Suppose a pure Si crystal has `5 xx 10^(28)` atoms `m^(-3)`. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that `n_(i) = 1.5 xx 10^(16) m^(-3)`.

Explanation

`n_(c) = 4.95 xx 10^(22), n_(h) = 4.75 xx 10^(9)`, n-type since `n_(e) gt gt n_(h)`
For charge neutrality `N_(D)-N_(A) = n_(e).n_(h) = n_(i)^(2)`
Solving these equations, `n_(e) = 1/2 [(N_(D)-N_(A)) + sqrt((N_(D)-N_(A))^(2) + 4n_(i)^(2))]`