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When displaced from equilibrium, the two liydrogen atoms in an \(\mathrm{H}_{2}\) molecule are acted on by a restoring force \(F_{x}=-k x\) with \(k=580 \mathrm{N} / \mathrm{m}\) . Calculate the oscillation frequency of the \(\mathrm{H}_{2}\) molecule. (Hint: The mass of a hydrogen atom is 1.008 atomic mass units, or 1 u; see Appendix E. As in Example 13.7 (Section \(13.4 ),\) use \(m / 2\) instead of \(m\) in the expression for \(f\) )

Short Answer

Expert verified
The oscillation frequency of the \(\mathrm{H}_{2}\) molecule is approximately \(1.32 \times 10^{14}\) Hz.

Step by step solution

01

Understand the Problem

We need to calculate the oscillation frequency of an \( \mathrm{H}_{2} \) molecule based on the restoring force given by \( F_x = -kx \) and the spring constant \( k = 580 \ \mathrm{N/m}\). The formula for oscillation frequency is derived from the harmonic oscillator model.
02

Use the Harmonic Oscillator Formula

The oscillation frequency \( f \) for a system acting like a spring is given by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]However, since the molecule consists of two atoms oscillating, we use \( \frac{m}{2} \) instead of \( m \).
03

Convert Mass to Kilograms

The atomic mass unit \( 1 \ \mathrm{u} \) is equivalent to \( 1.66053906660 \times 10^{-27} \ \mathrm{kg} \). Therefore, the mass of one hydrogen atom is \( 1.008 \ \mathrm{u} = 1.008 \times 1.66053906660 \times 10^{-27} \ \mathrm{kg} \).
04

Calculate the Effective Mass for the Molecule

In this scenario, use the effective mass for the pair of hydrogen atoms: \[ m_{\text{effective}} = \frac{1.008 \times 1.66053906660 \times 10^{-27}}{2} \ \mathrm{kg} \].Calculate the value of \( m_{\text{effective}} \).
05

Plug Values into the Frequency Formula

Calculate the frequency using the harmonic oscillator formula and the effective mass:\[ f = \frac{1}{2\pi} \sqrt{\frac{580}{\frac{1.008 \times 1.66053906660 \times 10^{-27}}{2}}} \].
06

Compute the Result

Perform the computation to find the value of \( f \). Ensure all mathematical operations are executed correctly, including taking the square root and division by \( 2\pi \).

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Key Concepts

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

Harmonic Oscillator
The harmonic oscillator is a fundamental concept in physics describing a system where a particle experiences a force proportional to its displacement from a position of equilibrium. This type of motion is common in springs, pendulums, and even molecules like the hydrogen molecule. The relation governing this motion is often represented by Hooke's law:
  • The force exerted by the spring or molecular bond is given by \( F = -kx \), where \( k \) is the spring constant, and \( x \) is the displacement from equilibrium.
  • The negative sign indicates that the force is a restoring force, acting in the opposite direction to the displacement.
The harmonic oscillator model helps in understanding the oscillation frequency of systems, which is the number of oscillations per unit time. The frequency formula is derived from the dynamic behavior of such systems, mathematically illustrated as a sinusoidal function:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]This formula is essential in calculating properties of molecular vibration such as those seen in an \(\mathrm{H}_2\) molecule.
Restoring Force
A restoring force is a critical component in understanding oscillatory systems, prominently featured in springs and molecular physics. It is called 'restoring' because it aims to bring the system back to equilibrium after displacement.
  • The classic example is Hookean springs, where the force is linearly proportional to displacement: \( F = -kx \).
  • The proportionality constant \( k \) defines the stiffness of the spring or bond; a higher \( k \) means a stronger force trying to restore equilibrium.
In the case of molecular systems like the hydrogen molecule, the concept scales down to atomic levels where chemical bonds exert similar restoring forces. These forces dictate the vibrational characteristics of molecules when evaluating things like molecular oscillation frequency. Understanding restoring forces helps us comprehend how molecular vibrations contribute to phenomena like infrared absorption and molecular bond strength.
Hydrogen Molecule
The hydrogen molecule \(\mathrm{H}_2\) is the simplest molecular system, consisting of two hydrogen atoms sharing one electron. The properties of this molecule make it an excellent subject for studying molecular vibrations and bond dynamics.
  • In the \( \mathrm{H}_2 \) molecule, two atoms oscillate about their center of mass due to the restoring force of their shared electron bond.
  • The atomic mass of a hydrogen atom is approximately \( 1.008 \ \mathrm{u} \), which must be converted to kilograms for use in physics calculations, using \( 1 \ \mathrm{u} = 1.66053906660 \times 10^{-27} \ \mathrm{kg} \).
  • Hydrogen molecules are often modeled as harmonic oscillators, with the atomic bond acting as the spring.
Studying \( \mathrm{H}_2 \) gives insight into fundamental molecular behavior and aids in understanding complex molecules. In physical chemistry, these models help determine the oscillation frequency crucial for characterizing molecular energy levels and interactions.

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Most popular questions from this chapter

A 175 -g glider ona horizontal, frictionless air trackis attached to a fixed ideal spring with force constant 155 \(\mathrm{N} / \mathrm{m}\) . At the instant you make measurements on the glider, it is moving at 0.815 \(\mathrm{m} / \mathrm{s}\) and is 3.00 \(\mathrm{cm}\) from its equilibrium point. Use energy conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

While on a visit to Minnesota (Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 \(\mathrm{cm} .\) The boat takes 3.5 \(\mathrm{s}\) to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 \(\mathrm{kg}\) ) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 \(\mathrm{cm}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

You pull a simple pendulum 0.240 \(\mathrm{m}\) long to the side through an angle of \(3.50^{\circ}\) and release it. (a) How much time does it take the pendulum bob to reach its highest speed? (b) How much time does it take if the pendulum is released at an angle of \(1.75^{\circ}\) instead of \(3.50^{\circ} ?\)

An object is undergoing SHM with period 1.200 s andamplitude 0.600 \(\mathrm{m}\) . At \(t=0\) the object is at \(x=0 .\) How far is the object from the equilibrium position when \(t=0.480\) s?

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

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