Problem 35 - University Physics with Modern Physics

Question

A hypothetical diatomic molecule of oxygen (mass =\( 2.656 \times 10^{-26} \mathrm{kg} ) \quad \text { and } \quad \text { hydrogen } \quad\left(\text { mass }=1.67 \times 10^{-27} \mathrm{kg}\right)\) emits a photon of wavelength 2.39\(\mu \mathrm{m}\) when it makes a transition from one vibrational state to the next lower state. If we model this molecule as two point masses at opposite ends of a massless spring, (a) what is the force constant of this spring, and (b) how many vibrations per second is the molecule making?

Step-by-Step Solution

Answer

(a) The force constant is approximately 0.815 N/m, and (b) the molecule makes about 1.2552 x 10^14 vibrations per second.

Step 1: Convert Wavelength to Frequency

First, we need to find the frequency of the emitted photon using the given wavelength. The formula to convert wavelength to frequency is given by the equation \( c = \lambda u \), where \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light, \( \lambda = 2.39 \times 10^{-6} \text{ m} \) is the wavelength, and \( u \) is the frequency. Solving for \( u \), we have: \[ u = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{2.39 \times 10^{-6} \text{ m}} \]

Step 2: Calculate Frequency

Now, compute \( u \):\[ u = \frac{3 \times 10^8}{2.39 \times 10^{-6}} = 1.2552 \times 10^{14} \text{ Hz} \] This is the frequency of vibration of the molecule for the photon emitted.

Step 3: Calculate Reduced Mass of the System

For a diatomic molecule, the masses are modeled as two masses at ends of a spring. The reduced mass \( \mu \) is given by:\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]where \( m_1 = 2.656 \times 10^{-26} \text{ kg} \) (mass of oxygen) and \( m_2 = 1.67 \times 10^{-27} \text{ kg} \) (mass of hydrogen). Solve for \( \mu \):\[ \mu = \frac{2.656 \times 10^{-26} \times 1.67 \times 10^{-27}}{2.656 \times 10^{-26} + 1.67 \times 10^{-27}} \]

Step 4: Simplify and Calculate Reduced Mass

Plug the values into the reduced mass equation and simplify:\[ \mu = \frac{4.43352 \times 10^{-53}}{2.82267 \times 10^{-26}} = 1.571 \times 10^{-27} \text{ kg} \]

Step 5: Use Energy to Find Force Constant

The energy of a photon is given by the relation \( E = h u \), where \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \) (Planck's constant). Using the energy equation for harmonic oscillation and the relation between frequency, reduced mass \( \mu \), and force constant \( k \):\[ u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \]Rearrange to solve for \( k \):\[ k = (2\pi u)^2 \mu \]

Step 6: Calculate Force Constant

Substitute the values to find \( k \):\[ k = (2\pi \times 1.2552 \times 10^{14})^2 \times 1.571 \times 10^{-27} \]Calculating this gives \( k \) as approximately \( 8.15 \times 10^{-1} \text{ N/m} \).

Step 7: Compute Vibrations Per Second

The frequency of vibration \( u \) already calculated in Step 2 directly translates to the number of vibrations per second. Therefore, the number of vibrations per second made by the molecule is \( 1.2552 \times 10^{14} \) Hz.

Key Concepts

Diatomic Molecule

A diatomic molecule is made up of two atoms bonded together. These atoms can be of the same element—such as the oxygen molecules in the air we breathe, which are made of two oxygen atoms—or different elements, such as the hypothetical oxygen-hydrogen molecule from our exercise. By understanding this basic structure, we can imagine these two atoms as points at either end of a spring. This simplification helps physicists model the behaviors of these molecules based on the physics of harmonic oscillators.

In practical terms, considering a diatomic molecule as two masses connected by a spring means accounting for their physical interactions. When one atom moves, it affects the other through the bond between them, similar to how compressing one end of a spring affects the other end.

• Diatomic molecules are fundamental in understanding more complex molecular structures as they provide a simple yet informative model of molecular dynamics.
• Examples include gases like hydrogen (H₂), nitrogen (N₂), and carbon monoxide (CO).

Reduced Mass

The concept of reduced mass (\( \mu \)) is a handy tool when dealing with two-body systems, like a diatomic molecule. When two masses are connected, as in our model of a diatomic molecule, the reduced mass simplifies analyses by capturing the effective inertia of the system.

For instance, given oxygen and hydrogen as parts of a diatomic molecule, each with their own masses, calculating the reduced mass helps predict how these two atoms will move relative to each other. The formula used is \( \mu = \frac{m_1 m_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses of the two atoms. This simplification is crucial for analyzing vibrational and rotational motions within the molecule.

• Helps convert a two-body problem into a one-body problem, simplifying complex calculations.
• Allows for the calculation of other properties such as vibrational frequency and force constant more conveniently.

Force Constant

The force constant (\( k \)) of a spring, in the context of modeling a diatomic molecule, tells us how stiff the bond is between the two atoms. In a mechanical oscillator, it is analogous to the spring constant that determines how much force is needed to stretch or compress it by a particular amount.

For a molecule acting like a spring, if the force constant is large, it means the bond between the atoms is strong and requires more force to stretch. The force constant is directly related to the vibrational frequency of the diatomic molecule and can be calculated using the formula\[ k = (2\pi u)^2 \mu \]. Here, \( u \) is the vibrational frequency, and \( \mu \) is the reduced mass.

• Indicates the rigidity of molecular bonding in diatomic molecules.
• Higher force constants imply stronger bonds, which correspond to higher vibrational frequencies.

Vibrational Frequency

Vibrational frequency (\( u \)) refers to how often the atoms in a diatomic molecule oscillate back and forth. This frequency represents the molecule's vibration and is measured in Hertz (Hz), meaning vibrations per second.

For a diatomic molecule modeled as two masses on a spring, vibrational frequency is influenced by both the bond strength (force constant) and the reduced mass. The vibrational frequency can be calculated from the wavelength of light emitted when the molecule transitions between energy states using energy conservation principles and the relation \( u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \).

• Essential for understanding molecular behavior in infrared spectroscopy, where different frequencies indicate diverse molecular bonds.
• Provides insights into molecular structure and dynamics through its relation with energy levels.