Question

When a diatomic molecule undergoes a transition from the \(l = 2\) to the \(l = 1\) rotational state, a photon with wavelength 54.3 \(\mu\)m is emitted. What is the moment of inertia of the molecule for an axis through its center of mass and perpendicular to the line connecting the nuclei?

Short answer

The moment of inertia for the molecule is about \(3.59 \times 10^{-47} \text{ kg} \cdot \text{m}^2\).

Step-by-step solution

Understanding Rotational Transitions

The molecule undergoes a transition from the rotational quantum state with angular momentum quantum number \( l = 2 \) to \( l = 1 \). Such a transition means the molecule releases a photon, and the energy difference between these states equates to the energy of this photon.

Finding Energy of Emitted Photon

The energy \( E \) of the emitted photon is related to its wavelength \( \lambda \) by the equation: \[E = \frac{hc}{\lambda}\]where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J}\cdot \text{s} \) and \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \). Substituting \( \lambda = 54.3 \times 10^{-6} \text{ m} \), we find \[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{54.3 \times 10^{-6}} \text{ J}\]

Calculate Change in Energy Levels

For a diatomic rotor, the rotational energy levels are given by:\[E_l = \frac{l(l+1)\hbar^2}{2I}\]where \( I \) is the moment of inertia and \( \hbar = \frac{h}{2\pi} \). The energy difference between \( l = 2 \) and \( l = 1 \) is: \[\Delta E = \frac{2(2+1)\hbar^2}{2I} - \frac{1(1+1)\hbar^2}{2I}\] which simplifies to \[\Delta E = \frac{6\hbar^2}{2I} - \frac{2\hbar^2}{2I} = \frac{4\hbar^2}{2I} = \frac{2\hbar^2}{I}\]

Equating Energy and Solving for moment of Inertia

Since the change in energy \( \Delta E \) is equal to the energy \( E \) of the emitted photon, we have:\[\frac{2\hbar^2}{I} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{54.3 \times 10^{-6}}\]Solving for \( I \), we get:\[I = \frac{2\hbar^2}{E}\]Substitute \( \hbar = \frac{6.626 \times 10^{-34}}{2\pi} \) and the calculated value of \( E \) into the equation to find \( I \).

Calculate I

We already derived that:\[I = \frac{2(1.054 \times 10^{-34})^2}{E}\]Substitute the value of \( E \) from Step 2 into this equation to find the precise value of the moment of inertia.

Key concepts

Diatomic Molecules

Diatomic molecules are molecules composed of only two atoms. These atoms can be of the same element, like oxygen (O2), or different elements, like carbon monoxide (CO). A distinct feature of diatomic molecules is their ability to rotate about an axis through their center of mass. This rotation is similar to the spinning of a disc. The diatomic molecule's rotation can be understood by its quantum states, particularly its rotational quantum states, which quantify this motion.Rotational quantum states are defined by a quantum number, typically denoted by \( l \). This quantum number represents the amount of angular momentum the molecule has during its rotational motion and unfortunately not the actual spinning frequency. These states are quantized, meaning that a diatomic molecule can only occupy specific energy levels corresponding to these quantum numbers. This foundational concept is crucial when analyzing diatomic molecules undergoing rotational transitions.

Rotational Transitions

Rotational transitions in diatomic molecules occur when a molecule changes from one rotational quantum state to another. Such changes are accompanied by either the absorption or emission of a photon. This photon carries away the energy difference between the two rotational states.Specifically, when a molecule transitions from a higher energy level (larger \( l \)) to a lower one (smaller \( l \)), it releases energy in the form of a photon. The rotational quantum states are crucial here, as they determine the allowable transitions. These transitions can only occur between adjacent quantum numbers (e.g., \( l = 2 \) to \( l = 1 \)), adhering to strict selection rules found in quantum mechanics. The emitted photon's energy can be calculated using Planck's constant and the speed of light along with its wavelength, as these factors define the connection between energy and wavelength in electromagnetic radiation.

Photon Energy

Photon energy is a fundamental concept in understanding rotational transitions. When molecules change states, they emit or absorb photons, which are electromagnetic radiation packets. The energy of a photon is inversely proportional to its wavelength, a relationship defined by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the photon energy, \( h \) is Planck's constant, and \( c \) is the speed of light.Applying this concept to a diatomic molecule, when a molecule undergoes a rotational transition, it releases a photon whose wavelength reveals the energy difference between the initial and final states. The shorter the wavelength, the higher the energy of the emitted photon. By calculating the photon energy, we can infer essential properties of molecular motion, including the molecule's moment of inertia, which is central to these rotational modes.

Rotational Quantum State

Rotational quantum states represent the allowable rotational energy levels of a diatomic molecule. These states are determined by the quantum number \( l \), which can take non-negative integer values starting from zero. Each value of \( l \) corresponds to a specific rotational energy level, calculated using the formula \( E_l = \frac{l(l+1)\hbar^2}{2I} \), where \( I \) is the moment of inertia, and \( \hbar = \frac{h}{2\pi} \).As a diatomic molecule transitions between these states, it moves to a different rotational quantum state, creating energy differences equivalent to the photon energy released or absorbed. Such transitions dictate the spectrum of frequencies or wavelengths a molecule can emit or absorb. Therefore, understanding rotational quantum states is key in exploring how molecules rotate and interact with radiation, providing insights into molecular structure and behavior.