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 e mu _e + n_h e mu_h = e ( n_e mu_e +n_h mu_h)`
As per given date, n is negligible as compared to n, so that we can write
` sigma = e n_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 S m^(-1)`
where S (siemen) stands for `Omega^(-1)`