= Solution
For equally nondegenerate levels, the <partition function> is $Z=e^{-\beta\epsilon}(1+e^{-\beta\Delta})$. The mean <energy> per particle is $U=\epsilon+\Delta/(1+e^{\beta\Delta})$, so the <Schottky anomaly> has <heat capacity>
$$
\boxed{C=k_B\left(\frac{\Delta}{k_BT}\right)^2
\frac{e^{\Delta/(k_BT)}}{[1+e^{\Delta/(k_BT)}]^2}.}
$$
At low <temperature>, $C\sim k_B(\Delta/k_BT)^2e^{-\Delta/(k_BT)}$ because excitation is suppressed. At high <temperature>, $C\sim\Delta^2/(4k_BT^2)$ because the two populations approach equality and the <energy> saturates. The peak is about $0.439k_B$ at $k_BT\simeq0.417\Delta$. The common offset $\epsilon$ does not affect <heat capacity>. Multiply by particle number for the total <heat capacity>, or divide by the corresponding <mass> for a mass-specific quantity.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-schottky-heat-capacity.png]
{title=Two-level Schottky heat capacity and its low- and high-temperature suppression}
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