= Solution
A substitutional group-V atom such as <phosphorus> has one more <valence electron> than <germanium> and acts as a <donor impurity>, giving an <n-type semiconductor>. Approximate its outer <electron> by a <hydrogenic donor state> with screened Coulomb attraction and <effective mass> $m_e^*$. The <ionization energy> is the positive binding-energy magnitude
$$
E_D=13.606\,\mathrm{eV}\,\frac{m_e^*/m_e}{\epsilon_r^2}
=13.606\frac{0.22}{15.8^2}\,\mathrm{eV},
\qquad\boxed{E_D\simeq0.0120\,\mathrm{eV}=12.0\,\mathrm{meV}.}
$$
Its <effective Bohr radius of a hydrogenic donor> is about $3.80\,\mathrm{nm}$, much larger than an atomic core, consistent with a shallow extended bound state.
At $300\,\mathrm K$, $k_BT=25.9\,\mathrm{meV}$, while the mean three-dimensional classical translational <kinetic energy> is $3k_BT/2=38.8\,\mathrm{meV}$. Both exceed the donor binding <energy>, so substantial thermal ionization is expected, unlike excitation across the much larger $0.661\,\mathrm{eV}$ host <band gap>. A precise ionized fraction also depends on the <chemical potential>, <dopant density> and <donor charge-state degeneracy>; comparing two energy scales alone does not prove complete ionization at arbitrary concentration.
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