Question

Germainium a 200.0 K has 1.16×10^10 free electrons per atom. If you wanted to have 2×10^3 as many electrons from arsenic doping as thermally free electrons in a Germainium semiconductor, how many arsenic atoms should there be per Germainium atom ?

Answer 1

$k=(\frac{n_{As}}{n_{Ge}})^{3/2},\implies$

$n_{As}=n_{Ge}k^{2/3}=1.16\cdot 10^{10}\cdot (2\cdot 10^3)^{2/3}=1.84\cdot 10^{12}.$