Electrons can be visualised as rattling around insde the metal, randomly changing direction when they collide with positive ions, much like a ball in a pinball machine. The average velocity and displacement of the electron along any direction over a period of time is zero because electrons are equally likely to be moving in any direction on their ‘random walk’.

Imagine that a conductor under investigation is part of an electrical circuit. Imagine closing a switch that connects a battery that applies an electric field E to the conductor in the positive x-direction. This exerts a force F = qE on each electron, which will accelerate according to Newton’s second law:

$$\begin{aligned} F \, &= qE \\ M_{e} a \, &= qE, \, \text{where} \, F = ma \, \text{and} \, M_{e} \, \text{is the mass of an electron}\\ a \, &= \frac{qE}{m_{e}} \, \rightarrow \text{Eqn 1} \end{aligned}$$

$$ \begin{aligned} v – u \, &= at, \, \text{where} \, u = 0 \, \text{as overall movement cancel out} \\ \bar{V_{f}} \, &= \alpha \tau, \, \text{where} \, \bar{V_{f}} \, \text{is the average velocity (final)} \\ \bar{V_{f}} \, &= a \frac{\lambda}{ < v >} \, \rightarrow \text{Eqn 2} \end{aligned}$$

, where

λ is the mean free path, (average distance that electrons travel between collisions)

τ is the average time interval between successive collisions,

< V > is the average speed of the electrons.

Typical Speed between collision (drift velocity) is: (by combining eqn 1 and 2)

$$\begin{aligned} V_{d} \, &= \bar{V_{f}} \\ V_{d} \, &= \frac{qE}{m} \left( \frac{\lambda}{ < v > } \right) \end{aligned}$$

THAANK YOU VERY MUCH! we have used so many things from you!!