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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

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 this3236

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