The hole and the free electron concentrations | vedclass

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(A) Given,the intrinsic carrier concentration $n_{i} = 1.4 	imes 10^{16} \ m^{-3}$.According to the law of mass action for semiconductors,the product

Vedclass · Accepted Answer

$0.49 	imes 10^{10} \ m^{-3}$

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(A) Given,the intrinsic carrier concentration $n_{i} = 1.4 	imes 10^{16} \ m^{-3}$.According to the law of mass action for semiconductors,the product of hole concentration $(n_{h})$ and electron concentration $(n_{e})$ is equal to the square of the intrinsic carrier concentration $(n_{i}^2)$:$n_{i}^2 = n_{h} 	imes n_{e}$Given $n_{h} = 4 	imes 10^{22} \ m^{-3}$,we need to find $n_{e}$.$n_{e} = \frac{n_{i}^2}{n_{h}}$$n_{e} = \frac{(1.4 	imes 10^{16})^2}{4 	imes 10^{22}}$$n_{e} = \frac{1.96 	imes 10^{32}}{4 	imes 10^{22}}$$n_{e} = 0.49 	imes 10^{10} \ m^{-3}$.

The hole and the free electron concentrations in a pure silicon at room temperature are given by $1.4 \times 10^{16} \ m^{-3}$ each under equilibrium. When it is doped with indium and the hole concentration is $n_{h} = 4 \times 10^{22} \ m^{-3}$,the electron concentration is

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