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Question 41.9: Modeling Iodine A number of times throughout this book, we h......

Modeling Iodine

A number of times throughout this book, we have modeled molecular bonds as tiny springs connecting atoms. (For examples, see Figures 5.15, 14.21, and 21.25.) In particular, a vibrating diatomic molecule may be modeled as a simple harmonic oscillator (Fig. 41.21). In this problem, we connect the quantum-mechanical simple harmonic model to the classical model we have used for a molecular bond. An iodine molecule consisting of two iodine atoms is observed to have equally spaced energy levels of  4.24 \times 10^{-21} J. Model the molecule as a particle-spring system, with one end of the spring fixed and the other attached to the oscillating particle. The particle’s mass is 1.053 \times 10^{-25} kg. (This the reduced or effective mass of the iodine molecule.) Find the spring constant.

5.15
14.21
21.25
41.21
Step-by-Step
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INTERPRET and ANTICIPATE
Modeling the iodine molecule as a quantum simple harmonic oscillator is reasonable because the observed energy spacing is uniform as predicted by Equation 41.26.

E_n=\left(n-\frac{1}{2}\right) h f \quad \quad (41.26)

The frequency of a classical object-spring oscillator (Section 16-5) depends on the mass and spring constant. So, making the connection between the classical and quantum models means finding the frequency of the quantum oscillator and from that frequency finding the spring constant.

SOLVE
The difference in adjacent energy levels depends on frequency.

\Delta E=h f \quad \quad (1)

The frequency of an object-spring oscillator is given by Equation 16.26 (modified by using ω = 2\pi f.)

f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \quad \quad 16.26

Combine Equations (1) and 16.26 to arrive at the spring constant.

\begin{aligned}& \frac{\Delta E}{h}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\& k=4 \pi^2 m\left(\frac{\Delta E}{h}\right)^2 \\& k=4 \pi^2\left(1.053 \times 10^{-25} kg \right)\left(\frac{4.24 \times 10^{-21} J }{6.626 \times 10^{-34} J \cdot s }\right)^2 \\ & k=1.70 \times 10^2 N / m\end{aligned}

CHECK and THINK
In this problem we have come full circle on our models of molecular bonds. We see why it was reasonable to model such bonds by tiny springs throughout this book. However, we must remember there is a major difference between a classical and quantum simple harmonic oscillator: In a classical oscillator the mechanical energy can have any value, but in a quantum oscillator the energy can only have certain discrete, evenly spaced values.

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