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 `=1.6x10^(-19)C`.

Explanation

`rho_(i)=(1)/(sigma _(i))=(1)/(n_(i)e(mu_(e)+mu_(h)))`
`n_(i)=(1)/(rho_(i)e(mu_(e)+mu_(h)))`
`=(1)/(3000xx1.6xx10^(-19)(0.12+0.025))`
`=1.4437xx10^(16)//m^(3)`
(a) when `10^(19)` atoms of phosphorus (donor atoms) are added per `m^(3)`, we have
`n_(e)gt gtn_(i)` or `n_(e)gt gtn_(h)`
`n_(e)=10^(19)`
`rho=(1)/(n_(e)emu_(e))=(1)/(10^(19)xx1.6xx10^(-19)xx0.12)`
`=5.21 ohm.m`
(b) when `2xx10^(19)` boron (accepter) atoms are also added per `m^(3)`
`n_(h)-n_(e)=n_(A)-n_(D)=2xx10^(19)-10^(19)`
`n_(h)=10^(19), since n_(e)lt lt n_(h)`
`rho=(1)/(n_(h)emu_(h))=(1)/(10^(19)xx1.6xx10^(-19)xx0.025)`
`=25 ohm.m`