Question

an n-type semiconductor is known to have an electron concentration of 5.63 x 1019 m-3. if the electron drift velocity is 113 m/s in an electric field of 510 v/m, calculate the conductivity [in (ω-m)-1] of this material.

Answer 1

Answer:

$1.99581248\ /\Omega m$

Explanation:

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Electron concentration = $5.63\times 10^{19}\ /m^{3}$

$v_d$ = Drift veloctiy = 113 m/s

E = Electric field = 510 V/m

Electron mobility is given by

$\mu=\dfrac{v_d}{E}\\\Rightarrow \mu=\dfrac{113}{510}\\\Rightarrow \mu=0.22156\ m^2/Vs$

Conductivity is given by

$\sigma=ne\mu\\\Rightarrow \sigma=5.63\times 10^{19}\times 1.6\times 10^{-19}\times 0.22156\\\Rightarrow \sigma=1.99581248\ /\Omega m$

The conductivity of this material is $1.99581248\ /\Omega m$