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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 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

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 and0597

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, 4P2536

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