Question

a. In the rotational spectrum of \(\mathrm{H}^{33} \mathrm{Cl}(I=2.65 \times\) \(\left.10^{-47} \mathrm{kg} \mathrm{m}^{2}\right),\) the transition corresponding to the \(J=4 \mathrm{to}\) \(J=5\) transition is the most intense. At what temperature was the spectrum obtained? b. At \(1000 .\) K, which rotational transition of \(\mathrm{H}^{35} \mathrm{Cl}\) would you expect to demonstrate the greatest intensity? c. Would you expect the answers for parts (a) and (b) to change if the spectrum were of \(\mathrm{H}^{37} \mathrm{Cl}\) ?

Short answer

a. The temperature at which the spectrum was obtained is approximately \(T = 23.8\) K. b. At 1000 K, the most intense rotational transition for HCl is the J=19 to J=20 transition. c. The answers for parts (a) and (b) would not change significantly if the spectrum were of HCl with a different isotope since the changes in the moment of inertia mostly cancel out in the intensity ratio.

Step-by-step solution

Boltzmann distribution for rotational levels

The probability (P) of a molecule being in a particular rotational energy level (E) is given by the Boltzmann distribution: \[P(E) = \frac{g(E)e^{-E/kT}}{Q}\] Where: - \(g(E)\) is the degeneracy of the energy level, - \(E\) is the energy, - \(k\) is the Boltzmann constant, - \(T\) is the temperature in Kelvin, and - \(Q\) is the partition function.

Degeneracy and energy of rotational levels

For linear molecules like HCl, the degeneracy \(g(J)\) is given by \(g(J) = 2J + 1\), and the energy of the rotational level is given by \(E_J = BJ(J + 1)\), where \(B\) is the rotational constant and \(J\) is the quantum number.

Find the intensity ratio

Since we know the most intense transition is from J=4 to J=5, we need to find the intensity ratio \(I_{4 \to 5} = \frac{P(E_5)}{P(E_4)}\). Due to cancellation in this ratio, the partition function is now irrelevant. Rearranging the Boltzmann distribution equation and plugging in the expressions for energy and degeneracy, we get: \[\frac{P(E_5)}{P(E_4)} = \frac{(2(5)+1)e^{-(B5(6))/kT}}{(2(4)+1)e^{-(B4(5))/kT}}\]

Solve for T

Solve the previous expression for the temperature \(T\): \[T = \frac{B(5)(6-4(5))}{k\ln{(\frac{11}{9})}}\] We are given the moment of inertia (I = \(2.65 \times 10^{-47}\) kg\(m^2\)) and can calculate the rotational constant \(B = \frac{\hbar^2}{2I}\), where \(\hbar\) is the reduced Planck constant. Substitute the values of \(B\) and \(k\) to obtain the temperature. #b. Most intense rotational transition at 1000 K#

Determine the most intense transition

To find which transition would have the greatest intensity at 1000 K, calculate the intensities of adjacent transitions: \(I_{0 \to 1}\), \(I_{1 \to 2}\), \(I_{2 \to 3}\), and so on, using the same procedure as in part (a) along with the given temperature. From step 3, we used the following expression: \[\frac{P(E_{J+1})}{P(E_J)} = \frac{(2(J+1)+1)e^{-(B(J+1)(J+2))/kT}}{(2J+1)e^{-(BJ(J+1))/kT}}\] When the expression begins decreasing, the previous transition had the greatest intensity. By examining the expression, we also can notice as \(J\) increases, the exponential term will eventually overpower the linear term (degeneracy term in the numerator) resulting in diminishing intensity. #c. Change of results for H^{37}Cl#

Compare isotopes

For different isotopes, the moment of inertia \(I\) will change, which in turn affects the rotational constant \(B\). However, due to the ratio in the Boltzmann distribution, the changes in moment of inertia and \(B\) will cancel out for the most part. Therefore, the answers for parts (a) and (b) should not differ significantly for different isotopes of HCl.

Key concepts

Boltzmann Distribution

The Boltzmann Distribution is a fundamental principle in statistical mechanics that explains how molecules distribute among different energy levels at thermal equilibrium. In relation to rotational spectra, it provides insights on how likely a molecule is to exist in a given rotational energy level. The probability of a molecule occupying an energy level, denoted as \(P(E)\), is governed by:- The degeneracy of the energy level, \(g(E)\)- The actual energy, \(E\)- Temperature, \(T\)- Boltzmann's constant, \(k\)The distribution is mathematically expressed as:\[P(E) = \frac{g(E)e^{-E/kT}}{Q}\]where \(Q\) is the partition function that ensures probabilities sum up to one. The equation describes how the probability decreases exponentially with increasing energy levels. The higher the temperature, the more likely molecules will occupy higher energy states. This principle helps to understand why certain transitions in rotational spectra show more intensity than others.

Rotational Energy Levels

In molecules, rotational energy levels represent the quantized states that a rotor can occupy as it rotates around an axis. For linear molecules like HCl, these states are determined by the rotational constant \(B\) and the quantum number \(J\). The energy associated with each level is defined by:- \(E_J = BJ(J + 1)\)Here, \(J\) is the rotational quantum number, and each level can be viewed as a rung on a ladder where molecules can "jump" between. Higher values of \(J\) correspond to higher rotational energies.Each rotational energy level is degenerate, meaning multiple states can have the same energy. The degeneracy is given by \(g(J) = 2J + 1\), indicating that as \(J\) increases, there are more possible orientations for the molecules. These energy levels and degeneracies are crucial for predicting the positions and intensities of spectral lines in a rotational spectrum.

Moment of Inertia

The Moment of Inertia, \(I\), is a measure of a molecule's resistance to rotational motion and plays a central role in determining its rotational spectrum. It depends on the mass distribution relative to the axis of rotation and is calculated for diatomic molecules using:- \(I = \mu r^2\)where:

• \(\mu\) is the reduced mass of the molecule
• \(r\) is the bond length

A larger moment of inertia indicates that a molecule can reach the same rotational energy level with lower angular velocity. Consequently, \(I\) influences the rotational constant \(B\), calculated as:\[B = \frac{\hbar^2}{2I}\]where \(\hbar\) is the reduced Planck's constant.In the context of rotational spectra, the moment of inertia determines the spacing between rotational energy levels. Variations in \(I\) for different isotopes affect the rotational constant, hence altering the observed spectral features. However, changes often cancel out in ratios involving Boltzmann distribution.

Quantum Transitions

Quantum Transitions describe the process of moving from one quantum state to another, fundamental to understanding spectroscopic observations. In rotational spectroscopy, these transitions occur between different rotational energy levels, which are quantized. When a molecule absorbs or emits radiation, it transitions between these levels, defined by changes in the quantum number \(J\).Key points about quantum transitions in the context of rotational spectra:

• The transition involves absorbing or emitting a photon with energy equal to the difference between two rotational energy levels.
• Allowed transitions follow the selection rule \(\Delta J = \pm 1\), meaning \(J\) increases or decreases by 1 across the transition.
• The spectra reveal discrete lines, each corresponding to a distinct transition, and the intensity of these lines is governed by the Boltzmann distribution and energy degeneracy.

Understanding quantum transitions helps predict and interpret the rotational spectral lines' intensities and positions, aiding in molecular identification and structural analysis.