Question

A semiconductor has equal electron and hole concentration of `6xx10^(8)//m^(3)`. On doping with certain impurity, electron concentration increases to `9xx10^(12)//m^(3)`. (i) Identify the new semiconductor obtained after doping. (ii) Calculate the new hole concentration. (iii) How does the energy gap vary with doping?

Answer 1

Here, `n_(i)=6xx10^(8)m^(-3) , n_(e)=9xx10^(12)m^(-3)`
`n_(h)=n_(i)^(2)/n_(e)=((6xx10^(8))^(2))/(9xx10^(12))=4xx10^(4)m^(-3)`
As, after doping, `n_(e)gtn_(h)`, so the new semiconductor is n-type. Energy gap decreases with doping.