For pure Ge,the intrinsic carrier density is | vedclass

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(A) The law of mass action for semiconductors states that $n_e \cdot n_h = n_i^2$,where $n_e$ is the electron density,$n_h$ is the hole density,and $n

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$10^{15}$

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(A) The law of mass action for semiconductors states that $n_e \cdot n_h = n_i^2$,where $n_e$ is the electron density,$n_h$ is the hole density,and $n_i$ is the intrinsic carrier density.Given: $n_i = 10^{19} \ m^{-3}$.When donor impurity is added,the electron density $n_e$ becomes approximately equal to the donor impurity density $N_D = 10^{23} \ m^{-3}$.Using the formula: $n_h = \frac{n_i^2}{n_e} = \frac{(10^{19})^2}{10^{23}}$.$n_h = \frac{10^{38}}{10^{23}} = 10^{15} \ m^{-3}$.

For pure $Ge$,the intrinsic carrier density is $10^{19} \ m^{-3}$. If donor impurity is added with a density of $10^{23} \ m^{-3}$,what will be the hole density (in $m^{-3}$)?

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