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Statistical thermodynamics question. Can somebody explain the low temperature Schottky anomaly for two level systems? Why is this anomaly not seen in all materials?

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The Schottky anomaly is a general feature of two levels systems. The heat capacity is defined as $\displaystyle C= \left( \frac{\partial U}{\partial T} \right)_\Delta$ where $U$ is the internal energy and the energy gap between the two levels is $k_B\Delta$, $k_B$ being the Boltzmann constant.

The gradient of the internal energy is thus important. At constant energy gap and at very low temperatures only the lowest level is populated and the internal energy is small, and heat capacity small as the internal energy increase is small, i.e gradient $U$ vs $T$ is small. However, as the temperature is increased the upper level begins to be populated and so the internal energy rapidly increases, then the gradient $U$ vs $T$ is large and so is heat capacity.

When both levels are almost populated equally, which occurs at very high temperature, no more energy can be absorbed and the gradient $dU/dT$ effectively becomes constant and so the heat capacity becomes small and tends to zero.

Similar behaviour is observed when there are three or four etc i.e. a small number of levels. In the 'normal' case of a molecule there is effectively no upper limit on the number of levels that can be populated as temperature increases, and the slope of $U$ vs $T$ has a constant finite value which leads to $3R$ as the heat capacity per mole at high temperatures.

schottsky

The figure shows the internal energy (in units of $k_B\Delta $) and heat capacity (in units of $k_B$) vs $T/\Delta$. To make the plots more general $T/\Delta$ is plotted as this ratio occurs in the final equation.

The equations are:

partition function $ Z=1+e^{-E/k_BT} = 1+e^{-\Delta/T} $ when energy gap $E=k_B\Delta $.

The Internal energy is $\displaystyle U=k_BT^2\left( \frac{\partial \ln(Z)}{\partial T} \right)_\Delta$.

The heat capacity is $C\displaystyle=\left( \frac{\partial U}{\partial T} \right)_\Delta= k_B\frac{ \Delta^2}{T^2}\frac{e^{\Delta/T} }{(1+e^{\Delta/T})^2}$

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