## Materials Science and Engineering: An Introduction

σ = 0.028 $(Ω*m)^{-1}$
Given: electron concentration = 5 x $10^{17}$ $m^{-3}$ electron drift velocity = 350 $\frac{m}{s}$ electric field = 1000 $\frac{V}{m}$ Required: conductivity (σ) Solution: To compute the conductivity of the material, Equation 18.16 will be used. σ = n|e|$µ_{e}$ Compute first the value of electron mobility using Equation 18.7: $µ_{e}$ = $\frac{V^{d}}{E}$ = $\frac{350 \frac{m}{s}}{1000 \frac{V}{m}}$ $µ_{e}$ = 0.35 $\frac{m^{2}}{V-s}$ Substituting the computed value in Equation 18.16; σ = $(5 \times 10^{17} m^{-3})(1.602 \times 10^{-19} C)(0.35 \frac{m^{2}}{V-s})$ σ = 0.028 $(Ω*m)^{-1}$