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A semiconductor has equal electron and hole concentration of 2 x 10⁸ m^-3. On doping with a certain impurity, the electron concentration increases to 4 x 10¹⁰ m^-3, then the new hole concentration of the semiconductor is

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10⁶ m^-3

10⁸ m^-3

10¹⁰ m^-3

10¹² m^-3

answer is A.

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Detailed Solution

$\begin{array}{l} n = \sqrt{n_{H} n_{e}} \\ \Rightarrow n_{e} n_{H} = n_{e}^{1} n_{H}^{1} \\ \Rightarrow n_{H}^{1} = \frac{2 \times 10^{8} \times 2 \times 10^{8}}{4 \times 10^{10}} = 10^{6} / m^{3} \end{array}$

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