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(C) For a semiconductor,the law of mass action states that $n_e n_h = n_i^2$,where $n_e$ is the electron concentration,$n_h$ is the hole concentration

Vedclass · Accepted Answer

$4 	imes 10^4 \, m^{-3}$

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(C) For a semiconductor,the law of mass action states that $n_e n_h = n_i^2$,where $n_e$ is the electron concentration,$n_h$ is the hole concentration,and $n_i$ is the intrinsic carrier concentration.Given that the initial intrinsic concentration $n_i = 6 	imes 10^8 \, m^{-3}$.After doping,the new electron concentration $n_e = 9 	imes 10^{12} \, m^{-3}$.Using the relation $n_h = \frac{n_i^2}{n_e}$,we substitute the values:$n_h = \frac{(6 	imes 10^8)^2}{9 	imes 10^{12}}$$n_h = \frac{36 	imes 10^{16}}{9 	imes 10^{12}}$$n_h = 4 	imes 10^4 \, m^{-3}$.

$A$ semiconductor has equal electron and hole concentration of $6 \times 10^8 \, m^{-3}$. On doping with a certain impurity,the electron concentration increases to $9 \times 10^{12} \, m^{-3}$. The new hole concentration is:

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