Charge neutrality of extrinsic semiconductors

Worked out. A pure silicon crystal with $4\times10^{28}$ atoms per cubic metre is doped with a 1 ppm (one part per million, i.e. $1/10^6$) concentration of antimony, a pentavalent donor impurity; the intrinsic carrier density is given as $n_i=1.2\times10^{16}\,m^{-3}$. The number of antimony atoms per cubic metre works out to $4\times10^{28}\times10^{-6}=4\times10^{22}\,m^{-3}$. Since every pentavalent donor atom contributes exactly one free electron to the crystal, the free-electron density becomes $n_e=4\times10^{22}\,m^{-3}$ (donor electrons vastly outnumber the negligible thermally-generated electrons). The hole density then follows from the mass-action law $n_h=n_i^2/n_e$, giving $n_h=(1.2\times10^{16})^2/(4\times10^{22})=3.6\times10^{9}\,m^{-3}$ -- many orders of magnitude smaller than $n_e$, confirming electr …

Worked out. A pure silicon crystal at 300 K has equal electron and hole densities $n_e=n_h=1.5\times10^{16}\,m^{-3}$ (so this is also the intrinsic carrier density $n_i$) before doping. The crystal is then doped with indium, a trivalent acceptor impurity, which raises the hole density to $n_h=4.5\times10^{22}\,m^{-3}$ (holes become the majority carrier, consistent with indium creating a p-type crystal). Using the mass-action law $n_e n_h=n_i^2$, the new (minority) electron density is $n_e=n_i^2/n_h=(1.5\times10^{16})^2/(4.5\times10^{22})=5\times10^{9}\,m^{-3}$ -- showing how heavily doping suppresses the minority-carrier density far below i …