Question:

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 \(\rho = \frac{m}{n e^2 \tau}\)
Updated On: May 19, 2025
  • \(\frac{m l}{n e^2 \tau A}\)
  • \(\frac{m \tau^2 A}{n e^2 l}\)
  • \(\frac{n e^2 \tau A}{2 m l}\)
  • \(\frac{n e^2 A}{2 m \tau l}\)
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The Correct Option is A

Approach Solution - 1

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:
\[ \rho = \frac{1}{\sigma} \]

Step 4: Substitute and simplify
\[ R = \frac{1}{\sigma} \frac{l}{A} = \frac{m}{n e^2 \tau} \frac{l}{A} = \frac{m l}{n e^2 \tau A} \]

Step 5: Final answer
The resistance of the wire is \(\frac{m l}{n e^2 \tau A}\).
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