The greenhouse-gas carbon dioxide molecule CO₂ strongly absorbs infrared radiation when its vibrational normal modes are excited by light at the normal-mode frequencies. CO₂ is a linear triatomic molecule, as shown in FIGURE CP15.82, with oxygen atoms of mass mo bonded to a central carbon atom of mass mc. You know from chemistry that the atomic masses of carbon and oxygen are, respectively, 12 and 16. Assume that the bond is an ideal spring with spring constant k. There are two normal modes of this system for which oscillations take place along the axis. (You can ignore additional bending modes.) In this problem, you will find the normal modes and then use experimental data to determine the bond spring constant.

g. The symmetric stretch frequency is known to be 4.00 X 10¹³ Hz. What is the spring constant of the C - O bond? Use 1 u = 1 atomic mass unit = 1.66 X 10⁻²⁷ kg to find the atomic masses in SI units. Interestingly, the spring constant is similar to that of springs you might use in the lab.

Question

The greenhouse-gas carbon dioxide molecule CO₂ strongly absorbs infrared radiation when its vibrational normal modes are excited by light at the normal-mode frequencies. CO₂ is a linear triatomic molecule, as shown in FIGURE CP15.82, with oxygen atoms of mass mo bonded to a central carbon atom of mass mc. You know from chemistry that the atomic masses of carbon and oxygen are, respectively, 12 and 16. Assume that the bond is an ideal spring with spring constant k. There are two normal modes of this system for which oscillations take place along the axis. (You can ignore additional bending modes.) In this problem, you will find the normal modes and then use experimental data to determine the bond spring constant.

g. The symmetric stretch frequency is known to be 4.00 X 10¹³ Hz. What is the spring constant of the C - O bond? Use 1 u = 1 atomic mass unit = 1.66 X 10⁻²⁷ kg to find the atomic masses in SI units. Interestingly, the spring constant is similar to that of springs you might use in the lab.

Accepted Answer

Hello, fellow physicist today were to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A young astrophysicist is studying the atmosphere of a newly discovered gas giant planet and is interested in creating a model of acetylene gas. Acetylene consists of two carbon capital C atoms and two hydrogen capital H atoms. As shown below the C dash C and H dash C bonds can be considered to be ideal springs with unknown spring constants K subscript C dash C and K subscript H dash C respectively. The symmetric sketch, the symmetric stretch frequency of the carbon dash carbon bond of acetylene is measured to be 9.94 multiplied by 10 to the power of 13 Hertz. If the atomic masses of carbon and hydrogen are known to be M subscript C to equal 12.0 at atomic mass units. And M subscript H equals 1.0 atomic mass units respectively, calculate the effective spring constant of the carbon dash carbon bond in acetylene. OK. So that's our end goal is to calculate the effective spring constant of the carbon dash, carbon bond and acetylene. Ok. So we're given some multiple choice answers. They're all in the same units of newtons per meters. So let's read them off to see what our final answer might be. A is 2.2 multiplied by 10 to the power of three B is 3.9 multiplied by 10 to the power of three C is 4.2 multiplied by 10 to the power of three and D is 5.2 multiplied by 10 to the power of three. OK. So first off, let us make the following assumptions that acetylene molecule behaves like a simple harmonic oscillator. That the motion is purely vibrational and does not involve any rotational or translational motion. And finally, that the molecular vibrations are not affected by the surrounding environment. In order to solve this problem, we must recall and use the relationship between symmetric stretch frequency, the reduced mass condition and the spring constant of the carbon dash carbon bond. The relationship can be derived from the classical harmonic oscillator equation which as we recall is omega is equal to the square root of K divided by mu. So to begin, let us determine the value for the reduced mass mu between the carbon dash carbon bond. Considering that the carbon dash carbon condition, we can write that mu is equal to M one multiplied by M two divided by M one plus M two which we can also say that mu is equal to the mass of carbon multiplied by the mass of carbon divided by the mass of carbon plus the mass of carbon. OK. So now at this stage, we can plug in our known variables to solve for a mute. So let's do that. So mu is equal to, so the carbon atom was determined to be. So MC M subscript C is 12.0 AM U atomic mass units. So am U so it's 12.0 AM U multiplied by 12.0 AM U divided by 12.0 AM U plus 12. AM U. So when we plug that into a calculator, we should get 6.0 atomic mass units, 6.0 AM U. So now we need to convert our atomic mass units to kilograms. So to do that, we take our which is equal to 6.0 AM U and we do a little bit of dimensional analysis to convert. So in 1 a.m. U there is 1.66 multiplied by 10 to the power of negative kg in one AM U in one atmosphere, I mean one atomic mass unit. OK. So those the AM us cancel out just leaving kilograms, which is what we want. So when we plug that into a calculator, we get 9.96 multiplied by 10 to the power of negative 27 kilograms. OK? So now we need to solve for the spring constant K. So essentially, we need to rearrange our equation that we wrote at the beginning, which was Omega is equal to the square root of K divided by mu. But we need to remember that Omega which is the angular frequency is equal to two pi multiplied by F which F is the frequency is equal to the square root of K divided by mu. So let me rearrange to solve for K. And this is for the K subscript C dash C. So the carbon to carbon bond is equal to two pi multiplied by F all squared multiplied by mu. So at this stage, we can plug in all of our known variables to solve for our spring constant K. So K subscripts C dash C so the carbon to carbon bond is equal to. Now let's start plugging in all of our known variables. Chew Pie multiplied by the frequency which the frequency is 9.94 multiplied by 10 to the power of 13 Hertz as given to us in the problem itself squared multiplied by our mu which we just found which was 9.96 multiplied by 10 to the power of negative kg. OK. So when we plug that into a calculator, we get that K our spring constant for the carbon to carbon bond is equal to 885. kilograms per seconds. Squared or 3885. newtons per meter. But we need to write it in scientific notation. So in scientific notation, when we round up is 3.9 multiplied by 10 to the power of three newtons per meter, which is our final answer. We did it. So let's go look at our multiple choice sensors to see what our final answer should be. The correct answer is the letter B 3.9 multiplied by 10 to the power of three newtons per meter. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Verified Solution

Key Concepts

Normal Modes of Vibration

Spring Constant (k)

Atomic Mass and SI Units