Question

Suppose a pure Si crystal has ` 5 xx 10^(28)` atoms `m^(-3)` . It is droped 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)`.

Answer 1

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) = ( 1.5 xx 10^(16))^(2) // 5 xx 10^(28) xx 16^(-6)`
`n_(h) =( 2.25 xx 10^(32)) //( 5xx 10^(22)) ~ 4.5xx 10^(9) m^(-3)`