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The mass of the deuterium molecule \(\left(\mathrm{D}_{2}\right)\) is twice that of the hydrogen molecule \(\left(\mathrm{H}_{2}\right) .\) If the vibrational frequency of \(\mathrm{H}_{2}\) is \(1.30 \times 10^{14} \mathrm{Hz},\) what is the vibrational frequency of \(\mathrm{D}_{2} ?\) Assume that the "spring constant" of attracting forces is the same for the two molecules.

Short Answer

Expert verified
The vibrational frequency of deuterium \(\mathrm{D}_{2}\) is approximately \(0.92 \times 10^{14}\) Hz

Step by step solution

01

Identify and denote given values

Denote the frequency of \(\mathrm{H}_{2}\) as \(F1 = 1.30 \times 10^{14} \mathrm{Hz}\), the mass of \(\mathrm{H}_{2}\) molecule as \(M1\), and the mass of deuterium \(\mathrm{D}_{2}\) molecule as \(M2 = 2M1\) as it is twice the mass of \(\mathrm{H}_{2}\). The frequency of \(\mathrm{D}_{2}\) can be denoted as \(F2\) and the spring constant for both is the same denoted as 'k'
02

Create the equation for \(\mathrm{D}_{2}\) using the harmonic oscillator formula

\[ F2 = \frac{1}{2\pi} \sqrt{\frac{k}{M2}}\]
03

Express k from the frequency formula for \(\mathrm{H}_{2}\)

From the first step, we know the spring constant (\(k\)) is the same for both molecules. We also know the formula for \(F1\), we can replace \(k\) with \(F1\) and \(M1\) in our equation for \(F2\): \[ k = (2 \pi F1)^2 \cdot M1\]
04

Substitute \(k\) in the \(\mathrm{D}_{2}\) frequency equation

Substitute k from step 3 into the equation for \(F2\): \[ F2 = \frac{1}{2\pi} \sqrt{\frac{(2\pi F1)^2 M1 }{M2} }\]
05

Simplify the equation

Since \(M2 = 2M1\), we have: \[ F2 = \frac{1}{2\pi} \cdot \frac{2\pi F1}{\sqrt{2}}\] So the final version of the equation is: \[ F2 = \frac{F1}{\sqrt{2}}\]
06

Calculate the frequency of \(\mathrm{D}_{2}\)

Calculate \(F2\) using \(F1\) value (vibrational frequency of \(\mathrm{H}_{2}\)) and substitution to the above obtained formula: \[ F2 = \frac{1.30 \times 10^{14}}{\sqrt{2}}\]

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Key Concepts

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

Vibrational Frequency
Vibrational frequency is a fundamental concept when studying molecules, particularly in the context of harmonic oscillators. It represents how fast a molecule vibrates in its lowest energy state. For simple diatomic molecules like hydrogen (\(\mathrm{H}_{2}\)) and deuterium (\(\mathrm{D}_{2}\)), this is crucial for understanding molecular behavior.
When we talk about vibrational frequency, we are referring to the number of times a molecule completes a vibration cycle per unit of time, typically measured in Hertz (Hz).
  • Low vibrational frequencies indicate slow molecular vibration.
  • High vibrational frequencies mean that the molecules are vibrating rapidly.
For the hydrogen molecule, the vibrational frequency is given as \(1.30 \times 10^{14} \mathrm{Hz}\). When considering molecules like deuterium with different masses, knowing the vibrational frequency helps to understand how these molecules will behave differently under the same conditions.
The vibrational frequency changes inversely with the square root of the reduced mass of the molecule. This means if the mass increases, the vibrational frequency decreases, provided the spring constant remains the same.
Deuterium
Deuterium is an isotope of hydrogen, symbolized as \(\mathrm{D}_{2}\) for its molecular form. It is sometimes referred to as heavy hydrogen due to its greater mass.
A standard hydrogen atom contains one proton in its nucleus, while a deuterium atom contains one proton and one neutron, effectively doubling its mass. As such, deuterium molecules are twice as heavy as hydrogen molecules, making \(\mathrm{D}_{2}\) two times heavier than \(\mathrm{H}_{2}\).
In our exercise, this distinction is crucial because mass directly affects vibrational frequency when the spring constant does not change.
  • Heavier molecules, like \(\mathrm{D}_{2}\), tend to vibrate more slowly than lighter molecules of the same spring constant.
  • This behavior is typical in systems analyzed by harmonic oscillator models.
Utilizing the mass of deuterium, we can calculate the adjusted vibrational frequency using derived formulas which take into account the increased mass. Thus, despite having the same spring constant, \(\mathrm{D}_{2}\) will have a lower vibrational frequency compared to \(\mathrm{H}_{2}\).
Spring Constant
The spring constant, symbolized often as \('k'\), is a measure of the stiffness of the bond between atoms in a molecule. In the harmonic oscillator model, it plays a crucial role in determining the vibrational frequency of a molecule.
Simply put, the spring constant represents how strongly the atoms are bonded together. For both \(\mathrm{H}_{2}\) and \(\mathrm{D}_{2}\), the exercise assumes that this spring constant remains unchanged.
  • A higher spring constant means the atoms are more tightly bound, leading to higher vibrational frequencies.
  • A lower spring constant means the atoms are more loosely bound, resulting in lower vibrational frequencies.
The relation between the spring constant and vibrational frequency can be expressed with the formula: \[ F = \frac{1}{2\pi} \sqrt{\frac{k}{M}} \]where \(M\) is the mass of the molecule. This formula shows that, with the spring constant held constant, changes in mass directly affect the vibrational frequency, making it central to problems that involve comparing different isotopes, as seen with hydrogen and deuterium.

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Most popular questions from this chapter

A \(1.00-\mathrm{kg}\) glider attached to a spring with a force constant of \(25.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless air track. At \(t=0\) the glider is released from rest at \(x=-3.00 \mathrm{cm} .\) (That is, the spring is compressed by \(3.00 \mathrm{cm} .)\) Find (a) the period of its motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as functions of time.

A \(7.00-\mathrm{kg}\) object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of \(2.60 \mathrm{s}\). Find the force constant of the spring.

A particle of mass \(m\) slides without friction inside a hemispherical bowl of radius \(R\). Show that, if it starts from rest with a small displacement from equilibrium, the particle moves in simple harmonic motion with an angular frequency equal to that of a simple pendulum of length \(R\) That is, \(\omega=\sqrt{g / R}\).

A \(2.00-\mathrm{kg}\) object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of \(20.0 \mathrm{N}\) is required to hold the object at rest when it is pulled \(0.200 \mathrm{m}\) from its equilibrium position (the origin of the \(x\) axis). The object is now released from rest with an initial position of \(x_{i}=0.200 \mathrm{m},\) and it subsequently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the maximum speed of the object. Where does this maximum speed occur? (d) Find the maximum acceleration of the object. Where does it occur? (e) Find the total energy of the oscillating system. Find (f) the speed and (g) the acceleration of the object when its position is equal to one third of the maximum value.

The angular position of a pendulum is represented by the equation \(\theta=(0.320 \mathrm{rad}) \cos \omega t,\) where \(\theta\) is in radians and \(\omega=4.43 \mathrm{rad} / \mathrm{s} .\) Determine the period and length of the pendulum.

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