If \(n\), \(e\), \(\tau\), and \(m\) represent the concentration, the charge, the relaxation time, and the mass of an electron in a metal, then the resistance of a wire made of the metal of length \(l\) and cross-sectional area \(A\) is:
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Resistance \(R = \rho \frac{l}{A}\), and Drude model gives \(\rho = \frac{m}{n e^2 \tau}\)
From the Drude model of electrical conduction, the resistivity \(\rho\) of a metal is given by:
\[
\rho = \frac{m}{n e^2 \tau}
\]
Where:
- \(m\): mass of the electron
- \(n\): number of free electrons per unit volume
- \(e\): charge of an electron
- \(\tau\): relaxation time
The resistance \(R\) of a wire of length \(l\) and cross-sectional area \(A\) is:
\[
R = \rho \cdot \frac{l}{A} = \left(\frac{m}{n e^2 \tau}\right) \cdot \frac{l}{A} = \frac{m l}{n e^2 \tau A}
\]
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Approach Solution -2
Step 1: Understand the given parameters
- \(n\) = concentration of free electrons per unit volume
- \(e\) = charge of an electron
- \(\tau\) = relaxation time (average time between collisions)
- \(m\) = mass of an electron
- \(l\) = length of the wire
- \(A\) = cross-sectional area of the wire
Step 2: Recall the expression for conductivity \(\sigma\)
From Drude’s model, conductivity is:
\[
\sigma = \frac{n e^2 \tau}{m}
\]
Step 3: Recall the relation between resistance and conductivity
Resistance \(R\) is related to resistivity \(\rho\) by:
\[
R = \rho \frac{l}{A}
\]
and resistivity is inverse of conductivity:
\[
\rho = \frac{1}{\sigma}
\]