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The vibration frequency of a hydrogen chloride molecule is \(8.66 \times 10^{13} \mathrm{Hz}\). How long does it take the molecule to complete one oscillation?

Short Answer

Expert verified
The hydrogen chloride molecule takes approximately \(1.15 \times 10^{-14}\) seconds to complete one oscillation.

Step by step solution

01

Identify the given parameters and the target

The given parameter is the frequency \(f = 8.66 \times 10^{13} Hz\). The target is to calculate the time taken for one oscillation, also known as the period \(T\).
02

Use the relation between frequency and period

The relation between frequency and period is \(T = 1/f\). Frequency \(f\) is the number of cycles per unit time and period \(T\) is the time taken for one cycle.
03

Substitute the given frequency into the formula

Substitute the given frequency \(f = 8.66 \times 10^{13} Hz\) into the formula \(T = 1/f\). Upon substituting the given frequency, the formula becomes \(T = 1 / (8.66 \times 10^{13} Hz)\)
04

Calculate the period

When we carry out the above division, we get \(T ≈ 1.15 \times 10^{-14} seconds\)

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Key Concepts

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

Oscillation Period Calculation
When you encounter a question about calculating the time it takes for an object to complete one cycle of oscillation—often referred to as the 'oscillation period'—it's actually very straightforward with the proper knowledge. In physics, we measure this time in seconds, and it provides valuable insights into the behavior of oscillating systems like molecules, pendulums, or even musical instruments.

Let's break down the steps needed for such a calculation, as the period (denoted as 'T') is the inverse of the frequency (represented by 'f'). If you know the frequency, as you do in the case with the hydrogen chloride molecule which vibrates at a frequency of 8.66 x 10^13 Hz, you simply take the reciprocal of this value to find the period.

Using the formula: \[T = \frac{1}{f}\],you plug in the frequency and perform the division. This simple calculation will give you the period in seconds, allowing you to understand how long one complete oscillation takes.
Relationship Between Frequency and Period
Understanding the relationship between frequency and period is fundamental in physics, especially when studying wave phenomena and harmonic motion. The key concept here is that frequency and period are inversely related. In other words, as the frequency of a vibrating system increases, the time it takes to complete one cycle—its period—decreases, and vice versa.

This inverse relationship can be expressed mathematically as:\[T = \frac{1}{f}\] and \[f = \frac{1}{T}\].

This tells us that if we're given the frequency of a system, we can find the period by simply taking the reciprocal of the frequency. Conversely, if we know the period, we can calculate the frequency by taking the reciprocal of the period. The units here are also important: frequency is measured in hertz (Hz), which equates to cycles per second, while period is measured in seconds. This inverse relationship is a cornerstone concept that helps us in analyzing the behavior of oscillatory systems.
Frequency of Hydrogen Chloride Molecule
In the realm of chemistry and physics, the frequency of a molecule's vibration can tell us a great deal about its energy and the bonds that hold it together. For the hydrogen chloride (HCl) molecule, a specified frequency of oscillation is provided, which is a staggering 8.66 x 10^13 Hz. This is a high frequency, indicative of the rapid oscillation between the hydrogen and chlorine atoms within the molecule.

Further implicating its significance, the frequency can relate to the infrared spectroscopy used in chemistry to analyze substances, where these vibration frequencies are detected and used to determine molecular identity and structure. Given this frequency, by using the inversely related formula for period calculation, one can deduce the duration of a single vibration cycle for the HCl molecule.

Understanding this frequency is not only practical for solving physics and chemistry problems, but it also provides a deeper insight into the energetic properties and behavior of the molecule at a quantum level.

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Most popular questions from this chapter

You're riding in a friend's \(1400-\mathrm{kg}\) car with bad shock absorbers, bouncing down the highway at \(20 \mathrm{m} / \mathrm{s}\) and executing vertical SHM with amplitude \(18 \mathrm{cm}\) and frequency \(0.67 \mathrm{Hz}\). Concerned about fuel efficiency, your friend wonders what percentage of the car's kinetic energy is tied up in this oscillation. Make an estimate, neglecting the wheels' rotational energy and the fact that not all of the car's mass participates in the oscillation.

Two mass-spring systems with the same mass are undergoing oscillatory motion with the same amplitudes. System 1 has twice the frequency of system \(2 .\) How do (a) their energies and (b) their maximum accelerations compare?

The human eye and muscles that hold it can be modeled as a mass-spring system with typical values \(m=7.5 \mathrm{g}\) and \(k=2.5 \mathrm{kN} / \mathrm{m} .\) What's the resonant frequency of this system? Shaking your head at this frequency blurs vision, as the eyeball undergoes resonant oscillations.

What happens to the frequency of a simple harmonic oscillator when the spring constant is doubled? When the mass is doubled?

You're working on the script of a movie whose plot involves a hole drilled straight through Earth's center and out the other side. You're asked to determine what will happen if a person falls into the hole. You find that the gravitational acceleration inside Earth points toward Earth's center, with magnitude given approximately by \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right),\) where \(g_{0}\) is the surface value, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. What do you report for the person's motion, including equations and values for any relevant parameters?

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