Problem 7 - An Introduction to Modern Astrophysics

Question

The rotational kinetic energy of a molecule is given by $$E_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}=\frac{L^{2}}{2 I}$$ where \(L\) is the molecule's angular momentum and \(I\) is its moment of inertia. The angular momentum is restricted by quantum mechanics to the discrete values where \(\ell=0,1,2, \ldots\) (a) For a diatomic molecule $$I=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}$$ where \(m_{1}\) and \(m_{2}\) are the masses of the individual atoms and \(r_{1}\) and \(r_{2}\) are their separations from the center of mass of the molecule. Show that \(I\) may be written as $$I=\mu r^{2}$$ where \(\mu\) is the reduced mass and \(r\) is the separation between the atoms in the molecule. (b) The separation between the carbon and oxygen atoms in CO is approximately \(0.12 \mathrm{nm},\) and the atomic masses of \(^{12} \mathrm{C},^{13} \mathrm{C},\) and \(^{16} \mathrm{O}\) are \(12.000 \mathrm{u}, 13.003 \mathrm{u},\) and \(15.995 \mathrm{u},\) respectively. Calculate the moments of inertia for \(^{12} \mathrm{CO}\) and \(^{13} \mathrm{CO}\) (c) What is the wavelength of the photon that is emitted by 12 CO during a transition between the rotational angular momentum states \(\ell=3\) and \(\ell=2 ?\) To which part of the electromagnetic spectrum does this correspond? (d) Repeat part (c) for \(^{13}\) CO. How do astronomers distinguish among different isotopes in the interstellar medium?

Step-by-Step Solution

Answer

Calculate rotational and vibrational states for each isotope and use the differences in emitted wavelengths to distinguish between them. Astronomers use unique spectral lines to identify isotopes.

Step 1: Breaking down the moment of inertia

For a diatomic molecule, the moment of inertia is given by the formula \( I = m_1 r_1^2 + m_2 r_2^2 \). From the center of mass concept, \( m_1 r_1 = m_2 r_2 \). Solving \( r_1 \) and \( r_2 \) from this equation gives \( r_1 = \frac{m_2}{m_1+m_2}r \) and \( r_2 = \frac{m_1}{m_1+m_2}r \). Substitute these into the inertia formula to show \( I = \mu r^2 \), with \( \mu = \frac{m_1 m_2}{m_1 + m_2} \).

Step 2: Calculate moment of inertia

For \( ^{12} \text{CO} \), use \( m_1 = 12.000 \text{u} \), \( m_2 = 15.995 \text{u} \), and \( r = 0.12 \text{ nm} \). Compute the reduced mass \( \mu = \frac{12.000 \times 15.995}{12.000 + 15.995} \). Substitute into \( I = \mu r^2 \) to find \( I \). Repeat similar calculations for \( ^{13} \text{CO} \). Convert atomic mass units to kilograms for accurate calculations.

Step 3: Finding energy difference between states

Use the rotational energy equation \( E_{\text{rot}} = \frac{L^2}{2I} \) and \( L = \hbar \sqrt{\ell (\ell + 1)} \). Calculate the energies for states \( \ell = 3 \) and \( \ell = 2 \): \( E_3 \) and \( E_2 \) respectively. The energy of the photon emitted is \( E_3 - E_2 \).

Step 4: Calculating photon's wavelength and spectrum

Use Planck's relation for wavelength \( \lambda = \frac{hc}{E_3 - E_2} \). Calculate \( \lambda \) using the energy difference found in the previous step. Check the wavelength against known electromagnetic spectrum ranges (e.g., infrared, microwave) to determine the type of radiation.

Step 5: Repeat for different isotope

Repeat the process for \( ^{13} \text{CO} \), calculating rotational energy levels, the differences in energies for the transition \( \ell = 3 \) to \( \ell = 2 \), and subsequently determining the wavelength emitted. Compare these results between isotopes to see the differences in emitted wavelengths.

Step 6: Conclusion on distinguishing isotopes

Astronomers distinguish different isotopes based on their unique spectral lines, which arise from differences in molecular rotational transitions. By identifying these spectral lines, astronomers can infer the presence of specific isotopes in the interstellar medium.

Key Concepts

Rotational Kinetic Energy

Rotational kinetic energy is a form of energy related to the spinning motion of molecules. Much like the kinetic energy of a moving object, rotational kinetic energy depends on specific physical properties—here represented by the moment of inertia and angular velocity. For molecules, it is calculated using the formula:

• \[ E_{\mathrm{rot}} = \frac{1}{2} I \omega^{2} = \frac{L^{2}}{2 I} \]
• Where \( I \) is the moment of inertia, \( \omega \) is the angular velocity, and \( L \) is the molecule's angular momentum.

In quantum mechanics, only specific, discrete angular momentum values are allowed. These are based on the principal quantum number \( \ell \), where \( \ell = 0, 1, 2, \ldots \). Each state change corresponds to an energy level transition, which is crucial for understanding molecular behaviors, especially during spectroscopic analysis.

Moment of Inertia in Molecules

Understanding the moment of inertia is key to predicting how molecules rotate. In diatomic molecules, the moment of inertia can be visualized as two atoms connected by a rigid bond. For such molecules, it's expressed as:

• \[ I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} \]
• Where \( m_{1} \) and \( m_{2} \) are atomic masses, and \( r_{1} \) and \( r_{2} \) are the distances from the center of mass.

Using the concept of reduced mass \( \mu \), the moment of inertia simplifies to \( I = \mu r^{2} \) with:

• \[ \mu = \frac{m_{1} m_{2}}{m_{1} + m_{2}} \]
• and \( r \) as the distance between the two atoms.

This reduced mass simplifies calculations and allows better predictions of molecular behavior, especially for molecules like carbon monoxide (CO), where atomic separations directly affect the molecule's rotational properties.

Photon Emission in Molecular Transitions

Photons are the particles of light released during energy state transitions within molecules. When a molecule transitions between rotational energy levels, such as from \( \ell = 3 \) to \( \ell = 2 \), it emits a photon with energy equivalent to the difference in energy levels. This emission is calculated using:

• \[ E_{\text{photon}} = E_3 - E_2 \]
• Where \( E_3 \) and \( E_2 \) are the rotational energies of the initial and final states.

The emitted photon's wavelength \( \lambda \) is derived using Planck's equation:

• \[ \lambda = \frac{hc}{E_{\text{photon}}} \]
• With \( h \) as Planck's constant and \( c \) as the speed of light.

These transitions often result in emission lines in the electromagnetic spectrum, aiding in the identification of molecular changes.

Astrophysical Spectral Analysis

Astrophysical spectral analysis allows scientists to decipher the chemical composition and physical conditions of stars, galaxies, and the interstellar medium. By analyzing the spectrum of light emitted or absorbed by substances, astronomers can determine the presence and quantities of specific isotopes.

• Spectral lines act as fingerprints for molecules and atoms, providing clues about various isotopic compositions and molecular transitions.
• For example, spectral lines arising from CO isotopes exhibit slight shifts due to differences in molecular mass. This is how astronomers identify different isotopes like \( ^{12} \text{CO} \) and \( ^{13} \text{CO} \).
• Spectral analysis also helps in detecting conditions such as temperature, density, and movement within astrophysical objects.

By studying the unique spectral signatures of each element or molecule, astronomers gain insights beyond what is visible, enhancing understanding of the universe's fundamental workings.