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?

Options

A. `2xx10^4` per `m^3`
B. `2xx10^2` per `m^3`
C. `4xx10^4` per `m^3`
D. `4xx10^2` per `m^3`

Answer

C

Explanation

As, `n_e n_h=n_i^2`
Here, `n_i=6xx10^8 per m^3 and n_e = 9xx10^12 per m^3`
`therefore n_h=n_i/n_e=((6xx10^8)^2)/(9xx10^12)=4xx10^4 per m^3`