Question

As per the Drude model of metals, the electrical resistance of a metallic wire of length \( L \) and cross-sectional area \( A \) is:
(Consider \( \tau \) as the relaxation time, \( m \) as electron mass, \( n \) as carrier concentration, and \( e \) as electronic charge)

• A. \( \frac{mL}{n e^2 A \tau} \)
• B. \( \frac{2mL}{n e^2 A \tau} \)
• C. \( \frac{mL}{2n e^2 A \tau} \)
• D. \( \frac{mL}{4n e^2 A \tau} \)

Hint:

For the Drude model of metals, the electrical resistance depends on the electron mass, the relaxation time, and the carrier concentration.

Answer 1

The Correct Option is A

To derive the electrical resistance using the Drude model, we use the following relationship: 1. The electrical resistance \( R \) is given by: \[ R = \frac{\rho L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire. 2. In the Drude model, the resistivity \( \rho \) is defined as: \[ \rho = \frac{m}{n e^2 \tau} \] where: - \( m \) is the electron mass, - \( n \) is the carrier concentration, - \( e \) is the electronic charge, and - \( \tau \) is the relaxation time (mean time between collisions). 3. Now substitute this expression for \( \rho \) into the resistance formula: \[ R = \frac{\frac{m}{n e^2 \tau} L}{A} \] 4. Simplifying the expression, we get: \[ R = \frac{m L}{n e^2 A \tau} \] Thus, the resistance of the metallic wire is \( \frac{m L}{n e^2 A \tau} \).