Question 705061

1. Energy and Degeneracy of Rotational States: The rotational energy for a state $j$ is given by: $E_j = \frac{\hbar^2}{2I}j(j+1)$

For each rotational energy level $j$, the degeneracy (number of allowed orientations or magnetic quantum numbers $m_j$) is: $g_j = 2j + 1$

2. Partition Function ($Z$): The rotational partition function is the sum over all states: $Z = \sum_{j=0}^{\infty} g_j e^{-\frac{E_j}{k_B T}}$ $Z = \sum_{j=0}^{\infty} (2j + 1) e^{-\frac{\hbar^2 j(j+1)}{2I k_B T}}$

3. High-Temperature Approximation: We are given the condition $k_B T \gg \frac{\hbar^2}{2I}$. Under this high-temperature limit, the energy spacing between adjacent rotational levels is very small compared to the thermal energy $k_B T$. Therefore, the rotational energy levels can be treated as a continuum, and we can replace the summation with an integral:

$Z \approx \int_{0}^{\infty} (2j + 1) e^{-\frac{\hbar^2 j(j+1)}{2I k_B T}} dj$

To solve this integral, we can use a simple substitution:

Let $x = j(j+1)$

Differentiating both sides with respect to $j$:

$dx = (2j + 1) dj$

The limits of integration remain from $0$ to $\infty$. Substituting $x$ and $dx$ into the integral:

$Z = \int_{0}^{\infty} e^{-\frac{\hbar^2}{2I k_B T} x} dx$

Evaluating the exponential integral:

$Z = \left[ \frac{-2I k_B T}{\hbar^2} e^{-\frac{\hbar^2}{2I k_B T} x} \right]_{0}^{\infty}$

$Z = \frac{-2I k_B T}{\hbar^2} \left( e^{-\infty} - e^0 \right)$

$Z = \frac{-2I k_B T}{\hbar^2} (0 - 1)$

$Z = \frac{2I k_B T}{\hbar^2}$

4. Probability of the Ground State: According to Boltzmann statistics, the probability $P_j$ of the molecule being in the $j$-th state is: $P_j = \frac{g_j e^{-\frac{E_j}{k_B T}}}{Z}$

For the ground state, $j = 0$:

Degeneracy $g_0 = 2(0) + 1 = 1$

Energy $E_0 = \frac{\hbar^2}{2I}(0)(1) = 0$

Substituting these ground state values into the probability equation: $P_0 = \frac{1 \cdot e^0}{Z}$

$P_0 = \frac{1}{Z}$

Now, substitute the value of the partition function $Z$ we calculated earlier:

$P_0 = \frac{1}{\left( \frac{2I k_B T}{\hbar^2} \right)}$

$P_0 = \frac{\hbar^2}{2I k_B T}$