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Two atoms of cesium (Cs) can form a Cs \(_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) molecule is \(0.447 \mathrm{nm} .\) Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is \(2.21 \times 10^{-25} \mathrm{~kg}\).

Short Answer

Expert verified
The moment of inertia is \(2.20507 \times 10^{-46} \, \text{kg·m}^2\).

Step by step solution

01

Understanding the System

In a Cs\(_2\) molecule, two cesium atoms are connected. The distance between the two atoms is given as 0.447 nm. We need to compute the moment of inertia about an axis that goes through the center of mass and is perpendicular to the line joining the atoms.
02

Convert Distance Units

Convert the given distance from nanometers (nm) to meters (m) for consistency in SI units. Since 1 nm = 10^{-9} m, the distance in meters is:\[d = 0.447 \times 10^{-9} = 0.447 \times 10^{-9} ext{ m}\]
03

Locate the Center of Mass

Since the two cesium atoms have the same mass, the center of mass is exactly halfway between them. Therefore, each atom is \(\frac{d}{2}\) meters from the center of mass.
04

Use Moment of Inertia Formula

The moment of inertia \(I\) for two point masses (each cesium atom) about an axis through the center of mass is given by:\[I = m \left( \frac{d}{2} \right)^2 + m \left( \frac{d}{2} \right)^2 = 2m \left( \frac{d}{2} \right)^2\]where \(m = 2.21 \times 10^{-25} \, \text{kg}\) is the mass of a cesium atom.
05

Plug in Values and Calculate I

Substitute the known values into the moment of inertia formula:\[I = 2 \times 2.21 \times 10^{-25} \times \left( \frac{0.447 \times 10^{-9}}{2} \right)^2\]Calculate \(I\) by evaluating this expression.
06

Final Calculation

Perform the arithmetic:\[\left(\frac{0.447 \times 10^{-9}}{2}\right)^2 = \left(0.2235 \times 10^{-9}\right)^2 = 0.04992225 \times 10^{-18}\]\[I = 2 \times 2.21 \times 10^{-25} \times 0.04992225 \times 10^{-18} = 2.20507 \times 10^{-46} \text{ kg·m}^2\]
07

Conclusion

Thus, the moment of inertia of the Cs\(_2\) molecule about the specified axis is \(2.20507 \times 10^{-46} \, \text{kg·m}^2\).

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Key Concepts

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

Cs2 Molecule
The Cs extsubscript{2} molecule is a simple diatomic molecule composed of two cesium (Cs) atoms. In chemistry, diatomic molecules are known for having two atoms which can be of the same or different chemical elements.
The cesium molecule, Cs extsubscript{2}, is formed when two cesium atoms bond together. This bonding is typically due to the sharing of electrons, creating a covalent bond. The molecule has a linear configuration because both atoms are identical and equally attract the shared electrons.
Understanding molecules like Cs extsubscript{2} can help in studying properties such as molecular orbitals, bond lengths, and moment of inertia, as seen in the given exercise.
Cesium Atom
A cesium atom is a soft, silvery-golden alkali metal found in the periodic table with the symbol Cs and atomic number 55. It is one of the most reactive and electropositive elements known. This reactivity makes cesium atoms interesting in numerous fields, ranging from physics to chemistry.
One unique property of cesium is its relatively large atomic mass and size compared to other alkali metals, which affects its physical and chemical behavior. The mass of the cesium atom is crucial in calculations involving molecular structures, such as determining the moment of inertia in the given exercise. This demonstrates how fundamental attributes of a cesium atom help determine the characteristics of larger structures like molecules.
Equilibrium Distance
In a molecule, the equilibrium distance is the distance where the forces of attraction and repulsion between atoms are balanced, resulting in a stable bonded state. For the Cs extsubscript{2} molecule, this equilibrium distance is given as 0.447 nm.
This measurement signifies the average distance between the nuclei of the cesium atoms when they are bonded in the Cs extsubscript{2} molecule.
Equilibrium distance is crucial for understanding molecular geometry and properties such as bond strength and molecular interactions. It's also pivotal in calculating molecular moment of inertia, as this distance partly defines the rotational dynamics of the molecule in question.
Mass of Cesium Atom
The mass of a cesium atom, given as 2.21 x 10 extsuperscript{-25} kg in the exercise, is a key parameter in physics and chemistry. Atomic mass, often expressed in terms of kilograms for scientific calculations, influences how atoms interact in chemical reactions and how they contribute to molecular properties.
In physics, the mass of components like cesium atoms is essential for calculations involving motion. For example, in calculating the moment of inertia for a molecule, the mass and distance from the axis of rotation are vital. By understanding how cesium's mass affects these calculations, we gain insight into the molecule's rotational behavior and dynamics.

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Most popular questions from this chapter

Consider a system of \(N\) free electrons within a volume \(V\). Even at absolute zero, such a system exerts a pressure \(p\) on its surroundings due to the motion of the electrons. To calculate this pressure, imagine that the volume increases by a small amount \(dV\). The electrons will do an amount of work \(p\) \(dV\) on their surroundings, which means that the total energy \(E_{tot}\) of the electrons will change by an amount \(dE_{tot} = -p dV\). Hence \(p = -dE_{tot}/dV\). (a) Show that the pressure of the electrons at absolute zero is \(p = \frac{3^{2/3}\pi^{4/3}\hbar^{2}}{5m} \lgroup \frac{N}{V}\ \rgroup^{5/3}\) (b) Evaluate this pressure for copper, which has a freeelectron concentration of \(8.45 \times 10^{28} m^{-3}\). Express your result in pascals and in atmospheres. (c) The pressure you found in part (b) is extremely high. Why, then, don't the electrons in a piece of copper simply explode out of the metal?

When an OH molecule undergoes a transition from the \(n = 0\) to the \(n = 1\) vibrational level, its internal vibrational energy increases by 0.463 eV. Calculate the frequency of vibration and the force constant for the interatomic force. (The mass of an oxygen atom is \(2.66 \times 10^{-26}\) kg, and the mass of a hydrogen atom is \(1.67 \times 10^{-27} kg.)\)

When a hypothetical diatomic molecule having atoms 0.8860 nm apart undergoes a rotational transition from the \(l\) = 2 state to the next lower state, it gives up a photon having energy 8.841 \(\times\) 10\(^-$$^4\) eV. When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up 0.2560 eV. Find the force constant of this molecule.

Our galaxy contains numerous \(molecular\) \(clouds\), regions many lightyears in extent in which the density is high enough and the temperature low enough for atoms to form into molecules. Most of the molecules are H\(_2\), but a small fraction of the molecules are carbon monoxide (CO). Such a molecular cloud in the constellation Orion is shown in Fig. P42.42. The upper image was made with an ordinary visiblelight telescope; the lower image shows the molecular cloud in Orion as imaged with a radio telescope tuned to a wavelength emitted by CO in a rotational transition. The different colors in the radio image indicate regions of the cloud that are moving either toward us (blue) or away from us (red) relative to the motion of the cloud as a whole, as determined by the Doppler shift of the radiation. (Since a molecular cloud has about 10,000 hydrogen molecules for each CO molecule, it might seem more reasonable to tune a radio telescope to emissions from H\(_2\) than to emissions from CO. Unfortunately, it turns out that the H\(_2\) molecules in molecular clouds do not radiate in either the radio or visible portions of the electromagnetic spectrum.) (a) Using the data in Example 42.2 (Section 42.2), calculate the energy and wavelength of the photon emitted by a CO molecule in an \(l\) \(=\) 1\(\rightarrow\) \(l\) \(=\) 0 rotational transition. (b) As a rule, molecules in a gas at temperature \(T\) will be found in a certain excited rotational energy level, provided the energy of that level is no higher than \(kT\) (see Problem 42.39). Use this rule to explain why astronomers can detect radiation from CO in molecular clouds even though the typical temperature of a molecular cloud is a very low 20 K.

If the energy of the \(H_{2}\) covalent bond is -4.48 eV, what wavelength of light is needed to break that molecule apart? In what part of the electromagnetic spectrum does this light lie?

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