CP The rotational energy levels of CO are calculated in Example 4...

Question

CP 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 = (1/2)Iω^2 , for the l = 1 level what are (b) the linear speed of each atom;

Accepted Answer

Hi everyone. In this practice problem, we're being asked to calculate the 02 molecules linear speed for L equals to rotation level. If the rotational energy of the 02 molecule is related to angular speed by the relation of E equals half I omega squared. Consider the rotational energy of the 02 molecule as 1.75 mili E V with a moment of inertia of 1.94 times 10 to the power of negative 46 kg meters squared about an axis passing through the center of mass and perpendicular to the line joining the two oxygen atoms. The options given for the molecules linear speed are a 1. m per second. B 80.6 m per second, C 86.0 m per second and D 60. m per second. So it is given in the problem statement and we want to recall that the rotational energy of the molecule is going to be given by the equation of E equals to half multiplied by I multiplied by omega squared where I is the moment of inertia about an axis passing through the center of mass and a perpendicular to the line joining the two oxygen atoms. And omega is the angular speed. It is given in the problem statement that the energy or E is going to equal to 1.75 Meli E V or an E V that will correspond to 1. 75 times 10 to the power of negative three E V. And want to convert this into Joel by multiplying this with 1.6 oh two times 10 to the power of negative 19 Joes for every one E V. And that will give us the energy in terms of Jos which will come up to a value of 17.222 times 10 to the power of negative 23 jos the moment of inertia or I is given the problem statement to be 1.94 times 10 to the power of negative 46 kg meters squared. And from the equation of the energy or E, we can rearrange that equation to get an equation for the angular speed or omega squared which will equal to two E divided by I. So let's substitute all of these values that we know two E will be equals to two multiplied by 17.222 times 10 to the power of negative 23 jos divide that with the moment of inertia which is 1.94 times 10 to the power of negative 46 kg meters square. And that will give us our omega squared value or squared off the angular speed to be equal to 1.775 times 10 to the power of 24 radium squared per second squared. Taking the square root of that, we will get our angular speed which will come out to be 1.332 times 10 to the power of 12 radians per seconds. So linear speed or V will be equal to R multiplied by angular speed or omega. So to get the R, we want to look at the distance between two oxygens, oxygen atoms in an 02 molecule which will be equal to R equals to 0. nanometer R is the cent the distance between the center of mass of 02 molecule and the center of oxygen atom. So this R or the 0.1 to one nanometer will have to be divided by two. So then our R will be equal to 0.6 oh five nanometer or in meter. This will be equal to 0.6 oh five times 10 to the power of negative nine m. Awesome. So then let's calculate the linear speed or V. So we want to input the I uh the R value which is 0.6 oh five times 10 to the power of negative nine m multiply that with the angler speed which is 1.332 times 10 to the power of 12 radiant per seconds. And then that will give us the linear speed of 80.6 m per second. So the linear speed oops the linear speed is going to be equal to 80.6 m per second, which will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

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CP 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 = (1/2)Iω^2 , for the l = 1 level what are (b) the linear speed of each atom;

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