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` =0.048m^2 V^(-1) s^(-1) `

Explanation

The conductivity of a semiconductor is the sum of the conductivities due to electrons and holes and is given by `sigma= sigma_(e)+ sigma_(h)=n_(e)emu_(e)+n_(h)emu_(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)`
`=(1.6xx10^(-19)C)(5.0xx10^(20)m^(-3))(0.048m^(2)V^(-1)s^(-1))`
`=3.84Omega^(-1)m^(-1)=3.84Sm^(-1)`
where S (siemen) stands for `Omega^(-1)`