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

Here, `rho=2500Omegam, mu_(e)=0.12m^(2)V^(-1)s_(-1)`,
`mu_(h)=0.45m^(2) V^(-1)s^(-1)`
(i)The resistivity of pure silicon is
`rho=1/sigma=(1)/(e(n_(e)mu_(e)+n_(h)mu_(h)))=(1)/(en_(i)(mu_(e)+mu_(h)))`
`( :' n_(e)=n_(h)=n_(i))`
or `n_(i)=(1)/(erho(mu_(e)+mu_(h)))`
`(1)/((1.6xx10^(-19))xx2500xx(0.12+0.45))`
`=1.52xx10^(16) m^(-3)`
When `10^(20)` atoms of phosphorous (i.e., donor ayoms of valence five) are added per m^(3), the semiconductor becomes n-type semiconductor. Resistivity,
`rho=(1)/(en_(e)mu_(e))=(1)/((1.6xx10^(-11))xx10^(20)xx0.12)`
`=52.1 Omegam`
Then `n_(e)-n_(h)=N_=10^(20)`
`( :' n_(h)=1.52xx10^(16)m^(-3))`
(ii) When 2xx20^(20) atoms of Boron (i.e., acceptor atoms of valence five) are added per m^(3), the semiconductor becomes p-type semiconductor. Then
`n_(h)-n_(e)=N_-N_=2xx10^(20)-10^(20)=10^(20)`
Since `n_(h)gt n_(e)`, hence `n_(h)=10_(20)`
Resistivity, `rho=(1)/(en_(h)mu_(h))
(1)/((1.6xx10^(-19))xx10^(20)xx0.045)`
`=138.9Omegam`