Problem 12 - University Physics with Modern Physics

Question

If a sodium chloride (NaCl) molecule could undergo an \(n \rightarrow n-1\) vibrational transition with no change in rotational quantum number, a photon with wavelength 20.0\(\mu \mathrm{m}\) would be emitted. The mass of a sodium atom is \(3.82 \times 10^{-26} \mathrm{kg},\) and the mass of a chlorine atom is \(5.81 \times 10^{-26} \mathrm{kg}\) . Calculate the force constant \(k^{\prime}\) for the interatomic force in NaCl.

Step-by-Step Solution

Answer

The force constant \(k'\) for NaCl is approximately 173 N/m.

Step 1: Derive the relationship for the force constant

In a diatomic molecule, the vibrational frequency \(u\) is related to the force constant \(k'\) by the formula:\[u = \frac{1}{2\pi} \, \sqrt{\frac{k'}{\mu}}\]where \(\mu\) is the reduced mass of the molecule. Our task is to determine \(k'\).

Step 2: Determine the reduced mass

The reduced mass \(\mu\) of the sodium chloride molecule is given by:\[\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\]Plugging in the given masses:\[\begin{align*}m_1 &= 3.82 \times 10^{-26} \text{ kg} \, (\text{mass of } \text{Na}) \m_2 &= 5.81 \times 10^{-26} \text{ kg} \, (\text{mass of } \text{Cl})\\mu &= \frac{(3.82 \times 10^{-26}) \cdot (5.81 \times 10^{-26})}{3.82 \times 10^{-26} + 5.81 \times 10^{-26}} \\mu &\approx 2.25 \times 10^{-26} \text{ kg}\end{align*}\]

Step 3: Calculate the frequency from the wavelength

The energy of the photon emitted is related to its frequency \(u\) by \(c/\lambda = u\), where \(c\) is the speed of light, and \(\lambda\) is the wavelength. We are given \(\lambda = 20.0 \underline{\phantom{xxx}} \mu\text{m} = 20.0 \times 10^{-6} \text{ m}\).Thus, the frequency \(u\) is:\[u = \frac{3.00 \times 10^8 \text{ m/s}}{20.0 \times 10^{-6} \text{ m}} = 1.50 \times 10^{13} \, \text{Hz}\]

Step 4: Calculate the force constant \(k'\)

We rearrange the earlier formula to solve for \(k'\):\[k' = (2\piu)^2 \cdot \mu\]Substituting the known values:\[k' = (2\pi \times 1.50 \times 10^{13})^2 \cdot 2.25 \times 10^{-26}\]\[k' \approx 1.73 \times 10^{2} \, \text{N/m}\]

Key Concepts

Force Constant Calculation

Understanding the force constant is key to analyzing vibrational transitions in molecules. The force constant, denoted as \(k'\), is a measure of how stiff the bond is between two atoms in a molecule. A higher force constant indicates a stronger, more difficult to stretch bond. This concept is vital in vibrational spectroscopy where molecular vibrations can be studied to gain insights about molecular bonds.

The force constant can be determined using the formula for vibrational frequency \(u\) derived from Hooke's Law for a harmonic oscillator:

• \(u = \frac{1}{2\pi} \, \sqrt{\frac{k'}{\mu}}\)

\(u\) represents the vibrational frequency, and \(\mu\) is the reduced mass of the molecule. Rearranging this formula to solve for the force constant, we get:

• \(k' = (2\pi u)^2 \cdot \mu\)

By substituting the correct values for \(u\) and \(\mu\), students can calculate the force constant, giving insight into the nature of the molecular bond in question.

Reduced Mass

Reduced mass is a crucial concept when dealing with systems of two interacting particles, like the sodium (Na) and chlorine (Cl) atoms forming a diatomic molecule here. It simplifies the math involved in dynamical problems by allowing us to treat two separate masses as one. This is especially helpful in vibrational analysis.

The formula to find the reduced mass \(\mu\) is:

• \(\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\)

where \(m_1\) and \(m_2\) are the masses of the sodium and chlorine atoms, respectively. This calculation helps reduce the complexity of the system from two masses to a single "effective" mass. Crucially, self-contained in the calculation of the vibrational frequency \(u\) and consequently the force constant \(k'\), the reduced mass ensures that the focus remains strictly on the vibrational characteristics of the molecule without getting the actual masses directly involved each time in the computation.

Photon Wavelength

Understanding the wavelength of a photon is fundamental in molecular spectroscopy, as it is directly related to the photon’s energy and frequency. When a molecule undergoes a vibrational transition, a photon with a specific wavelength is emitted or absorbed. This wavelength, \(\lambda\), can be used to determine the photon’s frequency \(u\) using the speed of light \(c\) as:

• \(u = \frac{c}{\lambda}\)

Given the wavelength in the exercise as \(20.0\,\mu \mathrm{m}\) or \(20.0 \times 10^{-6} \mathrm{m}\), plug this into the equation to find \(u\), understanding that \(c = 3.00 \times 10^8\, \mathrm{m/s}\).

A correct calculation of \(u\) is crucial because it serves as a bridge to finding other vital characteristics of the molecule, such as the force constant. By grasping how wavelength relates to frequency, students can deeply understand the enchanting interplay between light and matter.