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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 $$ 9 \times 10^{12} / m^3 $$. Calculate the new hole concentration.

Solution
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$$ n_e n_h = n^{2}_{i} $$ $$ \Rightarrow $$ $$ n_h = \dfrac{n^{2}_{i}}{n_e} = \dfrac{(6 \times 10^8)^2}{9 \times 10^{12}} = 4 \times 10^4 \, per \, m^3 $$.

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