m/s. This means that the mean free time and the mean free path are completely unaffected by the field, the size of the wire, etc. The same goes for the number of collisions per unit time and conduction electron. This means that if you double the sample, you get twice as many collisions.

Postcript: Consider the case without a field. There is a distribution function $f(\vec{v},\vec{x},t)$ for the electrons in the metal. The distribution function is isotropic, and the mean velocity and the current are zero. In a typical metal the distribution is close a sharp Fermi distribution, so that states below the Fermi velocity are occupied, and states above the Fermi velocity are empty. The Fermi edge is smeared out because of thermal effects. The mean (absolute) velocity of electrons that scatter is close to the Fermi velocity $v_F$. In a typical metal the Fermi velocity is large, on the order of $10^6$ m/s.

Now we turn on a field $E$. This shifts the equilibrium distribution $f\to f+\delta f$. The distribution $\delta f$ is anisotropic. In particular, the mean velocity is $\vec{v}_d$, called the drift velocity. The drift velocity is proportional to the electric field, $\vec{v}_d=\gamma/e \vec{E}$ ($\gamma$ is called the mobility). The mean current is also not zero, $\vec{\jmath}=\sigma \vec{E}$ ($\sigma$ is the conductivity).

The point is that $\delta f\ll f$, and $|\vec{v}_d|\ll v_F$, despite the fact that the total current is governed by $\vec{v}_d$. As a consequence the collision rate is determined by $f$, and is approximately idependent of $E$.

  • $\begingroup$ Can i get any calculation showing that doubling the volume doubles the number of collision? $\endgroup$ Jan 11, 2022 at 16:38
  • $\begingroup$ @PredakingAskboss I'm not sure that there is much of a calculation to be done. If the collision rate is independent of the current, the drift velocity, etc (only a function of the material, doping, temperature etc), then the total number of collisions is just proportional to the number of conduction electrons. Twice bigger sample, twice as many collisions. $\endgroup$
    – Thomas
    Jan 11, 2022 at 17:16
  • $\begingroup$ Why is drift veocity taken as $1m/s$? This velocity is dependent on magnitude of E and since $\tau$ is constant it should get double. $\endgroup$ Jan 31, 2022 at 22:17
  • $\begingroup$ @PredakingAskboss Both velocities are order of magnitude estimates. Yes, the drift velocity scales with E, but the point is that it is always much smaller than the Fermi velocity. The value of the drift velocity is irrelevant to the estimate of the collision rate, which is (to a very good approximation) independent of E. $\endgroup$
    – Thomas
    Feb 1, 2022 at 5:50
  • $\begingroup$ Well if drift velocity is fixed for a given wire then how current gets double when voltage provided also gets double? Assume that $v_d$ is the average velocity of all electrons in wire when Voltage is provided. $ I= v_d A n e$ .So when you double the voltage, what actually gets double? $A$ and $n$ are constant for a wire, also e is constant. The only thing that can change is $v_d$. $\endgroup$ Feb 3, 2022 at 22:05

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