A pure Ge specimen is doped with Al. The number density of acceptor atoms is approximately 10²¹ m⁻³. If the density of electron-hole pairs in an intrinsic semiconductor is approximately 10¹⁹ m⁻³, the number density of electrons in the specimen is

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(1) 10¹⁷ m⁻³. Given: acceptor density N_A≈10²¹ m⁻³, intrinsic density nᵢ≈10¹⁹ m⁻³. Idea: doping changes the two carrier densities but not their product. In any non-degenerate semiconductor nₑ nₕ=nᵢ² -- the mass-action law. Al is trivalent, so it is an acceptor and the specimen is p-type. At room temperature essentially every acceptor has taken an electron, so the hole density is set by the doping: nₕ≈ N_A=10²¹ m⁻³. Then nₑ=(nᵢ²)/(nₕ)=((10¹⁹)²)/(10²¹)=(10³⁸)/(10²¹). So the electron density is 10¹⁷ m⁻³: doping has pushed the minority carriers a hundred times below their intrinsic value.

A pure Ge specimen is doped with Al. The number density of acceptor atoms is approximately 10²¹ m⁻³. If the density of electron-hole pairs in an intrinsic semiconductor is approximately 10¹⁹ m⁻³, the number density of electrons in the specimen is

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