1.

Pure Si at 300 K has equal electron (n_(i)) concentrations of 1.5 xx 10^(16) m^(-3). Doping by indium increases n_(h) = 4.5 xx 10^(22) m^(-3). N_(e) in the doped Si is

Answer»

`5 xx 10^(9)`
`7 xx 10^(9)`
`9 xx 10^(9)`
`8 xx 10^(9)`

Solution :We KNOW that ofr a doped semiconductor in thermal equilibrium, we have `n_(e)n_(h)=n_(i)^(2)`
As PER GIVEN data, `n_(i?) =1.5 xx 10^(16) m^(-3)`
`n_(h) = 4.5 xx10^(22) m^(-3)`
Thus `n_(e)=(n_(i)^(2))/(n_(h)) =((1.5 xx10^(16))^(2)m^(-6))/(4.5 xx10^(22)m^(-3))=5.0 xx10^(9) m^(-3)`.


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