Question

Define drift velocity and mobility of electrons in a metallic conductor. The length of a conducting rod is 1 m and the potential difference between its ends is 4 volt. Electron density in the conductor is \( 5 \times 10^{24}~\text{m}^{-3} \) and its resistivity is \( 50 \times 10^{-8}~\Omega\)-m. Calculate the drift velocity of the electrons in the metal.

Hint:

Drift velocity \(v_d\) can be calculated using \(v_d = \frac{E}{n e \rho}\), where \(E\) is the electric field, \(n\) is the electron density, \(e\) is the electron charge, and \(\rho\) is the resistivity.

Answer 1

Drift Velocity and Mobility of Electrons:
- Drift velocity (\(v_d\)) is the average velocity of electrons in a conducting material under the influence of an electric field.
- Mobility (\(\mu\)) refers to the speed of an electron in a material when exposed to an electric field. It is given by the relation:
\[ v_d = \mu . E \] Where \(v_d\) is the drift velocity, \(\mu\) is the mobility, and \(E\) is the electric field.
For a conductor, the drift velocity can also be related to the current, electron density, and cross-sectional area.
Step 1: Calculate the Electric Field
The electric field \(E\) in the conductor is given by:
\[ E = \frac{V}{L} \] Where \(V\) is the potential difference and \(L\) is the length of the conductor.
\[ E = \frac{4~\text{V}}{1~\text{m}} = 4~\text{V/m} \] Step 2: Apply Ohm’s Law
The current \(I\) in the conductor is related to the resistivity (\(\rho\)), the current density (\(J\)), and the electric field (\(E\)) by:
\[ J = \sigma E = \frac{1}{\rho} E \] Where \(\sigma\) is the electrical conductivity and \(\rho\) is the resistivity of the material. The current density \(J\) is also related to the drift velocity and the electron density by:
\[ J = n e v_d \] Where: - \(n\) is the electron density (\(5 \times 10^{24}~\text{m}^{-3}\)), - \(e\) is the electron charge (\(1.6 \times 10^{-19}~\text{C}\)), - \(v_d\) is the drift velocity.
Equating both expressions for \(J\):
\[ \frac{E}{\rho} = n e v_d \] Step 3: Calculate the Drift Velocity
Rearranging the equation to solve for \(v_d\):
\[ v_d = \frac{E}{n e \rho} \] Substitute the known values: - \(E = 4~\text{V/m}\), - \(n = 5 \times 10^{24}~\text{m}^{-3}\), - \(e = 1.6 \times 10^{-19}~\text{C}\), - \(\rho = 50 \times 10^{-8}~\Omega\)-m.
\[ v_d = \frac{4}{(5 \times 10^{24}) (1.6 \times 10^{-19}) (50 \times 10^{-8})} \] \[ v_d = \frac{4}{(5 \times 1.6 \times 50 \times 10^0) \times 10^6} = \frac{4}{4 \times 10^6} = 10^{-6}~\text{m/s} \] Thus, the drift velocity of the electrons is:
\[ v_d = 1 \times 10^{-6}~\text{m/s} \]