For a pure Si crystal at 300 K,the electron | vedclass

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(D) In a semiconductor,the law of mass action states that $n_e 	imes n_h = n_i^2$,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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(D) In a semiconductor,the law of mass action states that $n_e 	imes n_h = n_i^2$,where $n_i$ is the intrinsic carrier concentration.Given,$n_i = 1.5 	imes 10^{16} \ m^{-3}$.After doping with Indium (a trivalent impurity),the hole concentration becomes $n_h = 4.5 	imes 10^{22} \ m^{-3}$.The new electron concentration $n_e$ is given by:$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}$.

For a pure $Si$ crystal at $300 \ K$,the electron $(n_e)$ and hole $(n_h)$ concentrations are equal,being $1.5 \times 10^{16} \ m^{-3}$. On doping with Indium,the hole concentration increases to $4.5 \times 10^{22} \ m^{-3}$. Calculate the new electron concentration $(n_e)$ in the doped $Si$.

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