 Donor binding energy, size and thermal ionisation

For a hydrogenic donor state, replace the bare Electron mass by $m^{\ast }=0.1m_e$ and the Coulomb permittivity by ${ϵ}_r{ϵ}_0$ with ${ϵ}_r=16$. Rescaling the Bohr model gives the positive ground-state binding magnitudeThe radius isusing the Bohr radius $a_0=5.29177\times 10^{-11}\mathrm{m}$. 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 $300\mathrm{K}$, $k_{B}T=25.9\mathrm{meV}$, about five times the binding energy. Thus 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 $k_{B}T$ alone does not determine it exactly. The large radius explains why a weakly bound donor extends across many unit cells.