A semiconductor has an electron concentration of 8 × 10¹³ per cm³ and a hole concentration of 5 × 10¹² per cm³. The electron mobility is 25000 cm²/Vs and the hole mobility is 100 cm²/Vs. Then,

ne=8 × 10¹³/cm³
nh=5 × 10¹²/cm³
μe=25000 cm²/(Vs)
μh=100 cm²/(Vs)
Since ne > nh, the semiconductor is n-type.
The conductivity of the semiconductor is
σ = e(neμe + nhμh) = 1.6 × 10⁻¹⁹((8 × 10¹³)(25000) + (5 × 10¹²)(100)) mho/cm = 0.32 mho/cm = 320 m mho/cm