Physical Chemistry Vibration-Rotation

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

Today, we are going to start on the next major broad topic of Physical Chemistry which is molecular spectroscopy.0004

We have done the thermodynamics, we have done the quantum mechanics, and now, we are going to bring0013

the quantum mechanics to bare and talk about probably the most important topic for practicality0016

is concerned for the chemist because pretty much everything you do as a chemist is going to be some spectroscopic technique.0024

We will be discussing the theory behind molecular spectroscopy, let us get started.0031

Molecular spectroscopy studies the interaction between a matter and electromagnetic radiation.0039

Molecular spectroscopy studies the interaction of molecules with electromagnetic radiation.0053

We get molecules with certain amount of radiation in different regions of the electromagnetic spectrum and we see what happens.0079

That information gives us most of the information that we have about what is happening inside the molecules and how molecules behave.0085

We will be concerned with three regions of the E and M spectrum.0095

We are going to be concerned with the microwave region.0115

Microwave region is related to rotations.0120

When a molecule is hit with microwave radiation, there are changes in the rotational quantum state of the molecule.0125

And we are also going to be concerned with the infrared, the infrared vibrations.0133

When a molecule is hit with some infrared radiation, changes in the vibration levels of the molecules take place.0138

We have already done a fair amount of IR spectroscopy from your course in organic chemistry.0147

And then of course, the last is visible and ultraviolet.0152

Visible and ultraviolet range electronic transitions, that is when it is in this range,0157

the energy of the visible and ultraviolet range, where electrons actually move to higher states themselves, electronic states.0165

Rotation vibration electronic, rotational spectroscopy, vibration spectroscopy, and electronic spectroscopy.0175

Let me see how I want to do this.0184

The absorption of microwave radiation, as we just said, we have rotational transitions.0202

Now in the IR, not only do we have vibrational transitions but accompany those vibrational transitions there are also rotational transitions.0212

I’m going to say vibrational + rotational transitions.0224

As you would expect in the visible UV range, we have not only electronic transitions0232

but there is much energy there that that energy causes an electronic transition,0240

it causes vibrational transitions and rotational transitions.0244

+ vibrational + rotational transition, all transitions take place when you are hitting it with energy and the visible UV range.0249

Most of the information that we get from spectroscopy, we actually get from electronic spectroscopy.0263

Electronic spectroscopy allows us to, it is difficult to analyze but everything that we need is there.0268

It gives us information on electronic states, on vibrational states, on rotational states.0275

The rotational spectra, the vibrational spectra tend to be easier, but they do not give is as much information.0280

Electronic spectra give us all the information that we want.0288

Let us see, the frequency of the absorb radiation comes from this δ E = H ν,0297

Planck's constant × the frequency is a change in energy from one energy state to another.0318

If we solve for the frequency, what you would end up with is the frequency of absorption is going to equal0324

the change in the energy between the 2 states divided by Planck's constant,0330

or we can say the energy of the upper state - the energy of the lower state divided by Planck's constant.0335

That gives us the actual frequency that we observe in the spectra.0345

That is what we are going to see on the spectrum.0348

When we see a line of the spectrum or a peak, it is this number right here.0350

That is what that is, it come from this relation.0356

What we are going to do, we would use the energies for the rigid rotator, the harmonic oscillator,0359

and electronic energies, to find this difference.0365

Because we want to find the equation for what is the absorption number.0369

In general, spectroscopy frequencies, I will put in parentheses.0376

Frequencies are listed in something called wave numbers, we have seen them before.0387

Frequencies are given in wave number.0391

That is what we would be working in,0394

are given in wave numbers which is just the inverse of the wavelength to inverse cm.0396

In general, it is going to be inverse cm.0404

The wave number, anything that is in a wave numbers is going to have a ~ over it.0408

That is equal to 1/ λ, or my preference it is equal to the actual frequency0412

that we got from the other equation just divided by the speed of light.0420

If you get a frequency, if you divide that frequency by the speed of light, you are going to end up getting your wave number.0424

We said transitions between vibrational states are accompanied by transitions and rotational states as well.0431

The following discussion is going to apply to diatomic molecules.0480

We will discuss polyatomic molecules but for right now,0484

we are just going to talk about diatomic molecules, homo nuclear and hetero nuclear.0487

The following discussion and for several more lessons, the following discussion applies to diatomic molecules.0495

Let us begin with the energy of the harmonic oscillator.0514

The energy of the harmonic oscillator HO is energy, that space on the quantum number R,0518

the vibrational quantum number is equal to H ν × R + ½, or R takes on the values 0, 1, 2, and so on.0532

When R is equal to 0, we have ½ H ν, that is the energy of the ground state.0545

When the quantum number is 0, put it in the equation for the energy which gives the ground state energy.0551

We see that there is always some vibrational energy, it is never a 0.0557

Where ν, this frequency, is equal to 2 π × the force constant of the molecule divided by the reduced mass ^½.0564

And this ν is called the fundamental vibration frequency.0583

You will sometimes see it as ν sub 0, something like that.0604

K is the force constant of the bond and how springy the bond is.0611

Is it really tight or is it really loose?0621

And μ is the reduced mass, we have seen the reduced mass before.0625

I just want to make sure that we understand what all of the parameters are.0632

There are selection rules for vibrational transition.0638

The selection rules for vibrational transitions and the selections rules are that δ R = + or -1.0642

In other words, if it is going to make a transition from one vibrational state to another,0660

it is going to go from 1 to 2, 2 to 3, 3 to 4, or 4 to 3, 3 to 2, 2 to 1.0664

In the case of emission, absorption up emission down, as far as the harmonic oscillator is concerned,0669

it is only going to go 1 or 2 steps.0677

I’m sorry, 1 step up or 1 step down.0681

Its not going to go from 1 to 5, from 1 to 4.0683

If it is going to make a transition from 1 to 5, it is going to pass to 2, 3, 4, and then to 5.0685

Those are the transitions that are allowed.0691

The other transition rule is the dipole moment of the molecule must vary during a vibration.0694

I would not worry too much about this idea of the selection rule.0715

Now the dipole moment is related to the selection rule, this is often called gross selection rule.0722

The name itself does not really matter.0728

The dipole moment of the molecule must vary during a vibrational transition.0729

This does not mean that the molecule has to have a permanent dipole.0737

It can or cannot have a permanent dipole, in the case of molecule like hydrogen chloride, it has a permanent dipole.0741

In the case of a molecule like N2, they are both the same.0748

It is homo nuclear so there is no permanent dipole.0752

Now, these are actually not the best examples to use but it has to change during the vibrations.0755

It does not have to have a permanent one.0764

But if the molecule, if somewhere during the vibration the dipole shows up then it is capable of a vibrational transition.0766

The one that we would be concert with most of all, is the changes in the quantum numbers itself.0779

In this case, δ R = + or -1, but it is good to know this because some of your classes depending on what they are going to cover,0783

what they are not going to cover, they may actually go into the mathematics behind this.0791

I do not know, we ourselves are not.0794

We are going to be concerned more with just the spectroscopic aspects.0795

As we said earlier, spectroscopy is conducted primarily in wave numbers.0801

Spectroscopy is conducted in inverse cm, in wave numbers.0817

We want to take this equation for energy and convert them to wave numbers.0825

For vibration, the symbol that we use is G.0832

G is a function of R, it is equal to the energy of R that we had from the previous page divided by HZ.0837

The energy that you have, the energy that is given in Joules divided by Planck's constant divided by the speed of light.0847

Any energy in Joules divided by those by HC is going to give you the number in inverse cm.0856

This is called the vibrational term and this is what you will see in the literature.0861

This G of R called the vibrational term and it is just a symbol for the vibrational energy expressed in terms of wave numbers,0868

expressed in inverse cm as opposed to joules or anything else.0882

G as a function of R is equal to, the E of R is H ν R + ½.0884

We will go ahead and divide by HC.0895

The H will cancel, what we end up with is this ν ~.0898

It is the frequency expressed in wave numbers, R + ½.0903

Everything else is the same, R is going to take on the values of 0, 1, 2, 3, and so on.0908

The reason is here we are left with N/ C.0916

That is just equal to N/ C.0921

G of R is equal to ν ~, the fundamental vibration frequency expressed in wave numbers, × R + ½,0930

where ν ~ is equal to this ν/ C, which is equal to 1/ 2 π C.0944

That is it, just a little bit of mathematical manipulation.0955

This is the important thing right here.0958

This is what we have, this gives us the energy of a particular vibrational state depending on the quantum number R.0961

Let me go ahead and put R = 0, 1, 2, and so n.0968

It gives the energy of that particular vibrational state expressed in inverse cm.0974

That is all this.0979

That is the energy of the harmonic oscillator.0983

Let us go ahead and talk about the energy of the rigid rotator.0986

The energy of the rigid rotator, in case you are wondering why we are talking about the harmonic oscillator and the rigid rotator,0994

these things are for spectroscopic movement, spectroscopic transitions.1004

The harmonic oscillator is this way, it is the molecule that is vibrating.1011

The mathematics behind this is what we are doing.1014

The rigid rotator is something that rotates like this.1018

When you have a diatomic molecule that is rotating, we modeled it with the mathematics of the rigid rotator.1021

That is all, that is what we are doing here.1026

The energy of the rigid rotator is, in terms of what it is that we studied earlier, EJ = H ̅² / 2I × J × J + 1,1028

where J is the rotational quantum number that takes on the values 0, 1, 2, and so on.1043

Here I is the rotational inertia of the molecule.1050

It is equal to the reduced mass × the equilibrium bond length, the radius length between the two nuclei.1054

This E stands for its equilibrium², that is it.1066

This is the moment of inertia.1069

This I is the moment of inertia or the rotational inertia of the molecule.1071

Our E is the equilibrium bond length.1080

One of the parameters that you will see when you look at a table of constants for spectroscopic data is you are going to see the R sub E.1092

You are going to see the equilibrium bond length for that molecule.1099

The rotational say this gives the energy of the rotational state, the degeneracy of each level.1104

In other words, the number of levels that actually has this energy.1111

The degeneracy is equal to 2 J + 1.1115

When we discussed the rigid rotator earlier, I do not think I explained why this degeneracy exists.1121

I may have, but I do not believe that I did.1144

I think I just throw it out there as a number.1146

I do not think I explained where this degeneracy in the rotational states comes from,1149

explained that nature of this degeneracy.1155

Here is what is going on, I will tell you and then I will write it all out.1170

Let us say J is 1, that is going to have certain energy.1175

It is going to rotate with certain energy.1182

However, for each quantum number, in this case 1, there are 2 × 1 + 1.1186

There are going to be three actual orientations in space, fundamental orientations where the rotation is going to have that energy.1194

That is what degeneracy means.1202

It is going to be a particular quantum state that has that same energy, that is what the degeneracy.1204

We know that by definition.1209

The orientation in space of the molecule does not affect its rotational energy.1211

In other words, if I have a certain molecule that is oriented this way and it is rotating like this,1215

or if it is this way rotating like this, or if it is this way rotating like this, they have the same energy.1221

That is what this degeneracy means.1228

In the case of J = 1, it can be this way, it can be this way, it can be this way.1230

If J = 2, that is 2 × 2 + 1.1238

It means there are going to be 5 fundamental orientations in space that give you that same energy.1241

It is going to be 1, 2, 3, 4, 5.1246

For 3, you are going to end up with 7 levels of degeneracy, 7 fundamental orientations in space that all have the same energy.1255

That is the nature of this degeneracy of the rotational states.1262

Let us write it all out and give you a little bit of a quantitative aspect of it.1265

In molecules orientation in space has no affect on this rotational energy.1274

J is the quantum number that represents the angular momentum of the rigid rotator.1298

In other words, we know that anything that rotates has an angular momentum.1325

That angular momentum is going to be perpendicular to the direction of rotation.1329

If the molecule is rotating like this, its angular momentum is that way.1332

The magnitude of that angular momentum that is what J is.1337

J² actually.1340

That is what it represents.1343

It represents the magnitude of the angular momentum.1345

If it is rotating this way, the angular momentum vector is pointing that way.1347

If it is rotating this way, it is pointing that way.1351

This is the Z axis, it can be rotating like this, like this like this.1356

Angular momentum pointing that way, angular momentum vector pointing this way,1363

angular momentum vector pointing this way.1366

For each value of J, there is a J sub Z.1369

It is the component of the angular momentum vector along the Z axis, J sub Z is, we already seen this before.1381

For angular momentum, for rotational angular, we have seen this for spin angular momentum already.1391

JZ is the component of the angular momentum vector along the Z axis.1395

JZ takes on the values 0, + or -1, + or -2, all way to + or - J.1422

If J is 1 then what we have is 1, 0, -1.1444

If J is 2, we have 2, 1, 0, -1, 2.1449

If J is 3, we have 3, 2, 1, 0, -1, -2, -3.1454

It is the projection of the angular momentum vector along the Z axis.1461

All of those orientations in space carry the same energy given by the,1466

Here is where 2J + 1 come from.1473

There are 2J + 1 value in J sub Z.1475

For each J, or each J, there are 2J + 1.1478

I will just call that fundamental orientation with the same energy.1490

When we discuss the hydrogen atom, we call the J we called it L, the angular momentum quantum number.1519

The rotational quantum number.1537

And we called this J sub Z, we call it M sub L.1544

This was the magnetic quantum number, that is all that is going on here.1551

Remember for each value of L, you have 0, + or -1, + or – 2, all the way to ± L.1558

That is the magnetic quantum number which takes on the values 0, + or -1, + or -2, all the way to + or – L.1567

We will do the rotational term.1589

We went ahead and expressed the vibrational energy in terms of wave numbers.1596

We are going to express the rotational energy in terms of wave numbers.1600

The rotational term is symbolized as F of J.1604

F of J is equal to E of J divided by HZ equal to H ̅² / 2I HZ × J × J + 1.1612

H ̅ is equal to H/ 2 π, that implies that H ̅² is equal to H²/ 4 π².1630

When we put all of this back in to here, we get that the rotational term,1644

in other words the rotational energy expressed in terms of wave numbers is going to be equal to H² / 8 I π² HZ × J × J + 1.1648

We get some cancellation with the H and one of these.1668

What we are left with is F of J is equal to H/ 8I π² ψ × J × J + 1.1671

And again, J takes on the value 0, 1, 2, 3, 4.1690

I will circle the whole thing not just the bottom part.1696

This whole thing this is called the rotational constant and is symbolized as a B with a ~ on it.1699

This is another one of the spectroscopic parameters that you find in a table of spectroscopic data.1722

Just like you see in the equilibrium bond length, you will also see the rotational constant.1730

And as we go on with the lessons, you will see that there are more and more constants that are actually tabulated.1734

Since that is symbolized that way, we will go ahead and write it as F of J is equal to B ~ × J × J + 1,1742

where J takes on the values of 0, 1, 2, and so on.1758

And B is of course what we just said F of J is now in inverse cm.1763

The selection rules for rotational transitions δ J = + or -1,1771

that means it can only go from one rotational state to the next, either up or down.1790

It is not going to jump 5 levels.1793

In this case, the molecule must have a permanent dipole.1795

In the case of a rotational transition, it has to have a permanent dipole.1807

In the case of the vibrational transition, there has to be a change in the dipole moment during the vibration, during the transition.1810

If the transition is to happen, it must have a permanent dipole.1821

That is the difference between the two.1825

We have the rigid rotator energy, we have the harmonic oscillator energy, therefore like we said,1828

the transitions in the infrared, the vibrational transitions are accompanied by rotational transitions.1835

The combined energy of the transition is going to be the energy of the rotation + the energy of the vibration.1841

The harmonic oscillator rigid rotator approximation for the energy of the molecule is1849

therefore, the sum of the vibrational rotational energy.1873

Therefore, the energy R J is equal to the vibrational term + the rotational term.1892

The vibrational energy + the rotational energy.1901

Let me make my J a little bit more clear so it is not connected like that.1903

E sub RJ is equal to, this one is ν prime R + ½ +, now we have B~ J × J + 1.1908

Here R takes on the values 0, 1, 2, and so on.1924

J takes on the values 0, 1, 2, and so on.1928

Notice, there are two quantum numbers here in the total expression for the energy.1932

Once again, this ν ~ is equal to 1/ 2 π Z × the force constant divided by the reduced mass ^½.1937

This is the fundamental frequency of the vibration.1949

B~ is equal to Planck's constant divided by 8I π² Z.1966

All the parameters are taken care of, this is the equation of the harmonic oscillator rigid rotator approximation.1976

The mathematical equation that approximates what we see when we look at vibration rotation spectra is this.1983

The energy level diagram looks like this.1997

Let me draw this one by hand, actually.2001

What we have is my harmonic oscillator R = 0, R = 1, R = 2.2004

Remember the spacing between energy levels is the same for the harmonic oscillator.2021

R = 3, this is R = 0.2027

R = 1, R = 2, R = 3.2030

Within each vibrational level, there is a series of rotational levels.2036

We have J = 0, J = 1, J = 2, J = 3, and so on.2042

Here we have this one, this one, this one, this one, this one.2053

For R = 2, for each vibrational level there is a series of rotational levels.2060

The spacing of the rotation levels is not the same, the energies.2072

Between each vibration level, or for each vibrational level2080

which is the vibrational quantum number R there is a progression of rotational states.2090

When a photon of infrared is absorbed,2119

not only does a vibrational transition take place from R to R + 12140

but several rotational transitions take place.2155

Several rotational transitions take place from J to either J + 1 or to J-1.2165

Right now, we are talking just about absorption.2182

When we are talking about absorption, we are going to go from R to R + 1.2184

If we are talking about emission, we would be talking about going from some level R to R -1.2189

For the sake of absorption, we are going up one vibrational level but the J level can actually go down or up.2194

You can go from J1 to J2 or J1 to J0, that is what this means.2202

But several rotation transitions take place when a photon of IR not only does the transition take place2211

from R to R + 1 but several rotational transitions take place from J to J + 1.2219

The selection rules, this is called vibration rotation.2230

When we to look a spectrum, it is called a vibration rotation spectrum.2242

We can get pure rotational spectra, we can get information of pure vibrational spectra2246

but when we run out of vibrational spectrum what we will get is a vibration rotation spectrum.2253

They are the combination of the R jump and the J jumps, that is what is happening.2256

The selection rules for vibration rotation are2267

δ R = + or -1, δ J = + or -1.2285

If R is + 1 that means it is going from a lower or to a higher state of absorption.2293

If R is -1, it means it is coming from a higher to a lower state, that is emission.2298

In the case of δ J + or -1, in the case of absorption, because you are going from one vibrational state to another vibrational state,2304

the rotational state might go up 1 or down 1.2316

But it is still absorption because it is actually going up an entire vibrational level.2322

You are still looking at a higher level.2331

Let us look at what the mathematics behind absorption.2337

What we want to do now is to derive an equation for the frequencies, for the spectra that we see.2341

Let us look at absorption, in the case of absorption δ R = + or -1.2352

Δ R is + 1, it is absorption, sorry about that.2365

Δ J is + or -1.2370

For R = + 1 and J = + 1, we would have two cases, 1 + 1 and 1 -1, + 1, the frequency of absorption.2374

The frequency of absorption is the difference between two energy levels.2389

The frequency of absorption is ν, what we observe.2399

It is the energy of the R + 1 J + 1 state - the lower state which is the energy of the R and J.2408

Let us go ahead and do the mathematics here.2434

This is going to equal, this energy term – this.2437

What we are going to have is prime × R + 1 + ½ + B~ × J + 1 × J + 2.2441

This is the upper state - the lower state which is ν~ × R + ½ + B ~ × J × J + 1.2455

I will not actually go through all the algebra here.2471

I think I actually will, I will do it for this one, that is not a problem.2481

This is equal to ν × R + 3/2 + B ~ × J² + 3J + 2 - ν × R + ½ - B ~ × J² + J,2484

I will distribute the - or both, what you end up with is ν ~ R + 3/2 ν ~ + B~ J² + 3 B ~ J + 2 B ~ – ν R - ½ ν - B ~ J.2511

This is just algebra, that is all it is.2538

It is always the worst part of mathematics.2540

Algebra has always been the worst and it will always be the worst.2542

Do not let it get to you.2546

A combined term, like for example I can cancel this one and this one.2548

I can cancel BJ² and that one I can combine the 3/2 ν - ½ ν.2553

When I'm left with is ν ~ + 2B ~ J + 2B ~.2559

Let me simplify this a little bit.2570

ν observed is going to equal ν + 2B.2572

I’m going to factor out the 2B ~, that is going to be J + 1, where J is going to take on the values 0, 1, 2, 3.2580

This is very important, J here is the value of the lower rotational state.2590

J is the value of the lower rotational state.2598

In other words, the smaller lower quantum number.2608

The quantum number of lower state.2611

It is the value of the quantum number in the lower state.2613

This is one of the equations.2637

This equation for the different values that J takes on.2640

What I'm going to see is the spectrum but I expect to see in the spectrum is this.2644

I expect to see a frequency at this number.2649

Its fundamental vibrational frequency + 2 × this rotational constant × whatever the J value happens to be in the lower state.2655

I expect to see a line there.2663

For the other case, for R = + 1.2668

This time J = -1, the observed frequency that I expect is going to be the energy of the R + 1 state J-1,2674

that is the upper state is J -1 – ERJ.2688

I go through the same algebra and what I end up with is, ν observe is going to equal ν -2 BJ.2697

Here, J takes on the values 1, 2, 3.2710

And again, J here is the value of the quantum number in the lower rotational state.2716

That is why there is a difference between these two.2725

Let us actually see what this looks like.2730

This is the other equation that I'm interested in right here.2731

This gives me one set of lines for different values of J.2739

This gives me another set of lines for different values of J.2742

Let us go ahead and go to a picture of the energy transitions to see what is happening first,2757

then we will take a look at the spectrum.2761

These right here, the blue lines, this blue level lower vibrational state,2765

Let me actually erase this.2771

Most books tend to use the symbol V or sort of a variation on ν as the vibrational quantum number.2774

I do not like that because it looks a lot like the frequency and it tends to get really confusing2782

which is why I use R for the vibrational quantum number.2788

This is R = 0, the 0 vibrational quantum state.2792

Here, this is the R = 1 vibrational quantum state.2795

Notice, within each vibration level, there are several rotational levels.2801

J = 0, J = 1, J = 2, J = 3, 4, 5.2806

And of course, in the upper level it has its own rotational quantum states 0, 1, 2, 3, 4, 5.2810

What we see in the spectrum is a series of lines.2817

For right now, let us not worry about with this Q branch is.2821

I will tell you in a second.2825

What is happening is R, the vibrational state is going up by 1.2827

We are jumping up from this vibrational state to this vibrational state.2832

Let us go to the R branch first, the R branch of the spectrum.2837

Here, the δ R = + 1 and the δ J = + 1.2845

That is fine, I will stick with blue.2856

I'm going from the J = 0, this one right here, this black line is J equal 0 to the J = 1.2859

The vibrational quantum state is going up by 1.2866

The rotational quantum state is going up by 1.2871

The molecules that are in the state of J = 1, they transition to the J = 2.2877

The 2 go up to 3, the 3 go up to 4, the 4 go up to 5.2885

That is the R branch.2894

The L branch represents δ R = + 1, we are going up a vibrational state.2896

But for the δ J = -1, in other words on the spectrum, I see a line for this transition,2905

a line for this transition, a line for this transition, and a line for this transition.2912

Now for the P branch, sorry about that.2918

It represents a vibrational state of going from a lower vibration to a higher vibration R = + 1,2926

but the rotational state drops by 1.2931

Here we are going from 1 to 0, J = 2 to J = 1.2933

J = 3 to J = 2, J = 4 to J = 3, J = 5 to J = 4.2942

There is a line for each of these transitions.2950

What we see is 1, 2, 3, 4, 5 and so on lines.2954

To the left we see 1, 2, 3, 4, 5, and so on lines.2958

This Q branch, notice where the lines go.2963

We are jumping up from a vibrational state to a vibrational state R = 0 to R = 1.2967

That is fine, that is not a problem.2971

The selection rule can handle that but we said that δ J has to be + or -1.2973

Δ J cannot be 0, that is a forbidden transition.2977

Notice this line right here, that is going from J = 0 to J = 0.2983

J = 1 to J = 1, because δ J = 0 is a forbidden transition, these transitions we do not see them in the spectrum.2988

What you see in the spectrum is this, no line at ν, the fundamental vibration frequency.2999

Remember what we had just a second ago, we had a couple of equations.3009

We said that for the R branch, we would end up seeing a bunch of lines at ν + 2 B × J + 1.3014

And for the P branch, we would see a bunch at ν – 2BJ.3028

There are going to be lines to the left, lines to the right of this number.3037

But we do not actually see a line in there because the transition from the δ J = 0 transition is forbidden.3042

There is no actual Q branch.3051

There are molecules with Q branch, they will show up3053

Your teacher may or may not decide not to talk about it.3056

We, ourselves, we will not talk about it.3060

It was not altogether that important, at least for what we are doing.3061

But know that there is a P branch, there is a Q branch, there is an R branch.3064

But in general, for the vibration rotation spectra of diatomic molecule,3068

because δ G = 0 is a forbidden transition, δ J has to equal + or -1, we do not see a Q branch.3072

It is just that nothing there.3079

Let us go ahead and take a look at the vibration rotation spectra.3082

This is a vibration rotation spectrum for HCL.3089

Do not worry about this, it should not be in here.3093

This is a vibration rotation of HCL.3097

You will sometimes see them this way, in terms of the peaks to be pointing down.3101

Other times you are going to see them when the peaks are pointing up.3107

The R branch here, right here, see these lines?3110

This right here, this little gap, this is the ν sub 0.3113

This is the fundamental frequency.3122

It to this point where the J = 0 to J = 0 transitions should happen.3123

But because they are forbidden, it is not going to happen, but that is where we see it.3128

If we know what ν 0 is, reading of the spectrum we just find that middle point and we go down and we mark it.3134

That is our ν sub 0, that is our fundamental vibration frequency.3143

The R branch we said is where you do,3148

Let us write it down.3156

It is equal to that, + 2B × J + 1.3159

Notice, it is where the absorption frequency is going to be increasing from the fundamental frequency.3168

The fundamental frequency + a certain amount.3175

The fundamental frequency + a certain amount.3178

The fundamental frequency + a certain amount + a certain amount, this is a second transition.3180

+ a certain amount, this is the third transition.3184

+ a certain amount, this is the fourth transition.3186

+ a certain amount, it is the fifth transition.3187

As the frequencies go up, that is the R branch.3190

This right here is the R branch.3193

Over here, the frequencies go down.3196

The fundamental frequency of the P branch is the fundamental frequency - some value as J increases.3199

- - -, these are going down.3206

This is the P branch.3209

Like I said, sometimes you see it the other way.3212

When you look at the spectra, you are not going to look at left and right.3214

You are looking at R and P.3220

R is for the frequencies that increase, P is for the frequencies that decrease from the fundamental frequency.3222

That is what is happening here.3228

Let us look at these lines, the fundamental frequency is here.3230

This one represents the transition from 0 to J = 1.3233

This one represents J = 1 to J = 2.3237

This one represents J = 2 to J = 3.3240

J = 3 to J = 4, and so on.3244

Here represents J = 1 to J = 0, J = 2 to J = 1, J = 3 to J = 2.3246

This is the δ J = + 1, here is the δ J = -1.3260

Let us look at the spectra that actually look the other way.3267

This is the spectrum for carbon monoxide.3271

Let me write it down here.3274

This is the spectrum from carbon monoxide.3276

In this particular case, the peaks are pointing up.3278

This is interesting.3280

The left is actually increasing, to the right it is actually decreasing.3288

Here is the fundamental frequency increase.3291

This is the R branch decreasing, that is the P branch.3294

That is all that is going on here.3299

Because the equation R ν + 2B × J + 1 and ν - 2 BJ, the difference in the spacing, this difference right here,3303

the difference between the lines is 2B.3322

The difference between them is 2B.3328

We see absorptions.3334

That is what is going on there.3340

Let us go ahead and see what else we have to say about this.3342

Notice there is a nonexistent Q branch here.3349

Let us talk about the intensities.3359

Notice that all of these lines they have different intensities.3361

Some are very very intense, some not so intense.3364

This is the last page.3370

Let me go ahead and write it over here and continue it down here.3374

The intensities of the transitions are related to the populations of each J level.3386

In other words, if I had a bunch of molecules, let us just say I have 100 molecules in level 2.3405

A hundred molecules in level 2, the more molecules that you have, the more transitions are going to take place.3413

The more molecules you have, each one of those molecules is going to make a transition.3428

Each one of those transitions is represented by, it is going to contribute to the size of the peak.3432

If you only have 10 molecules making the transition, let us say from level 3 to level 4,3437

it is going to have a certain height of a peak.3443

If you have 100 molecules making the transition from level 2 to level 3, that peak is going to be higher.3445

In this particular case, notice that the level 2 to level 3 to level 4, those tend to be the peaks of highest intensity.3452

That tells us that the rotational levels 2, 3, 4, those are the ones that are most highly populated at normal temperatures.3461

As far as the rotational state level molecule, most molecules are not in there,3472

0, 1, 2 rotational states at room temperature, most of them tend to be in the 3, 4, 5 rotational states.3476

That is where most of the molecules are.3483

Therefore, the transitions that are going to take place, they are going to have the ones of higher intensity.3485

Let me say that again.3491

The intensities of the transitions are related to the populations of each J level.3492

The greater the population of the level, the greater the number of transitions,3504

therefore, the more intense the line.3532

We see from these spectra that the J = 3, 4, 5, are the most populated.3542

Once again, notice that the Q branch does not exist because δ J = 0 is a forbidden transition.3574

Thank you so much for joining us here at www.educator.com.3582

We will see you next time for a continuation of molecular spectroscopy, bye.3584