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I was trying to understand n-type doping in diamonds. It mentions the low solid-state solubility for higher atomic radius potential dopants like P, As in diamonds. What does it mean?

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    $\begingroup$ That means you can't put many in without them wanting to precipitate into a secondary phase. Just like adding sugar to water, you can only put so much in. If the dopant isn't sitting nicely on a lattice site, it won't be a dopant. $\endgroup$
    – Jon Custer
    Aug 17, 2021 at 22:56

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To answer the question posed in the title:

Solid state solubility means this: it is possible for atoms of one kind to thermally diffuse into a lattice structure consisting of a different type of atom. In effect, those atoms that are diffusing in are dissolved within the structure of the lattice. They will experience a solubility limit in that lattice, beyond which the dissolved atoms will segregate out as a new structural phase dispersed in zones within that lattice.

Since all these processes are enabled by diffusion, they will obey a rate law that scales exponentially with temperature. Generally, this means greater solubility at higher temperatures and less at lower temperatures. If brought down quickly enough in temperature (a process called quenching), the structures formed at the elevated temperature will become frozen into the bulk structure and persist there in a metastable state at the lower temperature.

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