Lecture 15. Hydrogen Wavefunctions, Quantum Numbers, Term Symbols and Transitions
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Shaka, Athan J.
 UCI Chem 131A covers principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. 
 Hi, today in chemistry 131A we're going to talk about hydrogen wave functions, quantum numbers, term symbols, and transitions in atomic spectroscopy. Remember that the solution for the hydrogen atom gave us three quantum numbers, N that had to do with the energy, L that had to do with the square of the total angular momentum, M sub L which was the magnetic quantum number which was the projection of the angular momentum on the Z axis. And then there was another number, M sub S, that came from the intrinsic magnetic moment of the electron itself. And so if we have these four quantum numbers then we know everything that we can know about an electron in a hydrogen atom. However, once we have more than one electron then there's the well-known Pauli exclusion principle which you've probably learned from freshman chemistry. And that is no two electrons can have all four quantum numbers the same. So if N, L, and M sub L are the same, for example, the electrons in a 2PX orbital, then M sub S has to be different for the two electrons. One has to be so-called up and the other has to be down. In other words, they have to be paired. We'll see a little bit later what that actually means. And so as I said, electrons in the same spatial orbital have to be paired up magnetically with respect to the spin. Let's look at the possibilities for a neutral carbon atom with six electrons. The configuration from the periodic table for a carbon atom is 1S2, 2S2, 2P2 and that means that there's two electrons in the P shell that can occupy any of the three P orbitals. The other orbitals are filled. And the L equals one state for P, M sub L can be 1, 0 and minus 1. And we can figure out then the limits on L and S, which depend on each other because of the exclusion principle. Let's have a look at this. The first thing to note is that only the P electrons have any flexibility. The 1S shell is filled. That's out of commission. The 2S shell is filled. That's out of commission. And it's only the frontier orbitals that we're putting electrons into that have any flexibility as to how the electrons can orient. Whatever's already happened to 1S and 2S has happened and that's water under the bridge. For example then, here on slide 389, we have all these possible configurations of the electrons in these 3P orbitals are allowed. On the left, we've got two electrons in the M sub L equals plus 1 orbital and they have to be paired because of the exclusion principle. Or we could have one in the 1 orbital and one in the 0 and they could be also paired, one up, one down. And we could have one in 1 and one in the minus 1. And we could have them parallel as well, one in the 1 and one in the 0. All those possible configurations of the electrons in the P orbital shells are okay. They're all allowed. But on the bottom of the slide here, none of these are allowed. In the first one, we have M sub L equals plus 1. There are two electrons and they're both up. That means they have all four quantum numbers the same because N is 2, L is 1, M sub L is 1 and M sub S is plus 1 half for both of them, not allowed. Same thing with the others. Whenever we have two electrons in the same orbital, they can't have the same spin. Remember that we had a shorthand way to keep track of these various atomic levels which was the term symbol which gave us sort of a summary of how L and S would magnetically couple to give the total angular momentum J and would allow us to keep track of which transitions were allowed. Remember that the term symbol is 2S plus 1 on the left hand superscript, capital L which was SPD and so forth for the central letter. And then a value of J which is the way S and L couple and they can couple in various ways. They're various allowed values according to the Klebsch-Gordon series that we derived. 2S plus 1 is called the multiplicity but keep in mind that it is only really the number of states there if, in fact, L is greater than or equal to S. Because if L is less than or equal to S then there could be less states allowed. So if S is zero it's called a singlet. If S is a half it's a doublet. If S is one and a triplet. However, if we have 2S one half where L is zero then it's not really a doublet. There's only one state because there's only one level there for the electron to occupy. So it's not really a doublet in that case. Let's try another example and have a look at these then and see which values are allowed. So this is practice problem 20. What terms can arise from a neutral carbon atom with the electron configuration 1S2, 2S2, 2P2? Well, remember the way we approach these problems we have to figure out the possible values of L based on the values of little L for each electron. We couple those together. Then we figure out the possible values of big S based on spin one half for the electron. We couple those together. And then we use the Klebsch-Gordan series to get about possible values of J which naively go from L plus S down to the absolute value of L minus S. So if we can figure out these possible values of big L and big S we should be able to come up with the term symbols. But what we'll see in this case is that it's a little bit more complicated than it might first appear. And the reason why is that some of the values that we'd like to pick poly excludes from possibility. The possible values of big L with 2P electrons are 2, 1, and 0. Those are the three you can get from two values of little L equals plus 1. And the possible values of S, big S, the total spin angular momentum are 1 and 0. That's what you can get from two spin one half electrons. Therefore, if we just took the Klebsch-Gordan series it would seem that J could be well L plus S is 3 down to the absolute value of L minus S is 0. So it would seem that J could be 3, 2, 1, or 0. But we can't have J equals 3 when we have both electrons in the same shell. And the reason why is that if we had J equals 3 we'd have to have L equals 2. So let's put the two electrons in the plus 1 orbital. And then to get 3 we have to have S equals 1. So we have to have them parallel. That means that we have to have them both in the same orbital there. And then we have to have them with the same spin. But that's not allowed because if they're in the same orbital they have to have opposite spins. So we are allowed to take the maximum value in that case. If we had an excited carbon atom, and that's why some books give you these weird problems where it's an excited atom, where one of the electrons is in a 2P and one's in a 3P, then you just total them all up because there are different shells. And so the poly exclusion principle doesn't apply. They aren't in the same orbital. In this case, which is actually more important to work out because usually you want to work out the terms for the ground state of the atom to figure out what kind of spectroscopy you can see. You have to be quite a bit more careful how you do it. Therefore, if we have L equals 2, we have to have them paired. So we have S equals 0. The electrons have to have opposite spin. And the term that arises from that is singlet D. So if it's singlet D, then that means that J has to be 2 because S is 0. 2S plus 1 is 1. S is 0. D, so the series stops with J equals 2. And so the term that arises is singlet D2. For L equals 1, we're okay because then we can have one electron here and the other over here somewhere. And in that case, we get a triplet P term. And there are three values because now S is 1 and L is 1. So we get triplet P2, triplet P1, and triplet P0. And then the last term that we can have is L equals 0, S equals 0. And therefore, the only possible value of J is 0. And this would be called singlet S0. What you have to do then is you have to go back and say, what were the possibilities of the states I started with? Recall that we did that when we did this before with just coupling two electrons. We made sure that we accounted for all the possibilities that were there and then make sure that our terms cover all those and no extra ones. And like anything, like doing crosswords or Sudoku or anything, this takes some practice. But it can be quite a rewarding puzzle to work out to make sure that you've got it right, that you aren't including any non-allowed states in the terms. If you've got an open F shell with multiple electrons or an open D shell with multiple electrons, working out the allowed terms can really require a quiet room because it can get quite complicated to keep track of which ones are allowed and which ones are excluded because you can't have two electrons in the same orbital with the same spin. Suppose on the other hand we had five P electrons, say, flurring. If we took all five electrons and then we said, boy, this is a mess because all these orbitals are filled and so there's these millions of combinations that are no good because we can't have the electrons that way. It would be very, very difficult to do and it would take a long time. And so the trick we can do is we can treat the missing electron, the hole as if it were an electron and we can move it around and we can just treat it the same way we would if we had one electron. And that's because once it's more than half full, it's really the holes that are dictating which terms can arise. The electrons are filling all the orbitals. And it turns out that it's exactly the same and so you don't have to do it over. Therefore, the worst case is when you've got three P electrons or five D electrons. Once you've got more than that, it just reduces to a previous case where you have less electrons. So it's no harder. And so for flurring, excuse me, for oxygen, which has two P4 electrons, you get the same terms as carbon that had two P2 that we just did. And likewise, flurring gives rise to doublet P3 halves and doublet P1 half terms. Now I want to talk about the shapes of the hydrogen wave functions. We can plot solutions. We had the equations, but let's just look at the shapes of the wave functions themselves. Not the radial distribution function, but the angular part as well. And it's common practice to plot these then as a contour plot in which we pick some arbitrary contour in space and we say there's a certain likelihood that the electron is inside this contour. And then we can pick a color to code whether the phase of the wave function is plus or minus. Remember it's a wave, so it has a phase and it has a phase in 3D just like a sine wave has a phase in 1D. Sometimes before people used color, they'd mark a plus on one side of the wave function and a minus on the other. The problem with using plus and minus is that if you look at it and you aren't quite sure what it refers to, you might think it has something to do with charge, but it doesn't have anything to do with charge because electrons are always negative. And it's just whether the phase of the wave function is a positive or negative. And in between there's a node where the wave function has to be zero. The thing is we tend to avoid using the true eigenstates for the hydrogen atom and the reason why, which I referred to in a previous lecture, is that if we include I and we've got e to the im phi going around then that's difficult to figure out what you're going to plot because it's much easier to say well I'm going to plot if this number is equal to this thing I'm going to plot that surface where this number is equal to 0.9 or something like that. But if as you go around the phase is changing with respect to I then you'd have to figure out some way to code that. And you might be able to do that by coding in a rainbow, but on the other hand maybe that might be confusing when you include the phase. So as a dodge what tends to be done is instead of plotting the true eigenstates we plot linear combinations of plus and minus m sub l that either give the cosine part or the sine part. And recall from Euler's identity that e to the i theta is cos theta plus i sine theta. So we can pick linear combinations to plot those. And that's usually what's done. We pick then the real part, we throw away the i and we pick the real part and then we just plot the things. However, keep in mind that the plots depend a lot on the particular value of the contour that you pick. Do I pick that the electron has to have 99% probability of being inside this thing or 50% probability of being inside this thing or whatever? Usually 90% is a typical value, but as we saw for the hydrogen atom, if we go out just to the most likely radius of where the electron is, the Bohr radius, there's only a 32% probability that the electron is inside there. So perhaps you could argue that maybe you should pick a more conservative percentage to get a more realistic idea of the size of the spatial extent of the wave functions. So here are some wave functions then plotted. And red is positive phase in these plots and blue is negative phase. And the 1s is the little dot on the left. That's a little red thing. And it just is spherical as we would expect. And the 2s is a bigger sphere. There's a tiny red dot in the center. And then there's a bigger blue area. And that's because as we saw, the radial function for 2s has a node at some point where there is zero probability of finding the electron. And then we get these big fluffy orbitals, the 2pz, 2px, and 2py that look really quite big and distended. And that has to do of course with the particular contours that were chosen to plot these values. Recall that a node is a place where the wave function changes sign. Strictly speaking, it shouldn't just go to zero, but it should be plus on one side and minus on the other side. And a node in a 3D wave is a surface where the 3D wave has a value of zero. The radial nodes are in fact spheres. We saw that for estates there are nothing but radial nodes. And then the angular nodes are angles which turn out to be planes where the wave function is zero. A 1s function has zero nodes, a 2s has one radial node, and a 3s has two radial nodes. And in general, the number of nodes is n minus 1. If we look then at the 2pz orbital which I have here on slide 398, we see that there's an angular node, there's a plane in which the wave function vanishes. It's positive on one side, red on one side, blue on the other side, everywhere in between the wave function vanishes on the plane z is equal to zero. The d functions look like these. Here's 3Dxy looking like a big clover leaf. And you can clearly see that there's a disturbance building up in the angular variation of the electron as it assumes more quanta of angular momentum. That makes perfect sense. And so there are five 3D orbitals and we saw what they were. We wrote them all out but we didn't see exactly what they looked like so here's 3Dxy here with these four nodes. And unlike p, the blue are across from each other and the red are across from each other. p changes sign when you rotate it 180 degrees. This d changes sign if you rotate it 90 degrees. And that's because it's 2 quanta of angular momentum and not 1. Here's a different view of the d wave functions and how they may add up to give us a shell that has spherical symmetry. Four of them are pretty much like clover leafs. They're four things, red and blue. Here they're just gray. And here they're also plotted at a different level of the contour so they look a lot thinner than in the other plot. And then there's this funny one, 3Dz squared, which almost looks like a p orbital. And unlike the others where I can clearly see these planes where this thing is vanishing, this one, especially without color, is a little bit hard to see. But what we have with the 3Dz squared is we have a donut around the equator and then we have this thing that looks a little bit like a p orbital. But the difference is this thing is all the same color. It doesn't change sign. And this donut's all the same color. And you still look at it and you say, well, where are the angles where the nodes are? And the answer is they're at the magic angle. If you look back at the formula for this particular orbital, you'll find that there's an angle tilted where the wave function vanishes. And it vanishes all along that angle all the way out. And so that's how that one works. So it looks a little bit different, but in fact, it's exactly the same in terms of vanishing at values of an angle, just like the other ones do. Now let's talk about spin orbit coupling. We saw that we had these two lines for the sodium spectrum. And they were closely separated. And the two transitions were doublet p 3 halves goes to doublet s 1 half and doublet p 1 half goes to doublet s 1 half. Recall that delta J can be zero as long as J is not equal to zero. The energy levels then break down like this, as I've shown on this figure on slide 401. There are two levels, two excited states above the ground state, these p states, and they're split by an amount I've called delta in this slide called the spin orbit coupling. And what we'd like to do is explore this a little bit and figure out exactly where this comes from. The only difference between the two states is the orientation of the spin relative to the orbital angular momentum because they have the same value of s and l, but J is different. So s and l can add up like a triangle and make J the third leg of the triangle. Then J can be different because there can be several different values that work. And it must be then the energy difference is just whether this magnetic intrinsic bar magnet of the electron is aligned or against the magnetic field that it sees if from the perspective of the electron the nucleus is going around the other way and it creates a magnetic field that acts on the electron. If we know the measured optical frequency difference between the two levels, then we can estimate the magnetic field that the 3p electron and sodium appears to be experiencing from going around the nucleus of the sodium. The magnetic energy difference of an electron spin is delta E is equal to G, the so-called electron G factor which has a value about equal to 2 and then a conversion, the Bohr magneton that converts magnetic field to energy and then B, the value of the magnetic field that the spin is experiencing. That's the magnetic energy that this bar magnet that the electron has is going to feel. So let's do a practice problem and figure out what kind of field it experiences after we figure out exactly what the spin orbit coupling is. What's the spin orbit coupling in wave numbers, in electron volts and what is the apparent magnetic field that the electron in a sodium atom in the 2p orbital experiences? Okay, well the wavelengths of the two transitions are 589 nanometers and 589.6 nanometers and we know the energy difference is delta E is h nu 1 minus h nu 2 and that's hc times 1 over lambda 1 minus 1 over lambda 2 and we can convert that then to wave numbers as just hc times nu bar 1 minus nu bar 2. Those are the wave numbers in centimeters, inverse centimeters and that's, we can just write that as delta nu bar. So we can solve for delta nu bar and we just have to take the inverse of the wavelength which is in nanometers to the minus 1 times 10 to the 7 nanometers per centimeter and to be careful if you're doing an exam you may want to convert from nanometers to meters and meters to centimeters. It seems easy but it's easy to get also a factor of 10,000 off if you go too quickly and happen to go the wrong way with one of the conversions. If you do that then the spin orbit splitting in sodium is 17.28 wave numbers. If we convert that to EV which we do by converting to joules and then we just simply convert from joules to EV by multiplying by dividing rather by 1.6 times 10 to the minus 19 joules per EV, we get a very small value of 2.142 times 10 to the minus 3 electron volts. Keep in mind that the ionization energy of hydrogen to kick the electron completely out was about 13.6 electron volts and here we're talking 2 milli electron volts. So these magnetic interactions in these systems are very, very much smaller energy differences than the main electrostatic potential that the electron feels from the charge. And that's a common theme. Basically magnetic energies are always quite a bit smaller than electric energies. And in the hydrogen atom the spin orbit splitting is very, very tiny compared to sodium. And so it took a lot of detailed work to even work out that there was something there and then to explain it. And it's very interesting historically. It's too bad we don't have time to go through all the thought processes that went through finding the fine structure and hyper fine structure of these systems. But it is very interesting detective work from a science standpoint. Finally then, okay, we know this spin orbit splitting. Let's then use our formula for this, the energy difference of an electron in the magnetic field. We know the energy difference. We know the g value of the electron. We know the Bohr magneton. That's a constant. We look that up. Let's figure out then what the apparent magnetic field is that this electron is seeing. And what we get if we take delta e divided by g times the Bohr magneton is we get 18.5 Tesla. 18.5 Tesla is an absolutely huge magnetic field. If we tried to make a magnetic field like that in the laboratory, we would be very hard pressed. Even if we took tons and tons of wire and wound a gigantic electromagnet, we would not get such a big magnetic field. In fact, the very biggest NMR spectrometer at UCI has a magnetic field of 18.6 Tesla. But the magnet is 12 feet tall and it has several miles of superconducting wire wound into a gigantic solenoid to make this magnetic field, which is enormous. So you can see that because the particles are so close and things are moving so rapidly, that there are very large magnetic interactions on the electron, very large compared to what we can do in the laboratory. And so when we take atoms and we put them in even a pretty strong magnetic field, the energy levels do split and move and do things. But they split only a little bit compared to how big they're already split due to these internal magnetic fields from the nucleus. And that's good because that means that if we have two levels like this and then we turn on a magnetic field and they split, they go like that. And that makes it easy to keep track of what's going on. If they went like this and all over the place, it might be really hard to figure out what the spectrum was doing. If we look at other alkali metals like lithium, potassium, rubidium, and cesium, then because they have a single electron outside a closed shell, they all give rise to exactly the same terms as sodium. And so their spectra are very similar. They have the same allowed transitions as so on except for the absolute energy differences, which can be different because of the differences in the energy of the various orbitals, SP, and so forth. Why are they different? Well, they have different numbers of electrons and they're in different shells. So of course, they're going to be different. And it turns out for these systems, there's a slightly jiggered formula for the energy that works for these single electron systems when there's a single electron outside a closed shell. Sort of like hydrogen, you could think of a hydrogen as a single electron outside an empty shell. And in this case, we can write E, the energy of the state with respect to zero energy, just like with hydrogen, which depends on N and L now, not just N. As being, minus R, the Rydberg constant, divided by quantity N minus delta L squared. And delta L has to do with the other electrons being around changing the value, the apparent value of the energy. Delta L depends only on L, not on N. And that's important because since it doesn't depend on N, if we get a couple of N values, we can solve for delta L and then we can figure out, for example, we can estimate the ionization energy of an atom that's a non-hydrogen atom if it's an alkali metal by this means. Unfortunately, for other atoms, carbon or something else, no simple theory like this is even going to be close to the truth and so there are too many other problems. But at least for a single electron outside a closed shell, there's this slightly different formula. And this so-called quantum defect, as it has been called historically, has a value that is bigger for S orbitals. They have a bigger defect than P than D. And as you go out, it gets to be more and more ideal because as you increase the angular momentum, you're just out there. And so basically what you see is much more like a hydrogen atom because why all the other electrons cancel out with all the charge except the extra charge that makes the atom neutral with this electron out here way out there. But for S, even if you're in a big S shell, you can penetrate in. And so then you see that the energy is not nearly so ideal. So I've said that on this slide, the S electrons can penetrate. And here what I've drawn here is an attempt to try to rationalize why this is so, why we have to have this term. The potential when you get inside looks like a multiple charge because now you have other electrons outside you and so you actually see this gigantic nucleus and so you start diving down in energy. And that's this curve I've marked multiple charge. That's with some value that's not equal to 1. And then as you go way out and all the electrons are inside, then what you would expect is that it would look like a single charge. And on the interior part, I've drawn the same potential but with a single charge now, not with a multiple charge. And then the true potential has to look like the single potential far out but it has to look like the multiple charged one as you bury in and you're inside all the other electrons. So it has to sort of interpolate between the two curves that we've got and in fact that's exactly what that formula basically does for us. Let's look at the emission spectrum now of the neutral cesium atom as a practice problem. And we're going to look at it in some detail. In fact, we're going to look at it probably in enough detail that we won't be able to finish the whole thing in this lecture and that will be good because we'll have time to digest what we did and then come back and have another look at it. Practice problem 22 then is this. The strongest lines in the emission spectrum of the neutral cesium atom are at 11,178 and 11,732 wave numbers. And these lines are also seen in absorption. Wave numbers for related lines in the emission spectrum are 7357, 6803, then there's a semicolon, that's important. 3321, 2865, 2767, another semicolon, 11,411, 10,900, 10,857. One, construct an energy level diagram for the cesium atom and assign as many quantum numbers as you can. Two, could the ionization potential of cesium be estimated from these data? The first time you get a problem like this, especially if it's on an exam, is one horrible moment because you start to wonder if you ever had any experience with doing a calculation like this on a multi-electron system at all. And you might also wonder if I've got more than one electron, how do I know that it's only the outer electron that's monkeying about? What if I start to wonder suddenly, maybe one of the inner ones is doing something too and how can I figure out what's going on? The answer is it's always the outer one that's going and you never can have too excited because if you had too excited, the outer one would have ionized off before that and so you'd be doing spectroscopy on the cesium ion then. So you don't have to worry about that. It's just the outer 6S in this case, electron, that's doing the various transitions here. But we've got all these various lines and we've got to figure out what's going on. So we know that the ground state is 6S1, that that electron, the cesium's on the sixth row, it's in the first column. It's basically what the periodic table is telling us about the electron configuration. And we have those two strong lines. The strongest lines are called the resonance lines. The strongest lines in sodium are those two yellow lines. That's why the sodium lamp is yellow because those are doing all the emission. There are other transitions going on but they're less intense. The fact that those lines are seen in absorption, in absorption you have to start in the ground state. You don't start with the electron up here because if you have the electron up here and it's an absorption, the temperature would have to be, extremely high. And the temperature is not extremely high. The temperature is just whatever it is. And so we know that those two lines are it terminate in the ground state of the atom. And that's an important clue to help us assign them. Why are there two of them? Well, we go back to our term symbols. The term symbols for cesium are the same as sodium. So although this seems hard, it's actually not that hard because we know the ground state is doublet S1 half. There's a single electron outside a closed shell. And we know that the two excited states that are the first two there are doublet P3 halves and doublet P1 half. And we know that they're split. And they're split by the spin orbit coupling. And we know that if we've got a very big positive charge on the cesium atom that that spin orbit coupling should be big. And if we happen to remember, oftentimes you do if you're under pressure, that the sodium splitting was 17. And we expect some splitting that's going to be quite a bit bigger than 17 when we have this enormous cesium charge moving around. And that, my goodness, that could make an enormous magnetic field at the side of the electron. And therefore, these resonance lines give us the following picture which I've shown here on slide 410. We have the two doublet P3 halves, doublet P1 half states. And they are 6P. The other mistake you can make is you might mark them as 7P. They are not. They're 6P. The 6P and 6S and cesium have a huge energy difference because the 6S can penetrate the other levels. And the 6P has a node at the nucleus. So the 6P electron hardly ever sees the cesium nucleus because it has a zero there. And when you look at the probability, you have to square it. So it's far less likely. The S has a finite probability of being right at the nucleus. And in fact, in some cases, S electrons actually get gobbled up by the nucleus. And that turns out to be a mechanism for radioactive decay called electron capture which is just like what it sounds. So we can draw these two transitions. One of them is 11178 for the lower one. The other one is 11732 wave numbers. And the difference between them is 554 wave numbers which is a very, very huge value. But maybe not surprising because cesium has a lot more charge than the sodium atom. Okay, I'm going to leave it there. And we're going to digest this problem and come back to it in the next lecture. Meanwhile, we have to think why do we have groups of 2 and groups of 3? I think you can understand where the groups of 2 are going to be coming from. And then how can we figure out what these random collection of lines that we get in wave numbers actually means in terms of the energy level? So we have to have some way to kind of sort them. And what we're going to use to sort them, as you'll see in the next lecture, is we're going to use this value of 554. 554 is going to be our flashlight in the dark. We're going to keep looking for things that differ by 554. If they do, that means two things. That means that these two states, these two P states are involved in the thing. And that whatever level came down to these P states was allowed to make a transition to both of them. So we have a common level and it's going to these P states. So we'll leave it there and when we come back next time, what we'll try to do is fill in all the other energy levels for this atom and make all the assignments as we were asked to do and then we'll also try to see if we can estimate the ionization potential. Of course, if you get a problem like this on an exam, which I actually did, you don't say, well, no, you can't estimate the ionization potential. You assume you can and then you figure out how to do it. So we'll come back to that next time.
 Hi, today in chemistry 131A we're going to talk  about hydrogen wave functions, quantum numbers,  term symbols, and transitions in atomic spectroscopy.  Remember that the solution for the hydrogen atom gave us three  quantum numbers, N that had to do with the energy, L that had  to do with the square of the total angular momentum, M sub L  which was the magnetic quantum number which was the projection  of the angular momentum on the Z axis.  And then there was another number, M sub S,  that came from the intrinsic magnetic moment  of the electron itself.  And so if we have these four quantum numbers then we know  everything that we can know about an electron in a hydrogen atom.  However, once we have more than one electron then there's the  well-known Pauli exclusion principle which you've probably  learned from freshman chemistry.  And that is no two electrons can have all four quantum numbers  the same.  So if N, L, and M sub L are the same, for example,  the electrons in a 2PX orbital, then M sub S has to be different  for the two electrons.  One has to be so-called up and the other has to be down.  In other words, they have to be paired.  We'll see a little bit later what that actually means.  And so as I said, electrons in the same spatial orbital have  to be paired up magnetically with respect to the spin.  Let's look at the possibilities for a neutral carbon atom  with six electrons.  The configuration from the periodic table for a carbon atom is 1S2,  2S2, 2P2 and that means that there's two electrons in the P shell  that can occupy any of the three P orbitals.  The other orbitals are filled.  And the L equals one state for P, M sub L can be 1, 0 and minus 1.  And we can figure out then the limits on L and S,  which depend on each other because of the exclusion principle.  Let's have a look at this.  The first thing to note is that only the P electrons have any  flexibility.  The 1S shell is filled.  That's out of commission.  The 2S shell is filled.  That's out of commission.  And it's only the frontier orbitals that we're putting electrons  into that have any flexibility as to how the electrons can orient.  Whatever's already happened to 1S and 2S has happened  and that's water under the bridge.  For example then, here on slide 389, we have all these possible  configurations of the electrons in these 3P orbitals are allowed.  On the left, we've got two electrons in the M sub L equals plus 1 orbital  and they have to be paired because of the exclusion principle.  Or we could have one in the 1 orbital and one in the 0  and they could be also paired, one up, one down.  And we could have one in 1 and one in the minus 1.  And we could have them parallel as well, one in the 1 and one  in the 0.  All those possible configurations of the electrons  in the P orbital shells are okay.  They're all allowed.  But on the bottom of the slide here, none of these are allowed.  In the first one, we have M sub L equals plus 1.  There are two electrons and they're both up.  That means they have all four quantum numbers the same  because N is 2, L is 1, M sub L is 1 and M sub S is plus 1 half  for both of them, not allowed.  Same thing with the others.  Whenever we have two electrons in the same orbital,  they can't have the same spin.  Remember that we had a shorthand way to keep track  of these various atomic levels which was the term symbol  which gave us sort of a summary of how L and S would magnetically  couple to give the total angular momentum J and would allow us  to keep track of which transitions were allowed.  Remember that the term symbol is 2S plus 1  on the left hand superscript, capital L which was SPD  and so forth for the central letter.  And then a value of J which is the way S and L couple  and they can couple in various ways.  They're various allowed values according  to the Klebsch-Gordon series that we derived.  2S plus 1 is called the multiplicity but keep in mind  that it is only really the number of states there if,  in fact, L is greater than or equal to S. Because if L is less  than or equal to S then there could be less states allowed.  So if S is zero it's called a singlet.  If S is a half it's a doublet.  If S is one and a triplet.  However, if we have 2S one half where L is zero then it's not  really a doublet.  There's only one state because there's only one level there  for the electron to occupy.  So it's not really a doublet in that case.  Let's try another example and have a look at these then  and see which values are allowed.  So this is practice problem 20.  What terms can arise from a neutral carbon atom  with the electron configuration 1S2, 2S2, 2P2?  Well, remember the way we approach these problems we have  to figure out the possible values of L based on the values  of little L for each electron.  We couple those together.  Then we figure out the possible values of big S based  on spin one half for the electron.  We couple those together.  And then we use the Klebsch-Gordan series to get  about possible values of J which naively go from L plus S  down to the absolute value of L minus S. So if we can figure  out these possible values of big L and big S we should be able  to come up with the term symbols.  But what we'll see in this case is  that it's a little bit more complicated  than it might first appear.  And the reason why is that some of the values  that we'd like to pick poly excludes from possibility.  The possible values of big L with 2P electrons are 2, 1,  and 0. Those are the three you can get from two values  of little L equals plus 1.  And the possible values of S, big S,  the total spin angular momentum are 1 and 0.  That's what you can get from two spin one half electrons.  Therefore, if we just took the Klebsch-Gordan series it would  seem that J could be well L plus S is 3 down to the absolute  value of L minus S is 0.  So it would seem that J could be 3, 2, 1, or 0.  But we can't have J equals 3 when we have both electrons  in the same shell.  And the reason why is that if we had J equals 3 we'd have  to have L equals 2.  So let's put the two electrons in the plus 1 orbital.  And then to get 3 we have to have S equals 1.  So we have to have them parallel.  That means that we have to have them both  in the same orbital there.  And then we have to have them with the same spin.  But that's not allowed because if they're  in the same orbital they have to have opposite spins.  So we are allowed to take the maximum value in that case.  If we had an excited carbon atom, and that's why some books give  you these weird problems where it's an excited atom,  where one of the electrons is in a 2P and one's in a 3P,  then you just total them all up because there are different shells.  And so the poly exclusion principle doesn't apply.  They aren't in the same orbital.  In this case, which is actually more important to work out  because usually you want to work out the terms for the ground state  of the atom to figure out what kind  of spectroscopy you can see.  You have to be quite a bit more careful how you do it.  Therefore, if we have L equals 2, we have to have them paired.  So we have S equals 0.  The electrons have to have opposite spin.  And the term that arises from that is singlet D. So if it's  singlet D, then that means that J has to be 2 because S is 0.  2S plus 1 is 1.  S is 0. D, so the series stops with J equals 2.  And so the term that arises is singlet D2.  For L equals 1, we're okay because then we can have one electron here  and the other over here somewhere.  And in that case, we get a triplet P term.  And there are three values because now S is 1 and L is 1.  So we get triplet P2, triplet P1, and triplet P0.  And then the last term that we can have is L equals 0, S equals 0.  And therefore, the only possible value of J is 0.  And this would be called singlet S0.  What you have to do then is you have to go back and say,  what were the possibilities of the states I started with?  Recall that we did that when we did this before with just coupling  two electrons.  We made sure that we accounted for all the possibilities  that were there and then make sure that our terms cover all those  and no extra ones.  And like anything, like doing crosswords or Sudoku or anything,  this takes some practice.  But it can be quite a rewarding puzzle to work out to make sure  that you've got it right, that you aren't including any non-allowed  states in the terms.  If you've got an open F shell with multiple electrons  or an open D shell with multiple electrons,  working out the allowed terms can really require a quiet room  because it can get quite complicated to keep track  of which ones are allowed and which ones are excluded  because you can't have two electrons in the same orbital  with the same spin.  Suppose on the other hand we had five P electrons, say, flurring.  If we took all five electrons and then we said, boy,  this is a mess because all these orbitals are filled  and so there's these millions of combinations that are no good  because we can't have the electrons that way.  It would be very, very difficult to do  and it would take a long time.  And so the trick we can do is we can treat the missing electron,  the hole as if it were an electron and we can move it around  and we can just treat it the same way we would  if we had one electron.  And that's because once it's more than half full,  it's really the holes that are dictating which terms can arise.  The electrons are filling all the orbitals.  And it turns out that it's exactly the same  and so you don't have to do it over.  Therefore, the worst case is when you've got three P electrons  or five D electrons.  Once you've got more than that, it just reduces  to a previous case where you have less electrons.  So it's no harder.  And so for flurring, excuse me, for oxygen,  which has two P4 electrons, you get the same terms as carbon  that had two P2 that we just did.  And likewise, flurring gives rise to doublet P3 halves  and doublet P1 half terms.  Now I want to talk about the shapes  of the hydrogen wave functions.  We can plot solutions.  We had the equations, but let's just look at the shapes  of the wave functions themselves.  Not the radial distribution function,  but the angular part as well.  And it's common practice to plot these then as a contour plot  in which we pick some arbitrary contour in space  and we say there's a certain likelihood  that the electron is inside this contour.  And then we can pick a color to code whether the phase  of the wave function is plus or minus.  Remember it's a wave, so it has a phase and it has a phase  in 3D just like a sine wave has a phase in 1D.  Sometimes before people used color, they'd mark a plus  on one side of the wave function and a minus on the other.  The problem with using plus and minus is that if you look  at it and you aren't quite sure what it refers to,  you might think it has something to do with charge,  but it doesn't have anything to do with charge  because electrons are always negative.  And it's just whether the phase  of the wave function is a positive or negative.  And in between there's a node  where the wave function has to be zero.  The thing is we tend to avoid using the true eigenstates  for the hydrogen atom and the reason why,  which I referred to in a previous lecture,  is that if we include I and we've got e to the im phi going  around then that's difficult to figure  out what you're going to plot because it's much easier  to say well I'm going to plot if this number is equal  to this thing I'm going to plot that surface  where this number is equal to 0.9 or something like that.  But if as you go around the phase is changing with respect  to I then you'd have to figure out some way to code that.  And you might be able to do that by coding in a rainbow,  but on the other hand maybe that might be confusing  when you include the phase.  So as a dodge what tends to be done is instead  of plotting the true eigenstates we plot linear  combinations of plus and minus m sub l  that either give the cosine part or the sine part.  And recall from Euler's identity that e  to the i theta is cos theta plus i sine theta.  So we can pick linear combinations to plot those.  And that's usually what's done.  We pick then the real part, we throw away the i  and we pick the real part and then we just plot the things.  However, keep in mind that the plots depend a lot  on the particular value of the contour that you pick.  Do I pick that the electron has  to have 99% probability of being inside this thing  or 50% probability of being inside this thing or whatever?  Usually 90% is a typical value, but as we saw  for the hydrogen atom, if we go out just  to the most likely radius of where the electron is,  the Bohr radius, there's only a 32% probability  that the electron is inside there.  So perhaps you could argue  that maybe you should pick a more conservative percentage  to get a more realistic idea of the size  of the spatial extent of the wave functions.  So here are some wave functions then plotted.  And red is positive phase in these plots  and blue is negative phase.  And the 1s is the little dot on the left.  That's a little red thing.  And it just is spherical as we would expect.  And the 2s is a bigger sphere.  There's a tiny red dot in the center.  And then there's a bigger blue area.  And that's because as we saw, the radial function  for 2s has a node at some point  where there is zero probability of finding the electron.  And then we get these big fluffy orbitals, the 2pz,  2px, and 2py that look really quite big and distended.  And that has to do of course with the particular contours  that were chosen to plot these values.  Recall that a node is a place where the wave function changes  sign. Strictly speaking, it shouldn't just go to zero,  but it should be plus on one side and minus on the other side.  And a node in a 3D wave is a surface  where the 3D wave has a value of zero.  The radial nodes are in fact spheres.  We saw that for estates there are nothing but radial nodes.  And then the angular nodes are angles which turn out to be planes  where the wave function is zero.  A 1s function has zero nodes, a 2s has one radial node,  and a 3s has two radial nodes.  And in general, the number of nodes is n minus 1.  If we look then at the 2pz orbital which I have here  on slide 398, we see that there's an angular node,  there's a plane in which the wave function vanishes.  It's positive on one side, red on one side,  blue on the other side, everywhere in between the wave  function vanishes on the plane z is equal to zero.  The d functions look like these.  Here's 3Dxy looking like a big clover leaf.  And you can clearly see that there's a disturbance building  up in the angular variation of the electron  as it assumes more quanta of angular momentum.  That makes perfect sense.  And so there are five 3D orbitals and we saw what they were.  We wrote them all out but we didn't see exactly what they  looked like so here's 3Dxy here with these four nodes.  And unlike p, the blue are across from each other  and the red are across from each other.  p changes sign when you rotate it 180 degrees.  This d changes sign if you rotate it 90 degrees.  And that's because it's 2 quanta of angular momentum and not 1.  Here's a different view of the d wave functions  and how they may add up to give us a shell  that has spherical symmetry.  Four of them are pretty much like clover leafs.  They're four things, red and blue.  Here they're just gray.  And here they're also plotted at a different level  of the contour so they look a lot thinner  than in the other plot.  And then there's this funny one, 3Dz squared,  which almost looks like a p orbital.  And unlike the others where I can clearly see these planes  where this thing is vanishing, this one,  especially without color, is a little bit hard to see.  But what we have with the 3Dz squared is we have a donut  around the equator and then we have this thing  that looks a little bit like a p orbital.  But the difference is this thing is all the same color.  It doesn't change sign.  And this donut's all the same color.  And you still look at it and you say, well,  where are the angles where the nodes are?  And the answer is they're at the magic angle.  If you look back at the formula for this particular orbital,  you'll find that there's an angle tilted  where the wave function vanishes.  And it vanishes all along that angle all the way out.  And so that's how that one works.  So it looks a little bit different, but in fact,  it's exactly the same in terms of vanishing at values  of an angle, just like the other ones do.  Now let's talk about spin orbit coupling.  We saw that we had these two lines for the sodium spectrum.  And they were closely separated.  And the two transitions were doublet p 3 halves goes  to doublet s 1 half and doublet p 1 half goes to doublet s 1 half.  Recall that delta J can be zero as long  as J is not equal to zero.  The energy levels then break down like this,  as I've shown on this figure on slide 401.  There are two levels, two excited states above the ground state,  these p states, and they're split by an amount I've called delta  in this slide called the spin orbit coupling.  And what we'd like to do is explore this a little bit  and figure out exactly where this comes from.  The only difference between the two states is the orientation  of the spin relative to the orbital angular momentum  because they have the same value of s and l, but J is different.  So s and l can add up like a triangle and make J the third  leg of the triangle.  Then J can be different because there can be several different  values that work.  And it must be then the energy difference is just whether this  magnetic intrinsic bar magnet of the electron is aligned  or against the magnetic field that it sees if from the perspective  of the electron the nucleus is going around the other way  and it creates a magnetic field that acts on the electron.  If we know the measured optical frequency difference  between the two levels, then we can estimate the magnetic field  that the 3p electron and sodium appears to be experiencing  from going around the nucleus of the sodium.  The magnetic energy difference of an electron spin is delta E is  equal to G, the so-called electron G factor which has a value  about equal to 2 and then a conversion, the Bohr magneton  that converts magnetic field to energy and then B, the value  of the magnetic field that the spin is experiencing.  That's the magnetic energy that this bar magnet  that the electron has is going to feel.  So let's do a practice problem and figure  out what kind of field it experiences after we figure  out exactly what the spin orbit coupling is.  What's the spin orbit coupling in wave numbers, in electron volts  and what is the apparent magnetic field that the electron  in a sodium atom in the 2p orbital experiences?  Okay, well the wavelengths of the two transitions are 589 nanometers  and 589.6 nanometers and we know the energy difference is delta E  is h nu 1 minus h nu 2 and that's hc times 1 over lambda 1  minus 1 over lambda 2 and we can convert that then  to wave numbers as just hc times nu bar 1 minus nu bar 2.  Those are the wave numbers in centimeters,  inverse centimeters and that's, we can just write that  as delta nu bar.  So we can solve for delta nu bar and we just have  to take the inverse of the wavelength which is in nanometers  to the minus 1 times 10 to the 7 nanometers per centimeter  and to be careful if you're doing an exam you may want  to convert from nanometers to meters and meters to centimeters.  It seems easy but it's easy to get also a factor of 10,000 off  if you go too quickly and happen to go the wrong way with one  of the conversions.  If you do that then the spin orbit splitting  in sodium is 17.28 wave numbers.  If we convert that to EV which we do by converting to joules  and then we just simply convert from joules to EV by multiplying  by dividing rather by 1.6 times 10 to the minus 19 joules per EV,  we get a very small value of 2.142 times 10  to the minus 3 electron volts.  Keep in mind that the ionization energy of hydrogen  to kick the electron completely out was about 13.6 electron volts  and here we're talking 2 milli electron volts.  So these magnetic interactions in these systems are very,  very much smaller energy differences  than the main electrostatic potential  that the electron feels from the charge.  And that's a common theme.  Basically magnetic energies are always quite a bit smaller  than electric energies.  And in the hydrogen atom the spin orbit splitting is very,  very tiny compared to sodium.  And so it took a lot of detailed work to even work  out that there was something there and then to explain it.  And it's very interesting historically.  It's too bad we don't have time to go  through all the thought processes that went  through finding the fine structure and hyper fine structure  of these systems.  But it is very interesting detective work  from a science standpoint.  Finally then, okay, we know this spin orbit splitting.  Let's then use our formula for this, the energy difference  of an electron in the magnetic field.  We know the energy difference.  We know the g value of the electron.  We know the Bohr magneton.  That's a constant.  We look that up.  Let's figure out then what the apparent magnetic field is  that this electron is seeing.  And what we get if we take delta e divided by g times the  Bohr magneton is we get 18.5 Tesla.  18.5 Tesla is an absolutely huge magnetic field.  If we tried to make a magnetic field like that  in the laboratory, we would be very hard pressed.  Even if we took tons and tons of wire  and wound a gigantic electromagnet,  we would not get such a big magnetic field.  In fact, the very biggest NMR spectrometer at UCI has a  magnetic field of 18.6 Tesla.  But the magnet is 12 feet tall and it has several miles  of superconducting wire wound into a gigantic solenoid  to make this magnetic field, which is enormous.  So you can see that because the particles are so close  and things are moving so rapidly,  that there are very large magnetic interactions on the electron,  very large compared to what we can do in the laboratory.  And so when we take atoms and we put them  in even a pretty strong magnetic field,  the energy levels do split and move and do things.  But they split only a little bit compared  to how big they're already split due  to these internal magnetic fields from the nucleus.  And that's good because that means that if we have two levels  like this and then we turn on a magnetic field  and they split, they go like that.  And that makes it easy to keep track of what's going on.  If they went like this and all over the place,  it might be really hard to figure  out what the spectrum was doing.  If we look at other alkali metals like lithium,  potassium, rubidium, and cesium, then because they have a single  electron outside a closed shell, they all give rise  to exactly the same terms as sodium.  And so their spectra are very similar.  They have the same allowed transitions as so on except  for the absolute energy differences, which can be different  because of the differences in the energy  of the various orbitals, SP, and so forth.  Why are they different?  Well, they have different numbers of electrons  and they're in different shells.  So of course, they're going to be different.  And it turns out for these systems,  there's a slightly jiggered formula for the energy that works  for these single electron systems  when there's a single electron outside a closed shell.  Sort of like hydrogen, you could think of a hydrogen  as a single electron outside an empty shell.  And in this case, we can write E, the energy of the state  with respect to zero energy, just like with hydrogen,  which depends on N and L now, not just N. As being,  minus R, the Rydberg constant, divided by quantity N minus delta  L squared.  And delta L has to do with the other electrons being  around changing the value, the apparent value of the energy.  Delta L depends only on L, not on N. And that's important  because since it doesn't depend on N,  if we get a couple of N values, we can solve for delta L  and then we can figure out, for example,  we can estimate the ionization energy of an atom  that's a non-hydrogen atom if it's an alkali metal  by this means.  Unfortunately, for other atoms, carbon or something else,  no simple theory like this is even going to be close  to the truth and so there are too many other problems.  But at least for a single electron outside a closed shell,  there's this slightly different formula.  And this so-called quantum defect,  as it has been called historically, has a value  that is bigger for S orbitals.  They have a bigger defect than P than D.  And as you go out, it gets to be more and more ideal  because as you increase the angular momentum,  you're just out there.  And so basically what you see is much more  like a hydrogen atom because why all the other electrons cancel  out with all the charge except the extra charge  that makes the atom neutral with this electron  out here way out there.  But for S, even if you're in a big S shell,  you can penetrate in.  And so then you see that the energy is not nearly so ideal.  So I've said that on this slide, the S electrons can penetrate.  And here what I've drawn here is an attempt to try  to rationalize why this is so, why we have to have this term.  The potential when you get inside looks like a multiple charge  because now you have other electrons outside you  and so you actually see this gigantic nucleus  and so you start diving down in energy.  And that's this curve I've marked multiple charge.  That's with some value that's not equal to 1.  And then as you go way out and all the electrons are inside,  then what you would expect is that it would look  like a single charge.  And on the interior part, I've drawn the same potential  but with a single charge now, not with a multiple charge.  And then the true potential has to look  like the single potential far out but it has to look  like the multiple charged one as you bury in  and you're inside all the other electrons.  So it has to sort of interpolate between the two curves  that we've got and in fact that's exactly what  that formula basically does for us.  Let's look at the emission spectrum now  of the neutral cesium atom as a practice problem.  And we're going to look at it in some detail.  In fact, we're going to look at it probably in enough detail  that we won't be able to finish the whole thing  in this lecture and that will be good because we'll have time  to digest what we did and then come back  and have another look at it.  Practice problem 22 then is this.  The strongest lines in the emission spectrum  of the neutral cesium atom are at 11,178  and 11,732 wave numbers.  And these lines are also seen in absorption.  Wave numbers for related lines in the emission spectrum are 7357,  6803, then there's a semicolon, that's important.  3321, 2865, 2767, another semicolon, 11,411, 10,900, 10,857.  One, construct an energy level diagram for the cesium atom  and assign as many quantum numbers as you can.  Two, could the ionization potential  of cesium be estimated from these data?  The first time you get a problem like this,  especially if it's on an exam, is one horrible moment  because you start to wonder if you ever had any experience  with doing a calculation like this  on a multi-electron system at all.  And you might also wonder if I've got more than one electron,  how do I know that it's only the outer electron  that's monkeying about?  What if I start to wonder suddenly, maybe one  of the inner ones is doing something too  and how can I figure out what's going on?  The answer is it's always the outer one that's going  and you never can have too excited  because if you had too excited, the outer one would have ionized  off before that and so you'd be doing spectroscopy  on the cesium ion then.  So you don't have to worry about that.  It's just the outer 6S in this case, electron,  that's doing the various transitions here.  But we've got all these various lines and we've got  to figure out what's going on.  So we know that the ground state is 6S1,  that that electron, the cesium's on the sixth row,  it's in the first column.  It's basically what the periodic table is telling us  about the electron configuration.  And we have those two strong lines.  The strongest lines are called the resonance lines.  The strongest lines in sodium are those two yellow lines.  That's why the sodium lamp is yellow  because those are doing all the emission.  There are other transitions going on but they're less intense.  The fact that those lines are seen in absorption,  in absorption you have to start in the ground state.  You don't start with the electron up here  because if you have the electron up here  and it's an absorption, the temperature would have to be,  extremely high.  And the temperature is not extremely high.  The temperature is just whatever it is.  And so we know that those two lines are it terminate  in the ground state of the atom.  And that's an important clue to help us assign them.  Why are there two of them?  Well, we go back to our term symbols.  The term symbols for cesium are the same as sodium.  So although this seems hard, it's actually not that hard  because we know the ground state is doublet S1 half.  There's a single electron outside a closed shell.  And we know that the two excited states  that are the first two there are doublet P3 halves  and doublet P1 half.  And we know that they're split.  And they're split by the spin orbit coupling.  And we know that if we've got a very big positive charge  on the cesium atom that that spin orbit coupling should be big.  And if we happen to remember, oftentimes you do  if you're under pressure, that the sodium splitting was 17.  And we expect some splitting that's going  to be quite a bit bigger than 17  when we have this enormous cesium charge moving around.  And that, my goodness, that could make an enormous magnetic field  at the side of the electron.  And therefore, these resonance lines give us the following picture  which I've shown here on slide 410.  We have the two doublet P3 halves, doublet P1 half states.  And they are 6P.  The other mistake you can make is you might mark them as 7P.  They are not.  They're 6P. The 6P and 6S and cesium have a huge energy difference  because the 6S can penetrate the other levels.  And the 6P has a node at the nucleus.  So the 6P electron hardly ever sees the cesium nucleus  because it has a zero there.  And when you look at the probability, you have to square it.  So it's far less likely.  The S has a finite probability of being right at the nucleus.  And in fact, in some cases,  S electrons actually get gobbled up by the nucleus.  And that turns out to be a mechanism  for radioactive decay called electron capture  which is just like what it sounds.  So we can draw these two transitions.  One of them is 11178 for the lower one.  The other one is 11732 wave numbers.  And the difference between them is 554 wave numbers  which is a very, very huge value.  But maybe not surprising because cesium has a lot more charge  than the sodium atom.  Okay, I'm going to leave it there.  And we're going to digest this problem and come back to it  in the next lecture.  Meanwhile, we have to think why do we have groups of 2  and groups of 3?  I think you can understand where the groups  of 2 are going to be coming from.  And then how can we figure out what these random collection  of lines that we get in wave numbers actually means  in terms of the energy level?  So we have to have some way to kind of sort them.  And what we're going to use to sort them, as you'll see  in the next lecture, is we're going to use this value of 554.  554 is going to be our flashlight in the dark.  We're going to keep looking for things that differ by 554.  If they do, that means two things.  That means that these two states, these two P states are  involved in the thing.  And that whatever level came down to these P states was allowed  to make a transition to both of them.  So we have a common level and it's going to these P states.  So we'll leave it there and when we come back next time,  what we'll try to do is fill in all the other energy levels  for this atom and make all the assignments as we were asked  to do and then we'll also try to see  if we can estimate the ionization potential.  Of course, if you get a problem like this on an exam,  which I actually did, you don't say, well,  no, you can't estimate the ionization potential.  You assume you can and then you figure out how to do it.  So we'll come back to that next time.
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