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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: Calculate the Frequencies of the Transitions 0:09
• Example II: Specify Which Transitions are Allowed & Calculate the Frequencies of These Transitions 22:07
• Example III: Calculate the Vibrational State & Equilibrium Bond Length 34:31
• Example IV: Frequencies of the Overtones 49:28
• Example V: Vib-Rot Interaction, Centrifugal Distortion, & Anharmonicity 54:47

### Transcription: Example Problems II

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

Today, we are going to continue our example problems in molecular spectroscopy.0004

Let us jump right on in.0008

Our first example is the force constant for carbon monoxide is 1857 N/m.0010

The equilibrium bond length = 11.83 pm.0019

Using the rigid rotator harmonic oscillator approximation,0025

construct a table of energies for the first 5 rotational states for the vibrational levels R = 0 and R = 1.0029

The 5 rotational levels for R = 0, 5 rotational levels for R = 1.0039

Specify which transitions are allowed and calculate the frequencies of each transition,0043

associating each with its appropriate branch R or P.0049

Let us see what we can do.0054

Let me go ahead and work in blue today.0056

Under the rigid rotator harmonic oscillator approximation,0061

our energy equation is going to be energy RJ is equal to ν sub 0 R + ½ + this rotational constant × J × J + 1.0065

That is it, it is the vibrational energy + the rotational energy.0083

We have to find ν sub 0 that, and this, before we can actually start running J from 0 to 5.0088

Let us see what we can do.0107

Ν sub 0, the definition is 1/2 π C × the force constant divided by the reduced mass ^½.0109

We have to find the reduced mass.0124

The reduced mass is equal to the product of the weights divided by the sum of the weights.0128

12 × 16, 12 for carbon 16 for oxygen, divided by 12 + 16.0135

This is in atomic mass units, I’m going to multiply that by 1.661 × 10⁻²⁷ kg per atomic mass unit.0142

My atomic mass units cancel and I'm left with a reduced mass in kg.0153

When I do this calculation, I should get,0159

I hope that you are confirming my arithmetic, I am notorious for arithmetic mistakes.0161

The arithmetic is less important, the process is important.0167

The equations that you choose, that is what is important.0171

I may clear my throat a few more times than usual during these few lessons today.0176

× 10⁻²⁶ kg.0181

I want to keep writing these numbers over and over again.0189

When I go ahead and put this value of K into here, when I put this value ν into here and run this calculation,0192

I'm going to get a ν sub 0 equal to 214354.49.0201

The answer is actually going to be in inverse meters because this is N/ m.0212

When I convert that, I will just move that decimal over a couple of times,0218

I'm going to get 2143.54 inverse cm.0221

The problem is, the issues with spectroscopy are like many issues throughout quantum mechanics and thermodynamics.0228

They are just for conversion issues, just watch your units.0234

That is really all you have to watch out for.0239

We have that number.0242

Let us go ahead and find the rotational constant.0248

Let me go ahead and do that on the next page here.0251

Our rotational constant is equal to, the definition H/ 8 π² C × the rotational inertia.0255

We need I to put it into here.0267

I = that, that².0273

Therefore, I = R 1.139 × 10⁻²⁶ kg × RE is a 112.83 × 10⁻¹² m.0277

What we get is I = 1.45 × 10⁻⁴⁶ kg/ m², the unit of rotational inertia.0299

We put this value in here with all of the other values.0314

I will go ahead and write this one out, it is not a problem.0323

E = 6.626 × 10⁻³⁴ J-s divided by 8 π² × 2.9980328

× 10⁸ m/ s × 1.45 × 10⁻⁴⁶, that is the I.0344

When I do that, I will get 193.04 but this is going to be in inverse meters because a joule has meters.0355

That is going to give us, when I convert this, move the decimal over twice 1.9304 inverse cm.0367

Remember, inverse cm and inverse meters, it is the other way around.0376

It is going to be 100 inverse meters in 1 inverse cm, which is why this one goes to the left.0381

We have that and we have that.0390

We said that the energy was, let me go ahead and rewrite the equation.0392

The energy was E= sub RJ equal to ν sub 0 × R + ½.0397

This is the rigid rotator harmonic oscillator approximation + B ̃ J × J + 1.0406

Let me make my J a little more clear here.0417

For the R = 0 vibrational state, that is going to be E⁰ and J is going to run through these quantum numbers.0420

That is going to equal, when I put R and 0 into this equation, it becomes ½ ν sub 0 + B ̃ × J × J + 1.0435

And for the R = 1 vibrational state, I just put 1 into this equation and I get the E sub 1 J is equal to 3/2 ν sub 0 + B ̃ J × J + 1.0448

Let me double check that.0464

Therefore, our E0 J, when I actually put the ν 0 that I got and the B that I got in here,0469

I'm going to get equation that says 1071.77 + 1.9304, which is the rotational constant, × J × J + 1.0478

Now, I just set up a table of values.0490

I have J over here and I have my energy 0 J over here.0492

I’m going to take 0, 1, 2, 3, and 4.0498

I get 1071.8, I get 1075.6, I get 1083.4, I get 1094.9, I get 1110.4.0502

These are energies not frequencies of absorption or emission.0524

These are not spectral lines, these are not frequencies.0528

These are the energies of the actual rotational states in the R = 0 vibrational state.0530

Let us go ahead and do the energies for 1 vibrational state.0539

When I put the values in, I get 3, 2, 1, 5.0546

In other words 3/ 2, the ν sub 0.0549

3215.31 + 1.9304 which is the rotational constant, × J × J + 1.0551

Let us go ahead and set up my table of values again.0570

This is going to be E1 J.0573

I have 0, 1, 2, 3, and 4.0576

I have 3215.3, I have 3219.2, 3226.9, 3238.5, and 3253.9.0583

Let us go ahead and let me do it on one page here.0607

Let me see if I want to just go ahead and do this one in red.0614

I got J and I got E0 J.0617

Over here, I have J and I have E1 J.0623

I have 01234, 01234.0630

These are the first to 5 table of values.0636

I had 1071.8, 1075.6, 1083.4, 1094.9, and 1110.4.0639

Over here, I have 3215.3, 3219.2, 3226.9, 3238.5.0657

Again, I want them on the same page so I can see them together.0673

3253.9, the energies of the 0 vibrational state, the first 5 rotational for the 1 vibrational state.0677

The allowed transitions, δ J is + or -1.0690

The allowed transitions, the selection rule is δ J = + or -1.0699

Therefore, for the 0 state to the 1 state, the 0 vibrational state to the 1 vibarational state, I can go 0, 1, 2, 3, 4.0709

I have 0, 1, 2, 3, 4.0724

I can go from 0 to 1 that is + 1.0727

1 to 2, 2 to 3, and 3 to 4.0731

Let me do this one in black.0736

I can go from the lower 0, 1 to 0, to the 1, 3 to 2, 4 to 3.0738

Those are my allowed transitions.0746

The 0 state to 1 state, I can go 0 to 1, 1 to 2, 2 to 3, 3 to 4.0749

Or the 0 to 1 vibrational state, I can go rotational 1 to 0, 2 to 1, 3 to 2, and 4 to 3.0753

Those are my allowed transitions, the frequencies of those particular transitions.0762

The frequencies of these transitions, in other words the frequencies of the spectral lines of the R and P branches,0773

we have ν sub R branch is equal to ν sub 0 + 2B × J + 1, where J takes on the values 0, 1, 2, and so forth.0817

The P branch is going to be ν sub 0 -2B J.0835

Here, J takes on the values 1, 2, 3, and so on.0844

My ν sub R is going to be 2143.54 + 2 × 1.9304 × J + 1.0852

My ν sub P, those branches are going to be 2143.54 -2 × 1.9304 × J.0876

The R is 0 branch that represents the 0 to 1 transition, that was going to take place at 2147.4.0894

I will just put the J values in here.0904

J = 0, J = 1, J = 2, J = 3, J = 4.0905

That is the R0, let me be label it with the particular J value.0912

We put the R branch for J value 0 and it is going from 0 to 1.0919

When we do the P branch, it is actually going to be a P1 because it is going to be going from 1 to 0.0924

The R1, that represents the 1 to 2 transition.0932

The R2 represents the 2 to 3 transition.0937

The R3 represents the 3 to 4 transition.0941

These are the frequencies that we have.0946

We have 2151.3, we have 2155.1, and we have 2159.0948

We have the P1 transition, the P1 line, the P2 line, the P3 line, and the P4 line.0962

This one represents the 1 to 0 transition.0971

This one represents the 2 to 1 transition.0973

These are subscripts or the beginning energy level.0976

The energy level of the lower vibrational state.0979

For the lower rotational state, on the lower vibrational state.0984

In other words, the departure state not the arrival state.0991

I like that better, departure and arrival.0995

I think the mathematical terms are actually a lot better.0997

This one represents the 3 to 2 transition and this one represents the 4 to 3 transition.1003

These are 21392139.7, 2135.8.1012

These are going down, these are going up.1020

R branch increases, P branch decreases.1022

2132 and 2128.1027

There you have it, we found the energies, we found the allowed transitions.1035

These are the frequencies of the actual transitions that are allowed.1040

These are the labels for them representing this particular transition.1043

Let us go back to red here.1048

Be very careful to distinguish, especially on test.1054

On a test, you are going to be stressed out and your mind is going to be racing.1061

You need to really slow down when dealing with spectroscopy because they can be asking about energy,1065

Those are not the same.1072

Be careful to distinguish between the energy of a given level and the energy of transition,1074

the energy of transition between two energy levels.1113

The transition between two energy levels that is what we see in spectra are the transition energies from one level to another.1129

One final thing to notice, we found the energy levels for the E sub 0 J and E sub 1 J.1161

We used equations, the ν sub R, ν sub P to actually find those frequencies.1173

What you could have just done is take the difference between energies in the table.1179

Notice that the frequencies of transition,1184

instead of using equations for ν R branch and ν of the P branch,1203

we could have just taken δ E between the two vibrational states1235

for the appropriate J values.1262

For example, if I wanted the R1 line,1271

The R0 line that represents the transition from 0 to 1.1280

We have the energies already in the table.1289

We could have just taken the energy of 1 1 - the energy of 0 0.1290

This is the vibrational number, that is the vibration number.1304

Upper lower, the transition going from the 0 to 1 transition.1308

Upper lower, we could have just done that, just subtract the two values.1316

I will do that to the problem a little bit later on in this lesson.1319

Let us see what is next.1325

Using the table of parameters below and equations which correct for anharmonicity and vibration rotation interaction,1330

construct a table of energies for the first 5 rotational states of hydrogen iodide, for vibrational levels R = 0 and R = 1.1344

Specify which transitions are allowed and calculate the frequencies of these transitions,1353

associating each to its appropriate peak in the R and P branches.1356

The same sort of thing except now we need to make adjustment to the rigid rotator harmonic oscillator1361

and we are correcting for anharmonicity and vibration rotation interaction.1367

Notice, we are not correcting for centrifugal distortion.1371

We can always correct for all three, for two of those things, or for one of those things.1374

It just depends on what the problem is asking for.1378

We have our table of parameters, our rotational constant, our Α sub E,1381

our fundamental frequency and our correction for anharmonicity.1389

We said that the energy for the rigid rotator harmonic oscillator approximation1399

that was E sub RJ = ν sub 0 × R + ½ + B × J × J + 1.1405

This is the rigid rotator harmonic oscillator approximation.1417

It does not really matter.1422

The dependence of B on R, B sub R is equal to B sub E - Α sub E × R + ½.1424

This thing goes in where B is, that is the adjustment for vibration rotation interaction.1438

The adjustment for harmonicity, the energy is ν sub E × R + ½ - this X sub E ν sub E × R + ½².1446

This is the adjustment for anharmonicity.1465

Now when we put this together, when we put both of these into this,1469

we get a new equation for the energy which is RJ is equal to,1474

It gets a little complicated, a little long, no doubt about that.1482

R + ½ - X sub E ν sub E × R + ½² and the rotational term which is going to be B sub E - Α sub E × R + ½ × J × J + 1.1487

This is the ν for the two corrections that we have to make.1510

Notice again, this problem does not ask us to account for centrifugal distortion.1516

If it did, we have one more – that D thing, D × J² J + 1², but that does not matter here.1520

For R = 0, we get E of 0 J is equal to ½ ν sub E -1/4 X sub E ν sub E + this B sub E - Α sub E.1530

Again, these are just a bunch of parameters that we just have to account for.1551

It is not terribly a big deal.1554

R is 0 here so we can just go ahead and put the -1/2 Α sub E × J × J + 1.1557

This ends up being E sub 0 J is equal to 1154.1569

I will just put the numbers in.1577

In other words, I will put the ν sub E in, we put the X sub ν E in, the BE, the Α E, we just put it in.1579

That is it, nothing strange.1586

9.911 + 6.4275 × J × J + 1.1588

For the R = 1 vibrational state ,we get E of 1 J that is going to equal 3/2 ν sub E - 9/41601

X sub E ν sub E + B sub E – 3/2 Α sub E × J × J + 1.1614

And we end up as far as numerically is concerned, we end up with 3463.521 - 8189.199.1629

This is nothing more than just a bunch of tedious arithmetic, that really is L comes down to.1645

+ 6.2585, which is why I personally love theory as opposed to the tedium.1650

That is why we work with symbols, you do not want to do those arithmetic.1662

× J × J + 1.1665

This equation right here, let me go to blue, we use this equation and we use this equation.1669

We create a table of values.1676

We have J, J is going to take on the values 0, 1, 2, 3, 4, that is the first 5 states.1680

E sub 0 J, all the values are in inverse cm.1689

E1 J we have 1144.6, 1157.5, we have 1183.2, we have 1221.7,1695

we have 1273.1, we have 3374.3, we have 3386.8, we have 3411.9, 3449.1713

I completely understand, this is not something that you actually get used to.1735

After a long time doing this, you lose your mind from what are all these numbers.1739

3449.4 and this is going to be 3499.5.1751

The allowed transitions are δ J = + or -1.1760

We are going to have the 0 to 1, 1 to 2, 2 to 3, 3 to 4.1767

Or we are going to have the 1 to 0, 2 to 1.1773

Basically, all you have to do when you need to find the actual frequencies of the absorption,1777

the frequencies of the spectral lines, you are just going to take for example the 0 to 1 transition,1783

you are just going to take this number - this number.1788

The 1 to 2 transition, this number - this number.1793

You do not actually need the equations, which is why the energy equation is really the most important.1796

If you have that, you can get everything else.1801

The δ J = + or -1, when we do that, let us go ahead and create a new table of values here.1805

Once again, let us say δ J = + or -1.1815

The R0, R1, R2, R3, that represents the 0 to 1 transition, this is the 1 to 2 transition,1823

this is the 2 to 3 transition, this is the 3 to 4 transition.1834

This is going to be at 2242.2.1838

It is just the second number the second column - the first number the first column.1842

The energy of the 1 - the energy of the 0.1848

The energy of the 2 - the energy of the 1.1852

For the two vibrational states.1854

We do not necessarily need equations for Ν sub ρ and ν sub P.1858

2254.4, 2266.2, 2277.8, and we have the P branch.1863

This is going to be P1, P2, P3, P4.1875

This represents the 1 to 0 transition.1880

This represents the 2 to 1 transition.1883

This represents the 3 to 2 transition.1886

This represents the 4 to 3 transition.1888

This is going to be 2216.8.1893

This is going to be 2203.6.1897

This is going to be 2198.2.1901

And this is going to be 2176.3.1905

I really like the black.1918

If we wanted specific equations for ν sub ρ and ν sub P, basically the observed frequencies.1920

These numbers, we have to do some algebra, that is the problem.1934

We would have to do a lot of algebra actually.1945

We have to do a lot of algebra.1952

Essentially, what you have been doing is the following.1954

For ν sub R, you take the energy of the upper states and subtract the energy of the lower state.1959

That is what you are doing.1963

You are going to end up with, it is going to be the energy E R + 1 J + 1 - ERJ.1964

You have to do all that and come up with some final equation.1975

For the P branch, that is going to end up being E R + 1 J - 1 ERJ.1978

Based on the equation above, that long equation that we had for the energies,1990

you have to take one for the upper state then subtract that long equation, one for the lower state.1994

Go through all the algebra and come up with some equations, which is essentially what you see in your books.1999

Most of what you see in your books, the equations, the derivations of the spectroscopy section are just that.2004

They are just taking one energy and subtracting other energy.2010

It tends to look complicated but it is just a lot of algebra.2013

In order to come up with equations for the spectroscopic lines that we see,2016

based on whatever correction we are making it.2021

It is either going to be the harmonic oscillator rigid rotator approximation for the energy.2023

You can either make a correction for vibration rotation interaction.2029

You can make a correction for anharmonicity.2033

You are going to make a correction for centrifugal distortion.2035

Either 1, 2, or all 3 of those.2038

The equation is longer for every correction that you make.2041

You remember all this from the first lesson, the initial lesson that we did in spectroscopy2046

where all these things are coming from.2050

That it is not this ocean of equations that you are dealing with.2053

It is just the energies of the upper state - the energies of lower state that give you the frequency of the spectral line.2057

Let us do some more examples here.2068

I think we are on example number 3 now.2070

Wait, I think we are missing some data here.2078

We do not have this data but that is not a problem.2084

I actually have it right here with me, I will go ahead and write it.2086

For some odd reason, it did not end up on this page, my apologies.2088

Calculate B sub 0, B sub 1 of the equilibrium bond length for the R = 0 vibrational state and2091

the equilibrium bond length for R = 1 vibrational state, the B sub E and Α sub E for carbon monoxide.2100

I apologize, it did not show up on this slide.2118

Ν observed, we observed an R 0 line of 2173.81.2122

We observed R1 line at 2177.58.2132

We observed a P1 line at 2166.15.2140

And we observed a P2 line at 2162.27.2146

Given this data for two R lines and the first two lines of the P branch, how can I calculate all of these?2153

We see B0 and B1, that equations I want to deal with are going to only account for the vibration rotation interaction.2162

I do not see anything else here.2184

I do not see any X sub E, ν sub E.2185

I do not have to worry about the anharmonicity.2187

I do not see a D here so I do not have to worry about centrifugal distortion.2190

I see B sub 0 B sub 1, the equations that I’m going to be looking at,2193

I'm going to be looking at the rigid rotator harmonic oscillator corrected for vibration rotation interaction.2197

That is how you choose your equation.2203

You take a look at what constants are at your disposal, what they want, and I will tell you what to choose.2205

We see B0 and B1, we consider only the vibration rotation correction.2211

I always like to begin with the basic equation and then add to that.2234

It is just a nice way to keep sort of refreshing yourself and see that the equation over and over and over again.2238

Our harmonic oscillator rigid rotator equation E sub RJ, the energy is ν R + ½ + B × J × J + 1.2245

I do not need that parenthesis at J × J + 1.2263

The vibration rotation correction, that one says that this B, the rotational constant depends on R.2267

B sub R is equal to B sub E - Α sub E × R + ½.2279

This we put into here and get a new equation.2289

The corrected equation for the vibration rotation interaction is equal to,2297

Sometimes, I will put a tilde over the E, sometimes not.2302

It is in wave numbers, do not worry about it.2305

That is equal to, it is going to be ν R + ½2308

+ this thing B sub E - Α sub E R + ½ × J × J + 1.2323

Unless specifically stated otherwise,2338

always take the R = 0 to the R = 1 vibration transition.2357

That is the one that you want to take.2365

At normal temperatures, room temperatures, most molecules are actually going to be in their ground vibrational state.2368

They are also going to be in their ground electronic state.2374

However, at normal temperatures, it is the rotational states that molecules tend to be the higher states,2378

they cannot be the ground state.2384

J = 0, they can be anywhere from like J = 2 or 3, above that.2386

At normal temperatures, most molecules are in the vibrational ground state and electronic ground state.2393

It is only the rotational state that there are in excited states at normal temperatures.2401

We will talk more about that when we do statistical thermodynamics next.2405

When we look at this, our new observe, what we are looking for is this.2419

We need an equation that is going to be the energy of the upper state - the energy of lower state.2423

The energy being the equation that we just had.2433

Our R branch is going to be E of 1 J + 1 - E of 0 J.2435

Our ν sub P is going to be E of 1 J-1 E0 J.2448

When I do the algebra for these and I put the energy equation that I just had in the previous page,2459

one for the upper state and one for the lower state.2464

When I work out that algebra, here is what I get.2466

When we work out these differences, we get a rather daunting looking equation actually.2475

Ν sub 0 + B1 - B0 J² + 3 B1 - B0 J + 2 B1.2491

The J takes on the value starting from 0, 1, 2, and so on.2512

For the P branch, I get ν sub 0 + this same term B1 - B0 J².2518

The third term is different, this is going to be B1 - B0 × J.2529

Here, the J takes on the values of 1, 2, 3.2540

And again, for the R branch 0, so on.2543

The P branch is 1 and so on.2545

Let me go to blue here.2561

The R is 0 line that represents the J = 0, that is going to be ν sub 0 + 2 B1.2565

In other words, I put J = 0 into this equation and I see what I get.2577

I get ν sub 0 + 2 B1.2584

The data gave us the frequency of absorption, the ν sub 0, the R sub 0 line is at 2173.81.2588

The R1 line that is where J is equal to 1, that is ν sub 0 + 6 B1 – 2 B0 is equal to 2177.58.2602

The P sub 1 line that represents that J = 1.2622

I’m going to use this equation, the one for the ν sub P.2627

I get ν sub 0 -2 B0 = 2166.15 inverse cm.2631

For the P2 line that is the J = 2, I get ν sub 0.2642

I get + 2 B1 -6 B0.2647

And that one was observed at 2162.27.2654

These are the equations that I’m going to work with now.2660

I’m going to take the R1 equation - the P1 equation.2668

When I do that, I get 6 B1 = 11.43.2674

Therefore, B1 is equal to 1.905 inverse cm.2684

I found B1.2694

Now, I take the R0 line and I subtract from it the P2 line.2696

When I do that, I get 6 B0 = 11.54.2701

Therefore, B0 is equal to 1.923 inverse cm.2711

We found B1, we found B0.2722

BR, this B sub R that we just found is equal to H/ 8 π² C × I.2726

I is the reduced mass × this, whatever this is².2739

Now, we need to find the equilibrium bond length for each of these.2744

When I put the values in, I will get the following.2751

When I rearranged this for RE², R sub E vibrational state 1² is equal to 6.626 × 10⁻³⁴.2754

I’m not going to put the units in and I’m just going to put the numbers in.2771

8 π² 2.998 × 10¹⁰.2774

I went ahead and use cm directly.2782

1.139 × 10⁻²⁶ that was the reduced mass and then the 1.905 that is this number right here.2786

This goes down here, this comes up here.2797

When I solve for this, I get that it is equal to 113.6 pm.2800

The same for R0, when I do the same for R0, R sub E for the 0 state, I end up with a value of 113 pm.2812

We found RE 1, RE, we found B0, we found B1.2832

Let us go ahead and find for B sub E and Α sub E.2836

We said that b sub R is equal to B sub E – α sub E × R + ½.2850

We know what B0 is.2862

B0 is equal to B sub E – ½ Α sub E.2866

We know what B1 is, let us put that 0 into here, 1 into here for R.2874

We get B sub E - 3/2 Α sub E.2881

We know what B0 is, we already found it 1.923.2888

1.923 = B sub E – ½ Α sub E.2892

And we know that 1.905 which is B1, that = B sub E - 3/2 Α sub E.2903

Two equations and 2 unknowns.2914

Let us go ahead and take, this is equation 1 and this equation 2.2916

Let us go ahead and take equation 1 - equation 2.2921

When you do that, you end up with Α sub E is equal to 0.018 inverse cm.2928

Put this into one of these other equations and you end up with B sub E is equal to 1.932 inverse cm.2944

There you go, I hope that make sense.2957

Let us see what is next.2962

The IR spectrum of ML shows a fundamental line at 1877.62 inverse cm and the first overtone at 3728.66 inverse cm.2970

Find the values of ν sub E and X sub E ν sub E for ML.2983

Let us go ahead and work in black here.2993

The frequencies of the overtones for the anharmonic oscillator2998

are given by G of R – G of 0, just looking at vibrational transitions right now.3027

G of R to G of 0, the anharmonic oscillator.3037

Under normal temperatures, most molecules are in the ground vibrational states.3043

They are all going to start at the 0 vibrational state.3046

But δ R is no longer + or -1.3050

It can be + or -1, + or -2, + or -3.3053

The + or -1 is the fundamental.3056

The + or -2, those are the first overtone.3059

The first overtone, second overtone, third overtone.3062

The G of R equation that is equal to ν sub E × R + ½.3066

The anharmonic oscillator – X sub E ν sub E R + ½².3076

When we take the G sub R – G of 0, when we do the algebra,3078

what we get is the observe line equal to ν sub E × R – X sub E ν sub E × R × R + 1.3093

R1 runs from 1, 2, 3, and so on.3109

The fundamental transition is the transition from 0 to 1, that is the one that we see.3117

That is the brightest one that we see.3131

That is equal to, I put into this equation, fundamental 0 to 1.3135

I put R = 1, it is going to be ν sub E × 1 - X sub E Ν sub E × 1 × 1 + 1.3142

We end up with ν sub E - 2 X sub E ν sub E.3157

It tells us what that is already.3178

The fundamental is 1877.622.3179

1877.622 that is one of the equations that we are going to use.3184

The first overtone represents the transition from 0 to 2.3191

It is a weaker line.3198

That is going to be ν sub E × 2.3202

Just using this equation right here, taking care of the R = 1.3208

Now the first overtone R = 2 - X sub E ν sub E × 2 × 2 + 1.3214

What we end up with is 2 ν sub E – 6 X sub E ν sub E, it tells us what the first overtone is 3728.3227

3728.66 that is the second equation that we are going to use.3240

When you solve simultaneously, I will not go through the process.3247

I will let you go ahead and do that.3249

This equation and this equation, when you solve simultaneously,3250

you are going to get a value of ν sub E is equal to 1904.20 inverse cm and X sub E ν sub E is equal to 13.29 inverse cm.3253

That is it, nice and easy.3274

I know it is not nice and easy, believe me I do.3279

I have been there, the stuff still gives me grief too.3282

Let us take a look at our final example.3289

I’m actually wondering whether I should leave this example off because the first two lessons took care of it.3291

Of course, we had an example from the previous lesson that did this but we will see.3297

In examples 6 of the previous lesson, we asked you to calculate the frequencies of the first two lines of the R and P branches3304

for the vibration rotation spectrum of HBR under the harmonic oscillator rigid rotator approximation.3312

Here we ask you to repeat the problem.3320

But instead of the harmonic oscillator rigid rotator approximation,3323

we asked you to do so by accounting for all three corrections we made to the HIR equations.3326

We accounted for vibration rotation interaction for centrifugal distortion and for harmonicity.3335

Use the following spectroscopic parameters as necessary.3340

This is a great example because we get a chance to put everything together,3343

the HRR approximation and all of the corrections that we made.3350

I’m not going to write all the equations here.3367

I can do so in the last couple of lessons, basically what we are going to be doing is,3368

in this one you are going to take the harmonic oscillator approximation.3372

You are going to make the corrections for all three of these things.3378

The vibration rotation interaction, the centrifugal distortion, and the anharmonicity.3381

You got to find an equation for the total energy.3385

You are going to take the upper level - the lower level.3388

In other words, you are going to find δ E.3391

That is going to give you your equation.3395

You do not necessary have to go through the algebra, the equations are actually have been done for you.3398

They are in your books, I think we actually did in our lessons too.3402

When you do it, the equations that you come up with are the following.3404

Ν sub R, that is going to be ν sub 0 × 1 – 1.3411

2648, we are going to have ν sub E × 1 - 2 XE + 2 × BE × J + 1 - Α sub E × J + 1 × J + 3 - 4 D × J + 1³.3434

The J values are going to run from 0, 1, 2, and so on.3465

This is the equation that you get when you take the total energy,3471

the correction for all three things and the energy of the upper state – the energy of the lower state.3475

It is a lot of algebra but this is the equation you come up with.3482

This is the frequency of the spectral line that we should see.3484

The ν sub P branch that is going to be ν sub E × 1 - 2X sub E -2 B sub E × J - Α sub E × J × J - 2 + 4D J³.3492

Here J = 1, 2, and so on.3525

These are the two equations that we actually get.3529

Notice the equation for the spectral line changes depending on what corrections we are making.3532

In this case, we corrected for all three.3537

They give us ν sub E, we have that one.3542

We are going to need that parameter.3544

We are going to need this parameter.3546

We are going to need this parameter.3549

We do not need those others.3554

We also need the X sub E but they gave us N sub E ν sub E.3555

Actually, you have to take this.3559

We would need that.3561

We would have to take this parameter divided by that, in order to get just the X sub E.3561

Let us go ahead and do that first.3567

The data gives ν sub E and X sub E ν sub E.3570

Let us find X sub E alone.3586

The X sub E is equal to X sub E ν sub E divided by ν sub E.3591

We end up with 45.217 divided by 2648.975 and we end up with X sub E = 0.01707 inverse cm.3599

We put all these parameters into the equations.3617

What we end up with, ν of the R sub 0 line.3621

That is going to equal 2648.975 × 1 - 2 × 0.01707 + 23636

× 8. 465 × 0 + 1 - 0.2333 × 0 + 1 × 0 + 3 - 4 × 3.457 × 10⁻⁴ × 0 + 1³.3653

When I calculate that, I end up with 2574.768 inverse cm.3683

This is my R0 line.3693

I should see it there.3696

Notice, we expected something lower than what we get with the rigid rotator harmonic oscillator approximation.3699

Exactly what we got.3706

Now, when I do the same for the R1 line, I will write all this out.3708

What I end up with is 2590.522 inverse cm.3714

Notice the spacing between the lines.3724

The spacing between the spectral lines is this line - this line.3728

This line - this line, the absolute value there.3729

The spacing is ν of R1 - ν of R0, that is equal to 15.755 inverse cm.3735

2B E = 2 × 8.465 that is equal to 16.93 inverse cm.3750

You see that the spacing, the corrections that we have made.3763

On the actual spectrum, the spacing is less than 2B.3766

2B is 16.93.3769

The corrections that we made give us the spacing which is about less than 2B, which is exactly what we expect.3771

For the R branch, the spacing is smaller.3780

For the P branch, the spacing will actually be bigger than 2B.3783

The 2B is the rigid rotator harmonic oscillator approximation.3787

Let us go ahead and do the P branch.3792

I think I have one more page, If I’m not mistaken, yes I do.3794

I got ν of P1 and when I put into the equation for the P1, I end up with 2541.8.3801

Let me write this one, at least.3822

It is going to be 2648.975 × 1 - 2 × 0.01707 - 2 × 8.4653824

× 1 - 0.2333 × 1 × 1 - 2 + 4 × 3.457 × 10⁻⁴ × 1³.3844

This will give me 2541.843 inverse cm.3863

And when I do the same for the P2 line, I end up with 2524.690.3870

Now, the difference here, the δ ν.3885

In other words, one of them - the other, that is going to equal 17.153, which is greater than 2B,3891

which is exactly what we expect for the P branch.3903

That is it, thank you so much for joining us here at www.educator.com.3910

We will see you next time.3913