 So, we have the Boltzmann probability distribution, which tells us how to find the probabilities that a system is going to be in a particular state if we know the energies of the states. And in particular, we need to know the energies of all the states. In order to calculate the probabilities, we need to know Q. Q is the sum of these Boltzmann factors, the sum of e to the minus energy over kT for all the states the system can occupy. So if we know all the energies of all the states, then we're good. We can calculate Q. We can calculate the probabilities. That works fine for a system with a finite number of states, like butane, where there's an anti-state and two ghost states, or cyclohexane, where there's a boat state and a chair state. If we know what the states are and we know the energies of all of the states, then we're fine. But think for a second what might happen if we have a system like a hydrogen atom, where we know that the electron and the hydrogen atom might be in the 1s orbital, or the 2s orbital, or the 2p, or the 3s, or 3p, or 3d, or the 4spdf, and so on. So there's an infinite number. I can never run out of states to name. It might be in any one of those possible states. In order to calculate Q, I need to know the energies of all these states and what's an infinite list of states. So how do I go about calculating the probability that the molecule is in any one of these states if I can't stop writing down the infinite list of the states? I would need to be able to perform a sum over an infinite number of Boltzmann factors. And there are some tricks for doing infinite sums that we'll have occasion to use occasionally in the future. But for right now that sounds like a lot of work, writing down an infinite number of terms, summing up an infinite number of Boltzmann factors before we can even calculate Q to get started on the probabilities of each of these terms. So it turns out what we often want to do is when we don't know or can't know the energies of all the states, is to calculate the relative energies of two different states. So for example, let's use 1 and 2, well, we can use A and B. So in a trick, a lot like something we did a couple of video lectures ago, the probability of any particular state, state B is 1 over Q e to the minus energy B over KT using the Boltzmann distribution. Probability of some different state is 1 over Q e to the minus energy of that state divided by KT. If I divide these two things by each other, the ratio of PB to PA, that's just the Boltzmann term for B divided by the Boltzmann term for A. And now I can see that the Q's cancel. The ratio of these two exponentials, e to the minus eB over KT minus e to the minus EA over KT. That ratio of two exponentials is e to the difference between the two energies, or the two exponents, which I can also write as e to the minus delta e over KT, again, where delta e is the difference between the first state that I named the one in the numerator and the energy of the second state, the one in the denominator. So what that tells us is I've got narrative Q. I don't need to complete this infinite sum. I don't need to know what all these energies are. In order to calculate the relative probability of state B relative to state A, I only need to know the energies of those two states in particular A and B. So when we don't know all the energies, we can find the relative probabilities of two states if we know their energies. So that's an important result and one that we'll use frequently. I can use that to do an example. And let's stick with this hydrogen atom example. Electrons in a hydrogen atom in some particular orbital. The energy of the 1s orbital, that one I can call zero. That's my zero of energy. I'll draw an energy diagram. So here's the energies. There's the energy of the 1s orbital. The 2s orbital has a higher energy. That energy is a particular value in joules. That works out to be a very small number of joules, 10 to minus 18 joules, 1.63 times 10 to minus 18 joules. So that energy is higher for the 2s state than the 1s state. And those aren't the only two states. There's 3s and 3p and 3d and many other states in this system. But if those are the only two energies I know or the only two I want to work with, that still is enough information if I am asked how likely is it that I find an electron in the 2s state of a hydrogen atom relative to the 1s state at a particular temperature. And let's do 298 Kelvin. So we have enough information to solve that problem. We just need to do this calculation. The relative probabilities is e to the minus difference in energy over kt. So that's e to the minus difference in energies is 2s minus 1s over kt. So if I plug in numbers, that difference of energy, 2s is the upper state, has this much energy, 1.63 times 10 minus 18th joules we were given. Subtract from that 0, the energy of the lower state. So that's the difference is the same as the energy of the upper state. I need to divide that by Boltzmann's constant. And also divide it by the temperature, 298. Let's look at units before we go to the calculator. In this fraction in the exponent I've got a joules in the numerator, joules in the denominator. The denominator has a 1 over Kelvin and a Kelvin. Both of those cancel. So the exponent is unitless. And I can plug those into a calculator. If I do just the numbers in the exponent, I find that that's e to the negative 396. 10 to the minus 18 is quite a bit larger than 10 to the minus 23, even after the 10 to the minus 23 is multiplied by a few hundred Kelvin. So that ratio of numerator to denominator is about 400. Up in the exponent of an exponential, e to the minus 400, e to the minus 396, that's an almost ridiculously small number. That's something like 9 times 10 to the minus 173. So that's the numerical answer. How likely am I to be found in the 2s orbital of 100 and atom relative to the 1s to being found in the 1s orbital? Very unlikely. You find an electron in the 2s orbital of 100 and atom only one time out of 10 to the 172 or 3 times that you have 100 and atom. So almost never will you find at 298 Kelvin an electron in the 2s state of 100 and atom. Almost always you'll find it in the 1s state. And that matches what we expected from this problem. If I had asked you to recall what is the electron configuration of 100 and atom from general chemistry, you'd say that's 1s1. There's one electron found in the 1s orbital. We didn't teach you in general chemistry. Sometimes it's in the 1s. Sometimes it's in the 2s orbital. We taught you it's always in the 1s orbital. And we see that that's not literally true very, very rarely, but extremely rarely. You can find an electron in the 2s orbital at room temperature, but it's so rare that we don't really need to consider the possibility. So the Boltzmann distribution is telling us exactly how rare it is that we find electrons in the 2s orbital relative to the 1s orbital. So notice we've got a numerical answer to this question. We didn't need to know energies of anything other than the 1s and the 2s orbitals. We didn't need to know the energy of the 2p, 3s, and so on, all these other orbitals. Notice also that when we computed this answer, all that mattered was the difference in energy between the two orbitals. In the numbers I gave you, I said energy of the 1s orbital is 0, and energy of the 2s orbital was 1.63 times 10 to the minus 18th joules. The difference in energy is 1.63 times 10 to the minus 18th joules. In a different set of units, so in my way of looking at it, this was the 0 of energy. But from a different way of looking at it, one that's actually more common that we'll see later on in the semester, if we put the 0 of energy elsewhere and the energy of the 1s state in some different unit convention would be negative 2.18 times 10 to the minus 18th. The energy of the 2s state would be negative 0.55 times 10 to the minus 18th joules. Those energies, negative 2.18 times 10 to the minus 18th, that's not 0. Negative 0.55 is not 1.63. But still, the difference in energies between these two values is the same as the difference in energies between these two values. And when it comes to determining whether the molecule occupies the 1s orbital or the 2s orbital and with what probabilities, all that ends up mattering is the difference in energies. It doesn't matter where we put the 0 as long as the difference of energies is the same. So now we know how to calculate probabilities given the energies. The next question is going to be, where do those energies come from? If I give you the energies, you can now calculate the probabilities or the relative probabilities. But we'd also like to be able to predict what the energies are. If I say, for a hydrogen atom or some other atom, what are the energies, we can either go into the lab and measure them or we can predict them with physical chemistry. And in order to do that, we're going to have to use quantum mechanics. So that's what we'll start next, is learning a little bit about quantum mechanics.