## Materials Science and Engineering: An Introduction

$µ_{e} = 0.50 \frac{m^{2}}{V-s}$ and, $µ_{h} = 0.14 \frac{m^{2}}{V-s}$
Given: electrical characteristics of intrinsic and p-type extrinsic gallium antimonide (GaSb) (refer to the given Table) Required: electron mobility and hole mobility Solution: Using Equation 18.13, $σ = n|e|µ_{e} + p|e|µ_{h}$ Substituting the given values for intrinsic material, $8.9 \times 10^{4} (Ω-m)^-1$ = ($8.7 \times 10^{23} m^{-3}) (1.602 \times 10^{-19} C) µ_{e}$ + $(8.7 \times 10^{23} m^{-3}) \times (1.602 \times 10^{-19} C)µ_{h}$ Simplifying, $0.64 = µ_{e} + µ_{h}$ Applying Equation 18.13 for extrinsic GaSb, $2.3 \times 10^{5} (Ω-m)^{-1}$ = ($7.6 \times 10^{22} m^{-3}) (1.602 \times 10^{-19} C) µ_{e}$ + $(1.0 \times 10^{25} m^{-3}) \times (1.602 \times 10^{-19} C)µ_{h}$ Simplifying, $18.89 = µ_{e} + 131.58 µ_{h}$ Using these two equations simultaneously: $0.64 = µ_{e} + µ_{h}$ $18.89 = µ_{e} + 131.58 µ_{h}$ $µ_{e} = 0.50 \frac{m^{2}}{V-s}$ and, $µ_{h} = 0.14 \frac{m^{2}}{V-s}$