Ch 42: Molecules and Condensed Matter

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 42: Molecules and Condensed Matter

Problem 9a

Chapter 42, Problem 9a

The rotational energy levels of CO are calculated in Example $42.2$ . If the energy of the rotating molecule is described by the classical expression $K=($\frac$12)I$\omega$^2$ , for the $l = 1$ level, what is the angular speed of the rotating molecule?

Verified step by step guidance

Step 1: Begin by understanding the classical expression for rotational kinetic energy, $ K = \frac{1}{2} I \omega^2 $, where $ K $ is the rotational kinetic energy, $ I $ is the moment of inertia, and $ \omega $ is the angular speed.

Step 2: Recall that the energy levels for a rotating molecule are quantized and given by $ E_l = \frac{l(l+1)h^2}{8\pi^2I} $, where $ l $ is the rotational quantum number, $ h $ is Planck's constant, and $ I $ is the moment of inertia. For $ l = 1 $, substitute $ l = 1 $ into this formula to find the energy of the $ l = 1 $ level.

Step 3: Equate the quantized energy $ E_l $ for $ l = 1 $ to the classical expression for rotational kinetic energy $ K $. This gives $ \frac{l(l+1)h^2}{8\pi^2I} = \frac{1}{2}I\omega^2 $. Rearrange this equation to solve for $ \omega $, the angular speed.

Step 4: Substitute the known values for $ h $ (Planck's constant), $ I $ (moment of inertia, which can be calculated using $ I = \mu r^2 $, where $ \mu $ is the reduced mass and $ r $ is the bond length), and $ l = 1 $ into the equation derived in Step 3.

Step 5: Simplify the expression to isolate $ \omega $. This will give the angular speed of the rotating molecule for the $ l = 1 $ level. Ensure all units are consistent during substitution and simplification.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by an object due to its rotation. It is given by the formula K = (1/2)Iω^2, where I is the moment of inertia and ω is the angular speed. This concept is crucial for understanding how energy is distributed in rotating systems, particularly in molecular dynamics.

Moment of Inertia

The moment of inertia (I) is a measure of an object's resistance to changes in its rotation about an axis. It depends on the mass distribution relative to the axis of rotation. For molecules, the moment of inertia can be calculated based on the atomic masses and their distances from the center of mass, which is essential for determining the rotational energy levels.

Angular Speed

Angular speed (ω) is the rate at which an object rotates around an axis, typically measured in radians per second. In the context of molecular rotation, it describes how quickly the molecule is spinning at a given energy level. Understanding angular speed is vital for calculating the rotational kinetic energy and analyzing molecular behavior in quantum mechanics.

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Textbook Question

The rotational energy levels of CO are calculated in Example $42.2$ . If the energy of the rotating molecule is described by the classical expression $K=($\frac$12)I$\omega$^2$ , for the $l = 1$ level, what is the rotational period (the time for one rotation)?

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Textbook Question

The rotational energy levels of CO are calculated in Example $42.2$ . If the energy of the rotating molecule is described by the classical expression $K = (\frac{1}{2}) I ω^{2}$ , for the $l equals 1$ level, what is the linear speed of each atom?

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Textbook Question

During each of these processes, a photon of light is given up. In each process, what wavelength of light is given up, and in what part of the electromagnetic spectrum is that wavelength? A molecule decreases its vibrational energy by $0.198$ eV.

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Textbook Question

The average kinetic energy of an ideal-gas atom or molecule is $(\frac{3}{2}) k T$ , where $T$ is the Kelvin temperature (Chapter $18$ ). The rotational inertia of the H₂ molecule is $4.6 \times 10^{- 48}$ kg-m². What is the value of $T$ for which $(\frac{3}{2}) k T$ equals the energy separation between the $l equals 0$ and $l equals 1$ energy levels of H₂? What does this tell you about the number of H₂ molecules in the $l equals 1$ level at room temperature?

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Textbook Question

The H₂ molecule has a moment of inertia of $4.6 \times 10^{- 48}$ kg-m². What is the wavelength $l$ of the photon absorbed when H₂ makes a transition from the $l equals 3$ to the $l equals 4$ rotational level?

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Textbook Question

Two atoms of cesium (Cs) can form a $C s sub 2$ molecule. The equilibrium distance between the nuclei in a $C s sub 2$ molecule is $0.447$ 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 $2.21 \times 10^{- 25}$ kg.

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The rotational energy levels of CO are calculated in Example 42.242.242.2. If the energy of the rotating molecule is described by the classical expression K=(12)Iω2K=(\(\frac\)12)I\(\omega\)^2K=(21)Iω2, for the l=1 l = 1l=1 level, what is the angular speed of the rotating molecule?

Verified video answer for a similar problem:

Key Concepts

Rotational Kinetic Energy

Moment of Inertia

Angular Speed

The rotational energy levels of CO are calculated in Example $42.2$ . If the energy of the rotating molecule is described by the classical expression $K=(\(\frac\)12)I\(\omega\)^2$ , for the $l = 1$ level, what is the angular speed of the rotating molecule?