Answer
$µ_{e} = 0.50 \frac{m^{2}}{V-s}$ and,
$µ_{h} = 0.14 \frac{m^{2}}{V-s}$
Work Step by Step
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,
$µ_{h} = 0.14 \frac{m^{2}}{V-s}$
