 Open notes, open anything you can write in here, it doesn't matter, except for your phone. Okay, the certain atom has a three-fold degenerate ground state and non-degenerate electronic level. Excited level of 3,500 wave numbers and a three-fold degenerate level of 4,700 wave numbers. This is the last example we did on Wednesday, but we did it fast. So let's just do it again briefly, remind ourselves how this works. We want to use the form of the Boltzmann distribution law that we derive to figure out what the population of these different energy levels are. That population will depend on the degeneracy of the energy level. Here the degeneracy is 3. Here it's 1. Here it's 3. All right, we can factor this information into our answer. This is the version of the partition function that we want because it's the version that contains explicitly the degeneracy. This is the degeneracy of energy levels. All right, here's an energy level. Here's an energy level and here's an energy level. The degeneracy of this level is 3, 1, and 3. Okay, so this summation means that we have to sum over these three energy levels. There will be three terms in our summation. If there were five energy levels, there would be how many terms? Five is the correct answer. All right, three, that's the degeneracy exponential. What's that energy? It's zero, okay, and the first excited states at 3,500 wave numbers. The second excited states at 4,700 wave numbers. We need to work out what this is in units of joules. All right, we have our handy conversion factors. 8065.5 wave numbers for EV, 1.602 times 7, my second joules for EV, boom, there's the temperature that we care about, 1,900 Celsius, a Kelvin, or rather, and so this 2.649 is what? All right, it's the exponent here. There should be a minus sign in front of it and that's the exponent on this E. Okay, and so when I evaluate that E to the minus 2.649, I get 0.07069. Okay, and obviously this guy is just going to be three because that's zero. All right, so this exponential is just one. And so 3 plus that plus that is 3.156 and the first question, we always want to ask from recalculating a partition function is does, is the order of magnitude of this number of thought right or is it way up, all right? Have we made an unbelievably stupid mistake or could the answer be correct? All right, does it make sense? We always want to ask that question. What is the partition function? It's the number of thermally accessible states at the target temperature that we're talking about. Okay, and so if we want to figure out if this answer could possibly be right, we need to calculate what the thermal energy is in units of wave numbers. All right, let's calculate what the thermal energy is. What is the thermal energy? It's K times T. K is 1.381 in the semi-screen joules for Kelvin, 1900 Kelvin. 1900 degrees Kelvin, and so that product is 2.62 times 10 to the minus 20 joules as always any and certain joules is counterintuitive. It's always 10 to the minus 19, 20, 24 joules are not a unit that it's easy at least for me to grasp. So I always want to convert to EV or wave numbers, all right? In this case, we're just converting to wave numbers as we've done before, 1,321 wave numbers is what that turns out to be equal to. 1,321 wave numbers. Okay, so here's my energy level diagram. Here's my ground state, triply degenerate. Here's my first excited state. Here's my second excited state. Here is that 1,321 wave numbers. That's the amount of thermal energy that's present in the system, okay? And so of course these, so now we have to ask the question, what do we expect the partition function to be in this system at this temperature? Well, these ground states are definitely going to be thermally accessible, right? The ground states are always going to be thermally accessible when they're at zero energy, right? They would be accessible at 1 degrees Kelvin, all right? So we know that the partition function is going to be at least three. And then we can ask, are these excited states going to be populated as well? Well, you can see we don't have enough thermal energy to significantly populate these excited states and so we expect there to be a little bit of population of those states because what does that Boltzmann function look like? It's a decaying exponential, okay? So there's going to be some populate. This thermal energy doesn't cut off in a hard line here the way I've drawn. It sort of smears now. Yes. Yes. Did everyone hear that? Does that mean for this particular example the maximum that Q could be is seven? Yes is the answer. Okay. So we expect a little occupation of 3,500 and 4,700 states even though we're not close to them there's a tail on this thermal energy that we have to remember about, right? I'll draw it as a pink box but we know there isn't a hard cut off on that thermal energy. It extends exponentially at a higher energy. Okay, so qualitatively this number makes sense because here's three states that we know are going to be thermally accessible and then there will be some fraction of these four states here that are also thermally accessible. So 3.156 qualitatively that's what we expect here. If we had gotten four that should be a red flag. That sounds too high, all right? We shouldn't really have that much occupation of these excited states. And that level of intuition is only going to come from doing a bunch of problems, right? Now if you've done a bunch of problems it'll be clear to you that no, there's no way that 4 is the right answer and it's not the right answer. 2 can't be right, all right? We know there has to be a partition function of at least 3, okay? So look at the answer and ask yourself, we know the answer has to be between 3 and 7, all right? And 3 in a fraction is about what we expect. Let's do another one. These are the electronic states of carbon that turns out. Here are the energies, here are the degeneracies of those states. Now what we want to do is calculate the fractional population of every single one of these levels at 6,000 degrees Kelvin which is the temperature of the sun. We've got carbon atoms in the sun, what electronic states are they in? So we start with this equation that we derived on Monday, Wednesday. Okay, there's our partition function. What we want to evaluate is the fractional population, right? The number of atoms in the state divided by the total number of atoms in each one of those 4 states, okay? And so if I divide by n, that n just moves down here so this equation doesn't have an n in it anymore, all right? And this guy here, that's just our partition function q. Okay, in this case we will have 4 terms, yes, 1 for each, one of these 4 energy levels. The first one is going to go away with this level. The second one with this level right here, the third one here, this is really necessary. Okay, how do we calculate them, all right? Here are the 4 terms. The first number here is the degeneracy of each state, non-degenerate, triply degenerate, 5-fold, 5-fold, all right? 1, 3, 5, 5, then we've got the exponential. That contains a different energy for each state. That's the energy of the state, okay? So for this guy, the energy of that first state is at 16.4 wave numbers, pretty low, unbelievably low, all right? There it is. And here I'm just doing a unit conversion to get to joules again, all right? So I'm evaluating that term right there, all right? And of course there will be a minus sign in front of this 3.93124 when I put it in that exponent there. Put a minus sign on it, okay? And so it ends up being 2.99, this guy ends up being 4.95, this guy ends up being 0.43. I add them all together to get the partition function. I get 9.37 as the answer for the partition function, okay? So 9.37 of the 14 total states, there's 14 total states, 5, 5, 3, and 1, all right? Are thermally accessible at 6,000 degrees Kelvin? It's too small, right? 6,000 degrees is an enormous temperature, isn't it? So as always, does this make sense? The only way to know for sure is to calculate that thermal energy again, all right? 6,000 degrees Kelvin times K gives me this number in joules which once again is completely useless to me because I have no intuition about joules. It's always 10 to the minus 20 something, all right? But when I convert it to wave numbers rather, I get 4172 wave numbers, all right? That's a number that I do have a little bit of intuition about. All I've drawn here are the electronic states of carbon, right, from that table that I just showed you, right? Here's the ground state. It's non-degenerate. Here's the first excited state. It's only at 16.4 wave numbers, really, really close. First excited states at 43 wave numbers, then there's a discontinuity here, notice, on this energy axis, right? Because that's 5,000, that's the highest excited state. 10,103.7 wave numbers, okay? Where's the thermal energy on this diagram? Goes all the way up to 4,173, right? That's 6,000 degrees Kelvin. That's the thermal energy and it tails to higher energies, of course, okay? So what did we calculate for the partition function? 9.37. Does that make sense? 1, 2, 3, 4, 5, 6, 7, 8, 9. Does it make sense that all of those are thermally accessible? Yeah. 0.37, right? Hard to tell. All right, what that 0.37 is telling us is that there's some occupation of this guy, even though he's way the heck out there. Right, there's some occupation. It's a little surprising. That's a big energy gap, about 5,000 wave numbers, 6,000, okay? But we can use our intuition to tell us that this 9.37 is right about what we would expect, right, also close. Yes, 9.37 does make sense, okay? That's not what we were asked, by the way. We weren't asked to calculate the partition function. We were asked to calculate the relative populations of each one of those levels, okay? So we want to evaluate this. This is the number of atoms in the ground state divided by the total number of atoms. That's the relative population right there, okay? And so we take the first term. Here's our expression for the partition function. We take the first term, put that in the numerator, right? Divide by the partition function, that is the fractional population of the ground state. Now we take the second term, put that in the numerator for the N1 state, right? The first excited state, 2.99 divided by the partition function, that's the fractional occupation of that state that was at 14 wave numbers and so on, 4.95, 0.43, boom, boom, boom, boom, add them up and you should get 1. That's a check on whether you did this calculation correctly. Right, because obviously total population, right, once you've got it for all of the atoms and all of the states, it should equal 1, right? So 0.999 means there's a little rounding here, okay? Now we've used the Boltzmann equation before in N1, maybe in organic chemistry, but we've never, correct me if I'm wrong, been able to account for degeneracy in this way, right? This is the first time that you've ever been able to accurately calculate the population of energy levels like this. This is a super important thing to be able to do, okay? Very straightforward, very powerful. Now, it will be obvious to you, I hope. Here's a molecule, here's a molecule, here's a low temperature, there's not enough energy here to occupy very many of these excited vibrational energy levels that I've drawn here, right, we're almost all on the ground state, right? And at a higher temperature, we might occupy a bunch of different states in the same molecule. And if I calculated the number of molecules that have this configuration right here while there would be 4 and if I calculated how many have this configuration right here, there would be 24. So, qualitatively, we understand that energy is going to be correlated with W and with the partition function somehow. So let's see if we can figure out what the relationship is between the energy and the partition function, all right? We're going to keep coming back to this partition function in staff net. It's the central object, all right? So, this is the average energy, we're going to call it the average internal energy because it's exclusive of the translational energy. In other words, an atom might have some energy because it's zooming around. It's got some kinetic energy. This is the internal energy. This is the total energy to get the average energy. I divide by the total number of atoms or molecules, right? All right, so that makes sense. Total energy divided by the number of objects that have energy is the average energy per object, all right? This is my expression. What is this? This is the sum of the energy for every, this is the number of atoms or molecules that have a particular energy and that's the energy and so if I multiply those two things together and I sum over all the different energy levels, I should get the total energy and that's just it, okay? Now, we have an expression for n already from both of these distribution blocks. So, if I plug this expression in for that n right there, I think you can see immediately that this big n and that big n right there, they're going to cancel. Boom, okay? And so now, when I do that substitution, here's the equation that I get for the average energy, all I did is make a substitution here for that n, boom, okay? And now q is characteristic of this whole distribution of states and molecules at a particular temperature. So, I don't need to include it in this summation. It can go out front. It's a constant for every one of these states. It's not going to change. Okay? So, this is what my expression looks like. All I did is move 1 over q out front. Now, our definition for q is this. Look at something that I won't call interesting, but note something, if I take the derivative of q with respect to beta, right, the derivative of q with respect to beta, this is q, if I take the derivative with respect to beta, what are the rules for that? Remember, the rest of this guy's going to move out front, there it is, okay? And if I put a minus sign here, that minus sign is going to get extinguished, okay? So, does everybody agree that this is equal to this right here? All right, the reason that's interesting is because this right here is that, right? Dq d beta, minus dq d beta, minus the derivative of the partition function with respect to beta, it seems like a very abstract thing to be evaluating in the first place, all right? But my goodness, that gives us the average energy, all right? So, this is an equation in your book, I don't know. I should have laid it out, I don't remember which one. All right, but the average energy is just equal to minus 1 over q, dq d beta, there's that minus sign. That's what that minus sign is right there. And if I want to know what the energy is for n molecules, I put big n here. Often we want to know the energy per mole. Most of the time we want to calculate energy per mole, so that number is Avogadro's number for goodness sakes. Okay, you ready? Let's do an example. Here's a molecule, n, o. Two electronic states, 1 and 121.1 wave numbers, 1 and 0 at the ground state, both doubly degenerate. The analog is doubly degenerate, and it's 2, and 2, and 2, and 2, and 2, and 2, and 2. Calculate and plot the electronic partition function of NL from 0 to 1,000 k. Evaluate the term populations and the electronic contribution to the molar internal energy at 300 degrees Celsius. Which problem for next? Okay, 1,000 k, let's calibrate ourselves so we have some intuition about where we're going with this problem, where is that compared to 121 wave numbers, boo. 121 wave numbers is down here, 1,000 k turns out to be 695 wave numbers. Okay? So what are you going to predict? The partition function is going to be at that temperature. Yes? Yes, everyone here that insightful answer, 4 is the highest it can be. It can achieve 4 only at infinite temperature. So it's going to asymptotically approach 4 as the temperature increases, but it's never going to get there, right? Okay, let's see what we come up with. Here's the partition function, so hold that thought. Two levels this time, right? Both doubly degenerate, boo. There's the energy. We're not going to continue to write this unit conversion on the screen. The wave numbers to joules, I trust that you can do that now. And you can do it any way you want. You don't have to use my inversion factors. There's a million ways to do it. Oh, there it is again. All right, here's my plot of Q as a function of temperature all the way up to 1,000 degrees. Check it out. Does it look like it's approaching 4 asymptotically? It sort of does, right? What is it here? Maybe 3.7, 3.65, right? Intuition-wise, that's where we sort of expected it to be maybe a bit higher than that, because my goodness, we've got way more thermal energy than 121 wave numbers, don't we, right? So you might be a little surprised that it's not closer to 4, but what you'll find as you do more problems is, right, your intuition has to be calibrated a little bit, right? It's going to be more than 3.5, but less than 4, it's sort of in the right range of 1,000 degrees. Now, we want to calculate these term populations. The ground state is n0 divided by n. In the numerator, we have the first term of our part, here's our partition, here's our whole partition function, right, the ground state is the first term over the whole partition function, boom, 0.64. 64% of the molecules are in the ground state, even though, okay, well, we're talking about 300 degrees Kelvin here. Where's 300 degrees Kelvin on this darn? 300 degrees Kelvin is more than 121, but no more than 1,000, okay? Just to calibrate you, let me go back to where I was. Okay, so 0.64, 64% of the molecules are in the ground state, even though there's more thermal energy necessary than is necessary to populate that excited state, right? Most of the molecules are still in the ground state. That's interesting. And the excited state has, would have to have, of course, the other 36%, because there's only two states, right? And those two numbers better add up to 1, okay? I think this is a little counterintuitive because that number seems high when you've got almost a factor of two more thermal energy than you need to populate that excited state. You still have most of the molecules in the ground state. Okay, I think we're going to stop right there. We're not going to do the rest of this one. We'll do it on Monday. And you guys have a good weekend.