The number of silicon atoms per `m^(3) is 5xx10^(28)`. This is doped simultaneously with `5xx10^(22)` atoms per `m^(3)` of Arsenic and `5xx10^(20) per m^(3)` atoms of indium. Calculate the number of electrons and holes. Given that `n_(i)=1.5xx10^(16)m^(-3)`. Is the material n-type or p-type?

Explanation

Number of silicon atoms, `N = 5 xx 10^(28) " atoms"//m^(3)`
Number of arsenic atoms, `n_(As) = 5 xx 10^(22) " atoms"//m^(3)`
Number of indium atoms, `n_(In) = 5 xx 10^(20) " atoms"//m^(3)`
Number of thermally-generated electrons, `n_(i) = 1.5 xx 10^(16) "electrons"//m^(3)`
Number of electrons, `n_(e) = 5 xx 10^(22) - 1.5 xx 10^(16) approx 4.99 xx 10^(22)`
Number of holes `= n_(h)`
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
`n_(e)n_(h) = n_(i)^(2)`
`thereforen_(h)=(n_(i)^(2))/(n_(e))`
`=(1.5xx10^(16))^(2)/(4.99xx10^(22))approx4.51xx10^(9)`
Therefore, the number of electrons is approximately `4.99 xx 10^(22)` and the number of holes is about `4.51 xx 10^(9)`. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.