A pure Si crystal has 5×10²⁸ atoms m⁻³. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that nᵢ=1.5×10¹⁶ m⁻³.

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(3) 4.5×10⁹ m⁻³. Given: 5×10²⁸ Si atoms m⁻³, doped 1 ppm with pentavalent As, and nᵢ=1.5×10¹⁶ m⁻³. Idea: each pentavalent donor releases one electron, and whatever the doping the two densities obey nₑ nₕ=nᵢ². One part per million of the silicon atoms is replaced, so the donor density is n_D=5×10²⁸×10⁻⁶=5×10²² m⁻³. At room temperature every donor is ionised, and n_D swamps nᵢ, so the electron density is nₑ≈ n_D=5×10²² m⁻³. The holes then follow from the mass-action law: nₕ=(nᵢ²)/(nₑ)=((1.5×10¹⁶)²)/(5×10²²)=(2.25×10³²)/(5×10²²). So nₕ=4.5×10⁹ m⁻³ -- the density the options quote -- while nₑ=5×10²² m⁻³. Doping has raised the electrons by more than six powers of ten and pushed the holes down by nearly seven.

A pure Si crystal has 5×10²⁸ atoms m⁻³. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that nᵢ=1.5×10¹⁶ m⁻³.

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Why the other options are wrong