1.

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

Answer» `n_(e)=(n_(1)^(2))/(n_(h))=((1.5xx10^(16))^(2))/((4.5xx10^(22)))=5xx10^(9)m^(-3)`


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