 So we're now in a position where we can talk about the probability that each of the energy levels of a rigid rotor is occupied. The ground state, we might have expected would be most populated, but it turns out that's not quite true because if we look at that equation for the population of each one of these energy levels, it includes not just a Boltzmann factor for how high up the ladder of the energy state is, but also a degeneracy and the higher up this ladder we climb, the more degenerate, the higher the number of states there are at that level. So if we look a little more closely at this equation, so we've got constants all over the place in this equation, but this depends on the quantum number. If I want to know the probability as a function of the quantum number, the Ls show up in front of the exponential as just a linear term, an L, and then the exponential looks like e to the minus L squared, actually L squared plus L, but in a rough sense it looks like L times e to the minus L squared. That function, if I were to graph what L times e to the minus L squared looks like, or in fact what this piece of L function looks like as a function of L, at small values of L, this number is small, so the function is small, at large values of L, e to the minus L squared is certainly become very small, so the function becomes very small at high values of L. And what we have is a function that starts out small when L equals zero, goes through a maximum and comes back down. The key points about this are the degeneracy, this increase as I begin increasing the value of L, the probability of occupying the zero state, when I go to the first state or the second state, those probabilities are increasing and the reason they're increasing is because of the degeneracy. But then later on, so that happens at small values of L, at larger values of L, the function begins to decrease again, and that's Boltzmann telling us why that's true. The higher the value of L, the higher the energy of each energy level, the Boltzmann factor, either the full blown Boltzmann factor or this little sketch of the Boltzmann factor tells us that that Boltzmann factor becomes very small when L becomes very large, so for large values of L, the probability begins to decrease again. So there's some point after increasing and then decreasing, the function goes through a maximum at some point. So there's some particular value of L where this function takes on its highest possible value. That's interesting because it says the ground state is not the most occupied energy level. The L equals one state might be more occupied because even though each one of these individual states has a higher energy and is less occupied than the ground state, there's three of them, so the whole level might be more populated than the ground state. So somewhere up this ladder, you get to an energy level, maybe the L equals two or maybe the L equals three, that's most populated, and then the population starts to drop off as you continue to climb higher. So it's interesting to know what is that most populated rotational level? What can we expect a rotating diatomic molecule to do, whether it rotates with a little bit of energy in the L equals two level or a lot of energy in the L equals five level depends on the conditions. But we can learn something about what that maximum rotational level is or rotational level with the highest population. So the way we do that, of course, we're looking for the point on this graph that's the highest. We're looking for the point where the slope reaches zero. So since we have an equation for a piece of L, we can take its derivative and set it equal to zero. So the derivative is equal to zero when I'm at the most populated rotational level, L max. So let's go ahead and take the derivative of this expression with respect to L. The L's here show up both in the pre-factor as well as in the exponent, so we're going to have to use chain rule. I'm sorry, not chain rule, product rule. So if I take the derivative of this L, I've got one over q times two. That's the derivative of this quantity two L plus one. And since I've taken the derivative of this one, I leave the exponential alone e to the minus theta rotational over t L times L plus one. I need to add to that the derivative when I leave this L alone and take the derivative of the other ones. So leave the two L plus one alone. Take the derivative of this exponential. The derivative of an exponential is an exponential multiplied by the derivative of the exponent. So I've got constants in front of some L's, so this derivative is going to look like minus theta rotational over t. That's this minus theta rotational over t. And the derivative of L times L plus one, let's write this as L squared plus L. The derivative of L squared plus L is two L plus one. Notice that two L plus one is the same as this two L plus one. I can on the next line I'll write that as two L plus one quantity squared. So this long fairly ugly expression is the thing we want to set equal to zero, but luckily a lot of things factor out. So I've got a one over q in both terms. I've got the exponential in both terms. So let's go ahead and say one over q e to the minus theta over t L plus one. Those are the terms that those two terms have in common. Those all multiply. All that's left over in the first term is a two. All that's left over in the second term is a minus theta rotational over t and a two L plus one that shows up twice. So that whole thing we need to be equal to zero. One over q is not going to be zero. This exponential is not going to be zero. Only thing that can be zero is the quantity in brackets. So in the way to make that equal to zero is if the two is equal to the theta rotational over t times two L plus one quantity squared. So I need theta rotational over t times two L plus one squared to be equal to two. So I need to solve that expression. Remember what we're solving for. We're solving for the value of L that makes this function reach its max. So we're looking for this value of L at which the function reaches a maximum. So we just do some algebra to isolate this L. If I bring the t and the theta over to the right hand side I'll have a two L plus one squared on the left side. If I take the square root to get rid of the square I'll have this expression. I need to subtract one from both sides. So on the left side I've got square root of two t over theta minus one. That's equal to two L. And if I divide by two the last version of this equation will be L is equal to if I divide by two inside the square root it becomes a four. So two divided by four puts a two in the denominator and if I have minus one when I divide that by two it becomes minus one half. So I'll put that expression in a box it's one that we can reuse. That value of L that's our L max. That's the value of L at which the derivative of the function is equal to zero. So that tells us the most populated rotational level. So we can work an example or two and see how that works. So with our favorite molecules let's say we have carbon monoxide which has a rotational temperature of 2.77 Kelvin. If we have carbon monoxide molecules at a room temperature of 298 Kelvin then the most populated rotational level according to this expression we just arrived square root of temperature over twice the rotational temperature minus a half. So plug those numbers into our calculator 298 Kelvin divided by about five and a half Kelvin take the square root of that number subtract a half and we find that it's equal to six point eight. That's a little strange what does that mean? We've solved for the most populated rotational level rotational level is an integer that's a quantum number it has to be L equals zero, L equals one, L equals two, L equals three L equals where's L equals six point eight that's between six and seven so that's impossible there's no molecules that occupy the six point eight rotational level all that means is only zero or one or two only integer values of this function are acceptable so somewhere so six point eight is the max which is somewhere between the six value and the seven value so the way to interpret this result is to say maybe the sixth or maybe the seventh value is the most populated this one is closer to the L equals seventh level so if we wanted to be extra careful we could evaluate the L equals six and the L equals seven and see which one has a higher population using this expression or we can round off to seven and say the most populated rotational level is going to be the L equals seven level so not zero or one or two or three but remember because carbon monoxide behaved fairly classically somewhere way up multiple levels higher the L equals seven level is going to be the most populated one on the other hand if we take a molecule like HCl which has a rotational constant of fifteen point four Kelvin earlier when we asked some questions about that molecule at seventy seven that experimental temperature seventy seven Kelvin and decided it behaved somewhat quantum mechanically its most populated rotational level square root of temperature over twice the rotational temperature minus a half calculator tells us that that result comes out to be one point zero eight which again to round off to an integer rounds off to one so for HCl we decided that was a fairly quantum mechanical molecule because only about a handful of states are occupied the most populated state is this L equals one state there's more molecules in this L equals one energy level than are in the ground state but there's also more molecules here than there are up here so already the population has begun to fall off as we climb beyond L equals one this expression is is fairly useful it tells us something about the tension between the degeneracy that causes the states to be more occupied as I climb the ladder and the Boltzmann energy which causes them to be less occupied as I climb the ladder and somewhere in the middle there's a maximum and we can now calculate what that maximum is.