Question 19.11: CO2 vibration Within molecules, atoms interact with each oth......
\mathrm{\pmb{CO}}_2 vibration
Within molecules, atoms interact with each other via attractive and repulsive electric forces. These forces maintain a fairly consistent distance between the atoms. When the atoms get close to each other they repel; when they are pulled away they attract. Thus we can model the bond between the atoms as a spring. One of the vibrations in a \mathrm{CO}_2 molecule involves an asymmetric stretching and compression of the CO bonds between the carbon and oxygen atoms (see the figure at right). This bond has an effective spring constant of 1400 N/m. The vibration of the molecule has a natural frequency of 7.0 \times 10^{13} Hz. (a) Determine the effective mass of the \mathrm{CO}_2 molecule when vibrating in this way. (b) When a \mathrm{CO}_2 molecule in the atmosphere absorbs infrared energy emitted by Earth, the energy of the vibration is 4.7 \times 10^{-20} \mathrm{~J} . Estimate the amplitude of the vibration.
Sketch and translate The vibration is depicted in the figure above. We will use the expression for the vibration frequency of a cart with mass m and spring with spring constant k to estimate an effective mass of the vibrating molecule. We can use 1/2 kA² for the energy of vibration to estimate the amplitude.
Simplify and diagram Assume that the molecule can be modeled as a system undergoing SHM with a mass m at the end of a spring of spring constant k. We are also assuming that the physics ideas we are using apply to atomic-scale objects (which turns out not to be a very good assumption).
Represent mathematically The frequency of vibration of a mass at the end of a spring is
f=\frac{1}{2 \pi}\left(\frac{k}{m}\right)^{1 / 2}
The energy of vibration is
U_{\text {total }}=\frac{1}{2} k A^2
Solve and evaluate The effective mass of the vibrating system is
m=\frac{k}{4 \pi^2 f^2}=\frac{(1400 \mathrm{~N} / \mathrm{m})}{4 \pi^2\left(7.0 \times 10^{13} \mathrm{~s}^{-1}\right)^2}=7.2 \times 10^{-27} \mathrm{~kg}
The amplitude of vibration is
A=\left(\frac{2 U_{\text {total }}}{k}\right)^{1 / 2}=\left(\frac{2\left(4.7 \times 10^{-20} \mathrm{~J}\right)}{1400 \mathrm{~N} / \mathrm{m}}\right)^{1 / 2}
=8.2 \times 10^{-12} \mathrm{~m}
Are these numbers reasonable? The mass 7.2 \times 10^{-27} \mathrm{~kg} is approximately the mass of an oxygen atom. The amplitude is a few percent of the typical distance between atoms in a molecule. Both answers thus seem reasonable.
Try it yourself: Check the expressions used for each answer to see if they have the correct units.
Answer: m=\frac{k}{4 \pi^2 f^2} \text { has units } \frac{(\mathrm{N} / \mathrm{m})}{\left(\mathrm{s}^{-1}\right)^2}=\left(\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2}\right)\left(\frac{1}{\mathrm{~m}}\right)\left(\frac{1}{\mathrm{~s}^{-2}}\right)=\mathrm{kg} , the correct unit for mass.
A=\left(\frac{2 U_{\text {total }}}{k}\right)^{1 / 2} \text { has units }\left(\frac{\mathrm{J}}{\mathrm{N} / \mathrm{m}}\right)^{1 / 2}=\left(\frac{\mathrm{N} \cdot \mathrm{m}}{1} \cdot \frac{\mathrm{m}}{\mathrm{N}}\right)^{1 / 2}= \left(m^2\right)^{1 / 2}=m , the correct unit for amplitude.