Question

A semiconductor has an electron concentration of `0.45 xx 10^(12) m^(-3)` and a hole concentration of `5.0 xx 10^(20) m^(-3)`. Calculate its conductivity. Given electron mobility `= 0.135 m^(2) V^(-1) s^(-1)`, hole mobility `= 0.048 m^(2) V^(-1) s^(-1)`,

Answer 1

The conductivity of a semicon ductor is the sum of the conductivities due to electrons and holes and is given by
`sigma = sigma_(e) + sigma_(h) =n_(e)e mu_(e) +n_(h) e mu_(h) =e(n_(e)mu_(e)+n_(h)mu_(h))`
As per given date, `n_(e)` is negligible as compared to `n_(h)`, so that we can write
`sigma = en_(h) mu_(h)`
=`(1.6 xx 10^(-19) C) (5.0 xx 10^(20) m^(-3))(0.048 m^(2) V^(-1) s^(-1))`
=`3.84 Omega^(-1) m^(-1) =3.84 m^(-1)`.