Question

The resistivity of copper metal is \(1.7 \times 10^{-8} \, \Omega \, \text{m}\) and the atomic density of copper is \(8.5 \times 10^{28} \, \text{atoms/m}^3\). The collision time \(\tau\) for electrons in copper is:

• A. \(2.46 \times 10^{-14} \, \text{s}\)
• B. \(1.246 \times 10^{-14} \, \text{s}\)
• C. \(2.446 \times 10^{-14} \, \text{s}\)
• D. \(3.446 \times 10^{-14} \, \text{s}\)

Hint:

Use ρ =m/ne^2τ to relate resistivity, density, and collision time. Carefully substitute physical constants for accuracy.

Answer 1

The Correct Option is C

The collision time is calculated using:
\[\tau = \frac{m}{ne^2\rho}\]
where $m = 9.11 \times 10^{-31} \text{ kg}$ is the electron mass, $n = 8.5 \times 10^{28} \text{ atoms/m}^3$ is the electron density, $e = 1.6 \times 10^{-19} \text{ C}$ is the electron charge, and $\rho = 1.7 \times 10^{-8} \text{ }\Omega\text{m}$ is the resistivity.
Substituting:
\[\tau = \frac{9.11 \times 10^{-31}}{(8.5 \times 10^{28})(1.6 \times 10^{-19})^2(1.7 \times 10^{-8})} \approx 2.446 \times 10^{-14} \text{ s}\]