WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome to www.educator.com, welcome back to Physical Chemistry.
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Today, we are going to start our example problems for term symbols and atomic spectra.
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Let us jump right on in.
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What are the terms symbols for the NP1 configuration?
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The 2P1, 3P1, 4P1, the N part does not matter.
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It is actually just the electrons of the suborbital.
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This is really just true for P1.
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I'm going to solve this with a slightly different procedure than the one that I have used when I introduced terms symbols.
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Back when I introduced term symbols a few lessons back,
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we started by listing out all the microstates explicitly and then picking out those M sub L values and M sub S values.
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Then, I introduced this shorter procedure where we are actually making a table of the ML and MS values.
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Counting the largest ML, the largest MS, crossing them out until you come up with all of your term symbols.
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What I'm going to do is essentially just the reverse of that.
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I'm going to do this for this example problem and also the next several example problems.
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When we do the MP2 configuration and the 3 configuration, the D2 configuration goes through quite a few of these.
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Essentially, what we are doing is we are going to work with the ML and MS values directly
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and then generate the microstates for these term symbols as we go long, as opposed to the other way around.
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At some sense, it is the reverse.
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You can decide which one you like better.
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For this particular problem, I’m going to do both.
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I’m going to do this new way and then go back into the table way, then you will see which one you like better.
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They are essentially the same and at some point, you have to deal with the largest ML
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and largest MS and you are going to have to deal with the microstates.
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But this way with the numbers just tends to be a little bit faster.
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Let us see what we can do.
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Like everything else in science or anything for that matter, you have to do them a number of times in order to really understand.
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Passive understanding just by looking and seeing that you can follow that,
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that is one thing but being able to generate and it do it yourself, that is when you we only get your hands dirty,
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that is when you really understand what is happening.
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Let us see what we have got.
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We always want to begin by finding the number of viable microstates.
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Let me stick with black here.
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Always begin by finding the number of viable microstates and that was, we have this great name for that,
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it is the number of spin orbitals that we have available to occupy choosing the number of electrons.
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In this particular case, we are looking for the NP1 configuration.
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We have 1 electron and we are trying to distribute among the PX PY PZ.
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We have 2, 4, 6, we have 6 spin orbitals and we have 1 electron.
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It is going to be 6 choose 1, there are 6 microstates.
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6 different ways to arrange 1 electron in the P orbital.
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In this particular case, we already know which one they are.
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It is going to be up here, up here, up here, or down here, down here, down here.
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That takes care of it, but let us just go ahead and work it through.
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This is the important thing, always begin with that.
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Start by finding the number of viable microstates and as you find each term symbol,
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knock out the number of microstates and it will tell you how many you have left.
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We are also going to be finding the M sub L, that is equal to the m sub l1 + m sub l2 and so on.
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In this case it is only 1 electron.
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The M sub L, you remember that is the 1, 0, -1, in the case of the P orbital.
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For the D, it was 2, 1, 0, -1, -2, because there are 5 suborbital in the D orbital.
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We are going to be finding that number and we would be finding the M sub S.
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That is equal to the M sub S1 + M sub S2, that is the spin of the electron, +½ or - ½.
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We are going to add them all up for all of the electrons.
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We begin by looking for the largest ML possible.
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Here, because we are talking about 1 electron in the P.
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Here, the largest ML is going to equal 1.
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This particular thing, and the reason it is, is because we can put the 1 electron in the left most suborbital.
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Let us do that.
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Here, the ML value is 1 and ML value is 0, ML value is -1.
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We want the largest ML possible, for 1 electron it is over here.
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Here, ML = 1.
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The largest MS possible, in this particular case it is 1 electron, it is up spin ½ as opposed to downspin - ½.
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Given that ML is equal to 1, in this particular case, the largest MS possible is ½.
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If ML is equal to 1 that implies that L = 1.
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If MS is equal to ½ that implies that S is actually equal to ½, that implies that 2S + 1 is equal to 2 × ½ + 1 is equal to 2.
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When L is equal to 1 and S is equal to 2, that gives me a doublet P configuration, a doublet P term symbol.
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From here, we are going to calculate how many microstates are in this doublet P.
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L = 1 remember, when L = 0 that is S.
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When L = 1 that is P.
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When L = 2 that is D.
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When L = 3 that is F, and so on.
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S is ½, 2S + 1 that is the left most top left number.
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This is a doublet P configuration, we will worry about the J a little bit later.
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We found our basic term symbol.
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2L + 1 × 2S + 1, let us do it this way.
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The number of ML × the number of MS that is equal to 2L + 1 × 2S + 1.
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L is equal to 1, 2 × 1 + 1 is equal to 3.
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2S + 1 is equal 2, 3 × to = 6.
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There are 6 microstates in this level.
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There are 6 microstates that belong to this particular term symbol.
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These come from ML is equal to 1 or L is equal to 1, that means that ML is equal to 1, 0, -1.
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MS is equal to ½, S is equal to ½, that means that MS is equal to ½, and ½ -1 is – ½.
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There are 3, here is 2, 3 × 2 is 6.
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The states that you are going to get, the microstates come from 1, ½, 1 - ½, 0 ½ , 0 - ½, -1 ½, --1 ½.
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In this particular case, all 6 microstates are accounted for.
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Let us go ahead and generate these.
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The ML values takes on these values, MS takes on these values.
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Let us actually generate the microstates.
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Once again, doublet P we have ML = 1, 0, -1.
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We have MS = ½ - ½.
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We are going to have 1, ½, that is here.
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ML is 1, spin is ½, 1, 0, ½.
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ML is 0, spin is ½ - 1 ½.
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ML is -1, spin is ½.
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We will go 1 - ½ that is here, 0 - ½ -1 - ½, this accounts for all of the microstates in the doublet P.
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Let us go ahead and find the J values.
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The doublet P, we find the J values by taking L + S.
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In this particular case, L + S is going to be 1 + 0 = 1.
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And we find the absolute value of L – S.
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L + S is equal to L is 1, S is ½.
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This is going to be 3/2 the absolute value of L - S is equal to absolute value of 1 - ½ = ½.
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This is the upper value and we dropped by 1 until we get to that value.
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J is equal to 3/2 and is equal to ½.
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Therefore, our final term symbol we have the doublet P 3/2 and we have the doublet P ½.
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That is a complete term symbol for the MP1 configuration.
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2P1, 3P1, 4P1, it consists of 2 levels.
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There is going to be doublet P 3/2 term symbol, there is a doublet P ½ term symbol.
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Doublet P 3/2 energy state, doublet P ½ energy state.
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We will talk about the degeneracy.
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We know that the basic term symbol, the doublet P accounts for 6 microstates.
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The doublet P actually consists of doublet P 3/2 and the doublet P ½.
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The degeneracy, in other words the number of microstates in a given specific complete term symbol that is equal to 2 J + 1.
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In the case of doublet P 3/2, J is equal to 3/2.
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2 × 3/2 is 3 + 1 = 4.
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4 of these microstates belong to the doublet P 3/2 energy level.
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2 × ½ + 1 is 2, that means 2 of these microstates belong to the doublet P ½.
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Final is, which is the ground state?
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In other words, is it the doublet P 3/2 or the doublet P ½ the ground state?
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Hund’s rules say, whichever has the largest S.
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In this particular case, S2, S2.
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This is ½, this is ½, S is the same.
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Let us go ahead and put them out.
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We take the largest S, over here, S is the same.
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2, if S is the same, we take the largest L.
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In this case, we have P and P, L is the same.
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3, if that particular orbital is less than half filled which in this case it is less than half filled, it is only 1 electron in the P orbital.
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We are going to take the smallest J value, smallest J.
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This implies that we take the ½ which implies J = ½, which implies that the doublet P ½ is the particular ground state.
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It is the lowest energy, the most stable.
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Not necessarily the ground state, but of the two, that is the one that lower energy.
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Let us go ahead and do this.
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Recall that complimentary configurations have the same term symbols.
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The complement of P1 is P5.
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P5 also has a doublet P 3/2 and a doublet P ½ state.
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Let us do this with an actual table.
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The table version, this particular process that I just did.
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I will make a lot more sense when we see it for the next few configurations that we do.
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For the table, we have 1 electron, we are looking for the largest ML.
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We know the largest ML value is going to be 1.
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That means ML is equal to 1, 0, -1.
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The largest MS for 1 electron is also 1.
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MS is equal to 1, 0, -1.
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MS is actually ½, we have ½ and we have - ½.
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ML is going to be 1, 0, -1.
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MS is the upper row, M sub L the column.
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We have 1 electron, the sums of the ML values add up to these numbers.
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The sums of the MS values add up to these numbers.
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We get 1 + ½, 1 - ½, 0 + ½, 0 - ½.
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We get -1 + ½ - 1 - ½.
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We take the largest ML having a viable microstate in its row, 1.
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This one.
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We take the largest MS having a viable microstate in its column.
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Between these two, it is this one ½.
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We are essentially doing the same thing.
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We are just doing it backwards.
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The largest ML is equal to 1.
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The largest MS having a viable microstate in its column is equal to ½.
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ML = 1 implies that L = 1.
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MS = ½ implies that S = ½.
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It implies that 2 S + 1 is equal to 2.
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Therefore, we have a doublet P.
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We have ML is 1, 0, -1.
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And MS = ½ - ½.
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We are going to cross out 1 for each value of these.
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1, ½, 1 - ½, and we go to 0 1/2, 0 - ½ and then - ½ -1 - ½.
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We have accounted for all of them and from here, since we have the doublet P,
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we can go ahead and generate J value like we did before.
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That is all we are doing when we are doing the table.
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We are finding the largest ML and the largest MS, and then L is equal to 1.
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ML is equal to 1, 0, -1.
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S is equal to ½, MS is equal to ½ - ½.
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And we just combine these to knockout one for each.
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We are leftover with, we just continue on the same way.
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We pick the next largest ML, the next largest MS.
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I think an example will be a lot better and a lot more clear.
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What are the term symbols for the MP 2 configuration?
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We have 2 electrons and we still have the P orbitals.
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We have 6 spin orbitals but we have 2 electrons.
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MP2 means 6 choose 2, that is 6!/ 4! =2!.
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What we end up with is 15 viable microstates.
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We always want to begin with a number of viable microstates so that we know how many we are dealing with.
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We are looking for the microstate that is going to give the largest value of M sub L.
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The microstate giving the largest value of M sub L is going to be this one.
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We are going to put both of the electrons with 2 electrons.
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We are going to put both of the electrons in the m sub l into one.
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We have ML, M sub L for the first electron.
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This is getting confusing here.
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The first electron m sub l is equal to 1.
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The second electron m sub l is equal to 1.
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For the largest value that we can actually get is that putting them both into that m sub l to get a value of 2.
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Here, M sub L which is the sum of the m sub l is equal to 2.
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In this particular case, because I get the largest M sub L equal to 2,
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that I have to put them both in the same suborbital, they have to have opposite spin.
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In this particular case, the largest MS that I can generate from this configuration,
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this 1 configuration is going to be 0 because it is going to be m sub s = + ½.
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It is going to be m sub s = - ½.
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When I add these two together to get the M sub S, it is going to be 0.
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For this microstate, the largest M sub S is going to equal 0 because the 2 electrons must have opposite spin.
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We have ML is equal to 2 which implies the L is equal to 2.
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We have M sub S is equal to 0 which implies that S is equal to 0, which implies that 2 S + 1 is equal to 1.
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L is equal to 2 gives me D.
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2 S + 1 is 1 so I have a singlet D state.
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I need to find out how many microstates are in this particular energy level.
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In other words, the degeneracy of this basic term symbol.
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I'm going to do 2L + 1 × 2S + 1.
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2L + 1 = 5, 2S + 1 is equal to 1.
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5 × 1, there are 5 microstates in this level.
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Let us list the microstates, when you get the basic term symbol that is when you want to list those microstates.
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Let us list the microstates for the singlet D.
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Recall, we have L is equal to 2.
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L is equal to 2 which means the ML goes to 1, 0, -1, 2.
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MS is equal to 0 that means MS is equal to 0.
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We are going to have a microstate that is going to have the sum of the ML is 2, the sum of the MS is 0.
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2 0, 1 0, 0 0, -1 0, 2 0, those microstates are as follows.
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Up down, up down, up down, up down, there is a certain symmetry to this.
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I generate the largest ML with one configuration.
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The M sub L runs through all the values 2, 1, 0, -1, 2.
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MS =0.
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This is 2 0, this is 1 0, because ML is 1 and ML is 0, that give me a M of 1.
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Spin is +½, spin is -½ gives me a total spin of 0.
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Over here, m sub l is 1 and m sub l is -1.
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1 -1 is 0 so M sub L is 0.
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Up spin down spin, the total spin is 0.
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That make sense.
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These 5 microstates represent this term symbol.
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We will worry about that J values later.
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The next largest M sub L possible is ML is equal to 1.
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And this is achievable as I can do that.
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I can be do 2 up spins, the M sub L is equal to 1 because it is 1 + 0.
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M sub l is 1 and m sub l is 0.
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The total M sub L is 1, that is one possible way of getting ML is equal to 1 or I can do this and this.
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M sub l is 1, m sub l is 0.
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However, in this particular case, the two ways of actually achieving the largest M sub L is equal to 1 which one of these
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do I actually choose as my basis to decide what my M sub S is going to be.
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This one is accounted for already.
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If I look at the previous page where I listed the microstates,
00:28:37.100 --> 00:28:43.800
this microstate is already there where we have the first suborbital up spin, the second one suborbital with a downspin.
00:28:43.800 --> 00:28:47.900
Since this is accounted for, the one I pick is this one.
00:28:47.900 --> 00:28:56.900
For this particular one, the largest MS is going to be 1, both of them up spin.
00:28:56.900 --> 00:29:06.200
Here, M sub L is equal to 1 and M sub S is equal to 1, that implies that L = 1, that implies that S is equal to 1,
00:29:06.200 --> 00:29:09.300
that implies that 2S + 1 is equal to 3.
00:29:09.300 --> 00:29:16.600
L = to 1 is a P, 2S + 1 = 3, we end up with a triplet state.
00:29:16.600 --> 00:29:31.400
Let us go ahead and find the 2L + 1 × 2S + 1, 2L + 1, 2 × 1 + 1 is equal to 3.
00:29:31.400 --> 00:29:36.600
2S + 1 that is equal to 3, there are 9 microstates.
00:29:36.600 --> 00:29:42.400
There are 9 microstates in the triplet P basic term.
00:29:42.400 --> 00:29:46.500
Let us list what these microstates are.
00:29:46.500 --> 00:29:53.400
We have ML is equal to 1, 0, -1.
00:29:53.400 --> 00:29:56.600
MS is equal to 1, 0, -1.
00:29:56.600 --> 00:30:08.700
The microstates are going to represented such that we have 1 1, 1 0, 1 -1, 0 1, 0 0, 0 -1, -1 1, -1 0, -1 -1.
00:30:08.700 --> 00:30:11.100
There would be 9 of them.
00:30:11.100 --> 00:30:27.300
Here is what it would look like 123, 123, 123, 123, 123, 123, 123, 123, 123.
00:30:27.300 --> 00:30:42.700
This is the one that we chose, our basic so we start with that one.
00:30:42.700 --> 00:31:03.600
That is the 1 1, 0 1, -1 1, we are going to go to 0 1.
00:31:03.600 --> 00:31:08.400
That is what we just did the 1 1, 0 1, -1 1.
00:31:08.400 --> 00:31:13.700
We are going to do 1 0, 0 0, -1 0.
00:31:13.700 --> 00:31:20.000
The last 3 are going to be 1 -1, 0 -1, -1 -1.
00:31:20.000 --> 00:31:22.200
We are running through all of them.
00:31:22.200 --> 00:31:30.900
This first one, in order to pick the largest ML and the largest MS as possible given.
00:31:30.900 --> 00:31:34.900
The largest ML possible here was 1, that was achievable 2 ways.
00:31:34.900 --> 00:31:38.200
I can even go up spin up spin, down spin down spin.
00:31:38.200 --> 00:31:42.500
Which one do I choose to decide what my MS is going to be, what by largest MS?
00:31:42.500 --> 00:31:47.100
This was already accounted for in the D1 configuration, what we did in the previous page.
00:31:47.100 --> 00:31:50.600
Therefore, I choose this one.
00:31:50.600 --> 00:32:01.300
Here I get down up, down up, down up.
00:32:01.300 --> 00:32:10.300
I have down down, down down, down down, down down.
00:32:10.300 --> 00:32:18.700
These microstates all belong to the triplet P basic term symbol state.
00:32:18.700 --> 00:32:27.200
You basically just run through each combination, 1 1, 1 0, 1 -1.
00:32:27.200 --> 00:32:37.000
This is the 1 1 state, this is the 1 0 state, this is the 1 -1 state.
00:32:37.000 --> 00:32:40.300
M sub l is 1, m sub l is + 0.
00:32:40.300 --> 00:32:54.800
1 + 0 is 1 - ½ - ½ -1, I have to run for each possibility, that is all I'm doing.
00:32:54.800 --> 00:32:59.100
We have 9 microstates and we have 5 microstates previous.
00:32:59.100 --> 00:33:07.000
We have account for 14 of the microstates and we only have 1 left.
00:33:07.000 --> 00:33:18.600
We have accounted for 14 microstates that mean there is 1 left.
00:33:18.600 --> 00:33:37.700
The next possible largest ML is equal to 0.
00:33:37.700 --> 00:33:47.300
Lead me see, how can we actually get ML equal to 0?
00:33:47.300 --> 00:34:19.200
This is achievable as I can do that, that is one way or I can do that is one way, that is another way,
00:34:19.200 --> 00:34:22.400
here the MS is equal to 0.
00:34:22.400 --> 00:34:28.500
Here the MS is equal to 0, here the MS is equal to 0.
00:34:28.500 --> 00:34:33.100
I can go up up, here the MS is equal to 1.
00:34:33.100 --> 00:34:40.900
Here, I can go down down, MS is equal to 1.
00:34:40.900 --> 00:34:47.600
I'm looking for the next largest ML.
00:34:47.600 --> 00:34:52.000
Everything is already accounted for, the next largest M sub L is going to be 0.
00:34:52.000 --> 00:34:56.300
And the only way to get that is this one particular configuration or I can do this.
00:34:56.300 --> 00:35:00.400
Each of these configurations gives me an ML equal to 0.
00:35:00.400 --> 00:35:13.700
The problem is these of all that accounted for in the previous microstates, the previous 14.
00:35:13.700 --> 00:35:18.100
We have only one left with is this one.
00:35:18.100 --> 00:35:24.200
In this particular case, I have ML is equal to 0.
00:35:24.200 --> 00:35:30.800
I have MS is equal to 0 that means that L is equal to 0, that means that S is equal to 0,
00:35:30.800 --> 00:35:35.200
that means that 2S + 1 is equal to 1.
00:35:35.200 --> 00:35:43.600
L = 0 is S, 2S + 1 is 1, I have a singlet S state.
00:35:43.600 --> 00:35:52.600
In this particular case, 2L + 1 × to S + 1 is equal to 1 × 1, there is 1 microstate.
00:35:52.600 --> 00:35:55.500
We have accounted for all 15 microstates.
00:35:55.500 --> 00:36:05.300
In this particular case, this microstate ML is equal to 0, MS is equal to 0.
00:36:05.300 --> 00:36:11.100
The only microstate left is that one.
00:36:11.100 --> 00:36:26.500
There you go, we have singlet D state, we have a triplet P state, and we have a singlet S state.
00:36:26.500 --> 00:36:31.400
Let us go ahead and find the J values for this.
00:36:31.400 --> 00:36:39.000
For the singlet D state L + S is equal to 2 + 0 is equal to 2,
00:36:39.000 --> 00:36:47.100
the absolute value of L - S is equal to the absolute value of 2 -0 is equal to 2.
00:36:47.100 --> 00:36:53.800
J is equal to 2, we have a singlet D2 microstate.
00:36:53.800 --> 00:37:02.200
There we go, the degeneracy as we said before is 2J + 1.
00:37:02.200 --> 00:37:06.800
J is equal to 2, 2 × 2 is 4, 4 + 1 is 5.
00:37:06.800 --> 00:37:11.400
I will go ahead and put that in parentheses underneath.
00:37:11.400 --> 00:37:18.100
There are 5 microstates in that singlet D2.
00:37:18.100 --> 00:37:21.600
We have a triplet P.
00:37:21.600 --> 00:37:30.000
Our triplet P, our L + S is equal to 1 + 1 that is equal to 2.
00:37:30.000 --> 00:37:36.700
Our absolute value of L - S is equal to 1 -1 is equal to 0.
00:37:36.700 --> 00:37:41.100
J is equal to this, all the way down to this in increments of 1.
00:37:41.100 --> 00:37:50.800
2, 1, 0, we have a triplet P2, we have a triple a P1, we had a triple a P0.
00:37:50.800 --> 00:37:54.600
The degeneracy is 2J + 1.
00:37:54.600 --> 00:38:00.100
Of the 9 microstates belong to the triplet P state.
00:38:00.100 --> 00:38:14.000
All of those 9, 5 belong to the triplet P2 state, 3 belonged to the triplet P1 state and 1 belongs to the triplet P0 state.
00:38:14.000 --> 00:38:16.900
We have broken down even further.
00:38:16.900 --> 00:38:23.900
There are symbols all over the place.
00:38:23.900 --> 00:38:30.300
Singlet S, L + S = 0 + 0 = 0.
00:38:30.300 --> 00:38:39.500
The absolute value of L - S is equal to 0 -0 = 0, means that J is equal to 0.
00:38:39.500 --> 00:38:45.200
We have a singlet S0 and its degeneracy is going to be 2J + 1.
00:38:45.200 --> 00:38:50.700
It is going to be 1, 1 + 1 + 3 + 5 + 5 is equal to 15.
00:38:50.700 --> 00:38:53.900
All 15 microstates are accounted for.
00:38:53.900 --> 00:38:57.100
There is a singlet D level that has 5 microstates.
00:38:57.100 --> 00:38:59.700
There is a triplet P level that has 9 microstates.
00:38:59.700 --> 00:39:03.400
There is a singlet S level that has 1 microstate.
00:39:03.400 --> 00:39:08.400
The singlet D is actually a singlet D2 that contains all 5 microstates.
00:39:08.400 --> 00:39:15.900
The triplet P is broken up into 3 different energy levels, a triplet P2, triplet P1, triplet P0.
00:39:15.900 --> 00:39:19.700
Each having 5, 3, 1 microstate respectively.
00:39:19.700 --> 00:39:30.700
The final was here which is the ground state.
00:39:30.700 --> 00:39:34.500
When we say ground state, we mean which is the one with the lowest energy.
00:39:34.500 --> 00:39:48.400
We have 1, the first rule says largest S.
00:39:48.400 --> 00:39:58.800
The largest S is S = 1 which means that triplet P state.
00:39:58.800 --> 00:40:13.600
2, largest L, of these PPP, L is the same.
00:40:13.600 --> 00:40:16.800
3, less than half filled.
00:40:16.800 --> 00:40:20.300
It is less than half filled, we only have 2 electrons in there.
00:40:20.300 --> 00:40:29.900
Less than half filled means we are going to look for the smallest J, where the smallest J is the 0 value.
00:40:29.900 --> 00:40:42.000
Therefore, the triplet P0 is the ground state of all of these.
00:40:42.000 --> 00:40:49.500
Let us do another.
00:40:49.500 --> 00:40:53.300
What are the term symbols for the MP3 configuration?
00:40:53.300 --> 00:41:04.600
We have 3 electrons, we have 6 spin orbitals, 3 electrons, 6 choose 3.
00:41:04.600 --> 00:41:07.800
I will do that over here, I think.
00:41:07.800 --> 00:41:15.900
We have 6 choose 3 that is equal to 6!/ 3! is 3!.
00:41:15.900 --> 00:41:26.200
I will write it out, 6 × 5 × 4/ 3 × 2 × 1.
00:41:26.200 --> 00:41:29.400
One of these 3!s cancels the 3 × 2 × 1 up above.
00:41:29.400 --> 00:41:39.900
When you do that 6 × 5 × 4 divided by 6, 2 × 3 you are going to get 20 viable microstates.
00:41:39.900 --> 00:41:51.500
There are 20 possible ways of distributing 3 electrons in the spin orbitals,
00:41:51.500 --> 00:41:57.900
in the 3 suborbital of the P which has 6 spin suborbital.
00:41:57.900 --> 00:42:04.900
6 spin orbitals, there are 3 electrons, there are 20 different ways of actually arranging them.
00:42:04.900 --> 00:42:30.400
The largest ML possible is achievable as, if I put one there, one there, and one there.
00:42:30.400 --> 00:42:40.200
In this particular case, I have ML is 1, m sub l is 1, m sub l is 0.
00:42:40.200 --> 00:42:48.100
1 + 1 + 0 is equal to 2.
00:42:48.100 --> 00:42:52.700
The largest M sub L is equal to 2.
00:42:52.700 --> 00:42:56.800
In this particular arrangement, this and this spin they cancel out.
00:42:56.800 --> 00:43:12.400
Here, the largest M sub S is equal to ½.
00:43:12.400 --> 00:43:21.100
We have ML is equal to 2, we have M sub S equal to ½, that means that L is equal to 2,
00:43:21.100 --> 00:43:27.500
that means the S is equal to ½, that means that 2S + 1 is equal to 2.
00:43:27.500 --> 00:43:31.600
L = 2, that is a D, this is 2.
00:43:31.600 --> 00:43:38.600
We have a doublet D state.
00:43:38.600 --> 00:43:49.000
2L + 1 × 2S + 1 is going to be 5 × 2 is equal to 10.
00:43:49.000 --> 00:43:55.200
There are 10 microstates in this doublet D level.
00:43:55.200 --> 00:44:02.400
Let us list them.
00:44:02.400 --> 00:44:12.800
Let us list the 10 microstates, the one with the next largest ML.
00:44:12.800 --> 00:44:18.000
When we have a choice, we want to see which one of these microstates is accounted for.
00:44:18.000 --> 00:44:21.700
What we are doing this procedure is finding the basic term symbol and list the microstates for that term symbol.
00:44:21.700 --> 00:44:25.800
We are going to refer back to it, if we need to.
00:44:25.800 --> 00:44:28.100
Let us list the term well.
00:44:28.100 --> 00:44:37.100
ML, we said L is equal to 2 that means the ML is going to equal to 1, 0, -1, 2.
00:44:37.100 --> 00:44:42.000
M sub S or S is equal to ½ which means the M sub S is equal to ½ and -1/2.
00:44:42.000 --> 00:44:54.500
The microstates are going to be such that the total angular momentum is going to add up to 2.
00:44:54.500 --> 00:45:00.900
The total spin angular momentum is going to be ½ all the way down the line.
00:45:00.900 --> 00:45:09.000
We are going to have one with 2 ½, 1 ½, 0 ½ , -1 ½, 2 ½, and 2 – ½, 1 - ½, and so on.
00:45:09.000 --> 00:45:12.500
For total of 2 × 5, 10 microstates.
00:45:12.500 --> 00:45:19.400
Here is what they look like.
00:45:19.400 --> 00:45:33.600
We have 123, 123, 123, 123, 123, 123, 123, 123, 123.
00:45:33.600 --> 00:45:37.100
I will put the 10th one right down here, 123.
00:45:37.100 --> 00:45:40.000
We start with the one which we started with.
00:45:40.000 --> 00:45:56.800
Up down, up down, up up, up down, that is down electron.
00:45:56.800 --> 00:46:01.800
I just moved it once over here and once over here.
00:46:01.800 --> 00:46:14.000
We are just writing out the microstates that allow you to achieve a total angular of 2, total angular of half.
00:46:14.000 --> 00:46:18.600
Total angular 1 total angular half, total angular 0 total spin ½.
00:46:18.600 --> 00:46:24.700
Total angular -1 total spin ½, total angular -2 total spin ½.
00:46:24.700 --> 00:46:30.800
And then, total angular 2 total spin - ½, total angular 1 total spin - ½.
00:46:30.800 --> 00:46:40.400
Total angular 0 total spin - 1/2, total angular -1 total spin - ½, total angular -2 total spin - ½.
00:46:40.400 --> 00:46:43.500
You are just arranging them.
00:46:43.500 --> 00:46:48.200
It is going to be that, that and that.
00:46:48.200 --> 00:46:54.900
We have this, this and this.
00:46:54.900 --> 00:47:18.400
That that down, they have down down up, down up down, down down up, down down up.
00:47:18.400 --> 00:47:28.500
These are the 10 microstates that are in the doublet D level.
00:47:28.500 --> 00:47:36.000
Let us look for the next largest ML.
00:47:36.000 --> 00:47:46.200
The next largest ML is equal to 1 and achievable as follows.
00:47:46.200 --> 00:48:01.000
I can do this, this this.
00:48:01.000 --> 00:48:10.300
I could do this, this this, or I could do up down up.
00:48:10.300 --> 00:48:19.300
These are different ways that I can have a M sub L, a total angular momentum number of 1.
00:48:19.300 --> 00:48:27.400
This was has been account for in the previous set.
00:48:27.400 --> 00:48:30.000
That is why I listed them.
00:48:30.000 --> 00:48:38.700
I have listed list the microstates so that when I move to the next level, in this particular case the largest M sub L
00:48:38.700 --> 00:48:42.800
and I see the different ways that I can actually achieve that largest M sub L,
00:48:42.800 --> 00:48:47.400
I ignore the ones that were already accounted for.
00:48:47.400 --> 00:48:51.900
This one has not been accounted for, this is the one that I choose.
00:48:51.900 --> 00:48:58.600
In this particular case, the MS value for this one is ½.
00:48:58.600 --> 00:49:09.900
I have an ML equal to 1 and I have an MS = ½, that implies the L = 1, that implies that S = ½,
00:49:09.900 --> 00:49:14.800
that implies that to 2S + 1 is equal to 2.
00:49:14.800 --> 00:49:22.700
L = 1 is a P, this is a 2 so I have a doublet P state.
00:49:22.700 --> 00:49:36.600
2L + 1 × 2S + 1, 2L + 1 that is equal to 3 × 2S + 1 2, there are 6 microstates in the doublet P level.
00:49:36.600 --> 00:49:48.800
Let us go ahead and list those.
00:49:48.800 --> 00:49:55.100
L = 1, therefore M sub L = 1, 0, -1.
00:49:55.100 --> 00:50:00.700
M sub S, S is equal to ½, M sub S is equal to ½ - ½.
00:50:00.700 --> 00:50:03.100
2 × 3, there are 6.
00:50:03.100 --> 00:50:14.600
The one that we started with first, boom boom boom move is over one, leave out the same.
00:50:14.600 --> 00:50:23.600
Bring this here, I just have to make sure what is that I am writing, has noted that account for previously.
00:50:23.600 --> 00:50:25.700
That is all, that is all I’m doing them.
00:50:25.700 --> 00:50:42.800
I have that, that that, that, and this.
00:50:42.800 --> 00:50:48.600
This is why you want to have a list of the microstates that have been accounted for.
00:50:48.600 --> 00:50:57.100
If you are listing these microstates to satisfy these relations, if you list the one that are account for, just list another one.
00:50:57.100 --> 00:51:01.500
Find another way of doing it, there will always be another way of doing it.
00:51:01.500 --> 00:51:09.000
There has to be because that is 6 choose 3, there have to be 20 microstates available for you to do what you do,
00:51:09.000 --> 00:51:12.200
that is what all these numbers mean.
00:51:12.200 --> 00:51:15.400
If you end up listing something, as you are going through it here,
00:51:15.400 --> 00:51:23.700
listing the microstates to satisfy these conditions 1 ½, 0 ½, -1 ½, 1 - ½, 0 – ½, -1 – ½.
00:51:23.700 --> 00:51:32.900
If you list them and satisfy them up spin downspin, just make sure you have listed one that already been accounted for.
00:51:32.900 --> 00:51:35.500
If have, just list another one.
00:51:35.500 --> 00:51:39.000
Let me write that down.
00:51:39.000 --> 00:52:24.800
If you list a microstate that has been accounted for, just find another that satisfies the values of M sub L and M sub S.
00:52:24.800 --> 00:52:43.400
Find something that satisfied, just find another one that satisfies those values that is not been accounted for.
00:52:43.400 --> 00:52:55.300
There will always be one.
00:52:55.300 --> 00:53:03.600
Let me take a couple minutes here. And when you doing these problems, you are not probably going to be asked do more than a couple of these problems.
00:53:03.600 --> 00:53:08.600
There is a table in your book that lists with different term symbols are for each configuration,
00:53:08.600 --> 00:53:14.400
whether it is P1, P2, P3, whether it is D1, D2, D3, D4, D5.
00:53:14.400 --> 00:53:17.300
Whatever it is, it is already listed in your book.
00:53:17.300 --> 00:53:19.900
You are not going to be asked this is particular process.
00:53:19.900 --> 00:53:24.800
It can take a while, there is nothing simple about this.
00:53:24.800 --> 00:53:32.300
It is reasonably straightforward, you just have a lot of things to account for is have a lot of things on the paper.
00:53:32.300 --> 00:53:37.200
A lot of symbols floating around, a lot of things to check and doublet check again.
00:53:37.200 --> 00:53:43.000
You have clearly figured out by now, that is the nature of quantum mechanics.
00:53:43.000 --> 00:53:47.400
As we go deeper in science, things become more complex.
00:53:47.400 --> 00:53:52.000
Beautiful things emerge, symmetries start to emerge.
00:53:52.000 --> 00:53:56.100
Larger picture start to emerge but things become more complicated.
00:53:56.100 --> 00:54:03.300
This is the nature of the game and it is part of the excitement to be able to know that we can do something like this
00:54:03.300 --> 00:54:09.700
and achieve this magnificent, intellectual thing and know that we are right.
00:54:09.700 --> 00:54:15.500
Enough of that, let us go on here.
00:54:15.500 --> 00:54:22.200
We have taken care of ML is 1, ML is 2.
00:54:22.200 --> 00:54:59.300
The next largest ML, the next largest M sub L is equal to 0 and that is achievable as which is not been accounted for.
00:54:59.300 --> 00:55:04.800
We are good.
00:55:04.800 --> 00:55:13.500
ML = 0, here MS up, the largest MS is 3/2.
00:55:13.500 --> 00:55:23.900
This implies that L is equal 0, this implies that S = 3/2, this implies that 2S + 1 is equal to 4.
00:55:23.900 --> 00:55:27.100
L = 0 gives me an S term symbol.
00:55:27.100 --> 00:55:30.600
2S + 1 give me a spin multiplicity of 4.
00:55:30.600 --> 00:55:37.800
We have a quartet S or quadruplet S.
00:55:37.800 --> 00:55:49.100
2L + 1 × 2S + 1, 2L + 1 = 1 × 4 = 4 microstates.
00:55:49.100 --> 00:55:52.900
We have 10 microstates, 6 microstates, 4 microstates.
00:55:52.900 --> 00:56:03.600
We have accounted for all of them.
00:56:03.600 --> 00:56:20.700
ML is equal to 0 and M sub S is equal to 3/2, ½ - ½ -3/2.
00:56:20.700 --> 00:56:29.200
The total MS have to add the 0, total MS is ½ add to 3/2, and then add to ½ to - ½ to -3/2.
00:56:29.200 --> 00:56:36.100
123, 123, 123, 123, there are 4 microstates.
00:56:36.100 --> 00:56:42.200
The one that we chose as our basic one was that.
00:56:42.200 --> 00:56:51.800
I can go ahead and do this, I can go ahead and do this, I can go ahead and do this.
00:56:51.800 --> 00:56:58.200
From this one that has been accounted for, there are 3, I just switch them around
00:56:58.200 --> 00:57:01.100
until it is one that has not been accounted for.
00:57:01.100 --> 00:57:24.500
For the full symbols, we have a doublet D, we have a doublet P, and we have quartet S.
00:57:24.500 --> 00:57:39.100
For the doublet D, our L + S is equal to 2 + 0 is equal to 2.
00:57:39.100 --> 00:57:47.200
This is not right, 2S + 1 = 2, that means S = ½.
00:57:47.200 --> 00:57:58.000
L + S = 2 + ½ that =, 2 ½ that is equal to 5 ½.
00:57:58.000 --> 00:58:06.400
The absolute value of L - S is equal to 2 - ½, that is equal to 3/2.
00:58:06.400 --> 00:58:13.000
Therefore, J is equal to 5/2 down to 3/2.
00:58:13.000 --> 00:58:22.600
Therefore, we have a doublet D 5/2 and we have a doublet D D 3/2.
00:58:22.600 --> 00:58:37.200
The degeneracy of 2J + 1, 2 × 5/2 + 1, the degeneracy of this one is 6 and the degeneracy of this one is 4.
00:58:37.200 --> 00:58:39.900
Let us look at our doublet P state.
00:58:39.900 --> 00:59:05.400
Here L + S is equal to 1 + ½ which is equal to 3/2 and the absolute value of L - S is equal to 1 – ½ = ½.
00:59:05.400 --> 00:59:10.600
Therefore, J is equal to 3/2 and ½.
00:59:10.600 --> 00:59:17.500
We have a doublet P 3/2, we have a doublet P ½.
00:59:17.500 --> 00:59:24.500
The degeneracy is going to be 4 and 2.
00:59:24.500 --> 00:59:27.100
The doublet P have 6 microstates.
00:59:27.100 --> 00:59:33.800
Of all the 6 microstates, 4 of them belong to the doublet P 3/2 state, 2 of them belong to doublet P ½ state.
00:59:33.800 --> 00:59:58.400
The quadruplet S, the quadruplet S L + S is equal to 0 + 3/2 is equal to 3/2, L - S sequel to absolute value 0 -3/2 = 3/2.
00:59:58.400 --> 01:00:05.700
J is only equal to 3/2 which gives us a quadruplet S 3/2.
01:00:05.700 --> 01:00:11.800
The degeneracy of which is going to equal 4 2J + 1.
01:00:11.800 --> 01:00:18.200
I will go ahead and put that over here parentheses.
01:00:18.200 --> 01:00:36.400
The ground state, the largest S is this one, equal to 3/2.
01:00:36.400 --> 01:00:50.900
Our quadruplet S 3/2, that is the only one, that is the only term symbol for the largest S.
01:00:50.900 --> 01:00:54.700
That is the ground state, I do not have to look at the L, I do not have to look at the J.
01:00:54.700 --> 01:00:57.800
There we go.
01:00:57.800 --> 01:01:01.600
I know that it is tedious and I apologize for that.
01:01:01.600 --> 01:01:05.400
I hope that at least the procedure made a little bit of sense.
01:01:05.400 --> 01:01:10.100
If not, there still a couple of more in the next lesson of example problems and
01:01:10.100 --> 01:01:12.400
I'm going to be using the same procedure.
01:01:12.400 --> 01:01:17.300
Hopefully, just going and seeing it over and over will make a difference.
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Thank you so much for joining us here at www.educator.com.
01:01:19.100 --> 01:01:20.000
We will see you next time, bye.