For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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### Example Problems I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Example I: What are the Term Symbols for the np¹ Configuration? 0:10
- Example II: What are the Term Symbols for the np² Configuration? 20:38
- Example III: What are the Term Symbols for the np³ Configuration? 40:46

### Physical Chemistry Online Course

### Transcription: Example Problems I

*Hello, welcome to www.educator.com, welcome back to Physical Chemistry.*0000

*Today, we are going to start our example problems for term symbols and atomic spectra.*0004

*Let us jump right on in.*0009

*What are the terms symbols for the NP1 configuration?*0013

*The 2P1, 3P1, 4P1, the N part does not matter.*0019

*It is actually just the electrons of the suborbital.*0023

*This is really just true for P1.*0026

*I'm going to solve this with a slightly different procedure than the one that I have used when I introduced terms symbols.*0030

*Back when I introduced term symbols a few lessons back,*0038

*we started by listing out all the microstates explicitly and then picking out those M sub L values and M sub S values.*0041

*Then, I introduced this shorter procedure where we are actually making a table of the ML and MS values.*0049

*Counting the largest ML, the largest MS, crossing them out until you come up with all of your term symbols.*0056

*What I'm going to do is essentially just the reverse of that.*0064

*I'm going to do this for this example problem and also the next several example problems.*0067

*When we do the MP2 configuration and the 3 configuration, the D2 configuration goes through quite a few of these.*0074

*Essentially, what we are doing is we are going to work with the ML and MS values directly*0080

*and then generate the microstates for these term symbols as we go long, as opposed to the other way around.*0085

*At some sense, it is the reverse.*0093

*You can decide which one you like better.*0095

*For this particular problem, I’m going to do both.*0099

*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.*0101

*They are essentially the same and at some point, you have to deal with the largest ML*0107

*and largest MS and you are going to have to deal with the microstates.*0113

*But this way with the numbers just tends to be a little bit faster.*0116

*Let us see what we can do.*0120

*Like everything else in science or anything for that matter, you have to do them a number of times in order to really understand.*0123

*Passive understanding just by looking and seeing that you can follow that,*0133

*that is one thing but being able to generate and it do it yourself, that is when you we only get your hands dirty,*0137

*that is when you really understand what is happening.*0142

*Let us see what we have got.*0144

*We always want to begin by finding the number of viable microstates.*0146

*Let me stick with black here.*0151

*Always begin by finding the number of viable microstates and that was, we have this great name for that,*0153

*it is the number of spin orbitals that we have available to occupy choosing the number of electrons.*0176

*In this particular case, we are looking for the NP1 configuration.*0191

*We have 1 electron and we are trying to distribute among the PX PY PZ.*0195

*We have 2, 4, 6, we have 6 spin orbitals and we have 1 electron.*0202

*It is going to be 6 choose 1, there are 6 microstates.*0207

*6 different ways to arrange 1 electron in the P orbital.*0215

*In this particular case, we already know which one they are.*0219

*It is going to be up here, up here, up here, or down here, down here, down here.*0222

*That takes care of it, but let us just go ahead and work it through.*0226

*This is the important thing, always begin with that.*0228

*Start by finding the number of viable microstates and as you find each term symbol,*0231

*knock out the number of microstates and it will tell you how many you have left.*0235

*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.*0240

*In this case it is only 1 electron.*0258

*The M sub L, you remember that is the 1, 0, -1, in the case of the P orbital.*0260

*For the D, it was 2, 1, 0, -1, -2, because there are 5 suborbital in the D orbital.*0266

*We are going to be finding that number and we would be finding the M sub S.*0275

*That is equal to the M sub S1 + M sub S2, that is the spin of the electron, +½ or - ½.*0281

*We are going to add them all up for all of the electrons.*0292

*We begin by looking for the largest ML possible.*0297

*Here, because we are talking about 1 electron in the P.*0314

*Here, the largest ML is going to equal 1.*0323

*This particular thing, and the reason it is, is because we can put the 1 electron in the left most suborbital.*0334

*Let us do that.*0363

*Here, the ML value is 1 and ML value is 0, ML value is -1.*0365

*We want the largest ML possible, for 1 electron it is over here.*0371

*Here, ML = 1.*0376

*The largest MS possible, in this particular case it is 1 electron, it is up spin ½ as opposed to downspin - ½.*0381

*Given that ML is equal to 1, in this particular case, the largest MS possible is ½.*0395

*If ML is equal to 1 that implies that L = 1.*0406

*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.*0411

*When L is equal to 1 and S is equal to 2, that gives me a doublet P configuration, a doublet P term symbol.*0423

*From here, we are going to calculate how many microstates are in this doublet P.*0436

*L = 1 remember, when L = 0 that is S.*0442

*When L = 1 that is P.*0446

*When L = 2 that is D.*0448

*When L = 3 that is F, and so on.*0450

*S is ½, 2S + 1 that is the left most top left number.*0454

*This is a doublet P configuration, we will worry about the J a little bit later.*0462

*We found our basic term symbol.*0468

*2L + 1 × 2S + 1, let us do it this way.*0471

*The number of ML × the number of MS that is equal to 2L + 1 × 2S + 1.*0489

*L is equal to 1, 2 × 1 + 1 is equal to 3.*0501

*2S + 1 is equal 2, 3 × to = 6.*0505

*There are 6 microstates in this level.*0510

*There are 6 microstates that belong to this particular term symbol.*0517

*These come from ML is equal to 1 or L is equal to 1, that means that ML is equal to 1, 0, -1.*0521

*MS is equal to ½, S is equal to ½, that means that MS is equal to ½, and ½ -1 is – ½.*0537

*There are 3, here is 2, 3 × 2 is 6.*0547

*The states that you are going to get, the microstates come from 1, ½, 1 - ½, 0 ½ , 0 - ½, -1 ½, --1 ½.*0552

*In this particular case, all 6 microstates are accounted for.*0566

*Let us go ahead and generate these.*0574

*The ML values takes on these values, MS takes on these values.*0581

*Let us actually generate the microstates.*0586

*Once again, doublet P we have ML = 1, 0, -1.*0593

*We have MS = ½ - ½.*0607

*We are going to have 1, ½, that is here.*0612

*ML is 1, spin is ½, 1, 0, ½.*0625

*ML is 0, spin is ½ - 1 ½.*0632

*ML is -1, spin is ½.*0637

*We will go 1 - ½ that is here, 0 - ½ -1 - ½, this accounts for all of the microstates in the doublet P.*0640

*Let us go ahead and find the J values.*0654

*The doublet P, we find the J values by taking L + S.*0656

*In this particular case, L + S is going to be 1 + 0 = 1.*0661

*And we find the absolute value of L – S.*0667

*L + S is equal to L is 1, S is ½.*0677

*This is going to be 3/2 the absolute value of L - S is equal to absolute value of 1 - ½ = ½.*0683

*This is the upper value and we dropped by 1 until we get to that value.*0695

*J is equal to 3/2 and is equal to ½.*0699

*Therefore, our final term symbol we have the doublet P 3/2 and we have the doublet P ½.*0705

*That is a complete term symbol for the MP1 configuration.*0717

*2P1, 3P1, 4P1, it consists of 2 levels.*0724

*There is going to be doublet P 3/2 term symbol, there is a doublet P ½ term symbol.*0729

*Doublet P 3/2 energy state, doublet P ½ energy state.*0735

*We will talk about the degeneracy.*0750

*We know that the basic term symbol, the doublet P accounts for 6 microstates.*0755

*The doublet P actually consists of doublet P 3/2 and the doublet P ½.*0760

*The degeneracy, in other words the number of microstates in a given specific complete term symbol that is equal to 2 J + 1.*0765

*In the case of doublet P 3/2, J is equal to 3/2.*0774

*2 × 3/2 is 3 + 1 = 4.*0781

*4 of these microstates belong to the doublet P 3/2 energy level.*0785

*2 × ½ + 1 is 2, that means 2 of these microstates belong to the doublet P ½.*0791

*Final is, which is the ground state?*0805

*In other words, is it the doublet P 3/2 or the doublet P ½ the ground state?*0814

*Hund’s rules say, whichever has the largest S.*0824

*In this particular case, S2, S2.*0828

*This is ½, this is ½, S is the same.*0831

*Let us go ahead and put them out.*0837

*We take the largest S, over here, S is the same.*0840

*2, if S is the same, we take the largest L.*0849

*In this case, we have P and P, L is the same.*0855

*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.*0862

*We are going to take the smallest J value, smallest J.*0874

*This implies that we take the ½ which implies J = ½, which implies that the doublet P ½ is the particular ground state.*0879

*It is the lowest energy, the most stable.*0898

*Not necessarily the ground state, but of the two, that is the one that lower energy.*0902

*Let us go ahead and do this.*0916

*Recall that complimentary configurations have the same term symbols.*0918

*The complement of P1 is P5.*0950

*P5 also has a doublet P 3/2 and a doublet P ½ state.*0952

*Let us do this with an actual table.*0969

*The table version, this particular process that I just did.*0975

*I will make a lot more sense when we see it for the next few configurations that we do.*0980

*For the table, we have 1 electron, we are looking for the largest ML.*0987

*We know the largest ML value is going to be 1.*0996

*That means ML is equal to 1, 0, -1.*1001

*The largest MS for 1 electron is also 1.*1007

*MS is equal to 1, 0, -1.*1013

*MS is actually ½, we have ½ and we have - ½.*1020

*ML is going to be 1, 0, -1.*1037

*MS is the upper row, M sub L the column.*1041

*We have 1 electron, the sums of the ML values add up to these numbers.*1047

*The sums of the MS values add up to these numbers.*1056

*We get 1 + ½, 1 - ½, 0 + ½, 0 - ½.*1059

*We get -1 + ½ - 1 - ½.*1069

*We take the largest ML having a viable microstate in its row, 1.*1076

*This one.*1085

*We take the largest MS having a viable microstate in its column.*1089

*Between these two, it is this one ½.*1093

*We are essentially doing the same thing.*1098

*We are just doing it backwards.*1102

*The largest ML is equal to 1.*1104

*The largest MS having a viable microstate in its column is equal to ½.*1110

*ML = 1 implies that L = 1.*1121

*MS = ½ implies that S = ½.*1132

*It implies that 2 S + 1 is equal to 2.*1137

*Therefore, we have a doublet P.*1145

*We have ML is 1, 0, -1.*1154

*And MS = ½ - ½.*1163

*We are going to cross out 1 for each value of these.*1167

*1, ½, 1 - ½, and we go to 0 1/2, 0 - ½ and then - ½ -1 - ½.*1170

*We have accounted for all of them and from here, since we have the doublet P,*1185

*we can go ahead and generate J value like we did before.*1190

*That is all we are doing when we are doing the table.*1193

*We are finding the largest ML and the largest MS, and then L is equal to 1.*1196

*ML is equal to 1, 0, -1.*1206

*S is equal to ½, MS is equal to ½ - ½.*1209

*And we just combine these to knockout one for each.*1213

*We are leftover with, we just continue on the same way.*1217

*We pick the next largest ML, the next largest MS.*1221

*I think an example will be a lot better and a lot more clear.*1227

*What are the term symbols for the MP 2 configuration?*1242

*We have 2 electrons and we still have the P orbitals.*1248

*We have 6 spin orbitals but we have 2 electrons.*1252

*MP2 means 6 choose 2, that is 6!/ 4! =2!.*1258

*What we end up with is 15 viable microstates.*1272

*We always want to begin with a number of viable microstates so that we know how many we are dealing with.*1280

*We are looking for the microstate that is going to give the largest value of M sub L.*1287

*The microstate giving the largest value of M sub L is going to be this one.*1299

*We are going to put both of the electrons with 2 electrons.*1316

*We are going to put both of the electrons in the m sub l into one.*1318

*We have ML, M sub L for the first electron.*1326

*This is getting confusing here.*1330

*The first electron m sub l is equal to 1.*1334

*The second electron m sub l is equal to 1.*1338

*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.*1341

*Here, M sub L which is the sum of the m sub l is equal to 2.*1350

*In this particular case, because I get the largest M sub L equal to 2,*1359

*that I have to put them both in the same suborbital, they have to have opposite spin.*1366

*In this particular case, the largest MS that I can generate from this configuration,*1373

*this 1 configuration is going to be 0 because it is going to be m sub s = + ½.*1378

*It is going to be m sub s = - ½.*1387

*When I add these two together to get the M sub S, it is going to be 0.*1390

*For this microstate, the largest M sub S is going to equal 0 because the 2 electrons must have opposite spin.*1399

*We have ML is equal to 2 which implies the L is equal to 2.*1424

*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.*1434

*L is equal to 2 gives me D.*1444

*2 S + 1 is 1 so I have a singlet D state.*1446

*I need to find out how many microstates are in this particular energy level.*1450

*In other words, the degeneracy of this basic term symbol.*1458

*I'm going to do 2L + 1 × 2S + 1.*1462

*2L + 1 = 5, 2S + 1 is equal to 1.*1470

*5 × 1, there are 5 microstates in this level.*1475

*Let us list the microstates, when you get the basic term symbol that is when you want to list those microstates.*1485

*Let us list the microstates for the singlet D.*1494

*Recall, we have L is equal to 2.*1505

*L is equal to 2 which means the ML goes to 1, 0, -1, 2.*1510

*MS is equal to 0 that means MS is equal to 0.*1517

*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.*1522

*2 0, 1 0, 0 0, -1 0, 2 0, those microstates are as follows.*1529

*Up down, up down, up down, up down, there is a certain symmetry to this.*1545

*I generate the largest ML with one configuration.*1563

*The M sub L runs through all the values 2, 1, 0, -1, 2.*1568

*MS =0.*1573

*This is 2 0, this is 1 0, because ML is 1 and ML is 0, that give me a M of 1.*1576

*Spin is +½, spin is -½ gives me a total spin of 0.*1587

*Over here, m sub l is 1 and m sub l is -1.*1593

*1 -1 is 0 so M sub L is 0.*1599

*Up spin down spin, the total spin is 0.*1604

*That make sense.*1609

*These 5 microstates represent this term symbol.*1611

*We will worry about that J values later.*1617

*The next largest M sub L possible is ML is equal to 1.*1630

*And this is achievable as I can do that.*1655

*I can be do 2 up spins, the M sub L is equal to 1 because it is 1 + 0.*1676

*M sub l is 1 and m sub l is 0.*1683

*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.*1686

*M sub l is 1, m sub l is 0.*1695

*However, in this particular case, the two ways of actually achieving the largest M sub L is equal to 1 which one of these*1698

*do I actually choose as my basis to decide what my M sub S is going to be.*1705

*This one is accounted for already.*1710

*If I look at the previous page where I listed the microstates,*1714

*this microstate is already there where we have the first suborbital up spin, the second one suborbital with a downspin.*1717

*Since this is accounted for, the one I pick is this one.*1724

*For this particular one, the largest MS is going to be 1, both of them up spin.*1728

*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,*1737

*that implies that 2S + 1 is equal to 3.*1746

*L = to 1 is a P, 2S + 1 = 3, we end up with a triplet state.*1749

*Let us go ahead and find the 2L + 1 × 2S + 1, 2L + 1, 2 × 1 + 1 is equal to 3.*1756

*2S + 1 that is equal to 3, there are 9 microstates.*1771

*There are 9 microstates in the triplet P basic term.*1776

*Let us list what these microstates are.*1782

*We have ML is equal to 1, 0, -1.*1786

*MS is equal to 1, 0, -1.*1793

*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.*1796

*There would be 9 of them.*1809

*Here is what it would look like 123, 123, 123, 123, 123, 123, 123, 123, 123.*1811

*This is the one that we chose, our basic so we start with that one.*1827

*That is the 1 1, 0 1, -1 1, we are going to go to 0 1.*1843

*That is what we just did the 1 1, 0 1, -1 1.*1863

*We are going to do 1 0, 0 0, -1 0.*1868

*The last 3 are going to be 1 -1, 0 -1, -1 -1.*1874

*We are running through all of them.*1880

*This first one, in order to pick the largest ML and the largest MS as possible given.*1882

*The largest ML possible here was 1, that was achievable 2 ways.*1891

*I can even go up spin up spin, down spin down spin.*1895

*Which one do I choose to decide what my MS is going to be, what by largest MS?*1898

*This was already accounted for in the D1 configuration, what we did in the previous page.*1902

*Therefore, I choose this one.*1907

*Here I get down up, down up, down up.*1910

*I have down down, down down, down down, down down.*1921

*These microstates all belong to the triplet P basic term symbol state.*1930

*You basically just run through each combination, 1 1, 1 0, 1 -1.*1939

*This is the 1 1 state, this is the 1 0 state, this is the 1 -1 state.*1947

*M sub l is 1, m sub l is + 0.*1957

*1 + 0 is 1 - ½ - ½ -1, I have to run for each possibility, that is all I'm doing.*1960

*We have 9 microstates and we have 5 microstates previous.*1975

*We have account for 14 of the microstates and we only have 1 left.*1979

*We have accounted for 14 microstates that mean there is 1 left.*1987

*The next possible largest ML is equal to 0.*1998

*Lead me see, how can we actually get ML equal to 0?*2018

*This is achievable as I can do that, that is one way or I can do that is one way, that is another way,*2027

*here the MS is equal to 0.*2059

*Here the MS is equal to 0, here the MS is equal to 0.*2062

*I can go up up, here the MS is equal to 1.*2068

*Here, I can go down down, MS is equal to 1.*2073

*I'm looking for the next largest ML.*2081

*Everything is already accounted for, the next largest M sub L is going to be 0.*2087

*And the only way to get that is this one particular configuration or I can do this.*2092

*Each of these configurations gives me an ML equal to 0.*2096

*The problem is these of all that accounted for in the previous microstates, the previous 14.*2100

*We have only one left with is this one.*2114

*In this particular case, I have ML is equal to 0.*2118

*I have MS is equal to 0 that means that L is equal to 0, that means that S is equal to 0,*2124

*that means that 2S + 1 is equal to 1.*2131

*L = 0 is S, 2S + 1 is 1, I have a singlet S state.*2135

*In this particular case, 2L + 1 × to S + 1 is equal to 1 × 1, there is 1 microstate.*2143

*We have accounted for all 15 microstates.*2152

*In this particular case, this microstate ML is equal to 0, MS is equal to 0.*2155

*The only microstate left is that one.*2165

*There you go, we have singlet D state, we have a triplet P state, and we have a singlet S state.*2171

*Let us go ahead and find the J values for this.*2186

*For the singlet D state L + S is equal to 2 + 0 is equal to 2,*2191

*the absolute value of L - S is equal to the absolute value of 2 -0 is equal to 2.*2199

*J is equal to 2, we have a singlet D2 microstate.*2207

*There we go, the degeneracy as we said before is 2J + 1.*2214

*J is equal to 2, 2 × 2 is 4, 4 + 1 is 5.*2222

*I will go ahead and put that in parentheses underneath.*2227

*There are 5 microstates in that singlet D2.*2231

*We have a triplet P.*2238

*Our triplet P, our L + S is equal to 1 + 1 that is equal to 2.*2241

*Our absolute value of L - S is equal to 1 -1 is equal to 0.*2250

*J is equal to this, all the way down to this in increments of 1.*2257

*2, 1, 0, we have a triplet P2, we have a triple a P1, we had a triple a P0.*2261

*The degeneracy is 2J + 1.*2271

*Of the 9 microstates belong to the triplet P state.*2274

*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.*2280

*We have broken down even further.*2294

*There are symbols all over the place.*2297

*Singlet S, L + S = 0 + 0 = 0.*2304

*The absolute value of L - S is equal to 0 -0 = 0, means that J is equal to 0.*2310

*We have a singlet S0 and its degeneracy is going to be 2J + 1.*2319

*It is going to be 1, 1 + 1 + 3 + 5 + 5 is equal to 15.*2325

*All 15 microstates are accounted for.*2331

*There is a singlet D level that has 5 microstates.*2334

*There is a triplet P level that has 9 microstates.*2337

*There is a singlet S level that has 1 microstate.*2340

*The singlet D is actually a singlet D2 that contains all 5 microstates.*2343

*The triplet P is broken up into 3 different energy levels, a triplet P2, triplet P1, triplet P0.*2348

*Each having 5, 3, 1 microstate respectively.*2356

*The final was here which is the ground state.*2360

*When we say ground state, we mean which is the one with the lowest energy.*2371

*We have 1, the first rule says largest S.*2374

*The largest S is S = 1 which means that triplet P state.*2388

*2, largest L, of these PPP, L is the same.*2399

*3, less than half filled.*2413

*It is less than half filled, we only have 2 electrons in there.*2417

*Less than half filled means we are going to look for the smallest J, where the smallest J is the 0 value.*2420

*Therefore, the triplet P0 is the ground state of all of these.*2430

*Let us do another.*2442

*What are the term symbols for the MP3 configuration?*2449

*We have 3 electrons, we have 6 spin orbitals, 3 electrons, 6 choose 3.*2453

*I will do that over here, I think.*2464

*We have 6 choose 3 that is equal to 6!/ 3! is 3!.*2468

*I will write it out, 6 × 5 × 4/ 3 × 2 × 1.*2476

*One of these 3!s cancels the 3 × 2 × 1 up above.*2486

*When you do that 6 × 5 × 4 divided by 6, 2 × 3 you are going to get 20 viable microstates.*2489

*There are 20 possible ways of distributing 3 electrons in the spin orbitals,*2500

*in the 3 suborbital of the P which has 6 spin suborbital.*2511

*6 spin orbitals, there are 3 electrons, there are 20 different ways of actually arranging them.*2518

*The largest ML possible is achievable as, if I put one there, one there, and one there.*2525

*In this particular case, I have ML is 1, m sub l is 1, m sub l is 0.*2550

*1 + 1 + 0 is equal to 2.*2560

*The largest M sub L is equal to 2.*2568

*In this particular arrangement, this and this spin they cancel out.*2573

*Here, the largest M sub S is equal to ½.*2577

*We have ML is equal to 2, we have M sub S equal to ½, that means that L is equal to 2,*2592

*that means the S is equal to ½, that means that 2S + 1 is equal to 2.*2601

*L = 2, that is a D, this is 2.*2607

*We have a doublet D state.*2611

*2L + 1 × 2S + 1 is going to be 5 × 2 is equal to 10.*2618

*There are 10 microstates in this doublet D level.*2629

*Let us list them.*2635

*Let us list the 10 microstates, the one with the next largest ML.*2642

*When we have a choice, we want to see which one of these microstates is accounted for.*2653

*What we are doing this procedure is finding the basic term symbol and list the microstates for that term symbol.*2658

*We are going to refer back to it, if we need to.*2662

*Let us list the term well.*2666

*ML, we said L is equal to 2 that means the ML is going to equal to 1, 0, -1, 2.*2668

*M sub S or S is equal to ½ which means the M sub S is equal to ½ and -1/2.*2677

*The microstates are going to be such that the total angular momentum is going to add up to 2.*2682

*The total spin angular momentum is going to be ½ all the way down the line.*2694

*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.*2701

*For total of 2 × 5, 10 microstates.*2709

*Here is what they look like.*2712

*We have 123, 123, 123, 123, 123, 123, 123, 123, 123.*2719

*I will put the 10th one right down here, 123.*2733

*We start with the one which we started with.*2737

*Up down, up down, up up, up down, that is down electron.*2740

*I just moved it once over here and once over here.*2757

*We are just writing out the microstates that allow you to achieve a total angular of 2, total angular of half.*2762

*Total angular 1 total angular half, total angular 0 total spin ½.*2774

*Total angular -1 total spin ½, total angular -2 total spin ½.*2778

*And then, total angular 2 total spin - ½, total angular 1 total spin - ½.*2785

*Total angular 0 total spin - 1/2, total angular -1 total spin - ½, total angular -2 total spin - ½.*2791

*You are just arranging them.*2800

*It is going to be that, that and that.*2803

*We have this, this and this.*2808

*That that down, they have down down up, down up down, down down up, down down up.*2815

*These are the 10 microstates that are in the doublet D level.*2838

*Let us look for the next largest ML.*2848

*The next largest ML is equal to 1 and achievable as follows.*2856

*I can do this, this this.*2866

*I could do this, this this, or I could do up down up.*2881

*These are different ways that I can have a M sub L, a total angular momentum number of 1.*2890

*This was has been account for in the previous set.*2899

*That is why I listed them.*2907

*I have listed list the microstates so that when I move to the next level, in this particular case the largest M sub L*2910

*and I see the different ways that I can actually achieve that largest M sub L,*2919

*I ignore the ones that were already accounted for.*2923

*This one has not been accounted for, this is the one that I choose.*2927

*In this particular case, the MS value for this one is ½.*2932

*I have an ML equal to 1 and I have an MS = ½, that implies the L = 1, that implies that S = ½,*2938

*that implies that to 2S + 1 is equal to 2.*2950

*L = 1 is a P, this is a 2 so I have a doublet P state.*2955

*2L + 1 × 2S + 1, 2L + 1 that is equal to 3 × 2S + 1 2, there are 6 microstates in the doublet P level.*2963

*Let us go ahead and list those.*2976

*L = 1, therefore M sub L = 1, 0, -1.*2989

*M sub S, S is equal to ½, M sub S is equal to ½ - ½.*2995

*2 × 3, there are 6.*3001

*The one that we started with first, boom boom boom move is over one, leave out the same.*3003

*Bring this here, I just have to make sure what is that I am writing, has noted that account for previously.*3014

*That is all, that is all I’m doing them.*3023

*I have that, that that, that, and this.*3026

*This is why you want to have a list of the microstates that have been accounted for.*3043

*If you are listing these microstates to satisfy these relations, if you list the one that are account for, just list another one.*3048

*Find another way of doing it, there will always be another way of doing it.*3057

*There has to be because that is 6 choose 3, there have to be 20 microstates available for you to do what you do,*3061

*that is what all these numbers mean.*3069

*If you end up listing something, as you are going through it here,*3072

*listing the microstates to satisfy these conditions 1 ½, 0 ½, -1 ½, 1 - ½, 0 – ½, -1 – ½.*3075

*If you list them and satisfy them up spin downspin, just make sure you have listed one that already been accounted for.*3084

*If have, just list another one.*3093

*Let me write that down.*3095

*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.*3099

*Find something that satisfied, just find another one that satisfies those values that is not been accounted for.*3145

*There will always be one.*3163

*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.*3175

*There is a table in your book that lists with different term symbols are for each configuration,*3183

*whether it is P1, P2, P3, whether it is D1, D2, D3, D4, D5.*3188

*Whatever it is, it is already listed in your book.*3194

*You are not going to be asked this is particular process.*3197

*It can take a while, there is nothing simple about this.*3200

*It is reasonably straightforward, you just have a lot of things to account for is have a lot of things on the paper.*3205

*A lot of symbols floating around, a lot of things to check and doublet check again.*3212

*You have clearly figured out by now, that is the nature of quantum mechanics.*3217

*As we go deeper in science, things become more complex.*3223

*Beautiful things emerge, symmetries start to emerge.*3227

*Larger picture start to emerge but things become more complicated.*3232

*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*3236

*and achieve this magnificent, intellectual thing and know that we are right.*3243

*Enough of that, let us go on here.*3250

*We have taken care of ML is 1, ML is 2.*3255

*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.*3262

*We are good.*3299

*ML = 0, here MS up, the largest MS is 3/2.*3305

*This implies that L is equal 0, this implies that S = 3/2, this implies that 2S + 1 is equal to 4.*3313

*L = 0 gives me an S term symbol.*3324

*2S + 1 give me a spin multiplicity of 4.*3327

*We have a quartet S or quadruplet S.*3330

*2L + 1 × 2S + 1, 2L + 1 = 1 × 4 = 4 microstates.*3338

*We have 10 microstates, 6 microstates, 4 microstates.*3349

*We have accounted for all of them.*3353

*ML is equal to 0 and M sub S is equal to 3/2, ½ - ½ -3/2.*3363

*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.*3381

*123, 123, 123, 123, there are 4 microstates.*3389

*The one that we chose as our basic one was that.*3396

*I can go ahead and do this, I can go ahead and do this, I can go ahead and do this.*3402

*From this one that has been accounted for, there are 3, I just switch them around*3412

*until it is one that has not been accounted for.*3418

*For the full symbols, we have a doublet D, we have a doublet P, and we have quartet S.*3421

*For the doublet D, our L + S is equal to 2 + 0 is equal to 2.*3444

*This is not right, 2S + 1 = 2, that means S = ½.*3459

*L + S = 2 + ½ that =, 2 ½ that is equal to 5 ½.*3467

*The absolute value of L - S is equal to 2 - ½, that is equal to 3/2.*3478

*Therefore, J is equal to 5/2 down to 3/2.*3486

*Therefore, we have a doublet D 5/2 and we have a doublet D D 3/2.*3493

*The degeneracy of 2J + 1, 2 × 5/2 + 1, the degeneracy of this one is 6 and the degeneracy of this one is 4.*3502

*Let us look at our doublet P state.*3517

*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.*3520

*Therefore, J is equal to 3/2 and ½.*3545

*We have a doublet P 3/2, we have a doublet P ½.*3550

*The degeneracy is going to be 4 and 2.*3557

*The doublet P have 6 microstates.*3564

*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.*3567

*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.*3574

*J is only equal to 3/2 which gives us a quadruplet S 3/2.*3598

*The degeneracy of which is going to equal 4 2J + 1.*3606

*I will go ahead and put that over here parentheses.*3612

*The ground state, the largest S is this one, equal to 3/2.*3618

*Our quadruplet S 3/2, that is the only one, that is the only term symbol for the largest S.*3636

*That is the ground state, I do not have to look at the L, I do not have to look at the J.*3651

*There we go.*3655

*I know that it is tedious and I apologize for that.*3658

*I hope that at least the procedure made a little bit of sense.*3661

*If not, there still a couple of more in the next lesson of example problems and*3665

*I'm going to be using the same procedure.*3670

*Hopefully, just going and seeing it over and over will make a difference.*3672

*Thank you so much for joining us here at www.educator.com.*3677

*We will see you next time, bye.*3679

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