Suppose 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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Accepted Answer

(2) 4.5×10⁹ m⁻³. Given: 5×10²⁸ silicon atoms per cubic metre, one in a million replaced by pentavalent As, and nᵢ=1.5×10¹⁶ m⁻³. Idea: the donors fix the electron density, and the hole density then follows from nₑ nₕ=nᵢ², which holds whatever the doping. Donor density: 1 ppm is a factor 10⁻⁶, so n_D=5×10²⁸×10⁻⁶=5×10²² m⁻³. That is millions of times larger than nᵢ, so essentially every conduction electron came from an As atom: nₑ=5×10²² m⁻³. Holes: nₕ=(nᵢ²)/(nₑ), with nᵢ²=(1.5×10¹⁶)²=2.25×10³², and 2.25×10³²÷(5×10²²)=0.45×10¹⁰. So nₕ=4.5×10⁹ m⁻³, the density the options quote, alongside nₑ=5×10²² m⁻³.

Suppose 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.

Step-by-step solution

Why the other options are wrong