 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 potentialRescaling the hydrogen atom binding energy by the effective mass and dielectric screening gives the ionization energyThe 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 $a_B^{\ast }={ϵ}_r(m_e/m_e^{\ast })a_0\simeq 2.38\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.

At $1000\mathrm{K}$, $k_{B}T=0.08617\mathrm{eV}$, 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 $k_{B}T$ 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.