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_()=2.5xx10^(23)m^(-3)`
B. P– type having electron concentration `n_(e)=5xx10^(9)m^(-3)`
C. N– type with electron concentration `n_()=5xx10^(22)m^(-3)`
D. P– type with electron concentration `n_(e)=2.5xx10^(10)m^(-3)`

Answer

B

Explanation

`n_(i)=1.5xx10^(16)//m^(3)`
`n_(h)=N_(A)=4.5xx10^(22)//m^(3)`
Now, `n_(e)n_(h)=n_(i)^(2)`
`impliesn_(e)=((1.5xx10^(16))^(2))/(4.5xx10^(22))=5xx10^(9)//m^(3)`