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(D) For a semiconductor,the law of mass action states that the product of electron concentration $(n_{e})$ and hole concentration $(n_{h})$ is equal t

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$2 	imes 10^{9} \ m^{-3}$

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(D) For a semiconductor,the law of mass action states that the product of electron concentration $(n_{e})$ and hole concentration $(n_{h})$ is equal to the square of the intrinsic carrier concentration $(n_{i})$:$n_{e} n_{h} = n_{i}^{2}$Given that the intrinsic concentration $n_{i} = 10^{16} \ m^{-3}$ and the doped hole concentration $n_{h} = 5 	imes 10^{22} \ m^{-3}$,we can find $n_{e}$ as follows:$n_{e} = \frac{n_{i}^{2}}{n_{h}}$$n_{e} = \frac{(10^{16} \ m^{-3})^{2}}{5 	imes 10^{22} \ m^{-3}}$$n_{e} = \frac{10^{32}}{5 	imes 10^{22}} \ m^{-3}$$n_{e} = 0.2 	imes 10^{10} \ m^{-3} = 2 	imes 10^{9} \ m^{-3}$

$A$ pure semiconductor has equal electron and hole concentration of $10^{16} \ m^{-3}$. Doping by gallium increases $n_{h}$ to $5 \times 10^{22} \ m^{-3}$. Then,the value of $n_{e}$ in the doped semiconductor is

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