In pure silicon at 300 K temperature,the con | vedclass

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(A) For a semiconductor,the law of mass action states that $n_e \cdot n_h = n_i^2$,where $n_i$ is the intrinsic carrier concentration.Given,$n_i = 1.5

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$5 	imes 10^9 \ m^{-3}$

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(A) For a semiconductor,the law of mass action states that $n_e \cdot n_h = n_i^2$,where $n_i$ is the intrinsic carrier concentration.Given,$n_i = 1.5 	imes 10^{16} \ m^{-3}$ and the new hole concentration $n_h = 4.5 	imes 10^{22} \ m^{-3}$.Substituting these values into the formula:$n_e = \frac{n_i^2}{n_h} = \frac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{22}}$$n_e = \frac{2.25 	imes 10^{32}}{4.5 	imes 10^{22}}$$n_e = 0.5 	imes 10^{10} \ m^{-3} = 5 	imes 10^9 \ m^{-3}$.

In pure silicon at $300 \ K$ temperature,the concentration of electrons $n_e$ and holes $n_h$ is $1.5 \times 10^{16} \ m^{-3}$. By doping with Indium,$n_h$ is increased to $4.5 \times 10^{22} \ m^{-3}$. Calculate the concentration of electrons $n_e$ in the doped silicon.

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