Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is of

Options

A. n-type with electron concentration
`n_(e)=5xx10^(22)m^(-3)`
B. p-type with electron concentration
`n_(e)=2.5xx10^(10)m^(-3)`
C. n-type with electron concentration
`n_(e)=2.5xx10^(23)m^(-3)`
D. p-type with electron concentration
`n_(e)=5xx10^(9)m^(-3)`

Answer

D

Explanation

As `n_(i)^(2)=n_(e)xxn_(h)`
`(1.5xx10^(16))^(2)=n_(e)xx(4.5xx10^(22))`
`n_(e)=((1.5xx10^(16))^(2))/((4.5xx10^(22)))=5xx10^(9)`
As `n_(h) gt gt n_(e)`, so semiconductor is of p-type and
`n_(e)=5xx10^(9)`