Suppose a pure Si-crystal has `5xx10^(28) "atoms" m^(-3)`. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Give that `n_(i)=1.5xx10^(16)m^(-3)`.

Explanation

Note that thermally generated electrons `(n_(i) = 10^(16) m^(-3))` are negligibly small as compared to those produced by doping.
Therefore, `n_(e) ~~ N_(D)`.
Since, `n_(e)n_(h) = n_(i)^(2)`. The number of holes
`n_(h) = (2.25 xx 10^(32))//(5 xx 10^(22))- 4.5 xx 10^(9) m^(-3)`