= Donor binding energy, size and thermal ionisation
For a <hydrogenic donor state>, replace the bare <Electron> <mass> by $m^*=0.1m_e$ and the Coulomb permittivity by $\epsilon_r\epsilon_0$ with $\epsilon_r=16$. Rescaling the <Bohr model> gives the positive ground-state binding magnitude
$$
\boxed{E_B=13.6057\,\mathrm{eV}\frac{0.1}{16^2}=5.31\times10^{-3}\,\mathrm{eV}=5.31\,\mathrm{meV}}.
$$
The radius is
$$
\boxed{a_B^*=a_0\frac{16}{0.1}=8.47\times10^{-9}\,\mathrm m=8.47\,\mathrm{nm}},
$$
using the <Bohr radius> $a_0=5.29177\times10^{-11}\,\mathrm m$. The supplied positive hydrogen “<energy>” formula should be interpreted as a binding magnitude; the bound level's <energy> relative to the conduction-band edge is $-E_B$.
At $300\,\mathrm K$, $k_BT=25.9\,\mathrm{meV}$, about five times the binding <energy>. Thus \b[a substantial fraction, ordinarily most of the shallow donors in the dilute nondegenerate regime, will be ionised]. An exact fraction also depends on the <Fermi level>, donor concentration and degeneracies; comparing $E_B$ with $k_BT$ alone does not determine it exactly. The large radius explains why a weakly bound donor extends across many unit cells.
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