Diatomic Molecule as Rigid Rotor

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Question

Consider a molecule, such as Carbon Monoxide, which consists of two different atoms, one Carbon and one Oxygen, separated by a distance \(d\). Such a molecule can exist in quantum states of different orbital angular momentum. Each state has the energy \[ \epsilon_l={\hbar^2 \over 2I}l(l+1)\] where \(I=\mu d^2\) is the moment of inertia of the molecule about an axis through its centre of mass and \(\mu\) is the reduced mass defined by \(\frac 1\mu=\frac{1}{m_1} + \frac{1}{m_2}\). \(l=0, 1, 2, \ldots\) is the quantum number associated with the orbital angular momentum. Each energy level of the rotating molecule has the degeneracy \(g_l=2l+1\).

  1. Find the general expression for the canonical partition function \(Z\).
  2. Show that for high temperatures, \(Z\) can be approximated by an integral and calculate this integral.
  3. Evaluate the high temperature mean energy \(E\) and the heat capacity \(C_V\).
  4. Find the low-temperature approximations to the canonical partition function, the mean energy \(E\) and the heat capacity \(C_V\).

Answer

1. The generic partition function is given by
\begin{eqnarray}
Z&=&\sum_{j=0}^{\infty} g_j e^{-E_j \beta} \nonumber \\
&=&\sum_{l=0}^{\infty} (2l+1) e^{-l(l+1){\hbar^2 \over 2I}\beta} \nonumber
\end{eqnarray}

2. For high temperatures, the energy spacing between the energy levels is small compared to \(k_B T\), so the summation can be replaced by the integral
\begin{eqnarray}
Z&=&\int_0^\infty(2l+1)e^{-l(l+1){\hbar^2 \over 2I}\beta}dl \nonumber \\
&=&\int_0^\infty e^{{-\beta \hbar^2 l(l+1) \over 2I}}d(l(l+1)) \nonumber \\
&=&{2I \over \beta \hbar^2} \nonumber
\end{eqnarray}

3. Finding the energy in the high-temperature limit.
\begin{eqnarray}
\lt E \gt &=&-{\partial \over \partial \beta}\ln Z \nonumber \\
&=&-{\partial \over \partial \beta}\ln{2I\over \beta \hbar^2} \nonumber \\
&=&\frac 1\beta = k_B T \nonumber
\end{eqnarray}
And the heat capacity: \[C_V={\partial \lt E \gt \over \partial T} = k_B\]

4. For the low-temperature approximation, most of the particles will be in the ground state, so we can approximation the partition function by simply the first two terms like so:
\begin{eqnarray}
Z&=&\sum_{l=0}^{\infty} (2l+1) e^{-l(l+1){\hbar^2 \over 2I}\beta} \nonumber \\
&=&1+3e^{-\beta \hbar^2 / I} \nonumber
\end{eqnarray}
So the average energy again is
\begin{eqnarray}
\lt E \gt &=&-{\partial \over \partial \beta}\ln Z \nonumber \\
&=&-{\partial \over \partial \beta}\ln\bigg({1+3e^{-{\hbar^2 \over I} \beta}}\bigg) \nonumber \\
&=&-{3·\frac{-\hbar^2}{I}·e^{-\frac{\hbar^2}{I}\beta} \over 1+3e^{-\frac{\hbar^2}{I}\beta}} \nonumber \\
&=&{3\hbar^2 / I \over e^{\beta \hbar^2 / I}+3} \nonumber
\end{eqnarray}
For the heat capacity,
\begin{eqnarray}
C_V&=&{\partial \beta \over \partial T}{\partial \over \partial \beta}\lt E \gt \nonumber \\
&=&-{1 \over k_B T^2} {\partial \over \partial \beta} \bigg({3\hbar^2 / I \over e^{\beta \hbar^2 / I}+3}\bigg) \nonumber \\
&=&{3 \hbar^4 \over k_B T^2 I^2} {e^{\hbar^2 \beta / I} \over \big(e^{\beta \hbar^2 / I} + 3\big)^2} \nonumber \\
&\approx&3k_B\bigg({\hbar^2 \over I k_B T}\bigg)^2 \exp \bigg(-\hbar^2 / I k_B T\bigg) \nonumber
\end{eqnarray}