= Solution
Let two weakly coupled bodies exchange <energy> while their total <energy> $U$ is fixed. If body1 has $U_1$, the number of compatible states is $\Omega_1(U_1)\Omega_2(U-U_1)$. Its most probable equilibrium division maximizes the logarithm, giving
$$
\frac{\partial\ln\Omega_1}{\partial U_1}
=\frac{\partial\ln\Omega_2}{\partial U_2}.
$$
This equal intensive quantity labels <thermal equilibrium>. With $S=k_B\ln\Omega$ and thermodynamic $1/T=(\partial S/\partial U)_{V,N}$,
$$
\boxed{\left(\frac{\partial\ln\Omega}{\partial U}\right)_{V,N}=\frac1{k_BT}=\beta.}
$$
Ordinary positive <heat capacities> make the <entropy> maximum locally stable. Particle numbers and <volumes> are held fixed during the exchange.
For a subsystem coupled to a large <heat> bath, the exact level weight is proportional to $g_j\Omega_b(U-\epsilon_j)$. Expanding the bath <entropy> gives $\ln\Omega_b(U-\epsilon_j)=\ln\Omega_b(U)-\beta\epsilon_j+\cdots$. Normalization therefore gives
$$
\boxed{P(\epsilon_j)=\frac{g_je^{-\beta\epsilon_j}}Z,\qquad
Z=\sum_i g_ie^{-\beta\epsilon_i}.}
$$
The Boltzmann form uses the reservoir limit; an arbitrary small isolated pair instead retains its exact bath state count. The degeneracy counts distinct subsystem <microstates> at the same <energy>, rather than adding another Boltzmann factor for the whole level.
In the <two-level donor occupancy> model, set the single bound-state <energy> to zero and the conduction representative to $\Delta=0.04\,\mathrm{eV}$ with degeneracy $g$. Ignoring <spin>, its bound <probability> is
$$
P_b(T)=\frac1{1+g e^{-\Delta/(k_BT)}}.
$$
The measured model occupancy gives $g e^{-\Delta/(k_B150\,\mathrm K)}=9$, so
$$
\boxed{g=9e^{0.04/(150k_B)}\simeq1.99\times10^2\simeq200,}
$$
using $k_B=8.61733\times10^{-5}\,\mathrm{eV\,K^{-1}}$. The supplied occupancy and <energy> are rounded, so there is no reason to demand an exact nonrounded integer from that inference.
For $P_b=0.9$, the excited-to-bound weight must be $1/9$. Eliminate $g$ using the first point:
$$
\frac1T=\frac1{150\,\mathrm K}+\frac{2k_B\ln9}{\Delta},\qquad
\boxed{T\simeq62.0\,\mathrm K.}
$$
The <probability> decreases monotonically from1 as $T\to0$ toward $1/(1+g)\simeq0.005$ as $T\to\infty$. The nonzero high-temperature limit follows from the stipulated finite effective degeneracy, not from a full continuum conduction-band calculation. A realistic band with energy-dependent <density of states> need not retain this two-level limit.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-1-donor-occupancy.png]
{title=Donor occupancy}
{description=Two-level model with both inferred temperature points.}
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