Question

In copper at room temperature, the mobility of mobile electrons is about $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\raisebox{1ex}{$(m/s)$}\!\left/ \!\raisebox{-1ex}{$\mathbf{V}\mathbf{/}\mathbf{m}$}\right.$,and there are about$\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$mobile electrons per${\mathbf{m}}^{\mathbf{3}}$Calculate the conductivity$\mathit{\sigma }$and include the correct units. In actual practice, it is usually easier to measure the conductivity$\mathit{\sigma }$and deduce the mobility$\mathit{u}$from this measurement.

Short answer

$57.6\times {10}^6\mathrm{C}·{\mathrm{m}}^2/\mathrm{V}·\mathrm{s}$

Step-by-step solution

Given Data

Mobility of electrons, $\mu =4.5\times {10}^{-3}\raisebox{1ex}{$\left(\mathrm{m}/\mathrm{s}\right)$}\!\left/ \!\raisebox{-1ex}{$\left(\mathrm{V}/\mathrm{m}\right)$}\right.$

Number of electrons per${\mathrm{m}}^3,n=8\times {10}^8$

Concept

The conductivity is defined as the current flowing through a unit cross-sectional area.

Calculate the conductivity

Conductivity,

$\begin{array}{rcl}\sigma & =& \left(n\right)\left(e\right)\left(\mu \right)\left(\mathrm{where},e\mathrm{is}\mathrm{charge}\mathrm{of}\mathrm{electrons}(1.6\times {10}^{-19}\mathrm{C}\right)\\ & =& \left(8\times {10}^{28}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(4.5\times {10}^{-3}\right)\\ & =& 57.6\times {10}^6\mathrm{C}·{\mathrm{m}}^2/\mathrm{V}·\mathrm{s}\end{array}$

Hence, the conductivity is $57.6\times {10}^6\mathrm{C}·{\mathrm{m}}^2/\mathrm{V}·\mathrm{s}$