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# Lecture 16. Energy Level Diagrams, Spin-Orbit Coupling and the Pauli Principle

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all of the stuff that comes out but we don't keep track is just developing other things on a CCD camera if we assume that the 7 as than we have now this picture on slide 416 with the success 6 p splits into 2 levels the 7 as a single level again WAS ONE have we have those 2 lines 7 357 to the lower state and 68 all 3 to the upper stage now what we have by adding these numbers are as we have the energy difference between 6 hours and 7 minutes and that's how we're going estimate the ionization energy and a little better the other sets are grouped into 3 and what I've done then is I've just taken the 3 and I've taken all possible differences between 1 minus 2 1 minus 3 to minus-3 arranged the most positive numbers 33 21 minus 28 65 as 456 numbers I haven't seen it 33 21 minus 27 67 is 5 54 that I have seen so that's the hand that whatever is involved in these 3 is going to the same to there the central guys in this scheme that my linchpin for figuring out how this stuff works and then the other 1 28 65 minus 27 67 is 98 so that's another does as far as I'm concerned right now I don't know exactly what the
nuclear and then we can figure out the ionization potential from the six-state because then we can just put on any equals infinity for the state take the difference masters are over 6 minus the quantum defect quality square to take 6 minus 4 . 1 5 square and divide 1 0 9 7 3 7 by that and we get 32 thousand 63 where numbers that's our estimate for the ionization potential of cesium based on just 7 2 levels there and that's a little bit dicey because usually you like to have many more just to see what the trend now let's also try to get an estimate from the Mideast states for 1 thing that's going to confirm that the quantum numbers that we've assigned the corrected for another thing it's probably going to give us a better value because there's less distortion in the state's they more ideal but in order to do so we can't we got these 2 sets here at this month's but by 90 something the missus 40 we we needed to levels which are a sort of the unperturbed state energies before we have to stay in orbit couplings Internet to deal with that with the yesterday and the way you do that is you have to wait the 2 states by their multiplicity and you have to take the weighted mean the 3 halves has more sub-state them US excuse me the 5 have more substance than the 3 have and we get a ratio of 6 to 4 for the state's or 302 and therefore I shown on these 2 equations the difference between the 5 the state and the success today because I only have the difference I don't have the absolute is 2 times the difference to the lower level which is 1 1 1 7 8 plus 33 21 plus 3 times the difference to the upper level which is the same thing plus the 98 divided by 5 Parker taken 604 divided by 10 actually stay there for the energy difference that be the correct 0 of energy for the 550 versus the success is 14 thousand 557 . 8 numbers and if we do the same thing with the 60 versus the success we get a difference of 22 thousand 614 . 8 wave numbers and now what we can do is we can take the difference between those 2 numbers and then we have difference between 60 and 5 because whatever the differences that as drops out of the equation and using that difference of 80 57 we can then solved for the quantum defect in the deceased state by solving miners are times won over 6 -minus Delta sub D quantities square minus 1 over 5 and again we have a constraint that this solution be real that the positive and that is the lesson 5 and allowed value than of the quantum defect in that the state is 2 . 4 3 which is less than asthma recall said the quantum defect and gets smaller as you go out becomes more ideal using that we can get the ionization potential we 1st take are over 5 minus the quantum defects square but then what we have to do is we have to have the energy to to get out there from the success
so we now get an estimate from here from this formula and then we've got we gotta get up there so we take that lost 1 1 1 7 8 plus 33 21 and we get 31 thousand 198 wave numbers which is a little bit different than the best but closed very close considering the way we're doing this with this approximate formula and 2 levels and so really comforting if you get a result like this 1 accounts and you're going to get a great on in fact the literature values is 3 . 8 9 4 electron volts which Converse 231 thousand 324 wave numbers therefore the estimate from the G states it with their smaller bond defect is closer to the accepted value what I would say is that if you happen to have been trained to look at problems like this and then you just give a problem like this cold and you just know something about the hydrogen as a problem like this is pretty much nearly impossible and once you know how to do them and how to look for these common differences then pretty much you can get everything done although if it's not an alkali metal it's going to be much much harder and it just depends whether it would be able to estimate the ionization potential not you may need more data if there's probably had more data while we tried to do His we tried to organize the data rather than just solving 2 equations to and 1 equation 1 and rather than the and inserting ionization energy we try to make where the ionization energy would be interests and we will try to organize our variables so that the plot were linear and then what we do is we checked whether the plot is linear era 1st so that we know we're going to protecting and we will come soon and and then get the ionized section OK let's take a little bit more detailed look it's been orbit coupling this magnetic interaction we never included in the Hambletonian we had the kinetic energy but the electrons we factored out the nuclear motion the potential energy electrostatic energy and always said later there's this magnetic but how could we included if we want to include but we argued that the electron sees a magnetic field from the nucleus going around the other way and that's why there's only a spin orbit splitting and honest but we can do a little bit better than that because we know but the energy of a bar magnet for any died electric dipole on an electric field magnetic dipole in magnetic field is minus new be I or -minus million-dollar the and if insist that the magnetic moment is proportional to ask the spin and if the magnetic field is proportional to the the orbital angular momentum that we get energy term which has some number which I'm going to call data timescale .period because that should be the dark products of the 2 and energies are often .period products and forces are often cross products and that's the way things play out then we can cast the adopt product in terms of things that we actually know we want to get it in terms of quality numbers we can do a trick that we can take J . J that's a scalar product is the the total angular momentum of the atom with itself adjacent and substituting for JAL process in each place we get elsewhere which is held out but the word which is estimate close to know the answer and then we just saw that for all the which was going to be our energy terms and we get 1 J. squared minus elsewhere might successful and now we know the quantum numbers we just put in numbers for those things and more we don't have to worry about what the operators are we just put a number let's apply that to the double P 3 halves and double P 1 half states we know all the 1 numbers we know as we know well we know J. so we're going to take the expectation value of the spin orbit energy in the States well what we find that we put in the quantum numbers we get paid over 2 there are some things that has to do with how strong interaction is we'll get to that in a 2nd and we get further chased where J. little J times little J. plus 1 or at some other works -minus little Elton's little outpost 1 mines last was 1 we do that double P 3 have we simply get 15 over for minus to minus-3 over 4 data over to an end so we get paid over 2 times a are square and if we do it for a 381 have but 1st Excuse me Dublin P 1 half what we get is -minus beta age spots where you can see 1 of them is moved up by a certain amount and the other is moved down and that's why when we did the problem we took the weighted mean because they're moved by different amounts depending on the multiplicity so that the center of gravity remains the same the shares whatever it is this energy term data should depend on CD atomic number because after all we saw was cesium that was much much bigger than sodium and we would expect a bigger nucleus to generate a bigger magnetic field but it it's going to take us far too far afield to actually calculate from 1st principles with Thomas Fermi procession and all this other stuff that comes in and so I'm just going to quote the answer that age R square data the energy terms in front of the all-time best go sightseeing at the 4th and then a magnetic conversion factor and then the chief factor of the electron bore magnet on and then this term comes from doing some angular momentum Algebra 1 over and times a knock due to unstable times almost 1 and selfless 1 one-half that's quite a long and involved formula but what it shows is that as Annandale increased the splitting decreases we saw that again in the cesium we saw 1 of was nice the was 47 out so cents OK let's talk about 2 electron the we did hydrogen with alkali metals which is kind of a Dodger because they're just basically 1 electron and clouded charges
7 dimensional problem what I do is I put all the electrons into shells and orbitals like they would be in the hydrogen atoms
but with an appropriate value received and then I take 1 electron and I take all the others and ice near the mouth it's charge I just forget the fact that the real wave function there and I just stared amounted to some charge distributions and then I take the 1 electron I've got and get this weird charge distribution from all the other ones and the nucleus and I take this electron and I try to optimize its way function until it lowers the energy so that it's more ideas and then I freeze it and make it into the charge and I pick another electrons are always and minus 1 and I have 1 leftovers go round and round and round and when I can prove it any more than I say I've got a self-consistent solution doesn't mean it's correct in fact it's not correct because the problem and smearing it into a charge is pretty good but it's not the same as taking into account how things are actually moving so-called electron correlation and infect other theories and much better too but once we're done it was stated that go back to 2 electron animal forget the 27 national 1 if we go back and we've got to electrons Indiana we can treat the electron electron repulsion as per patient and then we can try to adjust the energy to see if we get something that's more acceptable than measures better with the spectroscopic lines which of course can be be measured too many many digits which is a great test between theory and experiment if you look back at the slides on perturbation theory remember that we started we we introduce this parameter that I called lambda and when land that was 0 0 we had our but undeterred solutions that we had the exact solution for and then when land is equal to 1 we have what we're trying to get and we've tried to connect pieced together 0 1 1 by taking a power series
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