What is the conductivity of a semiconductor if electron density = $5 × 10^{12}/cm^3$ and hole density = $8 × 10^{13}/cm^3$ $(μ_e = 2.3 m^2 V^{–1} s^{–1}, μ_h = 0.01 m^2V^{–1} s^{–1})$
Answer & explanation
Correct answer: option 2
Given: $μ_e = 2.3 m^2 V^{–1} s^{–1}$
$μ_h = 0.01 m^2V^{–1} s^{–1}, n_e=5 × 10^{12}/cm^3$
$=5 × 10^{18}n_h=8× 10^{13}/cm^3=8× 10^{19}/m^3$
Conductivity $σ=e[n_eμ_e + n_hμ_h]$
$=16× 10^{-19}[5 × 18^{18}×2.3+8×10^{19}×0.01]$
$=16× 10^{-1}[11.5+0.8]$
$=16× 10^{-1}×12.3=1.968Ω^{-1}m^{-1}$