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The allowed rotational energies \(E_{l}^{\text {rot }}\) of a diatomic molecule are given by $$E_{l}=\frac{l(l+1) \hbar^{2}}{2 I}$$ In this expression \(l\) is the rotational quantum number and can take the values \(l=0,1,2 \ldots ; I\) is the rotational inertia of the molecule about an axis through its center of mass; and \(\hbar=h / 2 \pi .\) The equilibrium separation of the two atoms in a diatomic hydrogen molecule \(H_{2}\) is about 0.074 nm. The mass of each hydrogen atom is about \(1.67 \times 10^{-27} \mathrm{kg}\) a. Show that the rotational inertia of the hydrogen molecule about an axis through its center of mass is about \(I=4.6 \times 10^{-48} \mathrm{kg} \cdot \mathrm{m}^{2}\) b. Find the difference in energy between the first excited rotational energy state and the ground rotational state. That is, find \(E_{1}^{\text {ot }}-E_{0}^{\text {fot }}\). Express the answer in both \(\mathrm{J}\) and \(\mathrm{eV}\). c. Find the relative likelihood \(P_{l=1} / P_{l=0}\) that a hydrogen molecule will be in its first excited rotational state in thermal equilibrium at room temperature, \(T=293 \mathrm{K}\) nore possible state degeneracies.

Step by step solution

01

Calculate Rotational Inertia I

For a diatomic molecule, the moment of inertia, \( I \), is given by \( I = \mu r^2 \). Here, \( \mu \) is the reduced mass, expressed as \( \mu = \frac{m_1 m_2}{m_1 + m_2} \), where both \( m_1 \) and \( m_2 \) are the masses of the hydrogen atoms. Since the molecule is symmetric, \( \mu = \frac{m m}{2m} = \frac{m}{2} \). With each hydrogen atom's mass \( m = 1.67 \times 10^{-27} \) kg, and \( r = 0.074 \) nm \( = 0.074 \times 10^{-9} \) m, the inertia \( I = \left(\frac{1.67 \times 10^{-27}}{2}\right) (0.074 \times 10^{-9})^2 \approx 4.6 \times 10^{-48} \text{ kg m}^2 \).

02

Determine Rotational Energy Levels

The energy levels for rotation are given by \( E_{l}=\frac{l(l+1)\hbar^{2}}{2I} \). Here, the Planck constant \( \hbar = \frac{h}{2\pi} \) and \( h = 6.626 \times 10^{-34} \text{ J s} \). For \( l = 0 \) (ground state), \( E_0 = \frac{0(0+1)\hbar^{2}}{2I} = 0 \). For \( l = 1 \) (first excited state), \( E_1 = \frac{1(1+1)\hbar^2}{2I} = \frac{2\hbar^2}{2I} = \frac{\hbar^2}{I} \).

03

Calculate Energy Difference E_1 - E_0

The energy difference is \( E_1 - E_0 = \frac{\hbar^2}{I} - 0 = \frac{\hbar^2}{I} \). Substitute the known values \( \hbar = \frac{6.626 \times 10^{-34}}{2\pi} \) and \( I = 4.6 \times 10^{-48} \text{ kg m}^2 \), we have \( E_1 - E_0 \approx \frac{(1.055 \times 10^{-34})^2}{4.6 \times 10^{-48}} \). Calculate to find \( E_1 - E_0 \).

04

Convert Energy Difference to eV

To convert the energy from Joules to electronvolts, use the conversion \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Divide the energy difference calculated in Step 3 by this conversion factor to find the energy in electronvolts.

05

Compute Relative Probability P(l=1)/P(l=0)

The relative likelihood is given by the Boltzmann distribution: \( \frac{P_{l=1}}{P_{l=0}} = \frac{g_1 e^{-{E_1}/{kT}}}{g_0 e^{-{E_0}/{kT}}} \), where \( g_l \) is the degeneracy (which is \( 2l+1 \)), \( k \) is Boltzmann's constant \( 1.38 \times 10^{-23} \text{ J/K} \), and \( T = 293 \text{ K} \). Here \( g_1 = 3 \), \( g_0 = 1 \), and since \( E_0 = 0 \), \( \frac{P_{l=1}}{P_{l=0}} = 3 e^{-E_1/kT} \). Compute this using the energy from Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diatomic Molecules

Diatomic molecules consist of two atoms that are covalently bonded. These are the simplest form of polyatomic molecules. Examples of such molecules include hydrogen (Hâ‚‚), oxygen (Oâ‚‚), and nitrogen (Nâ‚‚). Because of their simplicity, diatomic molecules are often the subject of studies in chemical bonding and molecular dynamics.
The rotation of these molecules around an axis perpendicular to the line joining the two atoms has quantized energy levels. This quantization arises due to the restrictions imposed by the uncertainty principle and quantum mechanics. Thus, diatomic molecules serve as a fundamental system to study rotational dynamics, providing essential insights into quantum mechanical behavior.
Understanding diatomic molecules involves knowing their bond lengths and masses, which are crucial for calculations like rotational inertia and energy levels.

Rotational Inertia

Rotational inertia, or moment of inertia, is pivotal for understanding the rotational dynamics of molecules. It describes how mass is distributed relative to an axis of rotation. For a diatomic molecule, rotational inertia (\(I\), is calculated using the formula: \(I = \mu r^2\), where \mu is the reduced mass and \r is the distance between the two atoms.
The reduced mass (\(\mu\),) is defined as \(\mu = \frac{m_1 m_2}{m_1 + m_2}\). In a diatomic molecule like hydrogen (\(H_2\)), both masses are equal; therefore, \(\mu = \frac{m}{2}\).
To further grasp rotational inertia, one can consider it analogous to linear inertia. Just as linear inertia is about an object resisting change in motion, rotational inertia is about an object resisting changes to its state of rotation. When you calculate the rotational inertia, you can better predict the necessary energy for rotational states, which directly ties into quantum mechanics theories.

Boltzmann Distribution

The Boltzmann distribution is key to understanding the equilibrium distribution of molecule rotational states at a given temperature. It predicts the relative probability of a molecule occupying different energy states. For rotational motions, this distribution is expressed as: \[\frac{P_{l}}{P_{0}} = \frac{g_{l}e^{-E_{l}/kT}}{g_{0}e^{-E_{0}/kT}}\]where \(P_{l}\) is the probability of the system in the l-th state, \(E_{l}\) is the energy of that state, \(k\) is Boltzmann's constant, \(T\) is temperature, and \(g_{l}\) denotes the degeneracy of the state.
As temperature rises, higher energy levels become more populated due to increasing thermal energy. The Boltzmann distribution therefore provides a tool for understanding molecular behavior under various thermal conditions, explaining how energy populations shift across different rotational levels.
This distribution highlights the dynamic nature of molecular systems in thermal environments, influencing fields from chemistry to statistical physics.

Quantum Mechanics

Quantum mechanics describes the physical properties of nature at the scale of atoms and subatomic particles. Its principles are crucial in comprehending rotational energy levels. In quantum mechanics, particles exhibit both wave-like and particle-like properties, which means that diatomic molecules may only exist in discrete energy levels for rotation.
The concept of quantized energy levels stems from the requirement that only certain rotational motions are possible. These quantized levels are determined by the rotational quantum number (\(l\)), and each level corresponds to a specific energy given by \(E_{l} = \frac{l(l+1)\hbar^2}{2I}\). This quantization results in fascinating phenomena, such as the lack of continuous energy emission and absorption, which classical physics cannot explain.
By combining principles of quantum mechanics with other tools such as rotational inertia and the Boltzmann distribution, we gain a full picture of a molecule's behavior, bridging the microscopic world of atoms to the macroscopic properties observed in materials.