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. Given that `n_(i)=1.5xx10^(16)m^(-3)`.

Explanation

Here, thermally generated electrons,
`n_(i)=1.5xx10^(16)m^(-3)`
1 ppm = 1 part per million `=10^(6)`
`therefore` Electron number density due to As atom,
`n_(D)=(5xx10^(28))/(10^(6))=5xx10^(22)m^(-3)`
Number of electron due to doping.
`n_(i)` is neglected compare to `n_(D)`
`therefore n_(e )=n_(D )=5xx10^(22)m^(-3)`
Now `n_(i)^(2)=n_(e )n_(h)`
`therefore n_(h)=(n_(i)^(2))/(n_(e ))`
`=((1.5xx10^(16))^(2))/(5xx10^(22))`
`therefore n_(h)=4.5xx10^(9) m^(-3)`