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(II) (a) Calculate the "characteristic rotational energy", \(\hbar^{2} / 2 I\) , for the \(\mathrm{O}_{2}\) molecule whose bond length is 0.121 \(\mathrm{nm}\) . (b) What are the energy and wavelength of photons emitted in an \(\ell=2\) to \(\ell=1\) transition?

Short Answer

Expert verified
The characteristic rotational energy is \(2.83 \times 10^{-22}\) J. The photon emitted has energy \(5.66 \times 10^{-22}\) J and wavelength 3520 nm.

Step by step solution

01

Understand the Characteristic Rotational Energy Formula

The characteristic rotational energy for a diatomic molecule is given as \( \frac{\hbar^2}{2I} \), where \( I \) is the moment of inertia. The moment of inertia for a diatomic molecule is \( I = \mu r^2 \) where \( \mu \) is the reduced mass and \( r \) is the bond length.
02

Calculate the Reduced Mass

The reduced mass \( \mu \) of the \( \mathrm{O}_{2} \) molecule can be calculated using \( \mu = \frac{m_1 m_2}{m_1 + m_2} \). The atomic mass of oxygen \( m_1 = m_2 = 16 \text{ u} = 2.66 \times 10^{-26} \text{ kg} \), so \( \mu = 1.33 \times 10^{-26} \text{ kg} \).
03

Calculate the Moment of Inertia

Substitute \( \mu = 1.33 \times 10^{-26} \text{ kg} \) and \( r = 0.121 \text{ nm} = 0.121 \times 10^{-9} \text{ m} \) into the formula: \[ I = \mu r^2 = 1.33 \times 10^{-26} \times (0.121 \times 10^{-9})^2 \, \text{kg} \cdot \text{m}^2 \] Calculating this gives \( I = 1.95 \times 10^{-46} \text{ kg}\cdot\text{m}^2 \).
04

Calculate Characteristic Rotational Energy

Substitute \( \hbar = 1.05 \times 10^{-34} \text{ J}\cdot\text{s} \) and \( I = 1.95 \times 10^{-46} \text{ kg}\cdot\text{m}^2 \) into the formula: \[ \text{Characteristic Energy} = \frac{\hbar^2}{2I} = \frac{(1.05 \times 10^{-34})^2}{2 \times 1.95 \times 10^{-46}} \text{ J} \] Calculating this gives \( 2.83 \times 10^{-22} \text{ J} \).
05

Calculate Energy Difference for Rotational Transition

The energy difference for a transition between rotational levels is given by \( E_{\ell} = \frac{\ell (\ell + 1)\hbar^2}{2I} \). For \( \Delta E = E_{2} - E_{1} \): \[ \Delta E = \left( \frac{2 \cdot 3 \cdot \hbar^2}{2I} \right) - \left( \frac{1 \cdot 2 \cdot \hbar^2}{2I} \right) = \frac{6\hbar^2}{2I} - \frac{2\hbar^2}{2I} = \frac{4\hbar^2}{2I} \] Substituting values gives \( \Delta E = 5.66 \times 10^{-22} \text{ J} \).
06

Calculate Wavelength of Emitted Photon

Use the energy-wavelength relation \( \Delta E = \frac{hc}{\lambda} \) to find \( \lambda \). Substituting \( \Delta E = 5.66 \times 10^{-22} \text{ J} \), \( h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s} \), and \( c = 3.00 \times 10^{8} \text{ m/s} \) gives: \[ \lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \cdot 3.00 \times 10^{8}}{5.66 \times 10^{-22}} \text{ m} \] This gives \( \lambda = 3.52 \times 10^{-6} \text{ m} \), or \( 3520 \text{ nm} \).

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Key Concepts

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

Diatomic Molecule
A diatomic molecule is a molecule composed of only two atoms. These atoms can either be of the same element or of different elements. An example of a diatomic molecule involving the same element is oxygen (\(\mathrm{O}_{2}\)), while carbon monoxide (\(\mathrm{CO}\)) is an example involving different elements. Diatomic molecules are key to understanding molecular physics because their simplicity allows us to study their rotational and vibrational dynamics effectively.
In the context of rotational energy, we specifically look at how these molecules rotate around an axis. For example, when analyzing the rotational energy of \(\mathrm{O}_{2}\), we focus on how the bond length and molecular mass contribute to determining the molecule’s moment of inertia and subsequent rotational energy levels. Understanding diatomic molecules offers insight into complex molecular behavior in interactions and reactions.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation. In the case of a diatomic molecule like \(\mathrm{O}_{2}\), the moment of inertia can be calculated using the formula: \[ I = \mu r^2 \] where \(\mu\) is the reduced mass and \(r\) is the bond length.
Calculating the moment of inertia for a molecule involves knowing these variables:
  • Reduced Mass: Derived from the masses of the two individual atoms within the molecule.
  • Bond Length: The distance between the two atomic nuclei in a diatomic molecule.
  • Unit: The unit of moment of inertia in scientific calculations is kg·m².
For \(\mathrm{O}_{2}\), using the known bond length and reduced mass, we can compute the moment of inertia to determine how easily the molecule can rotate. This is crucial for calculating the characteristic rotational energy, further getting insights into molecule behavior.
Reduced Mass
The reduced mass is an important concept used in calculating the moment of inertia and rotational energy of a diatomic molecule. It is effectively the "effective mass" that accounts for both objects in a two-body problem and ensures symmetry in the equations of motion.
The reduced mass \(\mu\) is given by the formula: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] where \(m_1\) and \(m_2\) are the masses of the two atoms forming the diatomic molecule.
In the example of an oxygen molecule \(\mathrm{O}_{2}\):
  • Both atoms have identical mass.
  • Hence, the reduced mass is a simple fraction of the individual atomic mass.
Understanding and accurately calculating the reduced mass allows for precise determination of other properties such as the moment of inertia, which subsequently affects rotational energy calculations. This key measure provides deeper insights into how molecular dynamics unfold.
Photon Wavelength
In the context of rotational transitions in diatomic molecules, photon wavelength is a crucial concept. When a molecule transitions between rotational energy levels, it either absorbs or emits a photon. This process is accompanied by a change in energy corresponding to the difference in these levels.
According to the energy-wavelength relation given by: \[ \Delta E = \frac{hc}{\lambda} \] where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
When the emitted photon's wavelength is calculated, it allows us to understand the energy changes involved in molecular rotational transitions. For example:
  • Photon emission during a transition from \(\ell=2\) to \(\ell=1\) in \(\mathrm{O}_{2}\) results in identifying the specific wavelength of light associated with this energy change.
  • This process is crucial in spectroscopy, where different wavelengths help identify molecular structures and energy distribution.
By comprehending photon wavelength calculations, we better understand molecular transitions and the resultant electromagnetic interactions.

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