 So the energy levels for a rigid rotor model are given by this expression. They depend on the quantum number L and this formula can be used. We've calculated energy levels for the first few of these levels and they get further and further apart as they go up. What we'll do now is take a closer look at this expression and point out a few things. First of all, this quantity, if you are particularly familiar with your probably freshman year physics, this quantity might sound familiar. Mu times r squared, that's a quantity that we can give a name to and that's called the moment of inertia and you don't necessarily need to remember that term from physics or need to have seen it before in physics but what the moment of inertia means is for rotational motion what matters for how difficult something is to rotate is not just its mass but how far away it is from the axis that it's rotating. So for example, if I want to rotate this pen it's a lot easier to rotate it on an axis close to my body than it is it's harder to rotate it on an axis that's far from my body. So the moment of inertia tells us how to combine both the mass of the object and the distance away from the rotational axis for this rotating diatomic molecule that we're treating as a rigid rotor. Remember what we've done is instead of having atoms with masses m1 and m2 spinning around some center of mass instead we've chosen to represent those as an object of mass mu rotating at some distance r, the bond length of the molecule away from the center of mass. So the mass of this rotating object is mu the distance away from the center of rotation is r and mu r squared gives us the moment of inertia. So regardless of whether that concept from a physics class you may have taken helps you understand this equation a little better or whether you think of it as just a way of cleaning up the notation we can write the energy levels for a rigid rotor in this form instead where we've just combined the mu r squared into this variable i as a shorthand and sometimes you'll see these energy levels written in that way sometimes you'll see them in fact written in a different formula so this simplification used moment of inertia for another simplification I'll remind you that Planck's constant shows up in a couple different forms either as Planck's constant or as this reduced Planck's constant h bar which is the ordinary Planck's constant divided by 2 pi so h bar squared would be h squared over 4 pi squared that looks an awful lot like this h squared over 8 pi squared we just have an extra factor of 2 in the denominator so if we want to simplify things even more instead of writing h squared over 8 pi squared oftentimes we write h bar squared leaving that extra factor of 2 in the denominator there's still an i at the moment of inertia in the denominator and I've still got an l times an l plus 1 that I haven't done anything with that equation's gotten a little simpler still so sometimes you'll see the expression written that way it's simpler written in terms of h bar than an h I'll continue mostly writing it with an h but you'll often see it in the h bar form and a lot of people like this expression because this h bar squared divided by twice a quantity that behaves a little bit like a mass but isn't quite like a mass is very similar to the form that energies take in other systems like we've seen in the particle in a box that have h bar squared divided by twice something that feels a little bit like a mass and if we want to continue this process making this equation as simple looking as we can just by hiding different constants and variables inside of other redefined constants we can take it all the way to an extreme I'll define this quantity b sub b which I'll call the rotational constant let's go ahead and take all these constants so everything that multiplies the l and the l plus 1 let's go ahead and lump all those into one large constant so let's let h squared over 8 pi squared mu r squared be this thing that we call the rotational constant so with that definition so whether we call it h squared over 8 pi squared mu r squared or whether we use an i h squared over 8 pi squared i or we could even call it h bar squared over 2i those are all different definitions different equivalent definitions of this rotational constant with those now all of the constants have been lumped into one and I can say that the energy levels for this rigid rotor molecule are just the rotational constant times l times l plus 1 so rather than constantly having to divide h squared by 8 pi squared and so on if we know the value of the rotational constant then that makes the math a little bit simpler so let's see how that works let's do an example let's take a molecule like carbon monoxide so it's a diatomic molecule different masses on one side than the other that molecule can rotate if we treat that rotating molecule with the rigid rotor model that we've developed these wave functions and energy levels for I can tell you that the bond length of a carbon monoxide molecule is 1.13 angstroms and maybe I would like to know what is the rotational constant for that carbon monoxide molecule so notice I have to ask not just what is the rotational constant in general for all molecules everywhere that rotational constant depends on the moment of inertia depends on the reduced mass depends on the bond length of the molecule so every different molecule a carbon monoxide molecule, a nitrogen molecule they all have different masses they all have different bond lengths so they're going to have different values of the rotational constant that's going to be different for each individual species so we can calculate well that's actually in order to calculate the rotational constant I'm going to need to know not just the bond length which I've given you but the reduced mass of the molecule so we have to calculate that first the reduced mass for carbon monoxide remember the definition of the reduced mass is mass of atom one times the mass of atom two divided by the sum of the two masses for us, atom one is a carbon atom atom two is an oxygen atom or vice versa so if we calculate atomic masses 12 grams per mole for carbon took a couple sig figs at least 16 for oxygen 12 plus 16 would be 28 grams per mole for the carbon monoxide molecule that's going to give us calculator says 6.86 grams per mole so I'll stop there for a second and talk about that value so what does that mean that the reduced mass of a carbon monoxide molecule is 6.86 grams per mole that's not the same as the mass of a carbon atom it's not the same as the mass of an oxygen atom it's lower in fact than either one of those that's part of the reason we call it the reduced mass is this mass is smaller than the masses of the atoms that I'm combining to make this molecule remember what the reduced mass means this rotating carbon monoxide molecule that spins about its center of mass is equivalent in a physics kinetic energy angular momentum sort of sense it's equivalent to a single point with mass 6.86 rotating around a center the origin with a bond length equal to the length equal to the bond length of the molecule so it's at a larger moment arm it's a larger distance away from the center of mass than either the carbon or the oxidase so in order to rotate equivalently it has a lighter mass and a longer rotational arm but mathematically the reduced mass works out to this value of 6.86 grams per mole we're going to need that value not in grams per mole but in kilograms so grams cancel leave me kilograms I don't want moles so I'll use Avogadro's number to get rid of moles so grams cancel moles cancel and in units of kilograms the reduced mass of this molecule works out to be of course a very small number of kilograms 1.14 times 10 minus 26 kilograms that's the reduced mass of a carbon monoxide molecule we can use that to calculate the rotational constant of the carbon monoxide molecule so now I'll use this formula h squared over 8 pi squared mu r squared we know the values of all those quantities now Planck's constant 8 pi squared the reduced mass we've just calculated I'll write all that out just so we can double check how the units work and then the bond length 1.13 angstroms I've given it to you in angstroms but an angstrom is 10 minus 10 meters so I'll convert that to SI units and don't forget to square the bond length so there's the quantities we've written all in SI units so units should cancel in fact we have a kilogram meters squared underneath the second squared that's one kilogram meters squared per second squared is a joule so I've got joules in the denominator that composite unit of joule is one of these joules in the numerator of which there are two leaving me with when I'm done one unit of joules left in the numerator and that makes sense because the unit on this quantity that we're calling the rotational constants that should be in units of energy because if I just multiply that by some integers I'll learn the energy of my rotating molecule so if we ask our calculator what that math works out to be we find it's 3.82 times 10 to the minus 23rd joules so that's a number looks like a very small number what that number tells us now that we have this rotational constant if we want to know the energy levels for this molecule the math once we have this number has gotten a lot simpler remember the ground state energy has the ground state energy level has an energy zero the first state constant times 1 times 2 now we can say in our new notation that energy level is just 2 times the rotational constant the next one up 2 times 3 is 6 times the rotational constant and then 12 times the rotational constant and so on so if we want to know any one of these energy levels we just take this rotational constant and multiply it by 0 or 2 or 6 or 12 or 20 or whichever level we're interested in one advantage of knowing the rotational constant is now we have an easy way to calculate the energy levels themselves we can also now ask ourselves these gaps between the energy levels how big are they? 3.82 times 10 to the minus 23rd joules is that a big number? Is that a small number? remember the way to answer that question is to compare that energy to kT to use kT as our benchmark for energies so in order to understand whether this gap is large or small compared to kT in other words whether these energy levels are easily populated at the temperature we're interested in or not easily populated we have to compare those energies to kT and that's what we'll tackle next