= Solution
A substitutional phosphorus atom supplies one more valence electron than the silicon atom it replaces. In the shallow <hydrogenic donor state> model, the remaining positive impurity attracts that extra electron through the screened <Coulomb potential>
$$
V(r)=-\frac{e^2}{4\pi\epsilon_0\epsilon_r r}.
$$
Rescaling the <hydrogen atom> binding energy by the <effective mass> and dielectric screening gives the ionization energy
$$
\boxed{E_D^{\mathrm{bind}}=13.6057\,\mathrm{eV}\,
\frac{m_e^*/m_e}{\epsilon_r^2}
=13.6057\frac{0.26}{11.7^2}\,\mathrm{eV}
=0.02584\,\mathrm{eV}.}
$$
The donor level lies at $E_c-E_D^{\mathrm{bind}}$; its binding magnitude is positive. The associated <effective Bohr radius of a hydrogenic donor> is $a_B^*=\epsilon_r(m_e/m_e^*)a_0\simeq2.38\,\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.
At $1000\,\mathrm K$, $k_BT=0.08617\,\mathrm{eV}$, approximately $3.3$ times this binding energy. Dilute donors are consequently expected to be largely ionized, rather than frozen out. The actual ionized fraction also depends on the <chemical potential>, available conduction states and <donor charge-state degeneracy>; comparing $k_BT$ to the binding energy alone is not a normalized ionization probability. At sufficiently high <temperature>, intrinsic electron-hole production can dominate the total carriers even though donors are ionized. Band anisotropy and short-range impurity corrections limit the numerical hydrogenic estimate.
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