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Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is \(f=7 \times 10^{13} \mathrm{~Hz}\). The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to be about \(5 \times 10^{-20} \mathrm{~J}\). In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about \(1.6 \times 10^{-10} \mathrm{~m} ?\)

Short Answer

Expert verified
The spring constant k is \(4.8 \times 10^{-2} N / m\), the maximum speed of the H atom is \(1.3 \times 10^3 m / s\), and the amplitude of the vibrational motion is \(1.0 \times 10^{-10} m\). The amplitude is smaller than the equilibrium distance between the Hydrogen and Iodine atoms.

Step by step solution

01

Calculate the spring constant.

From Hooke's law, we know that \( F = -kx \), where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. However, since we are dealing with simple harmonic motion, we can also use the equation \( f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \) where f is the frequency and m is the mass of the object in motion. In this case, we have the frequency and the mass (mass of a proton, approximately \(1.67 \times 10^{-27} kg\)), meaning we can isolate k in the equation and solve: \( k = (2\pi f)^2 \times m \).
02

Find the maximum speed of the H atom.

In the case of SHM, the maximum speed of the atom occurs when it passes through the equilibrium position, where all of the energy is kinetic. We know that the kinetic energy can be found using the equation \( \frac {1}{2}mv^2 \), and that the total vibrational energy would be equal to the maximum kinetic energy. Therefore, we can rewrite the equation as \( v = \sqrt{\frac{2E}{m}} \), where E is the vibrational energy and m is the mass of the hydrogen atom.
03

Determine the amplitude.

The energy in a harmonic oscillator can be represented as \( E = \frac {1}{2}ka^2 \), where k is the spring constant, and a is the amplitude of the motion. We want to solve for a, meaning that we can rewrite the equation as \( a = \sqrt {\frac {2E}{k}} \). Once we have calculated a, we can compare it to the given equilibrium distance between the two atoms.

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Key Concepts

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

Quantum Mechanics in Vibrational Motion
Quantum mechanics plays a crucial role in understanding the vibrational motion of molecules. In the realm of tiny particles like atoms and molecules, quantum mechanics provides a much more accurate description than classical physics. Vibrations in molecules are quantized, meaning they have specific, fixed energy levels. These energy levels result from the vibrational states of the molecule.

In a molecular vibration, the nuclei of the atoms involved vibrate around a point, much like masses connected by springs. However, unlike classical spring-mass systems, molecular vibrations are governed by quantum rules. This leads to phenomena such as zero-point energy, which is the lowest possible energy a quantum mechanical physical system may have, and it is not zero due to quantum fluctuations.
  • Quantum mechanics provides the framework to calculate these discrete vibrational energy levels.
  • It explains why molecules can absorb or emit photons when transitioning between different vibrational states.
The vibrational motion of molecules has significant implications in spectroscopy, where understanding these discrete energy levels helps identify substances and study molecular structures.
Classical Physics and Simple Harmonic Motion
Classical physics helps simplify the analysis of vibrational motion by modeling it as simple harmonic motion (SHM). In SHM, an object oscillates around an equilibrium position under a restoring force proportional to displacement. This makes it easier to describe the motion mathematically, even for complex systems like molecules.

In the classical approach, we consider molecules as tiny masses connected by springs, representing the bonds. This analogy provides a valuable method to estimate vibrational properties like frequency and amplitude using simple equations.
  • In SHM, variables such as frequency and spring constant are essential.
  • The equations of SHM help determine behaviors like maximum speed and amplitude.
While classical models may not explain all aspects, they are particularly useful in predicting general behavior and initial estimates before diving into exact quantum mechanical calculations.
Understanding Vibrational Frequency
The vibrational frequency of a molecular system is a key concept in both classical physics and quantum mechanics. It refers to how often the atoms in a molecule oscillate back and forth in a second. A higher frequency indicates faster oscillations, which often correlate with stronger molecular bonds.

In the classical model, vibrational frequency can be calculated using the relationship between the spring constant and the mass of the system:\[ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\]where \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass of the atom in motion.
  • This formula shows how frequency depends on the mass of the moving atom and the stiffness of the bond.
  • More massive atoms tend to have lower vibrational frequencies.
Understanding vibrational frequency is crucial for interpreting vibrational spectra in analytical chemistry and material science.
The Role of Spring Constant
The spring constant, often denoted as \( k \), is a measure of the stiffness of the spring, or in molecular terms, the stiffness of the bond linking the atoms. A larger spring constant means a stiffer bond that requires more force to be displaced.

In simple harmonic motion, the spring constant is directly related to the force exerted by the spring and can be determined by:\[ k = (2\pi f)^2 \times m\]where \( f \) is the frequency, and \( m \) is the mass of the moving atom.
  • The spring constant is essential for determining the system's energy dynamics.
  • It plays a critical role in calculating the amplitude and speed of vibrational motion.
This parameter helps in understanding how tightly atoms are bonded in a molecule, which is vital for exploring mechanical properties and stability in materials. In the context of molecular physics, it is instrumental in predicting the behavior of atoms during vibrations and informs the formation and breaking of chemical bonds.

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Most popular questions from this chapter

A \(10.0 \mathrm{~kg}\) mass is traveling to the right with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second \(10.0 \mathrm{~kg}\) mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant \(170.0 \mathrm{~N} / \mathrm{m}\). The other end of the spring is fixed to a wall to the right of the second mass. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

An object is undergoing SHM with period \(1.200 \mathrm{~s}\) and amplitude \(0.600 \mathrm{~m} .\) At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{~s} ?\)

A slender rod of length \(80.0 \mathrm{~cm}\) and mass \(0.400 \mathrm{~kg}\) has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia \(I_{\mathrm{cm}}\) of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate \(I_{\mathrm{cm}}\). Therefore, you make the following measurements: You suspend the rod about an axis that is a distance \(d\) (measured in meters) above the center of the rod and measure the period \(T\) (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of \(d\). When you plot your data as \(T^{2}-4 \pi^{2} d / g\) versus \(1 / d\), the data lie close to a straight line that has slope \(0.320 \mathrm{~m} \cdot \mathrm{s}^{2} .\) What is the value of \(I_{\mathrm{cm}}\) for the rod?

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is \(0.165 \mathrm{~m}\). The maximum speed of the block is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the maximum magnitude of the acceleration of the block?

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