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

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

### Example Problems II

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
- Example I: Find the Term Symbols for the nd² Configuration
- Example II: Find the Term Symbols for the 1s¹2p¹ Configuration
- Example III: Calculate the Separation Between the Doublets in the Lyman Series for Atomic Hydrogen
- Example IV: Calculate the Frequencies of the Lines for the (4d) ²D → (3p) ²P Transition

- Intro 0:00
- Example I: Find the Term Symbols for the nd² Configuration 0:11
- Example II: Find the Term Symbols for the 1s¹2p¹ Configuration 27:02
- Example III: Calculate the Separation Between the Doublets in the Lyman Series for Atomic Hydrogen 41:41
- Example IV: Calculate the Frequencies of the Lines for the (4d) ²D → (3p) ²P Transition 48:53

### Physical Chemistry Online Course

### Transcription: Example Problems II

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

*Today, we are going to do another set of example problems for the term symbols and for atomic spectra.*0004

*Let us get started.*0011

*This next one, find the term symbols for the ND2 configuration.*0014

*We are doing the same thing that we did before, just more practice.*0020

*ND2, we know that D orbital has 5 suborbitals.*0024

*There are 2 spin orbitals for every suborbital, we had a total of 10.*0035

*ND2 that means 10 and we have 2 electrons, that is going to be 10 choose 2 viable microstates.*0041

*That is equal to 10!/ 8! is 2!, what you end up with is 90/ 2 = 45.*0051

*There are 45 microstates, 45 viable microstates.*0061

*This tends to get very complex very quickly.*0066

*45 viable microstates but again it is reasonably easy to handle.*0071

*It is tedious but it is reasonably easy to handle.*0077

*We have the D which is 12345, m sub l is 2, 1, 0, -1, -2.*0080

*Remember the quantum numbers represent the locations.*0095

*The largest ML achievable, we start with the same way.*0098

*Find the viable microstates and then find the largest ML.*0103

*The largest ML achievable is ML is equal to 4.*0108

*1, 2, 3, 4, 5, that is achievable if we put both electrons in the m sub l = 2.*0117

*Here, because we put them both for the same suborbital, they have to have opposite spin, ½ and -1/2. *0125

*Here, the largest MS, the largest M sub S is going to equal 0.*0137

*We have M sub L = 4, M sub S = 0, this implies that L = 4, this implies that S = 0.*0147

*It implies that 2S + 1 which is the spin multiplicity is equal to 1.*0159

*The 4 gives us a G term symbol and the 1.*0165

*We have a singlet G state.*0168

*The singlet G state, let us go ahead and do our 2L + 1 × 2S + 1.*0173

*2L + 1 is equal to, L is 4, 2 × 4 is 8, 8 + 1 is 9, 9 × 1.*0198

*9 × 1, we have 9 microstates that belong to the singlet G state.*0204

*ML is equal to 4, 3, 2, 1, 0, -1, -2, -3, -4 and MS is equal to 0.*0212

*We have 4 0, 3 0, 2 0, 1 0, 0 0.*0225

*That that, that that, that accounts for the 9 microstates.*0229

*Let us explicitly list what those microstates are.*0232

*12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, there we go.*0236

*We started off with this and this, let us just run through this entire.*0258

*This is 4 0, we need to go 3 0, 2 0, 1 0, 0 0, -1 0, -2 0, -3 0, -4 0.*0265

*We run through them, this is just the beginning.*0274

*That is there, that is that one, that is that one, that is that one.*0278

*For example, if I take this one, here the m sub l is 2.*0290

*Here, the m sub l is -1.*0296

*2 -1 is 1, up spin downspin is 0.*0299

*This microstate represents the 1 0 combination.*0303

*We work our way down this way which is going to be this this, this this, this this, and this this.*0312

*These are the 10 microstates that have the same energy.*0323

*The degeneracy of this singlet G is 10.*0331

*The next largest ML, the next largest M sub L is equal to 3.*0340

*Mind you, just because I have the next largest ML = 4, the next largest ML = 3, I just do not work my way down.*0351

*I do, but I still have to be careful because I still might have another state where the ML can be the same.*0361

*And it is the same if there is another configuration that I can come up with that has not been accounted for.*0369

*That still gives me that same ML.*0375

*The next largest ML is equal to 3, achievable as 1, 2, 3, 4, 5.*0378

*This is achievable as this, 2, 1, 3.*0397

*This particular microstate has not been accounted for.*0406

*It was not in the previous 10 that I listed, this is fine.*0410

*Here, the MS for this is equal to 1 +½ +½.*0413

*I have an ML equal to 3 and have an MS equal to 1, that implies that L = 3, *0420

*this implies that S = 1 and implies that 2S + 1 which is the spin multiplicity is equal to 3.*0431

*L = 3 gives me an F.*0439

*That was the triplet S state.*0442

*2L + 1 × 2S + 1, 2 × 3 is 6, 6 + 1 = 7, 2S + 1 that is this, × 3 we have 21 microstates.*0446

*21 microstates for this level.*0463

*We have the ML values, ML3 which means 3, 2, 1, 0, -1, -2, -3.*0474

*MS 1, 0, -1, 3 × 7 = 21.*0484

*I need to arrange, I need to list 21 microstates that give me *0491

*all possible combinations 3 1, 2 1, 1 1, 0 1, all the way down 3 0, 2 0, 1 0, all the way down.*0496

*Here is how it looks.*0510

*We have 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 7.*0515

*12345, 12345, 12345, 6 7.*0537

*You see the fact of the matter is one procedure is not necessarily better than another.*0558

*At some point, you have to list the microstates.*0563

*I prefer this particular procedure working with the largest ML *0565

*and with each term symbol that I find will listing the microstates, because I like the list microstates explicitly like this.*0568

*I do not like that whole 1 + 0 – symbolism, that confuses me after all these years, I still get confused.*0577

*I’m going to do the next problem like that one because the problem is going to be slightly different.*0584

*That is a problem that you often see and you probably want to do one of them.*0590

*Your actual homework assignment but in general I really prefer this particular method.*0592

*Finding the term symbol based on the largest ML and the largest MS from a basic configuration and*0597

*then using that basic configuration to generate all the microstates for that particular term symbol.*0603

*I just think it works out better.*0608

*It is tedious.*0610

*12345, 12345, 12345, this is our basic.*0614

*I think I’m going to go to red here.*0627

*Let us go up and up, that is that one, that is that one.*0633

*When we go down up, down up, down up, down up, down up, down up, down up.*0649

*There is a symmetry when you are listing these microstates, find a pattern, follow that pattern.*0663

*We have down and down, down and down, down and down, down down, down down, down down, down and down.*0670

*There we go, those are our 21 microstates here.*0684

*Let us go back to black.*0691

*The next largest value of ML is equal to 2 and this is achievable as I can either do a 2 and 0, both up spin.*0705

*Or I can do a 1 and 1 up spin down spin.*0734

*This one has been accounted for already.*0739

*I just need to find one that has not been accounted for as my basis and *0744

*then use my ML and MS for that one to generate my term symbol.*0750

*This one has not been accounted for, that is the one that I'm going to use.*0755

*In this particular case, here MS is equal to 0.*0760

*I have an ML = 2, I will go ahead and write the MS.*0765

*MS is equal to 0 and ML = 2 implies the L = 2, implies that S = 0, implies that 2S + 1 is equal to 1.*0769

*L = 2 gives me the D, the 1.*0780

*I have a singlet D state.*0783

*Therefore, my 2L + 1, my degeneracy for this basic term symbol 2S + 1 is equal to 5 × 1 *0786

*which is 5 microstates in the singlet D.*0796

*ML = 2, 1, 0, -1, -2, I hope by now, the basic pattern that I’m doing is what is that *0801

*we are doing is starting to come together.*0810

*MS = 0, I'm going to have 12345, 12345, 12345, 12345, 12345.*0814

*Let us go ahead and go to red.*0829

*We picked this as our standard, we will go here.*0833

*The next one is going to go, I will leave 1 and electron here.*0837

*I will leave an electron here.*0841

*Notice, in this case my basic is here.*0853

*I have come over here already, I'm not occupied with the left side anymore and not occupied here.*0857

*I do not take this over to this, I bring this here.*0864

*These are the 5 microstates.*0873

*The individual M sub L is needed to add to these numbers and the individual M sub S need to add to this number.*0876

*That takes care of that, let us go back to black.*0887

*The next value of M sub L can take, if you make a mistake in any of this, *0897

*you will find out because what is going to end up happening is you are going to end up *0922

*using all your microstates before your term symbols run out.*0928

*You are going to end up using, you are going to run out of microstates.*0933

*Where you have too many or not enough, believe me, if you have made a mistake it will show up.*0937

*It may not show up right away but it will show up eventually.*0943

*You are going to end up short on either terms symbols or in microstates, that is what is going to happen.*0946

*Let us see, the next largest value ML can take is ML is equal to 1.*0954

*And that is achievable 12345, this or,*0964

*this is achievable as I can put on both up spin like that or I can do up spin down spin like that.*0973

*This one has already been accounted for.*0982

*It is already taken care of.*0986

*This one has not been accounted for, that is the one that I use.*0987

*Here, MS is equal to 1.*0991

*I have got ML is equal to 1, I have got MS is equal to 1.*0993

*That implies that L = 1, that implies that S = 1, that implies that 2S + 1 which is the spin multiplicity = 3.*0997

*That is going to give me a triplet P state.*1012

*2L + 1 × 2S + 1 is equal to 3 × 3, I have 9 microstates.*1017

*ML = 1, 0, -1.*1031

*MS = 1, 0, -1, 3 × 3 is equal to 9 microstates.*1034

*I have got 12345, 12345, and 12345, 12345, 12345, 12345, 12345, 12345, 12345.*1040

*The state that I picked was this and this.*1064

*I have this this, I have this, I have this.*1069

*I can go this way, that way, this way, that way, this way, that way.*1074

*I can go down, this way that way, this way that way, this way this way.*1082

*M sub L is 1, 1 + 0 is 1.*1096

*Up spin up spin is 1.*1101

*1 + 0 is 1 downspin up spin 0.*1108

*1 + 0 is 1 down spin down spin -1.*1114

*1 + -1 is 0 up spin up spin 1.*1122

*1 + 1 m sub l + m sub l.*1129

*1 + -1 is 0 down spin up spin 0, that is 0 0.*1133

*0 -1 -1 downspin up spin 0, that is it.*1142

*That is all that is going on here.*1148

*Let us see what we have got.*1153

*We just have taken care of 9 + 21 + 5 + 9 = 44 microstates.*1158

*We have accounted for 44 of the 45 microstates.*1171

*There is 1 microstate left.*1174

*The next largest ML is equal to 0.*1181

*Achievable as that, this state is not an accounted for.*1191

*Yes, this is the one that we use.*1200

*And the MS here is equal to 0.*1202

*We have ML is equal to 0, we have MS is equal to 0.*1205

*It implies that L is equal to 0, it implies that S is equal to 0.*1209

*It implies that 2S + 1 is equal to 1.*1215

*2L + 1 × 2S + 1 is equal to 1 × 1 = to 1 microstate.*1224

*It corresponds and this particular microstate, we have ML = 0, we have MS = 0.*1231

*The only way that this is achievable is that.*1239

*We have singlet G, triplet F, singlet D, triplet P, singlet S.*1247

*Our singlet G, L + S = 4 + 0 = 4.*1264

*The absolute value of L - S = the absolute value of 4 -0 = 4.*1273

*J is equal to 4.*1281

*What we have is a singlet G 4 state.*1284

*Let me go ahead and put degeneracy in there.*1293

*Once again, the degeneracy of the complete term symbol is equal to 2J + 1.*1297

*This 2L + 1 × the 2S + 1 that is equal to the degeneracy of the basic term symbol without the J value.*1305

*The degeneracy of the actual full term symbol is 2J + 1.*1313

*In this particular case, I will put a parentheses, it is equal to 9.*1318

*Triplet S state, L + S = 3 + 1 = 4.*1336

*The absolute value of L - S = absolute value of 3 -1 is equal to 2.*1345

*Therefore, J = 4, 3, 2.*1353

*Start here and work your way down there by increments of 1.*1357

*We have a triplet F 4, we have a triplet F 3, and we have a triplet F 2.*1361

*The degeneracy are 9, 7, and 5.*1370

*9 + 7 + 5 is 21.*1376

*We have a singlet D state.*1387

*L + S is equal to 2 + 0 is equal to 2.*1391

*The absolute value of L – S, the absolute value of 2 -0 is equal 2.*1395

*Therefore, the J is equal to 2 so we have a singlet D2 state degeneracy of 5.*1402

*And we have our triplet P.*1411

*Let me go ahead do on the next page here.*1418

*I have a triplet P, L + S = 1 + 1 is equal 2.*1426

*The absolute value of L - S = 1 -1 = 0.*1444

*Therefore, J is equal to 2, 1, 0.*1451

*I have a triplet P2, I have a triplet P1, and I have a triplet P0 state.*1455

*The degeneracy are 5, 3, and 1 for total of 9.*1462

*Remember, the total degeneracy, the basic term symbol.*1470

*The triplet P was 9, these are individual breakdowns.*1472

*5 microstates in the triplet P2, 3 microstates in the triplet P1, 1 microstate in the triplet P0 state.*1475

*Final state here, we have singlet S.*1485

*L +S is equal to 0 + 0 is equal to 0.*1492

*The absolute of L - S = absolute 0 -0 = 0.*1498

*J = 0, we have a singlet S0 and the degeneracy is 1.*1503

*Ground state, we are looking for the one with the largest S value.*1512

*The largest S is equal to 3.*1523

*Let us try this again.*1537

*We have a singlet G, we had a triplet F, we have a singlet D, we had a triplet P, and we had a singlet S.*1545

*The largest S is going to be 3.*1559

*The largest S is equal to 3 that means that triplet F or the triplet P.*1563

*The next is the largest L value.*1574

*The largest L between F and the P is the F, the triplet F.*1577

*Because we are talking about the D2 which is less than half filled, we are going to take the smallest J value.*1583

*Less than half means the smallest J value.*1593

*We had a triplet F4, triplet F3, triplet F2.*1598

*Our triplet F2 is our ground state for that particular electron configuration of the D2.*1605

*Let us see what is next.*1623

*Find the term symbols for the NS1 ND1 configuration.*1625

*I am a little bit of a typo here, this should actually the NP1 configuration not ND1 configuration.*1639

*NS1 NP1 configuration, this is an excited configuration.*1650

*You have 1 electron in the S and 1 electron in the P.*1655

*I d o not want to write this.*1670

*I do not think I wrote this the way that I actually wanted to write this.*1671

*Let me do this.*1674

*I’m going to be real specific about it.*1679

*Let me go ahead and do it this way.*1690

*Let me go ahead and rewrite this configuration.*1692

*Let us do it as NS1 and we will do N prime.*1695

*This is crazy, I’m going to pick a specific one.*1705

*The 1S1 and 2P1.*1709

*We had the 1S2 excited, when the electron has been excited to the P level.*1714

*The primary quantum numbers do not really matter.*1719

*Ultimately, what we are concerned with is this S1 P1, that is what really matters.*1721

*For excited states where electrons are in different suborbital, I think it is best to do a table or write out the configuration explicitly.*1727

*What I have done here, I have actually done a table.*1781

*However, like I said I'm not a big fan of tables.*1797

*If I'm going to do the microstates, I much rather just write them all up explicitly.*1803

*At least that way I can see what it is that I'm actually dealing with.*1808

*I’m going to go ahead and do that.*1811

*I’m going to modify what it is that I have done.*1811

*I’m going to write out the possible microstates, the 1S1, 2P1. *1814

*First of all, we are going to do the same thing and find out the number of viable microstates here.*1821

*First of all, let us see what we are looking at.*1828

*Let me go ahead and do this in blue.*1829

*We are looking at this, the S and then some P.*1832

*Because they are in different orbitals, here you have the 1 and here you have the 2, basically the S and P,*1837

*because they are in separate, you have to treat them differently.*1852

*Here you have 1 electron has two different ways of being in this S orbital.*1855

*1 electron in the P has 6 different ways.*1861

*Basically, here you have 2 choose 1.*1864

*We are going to multiply that by the 6 choose 1.*1867

*It is going to be 2 × 6, we are going to end up with 12 viable microstates.*1871

*Because the electrons are in different orbitals S and P, *1877

*I have to take the number of ways of finding the electron in one orbital × the number of ways of finding the P in the other orbital, *1883

*and I have to multiply those numbers.*1890

*There are 12 viable microstates.*1893

*I’m going to go ahead and list out those 12 microstates.*1897

*Maybe I should just use a table.*1902

*I’m going to go ahead and use a table here.*1904

*For excited states where electrons are in different suborbitals, the largest ML, *1910

*the largest M sub L achievable is if you put one here.*1923

*Here the ML is, the m sub l is 0, here the m sub l is 1, 0, -1.*1935

*If I do this, the largest M sub L possible is if an electron goes into this orbital and this orbital, that is going to equal 1.*1941

*The largest M sub S, if they both have an up spin ½ and ½, it is going to be that.*1952

*Our table, the ML is going to be 1, 0, -1.*1960

*The MS is going to be 1, 0, -1.*1963

*We are going to have 1, 0, -1.*1968

*We would have 1, 0, -1.*1971

*This is M sub S, this is M sub L.*1978

*The different ways of getting an ML value of 1 is 0, 1.*1986

*This is going to be 0 0, this is going to be 0 -1.*1995

*A spin of ½ are both going to be + ½.*1999

*This is going to be both + ½ and both + ½.*2002

*Here, this is going to be 0 0.*2011

*It is going to be 1, it is going to be 0 1 and this is going to be + and this is going to be -.*2018

*I can also do it with 0 - and 1 +.*2031

*I can have an M sub L value of 0 and 1.*2036

*0 and 1 add to 1, + and - add to 0.*2039

*0 and 1 add to 1, - and + add to 0.*2042

*This is going to be 0 + 0 - or 0 -0 +, these are not the same.*2048

*These are not the same because we are in different orbitals.*2056

*Because the electrons are in different orbital.*2060

*1 is in the S and one is in the P, very different.*2062

*These are not equivalent orbitals.*2066

*And this is going to be 0 -1 + -, 0 -1 - +.*2069

*And over here, I will do -1, this is going to be 0 -1, 0 -1.*2077

*0 0, 0 1, 0 – 1, this is going to be - -, - -, - -.*2093

*There we go, these are 9 microstates.*2105

*Yes, these are all the possible microstates that are available.*2111

*The largest ML is 1 which means 1, 0, -1 and the largest MS that has a viable microstate is 1.*2115

*It is going to be 1, 0, -1.*2123

*ML is going to be 1, 0, -1.*2131

*MS is 1, 0, -1.*2135

*For every combination, I will knock one out.*2138

*1 1 knock one out , 1 0 knock one out, 1 -1 knock one out, 0 1, 0 0, 0 -1, 0 1, 0 0, 0 -1 knock them out.*2141

*And it does not matter which one I choose.*2165

*-1 1, -1 0, -1 -1, -1 1, -1 0, -1 -1 knock one out.*2168

*For this ML = 1 MS = 1 that implies that L = 1, that implies that S = 1.*2179

*It implies that 2S + 1 = 3.*2189

*Here, I am looking at the triplet P state, I think.*2195

*Yes, triplet P and that has 9 microstates.*2201

*I know that already, 9 microstates.*2207

*And I do not know that is the 2L + 1.*2211

*Let me go ahead and stick with the same procedure.*2212

*2L + 1 × 2S + 1 should be 6 microstates.*2217

*123456789, 9 microstates.*2222

*2 L + 1 = 3 × 3 is equal to 9 microstates and here they are.*2225

*The only difference is this is 0 1 symbolism instead of the actual writing out the specific explicit orbital configuration.*2236

*The triplet P state, after I have done this, the table I'm left with is going to be the following.*2249

*I still have the 1, 0, and the -1 and I have the 1, 0, and -1, except all I have left now is 0, 1.*2256

*This is - this is +, this is 0 0 - +, and this is 0 -1 - +.*2271

*This is what is left over.*2280

*Here, the largest M sub L with a viable microstate is again, notice 1.*2286

*Here the largest M sub L is 1.*2293

*The largest M sub S is 0.*2297

*You notice in both cases, I ended up with a largest ML possible of 1.*2302

*I did not just go down to 0.*2308

*In this case, we have ML = 1, MS = 0.*2311

*That implies that L = 1, that implies that S = 0, that implies that 2S + 1 is equal to 1.*2319

*And this is going to be a single P state.*2327

*2L + 1 × 2S + 1, that is equal to 3 × 1 = 3 microstates.*2335

*These are your 3 microstates, we knock them out because ML = 1, 0, -1 and MS is equal to 0.*2348

*1 0, 0 0, -1 0, we knock them out.*2362

*We are left with that.*2373

*Let us go ahead and deal with our triplet P state.*2375

*Triplet P, L + S is equal to 1 + 1 is equal 2.*2378

*The absolute value of L - S is equal to 1 -1 is equal to 0.*2388

*Therefore, J is equal to 2, 1, 0.*2394

*I have a triplet P2, I have a triplet P1, I have a triplet P0.*2398

*The degeneracies are 5, 3, and 1.*2405

*I have a singlet P, it is going to be L + S = 1 + 0 is equal to 1.*2416

*L – S, absolute value is equal to 1 -0 = 1.*2427

*Therefore, J is equal to 1.*2431

*Therefore, I have a singlet P1 state and its degeneracy is equal to 3.*2434

*Which is the ground state?*2444

*The largest S between the triplet P and the singlet P, it is going to be the triplet P.*2450

*And then I have the largest L value, that does not matter because they are all P.*2459

*Triplet P2, triplet P1, triplet P0 and it is less than half filled.*2467

*Therefore, we are going to pick the smallest J which means we are going to pick the triplet P0 as the ground state.*2479

*That really is exhausting.*2494

*Let us do a little bit some of spectra.*2500

*Use the table below to calculate the separation between the doublet in the Lyman series for atomic hydrogen go up to N = 4.*2503

*You remember, our discussion a couple of lessons ago *2520

*when we are talking about the Lyman series and we are talking about the fine structure.*2523

*We said that the fine structure consist of doublets.*2530

*Each of those lines actually split into two because the P level, the 2P, 3P, 4P*2536

*it consists of a doublet P 3/2 and a doublet P ½.*2542

*We see them right here.*2549

*For the 2P, we have a doublet P 3/2 so that is a double line, that is a double line, that is a double line.*2551

*And where is the 4P, that is the double line and so on.*2565

*I want to go to N = 4.*2575

*We are going to do N = 2, N = 3, N = 4.*2577

*I do not want you to find a separation between the doublets.*2580

*The separation between those doublet, one of those is going to be a transition from here to here.*2583

*It is going to be a transition from here to here.*2591

*The separation between those is just the difference in energy between this and this.*2595

*The difference in energy between this and this.*2601

*The difference in energy this and this.*2605

*This can take a higher number and subtract the lower number, that is all.*2607

*Let us go ahead and write all this out.*2614

*Recall the Lyman series is for the NP 1 down to 1S1 transition.*2623

*The table shows NP1 with 2 states, doublet P 3/2 and a doublet P ½.*2651

*Again, the problem is to find the separation between the two.*2676

*We just take the difference between energy is higher - the lower.*2678

*I will write everything.*2687

*The problem asks us to find a separation between the lines for NP1 *2692

*and doublet P 3/2 down to 1S1 which is a doublet S 1/2 and NP1 doublet P 1/2 down to 1S1 doublet S ½.*2726

*We need to find the difference in energy.*2750

*For N = 2, we have 82259.28501 - 82258.9191, the difference in energy is 0.3659 wave numbers.*2757

*For N =3, we have 97.*2795

*We are just taking that – that, that is all we are doing.*2800

*97492.3196 - 97482.2112 the δ E = 0.1084 inverse cm.*2805

*N = 4, we have 102823.8942 - 102823.8486 δ E = 00.0457 inverse cm.*2834

*This probably is something that I should have told you a little bit earlier, I will go ahead and tell you now, I apologize.*2863

*I decided to use energy tables that I got off of the NIST web site.*2869

*They are the most up to date, as far as I know.*2873

*Very recent, the spectra and energy values were calculated within the last 45 years, if I'm not mistaken the last time I checked.*2876

*Your book making a different tables, tables that have different values.*2887

*The values are very close to these, but again as the years go on, as our techniques improve, our values improve.*2892

*Do not worry about if you will get 0.0457, if you are using different values from different tables.*2899

*Perhaps, your book was printed a long time ago.*2905

*You are going to be using table that have different values.*2909

*Do not worry about it, go with the values in the table you are given.*2911

*The table itself is not that important.*2915

*It is the process that is important.*2916

*I just wan to you to know that.*2918

*You do not have to get that same numbers I'm getting, you are probably using older tables.*2920

*The ones that are in your book, that is not a problem at all.*2924

*They are very minor differences.*2929

*Let us go ahead and round this out.*2933

*Using the table below for atomic sodium, this time it is atomic sodium.*2936

*Calculate the frequencies in inverse cm, all the lines for the 4D transitions.*2939

*This time we are going to calculate the frequencies of the spectral lines for the transition from the 4D to 3P level.*2951

*The 4D has a doublet D and the 3P has doublet P.*2961

*Let us go ahead and we will see what we have got.*2968

*The 4D, let us look at 4D here for sodium.*2970

*For sodium, you are looking at 1S2, 2S2, 2P6, 3S1 configuration.*2975

*This is the grounds state right here, the 2P6 3S1 configuration.*2983

*Its term symbol is a doublet S ½.*2988

*We know that, the S1 configuration is always a doublet S ½.*2993

*The primary quantum number is irrelevant.*2999

*We are going up to, we are going from 4D to 3P.*3003

*First of all, let us find the 3P if we can, it is right here.*3007

*The 3P configuration, we are going to have those two.*3013

*Let us look for the 4D configuration.*3022

*For D, we have that.*3024

*We have two possible levels in the D.*3027

*We have a doublet D 5/2 and a doublet P 3/2 and it could go down to a doublet a P 1/2 and doublet a P 3/2.*3030

*This is the transition is taking place.*3039

*The doublet D to singlet P represents one line.*3042

*The fine structure of that line shows this one going to this one.*3047

*The doublet D 5/2 to the doublet P ½.*3053

*The doublet P 5/2 to the doublet P 3/2.*3058

*The doublet P 3/2 to the doublet P ½.*3063

*The doublet P 3/2 to the doublet P 3/2.*3066

*There are 4 possible transitions.*3068

*The fine structure could show up to 4 possible lines.*3070

*Let us see which one of those lines, which of those transitions is actually allowed and which is not allowed.*3073

*Let us go ahead and do this one.*3082

*Our doublet D has doublet D 5/2 and it has a doublet D 3/2.*3088

*Our doublet P has a doublet P 3/2 and a doublet P ½.*3101

*There are 4 possible transitions.*3110

*Here to here, here to here, here to here, and here to here.*3119

*We have the doublet D 5/2 down to the doublet P 3/2.*3126

*I have got the doublet D 5/2 down to the doublet P ½.*3137

*I have got the doublet D 3/2 down to the doublet P 3/2.*3147

*Let me go ahead and put the other one right here.*3158

*I have got the doublet D 3/2 down to the doublet P ½.*3161

*Let us calculate some values here.*3166

*Here our δ L, for this one right here, our δ L = -1 DP.*3169

*Our δ S, doublet and doublet = 0.*3179

*Our δ J, 5/2 to 3/2 = -1.*3182

*This one is allowed.*3187

*Here we have a δ L = -1, we have a δ S = 0, we have δ J 5/2 to ½, it is -2.*3190

*This is not allowed.*3199

*Here we have δ L is equal to -1, δ S = 0 and we have δ J is equal to 0.*3202

*Yes, this one is allowed.*3212

*Here we have δ L = D to P-1, δ S = 0, and δ J = 0.*3213

*This one is allowed.*3222

*The 4 possibilities that is allowed, that is allowed, and that is allowed.*3224

*This transition is not allowed.*3231

*We are going to see 3 lines in the fine structure of this particular transition.*3233

*The rest is just working out the numbers.*3239

*I'm going to go ahead and do the doublet D 5/2 down to the doublet P 3/2.*3243

*That is going to be 34548.729 - 16973.366 = 17575.363 inverse cm.*3254

*That is where I’m going to see one of my lines.*3276

*The doublet P 3/2 down to the doublet P 3/2, we are going to have 34548.764 - 16973.366 = 17575.398 inverse cm.*3281

*That is going to be the second of the 3 lines.*3318

*Of course, the last one is doublet D 3/2 down to a doublet P ½.*3323

*That is going to be 34548.764 - 16956.170.*3332

*I think I hate most about quantum mechanics is the computational part.*3344

*The theoretical part in quantum mechanics is great, all the symbols not these tedious calculation.*3348

*But it is a necessity.*3354

*= 17592.594 inverse cm, this is that line.*3357

*The fine structure of this particular transition will show 3 unevenly spaced lines, the differences between these numbers.*3369

*This and this, and this and this, it is not even.*3378

*You will get something that looks like that, the separation.*3383

*That is it, there we go.*3388

*Thank you so much for joining us here at www.educator.com.*3391

*We will see you next time.*3393

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