 So we've seen a lot of equations for harmonic oscillators, and we've managed to get this far without plugging in any actual numbers for actual real-world diatomic molecules that are the molecules we're supposed to be treating with this harmonic oscillator model. So that's what we'll do next is see what the properties are of these harmonic oscillators when we're using them to describe actual real-world molecules. And the real question is going to be how does the difference in energy between these harmonic oscillator states, remember every pair of successive states is separated by the same amount of energy, h times the vibrational frequency, regardless of whether it's the zero to the first, or the first to the second, or the second to the third, that delta e difference of h nu, how does that compare to kT? If our temperature is hot compared to that, kT is large compared to that difference in energy, then thermal energy is enough to excite many of these energy levels. If kT is small compared to that difference in energy, then we're stuck down in the ground state. So that's going to make an important difference in the properties of these molecules. So we could ask whether h nu is large compared to kT. Looking at the partition function, that's going to affect the size of this exponent is h nu large compared to kT, is this h nu large compared to 2 kT, that's going to affect the size of the partition function. Equivalently, instead of asking whether h nu is large compared to kT, if we divide both sides of this expression by k, then we're asking whether h nu over k is large compared to the temperature. So all I've done is rearranged that collection of variables so that instead of asking a question about this difference in energies compared to kT, which is an energy, I'm asking a question about this collection of constants compared to the temperature. So this collection of constants has units of temperature, and in fact we call this collection of constants the vibrational temperature. And you can compare that to a constant, a similar constant we define for the rigid rotor model where we define the rotational temperature, which was just a collection of constants that had the units of temperature that told us whether the rotational properties of the molecule were large or small compared to the temperature we're interested in at a given set of conditions. So this is the same idea for the harmonic oscillator, a collection of constants. This energy difference divided by Boltzmann's constant tells us something about the difference in energy levels or tells us in terms that are more easy to compare, whether that collection of constants with units of temperature appears large or small compared to the physical temperature of the system. So let's go ahead and work an example and see how that works out. And our main goal is going to be to find out how that vibrational temperature for a real world molecule compares to real world temperatures. So let's take the carbon monoxide molecule, a diatomic molecule that we've considered before. We know that the reduced mass for that molecule, we've calculated that quantity before that's mass of a carbon times mass of an oxygen divided by the two masses summed together. That's 1.14 times 10 minus 26th kilograms. A number that we haven't seen before that I'll tell you for the first time, the spring constant. So notice to use these expressions, we're gonna have Ks of two different types. We have K that is Boltzmann's constant. Every time K shows up as a Kt, that's Boltzmann's constant. But we have a different K, this is not Boltzmann's constant. This is the spring constant. So this spring constant, remember, in the potential energy one half Kx squared, that tells us how quickly or how slowly the potential energy increases as I stretch the bond in this diatomic molecule when I treat it as a hooks law spring. So that spring constant for a carbon monoxide molecule, spring constant will have units of energy over distance squared. Because if I take one half Kx squared, if I multiply by distance squared, I get something that's a potential energy. So that is the spring constant for a diatomic molecule. If I were to stretch the bond on a carbon monoxide molecule by a full meter, it would cost me 1,880 joules. If it were a hooks law spring, meaning if that bond didn't break, if I stretch the bond by a full meter, which of course it would. So this is only for the harmonic oscillator approximation. And in practice, we stretch the bond by only fractions of a nanometer and the energy goes up by tiny fractions of a joule. But this is telling us the spring constant for the stretches of that bond. We can, with that information, first of all calculate what is the vibrational frequency of this carbon monoxide molecule? We have seen this expression, the vibrational frequency if we know the spring constant and the reduced mass, the vibrational frequency is 1 over 2 pi square root of spring constant divided by reduced mass. So just plug those numbers in, which we'll do mainly to see how the units work out. So if I insert the spring constant in the reduced mass, remembering that a joule, remember that a joule is a, in the numerator, I've got units of kilogram meters squared per second squared. That's a joule divided by an extra factor of meters squared. So that cancels the meters squareds. In the denominator of this large fraction, I've got a kilogram. That's this term here, so the kilograms cancel. And what I'm left with is the units inside this square root are 1 over second squared. So once I take the square root, the units are going to be 1 over seconds. And if I do the arithmetic, 1 over 2 pi times the square root of these two quantities, that works out to be 6.5 times 10 to the 13th per second. So 6.5 times 10 to the 13th hurts. What that means is that if we were to treat this carbon monoxide molecule with this spring constant and this mass as a classical vibrating Hooke's law spring, then it's going to oscillate with a frequency of 6.5, 65, what is that? Million, billion, trillion, 65 trillion times per second. So that's telling us how quickly that molecule oscillates. We're not so much interested in the vibrational frequency as we are in the vibrational temperature. That will be a much more intuitive feeling number. So the reason we calculated the vibrational frequency is so we get inserted into this expression to calculate the vibrational temperature. So vibrational temperature, h nu over k. Again, just making sure the units will work out. I'll insert Planck's constant. I'll multiply that by this vibrational frequency, 6.5 times 10 to the 13th per second. Those are both multiplied together in the numerator. And if I divide by k, this k, remember which came from a kt, is Boltzmann's constant. Taking a quick look at the units. Planck's constant has joules time seconds. The frequency involves one over seconds. So those units of seconds cancel. I have a joules in the numerator, a joules in the denominator that cancel. And my units are one over Kelvin, or total units of Kelvin. And if I do the arithmetic in this expression, I find that the vibrational temperature of a carbon monoxide molecule works out to be 3,100 Kelvin. So what does that mean? Is that a large number or a small number? Unlike this frequency, or unlike the difference of energy if we get calculated in that in joules, that number immediately gives you an impression of whether that's a large number or a small number, compared to temperatures that we're used to dealing with, 300 Kelvin room temperature. That clearly appears to be a pretty large temperature. And in general, that's going to be true for any real-world diatomic molecule. Typical reduced masses, typical spring constants for diatomic molecules will result in vibrational temperatures that are quite a bit larger than room temperature. So if it's true for carbon monoxide, it's also generally true for most diatomics. The vibrational temperature will typically be larger than room temperature by, in this case, by a full order of magnitude, sometimes by a little more, sometimes by a little less. And again, this is true at 300 Kelvin. It's not true if I were to do an experiment at 3,000 Kelvin, of course. But for typical temperatures, typical molecules, the vibrational temperature ends up larger than the temperature. Thinking back to what that means, if the vibrational temperature is larger than the temperature, that's equivalent to saying H nu is larger than kT. The difference in energy levels, difference in energy between two energy levels is larger than kT. This difference is large compared to kT. So if I, to scale, draw something 1-tenth is big. So kT is only this big. The difference in energy levels is this big. The thermal energy we have accessible to us at room temperature of 300 Kelvin can only excite energy levels that are within this distance or a few multiples of that distance above the ground state. So what that means is few of these vibrational states are occupied. At 300 Kelvin, we don't have enough thermal energy to be populating states that are 3,100 Kelvin or 6,200 Kelvin above the ground state in units of temperature. So very few of those states end up occupied. And what that means is that the partition function, if we were to calculate a numerical value for the partition function, that's going to end up fairly small. Again, where a small partition function means that relatively few of the states are occupied. So while we've only worked the example out for one molecule, carbon monoxide, those trends will be generally the same for all diatomic molecules. And it's an important feature of diatomic molecules that their vibrational states are not very well populated at room temperature. So we can do some examples to calculate specifically how populated those excited states are. And that's coming up next.