If resistivity of pure silicon is `3000 Omega meter`, and the electron and hole mobilities are `0.12 m^(2)V^(-1)s^(-1)` and `0.045m^(2)V^(-1)s^(-1)` respectively, determine the resistivity of a specimen of the material when `10^(19)` atoms of phosphorous are added per `m^(3)` are also added. Given charge on electron `=1.6xx10^(-19)C`.

Explanation

The resistivity of pure Si is given by `rho = (1)/(sigma)=(1)/( e(n_(e)u_(e)+n_(h)u_(h)))=(1)/(en_(i)(u_(e)+u_(h))` or `n_(i) =(1)/(e_(rho)(u_(e)+u_(h))` =`(1)/(1.6xx10^(-19)xx3000(0.12+0.045)` =`1.26xx10^(16) m^(-3)` when `10^(19)` atoms of phosphorus (donor atoms of valence five)are added per `m^(3)`, the semiconductor becomes n-type semiconductor.'therefore `n_(e)-n_(h)` approx `n_(e)=N_(d)` because `n_(h)=1.26xx10^(16)`Resistivity `rho =(1)/(n_(e)u_(e)e)=(1)/(1.6xx10^(-19)xx10^(19)xx0.12) = 5.21omega m`.