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?

Options

A. `3.24xx10^(6)m^(-3)`
B. `6.24xx10^(8)m^(-3)`
C. `4.54xx10^(9)m^(-3)`
D. None of these

Answer

C

Explanation

For each atom doped of arsenic, one free electron is received similarly of each atom doped of indium, a vecancy is created. So, number of free electrons introduced by pentavalent impurity is
`N_(AS) = 5 xx 10^(22)m^(-3)`
The number of holes introduced by trivalent impurity added is
`N_(I)=5xx10^(20)m^(-3)`
So, net number of electrons added is
`n_(e)=N_(AS)-N_(I)`
`=5 xx10^(22)-5xx10^(20)`
`=4.95xx10^(22)m^(-3)`
Now, by the law of mass action.
`n_(e)n_(h) =n_(i)^(2)`
So, `n_(h) =(n_(i)^(2))/(n_(e))=((1.5xx10^(16))^(2))/(4.95xx10^(22))`
`rArr n_(h)=4.54xx10^(9)m^(-3)`
As, `n_(e) gt n_(h)` ( number of hles ). So, the material is n-type semiconductor.