Question
A semiconductor has equal electron and hole concentration of $6 \times 10^8/m^3.$ On doping with certain impurity, electron concentration increases to $9 \times 10^{12}/m^3.$
  1. Identify the new semiconductor obtained after doping.
  2. alculate the new hole concentration.

Answer

  1. The doped semiconductor is $n-$type.
  2. $\text{n}_\text{e}\text{n}_\text{h}=\text{n}^2_\text{i}$
  3. $\Rightarrow\text{n}_\text{h}=\frac{\text{n}^2_\text{i}}{\text{n}_\text{e}}$
    $=\frac{(6\times10^8)^2}{9\times10^{12}}=4\times 10^4\text{per m}^3$

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