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

`1.99581248\ /\Omega m`

Step-by-step explanation:

e = Charge of electron =
`1.6* 10^(-19)\ C`

n = Electron concentration =
`5.63* 10^(19)\ /m^(3)`

`v_d` = Drift veloctiy = 113 m/s

E = Electric field = 510 V/m

Electron mobility is given by

`\mu=(v_d)/(E)\\\Rightarrow \mu=(113)/(510)\\\Rightarrow \mu=0.22156\ m^2/Vs`

Conductivity is given by

`\sigma=ne\mu\\\Rightarrow \sigma=5.63* 10^(19)* 1.6* 10^(-19)* 0.22156\\\Rightarrow \sigma=1.99581248\ /\Omega m`

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