Question

What do you understand by free electron and drift velocity? Deduce Ohm's law on the basis of the principle of drift velocity.

Hint:

Free electrons drift with \(v_d = eE\tau/m\); use \(I = neAv_d\) and \(E=V/l\) to reach \(V=IR\) with \(R\) a constant.

Answer 1

Step 1: Free electron.
In a metal the outermost (valence) electrons of the atoms are very loosely bound. They break away from their parent atoms and move randomly throughout the whole metal like the molecules of a gas. These loosely bound, freely moving electrons are called free electrons (or conduction electrons). Their number density \(n\) is very large, of the order of \(10^{28}\) to \(10^{29}\) per cubic metre.

Step 2: Drift velocity.
Without a field the free electrons move randomly, so their average velocity is zero. When a potential difference is applied, an electric field \(E\) acts on them and they gain a small steady average velocity opposite to the field. This small average velocity acquired by the free electrons under the applied electric field is called the drift velocity \(v_d\).

Step 3: Expression for drift velocity.
Force on an electron \(= eE\), so its acceleration \(a = \dfrac{eE}{m}\). If \(\tau\) is the average time between two successive collisions (relaxation time), then
\[ v_d = a\tau = \frac{eE\tau}{m} \]

Step 4: Relate current to drift velocity.
For a conductor of area of cross-section \(A\), the current is
\[ I = neAv_d \]
Substituting \(v_d\):
\[ I = neA\left(\frac{eE\tau}{m}\right) = \frac{ne^2 A \tau}{m}\,E \]

Step 5: Introduce the potential difference and deduce Ohm's law.
For a wire of length \(l\), \(E = \dfrac{V}{l}\). Then
\[ I = \frac{ne^2 A \tau}{m}\cdot\frac{V}{l} \]
Rearranging,
\[ \frac{V}{I} = \frac{m\,l}{ne^2 \tau A} = R \]
At constant temperature \(n, \tau, m, l, A\) are all constant, so \(R\) is a constant. Hence
\[ V = IR \quad\Rightarrow\quad V \propto I \]
which is Ohm's law.

\[\boxed{V = IR,\quad R = \dfrac{m\,l}{ne^2 \tau A}}\]