Raffi Hovasapian

Vibration-Rotation Interaction

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Section 1: Classical Thermodynamics Preliminaries

The Ideal Gas Law

46m 5s

Intro

0:00

Course Overview

0:16

Thermodynamics & Classical Thermodynamics

0:17

Structure of the Course

1:30

The Ideal Gas Law

3:06

Ideal Gas Law: PV=nRT

3:07

Units of Pressure

4:51

Manipulating Units

5:52

Atmosphere : atm

8:15

Millimeter of Mercury: mm Hg

8:48

SI Unit of Volume

9:32

SI Unit of Temperature

10:32

Value of R (Gas Constant): Pv = nRT

10:51

Extensive and Intensive Variables (Properties)

15:23

Intensive Property

15:52

Extensive Property

16:30

Example: Extensive and Intensive Variables

18:20

Ideal Gas Law

19:24

Ideal Gas Law with Intensive Variables

19:25

Graphing Equations

23:51

Hold T Constant & Graph P vs. V

23:52

Hold P Constant & Graph V vs. T

31:08

Hold V Constant & Graph P vs. T

34:38

Isochores or Isometrics

37:08

Physical Chemistry Vibration-Rotation Interaction

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Vibration-Rotation Interaction

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Intro

Vibration-Rotation Interaction

Example I: HCl Vibration-Rotation Spectrum

Intro 0:00
Vibration-Rotation Interaction 0:13

Vibration-Rotation Spectrum: HCl
Bond Length & Vibrational State
Vibration Rotation Interaction
Case 1
Case 2

Example I: HCl Vibration-Rotation Spectrum 22:58

Rotational Constant for the 0 & 1 Vibrational State
Equilibrium Bond Length for the 1 Vibrational State
Equilibrium Bond Length for the 0 Vibrational State
Bₑ & αₑ

Physical Chemistry Online Course

Section 1: Classical Thermodynamics Preliminaries
	The Ideal Gas Law	46:05
	Math Lesson 1: Partial Differentiation	46:02
Section 2: Energy
	Energy & the First Law I	1:06:45
	Energy & the First Law II	1:06:33
	Energy & the First Law III	1:02:17
	Changes in Energy & State: Constant Volume	1:04:39
	Joule's Experiment	16:50
	Changes in Energy & State: Constant Pressure	43:40
	The Relationship Between Cp & Cv	32:23
	The Joule Thompson Experiment	39:15
	Adiabatic Changes of State	35:52
Section 3: Energy Example Problems
	1st Law Example Problems I	42:40
	1st Law Example Problems II	1:00:23
	1st Law Example Problems III	44:34
	Thermochemistry Example Problems	59:07
Section 4: Entropy
	Entropy	49:16
	Math Lesson II	33:59
	Entropy As a Function of Temperature & Volume	54:37
	Entropy as a Function of Temperature & Pressure	31:18
	Summary of Entropy So Far	23:06
	Entropy Changes for an Ideal Gas	25:42
Section 5: Entropy Example Problems
	Entropy Example Problems I	43:39
	Entropy Example Problems II	56:44
	Entropy Example Problems III	57:06
Section 6: Entropy and Probability
	Entropy & Probability I	54:35
	Entropy & Probability II	35:05
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
	Spontaneity & Equilibrium I	28:42
	Spontaneity & Equilibrium II	34:38
	The Fundamental Equations of Thermodynamics	30:50
	The General Thermodynamic Equations of State	34:06
	Properties of the Helmholtz & Gibbs Energies	39:18
	The Entropy of the Universe & the Surroundings	19:40
Section 8: Free Energy Example Problems
	Free Energy Example Problems I	54:16
	Free Energy Example Problems II	31:17
	Free Energy Example Problems III	45:00
	Three Miscellaneous Example Problems	58:05
Section 9: Equation Review for Thermodynamics
	Looking Back Over Everything: All the Equations in One Place	25:20
Section 10: Quantum Mechanics Preliminaries
	Complex Numbers	34:25
	Probability & Statistics	59:57
	SchrÓ§dinger Equation & Operators	42:05
	SchrÓ§dinger Equation as an Eigenvalue Problem	30:26
	The Plausibility of the SchrÓ§dinger Equation	21:34
Section 11: The Particle in a Box
	The Particle in a Box Part I	56:22
	The Particle in a Box Part II	45:24
	The Particle in a Box Part III	48:43
Section 12: Postulates and Principles of Quantum Mechanics
	The Postulates & Principles of Quantum Mechanics, Part I	46:18
	The Postulates & Principles of Quantum Mechanics, Part II	39:28
	The Postulates & Principles of Quantum Mechanics, Part III	35:32
	The Postulates & Principles of Quantum Mechanics, Part IV	29:55
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
	Example Problems I	54:25
	Example Problems II	46:58
	Example Problems III	44:11
Section 14: The Harmonic Oscillator
	The Harmonic Oscillator I	35:33
	The Harmonic Oscillator II	43:04
	The Harmonic Oscillator III	26:30
Section 15: The Rigid Rotator
	The Rigid Rotator I	41:10
	The Rigid Rotator II	30:32
	The Rigid Rotator III	35:19
Section 16: Oscillator and Rotator Example Problems
	Example Problems I	33:48
	Example Problems II	46:43
Section 17: The Hydrogen Atom
	The Hydrogen Atom I	40:00
	The Hydrogen Atom II	35:58
	The Hydrogen Atom III	36:18
	The Hydrogen Atom IV	33:55
	The Hydrogen Atom V: Where We Are	51:53
	The Hydrogen Atom VI	51:53
	The Hydrogen Atom VII	34:29
Section 18: Hydrogen Atom Example Problems
	Hydrogen Atom Example Problems I	43:49
	The Hydrogen Atom Example Problems II	1:01:57
	The Hydrogen Atom Example Problems III	48:33
	The Hydrogen Atom Example Problems IV	48:33
	The Hydrogen Atom Example Problems V	48:33
Section 19: Spin Quantum Number and Atomic Term Symbols
	Spin Quantum Number: Term Symbols I	59:18
	Spin Quantum Number: Term Symbols II	34:54
	Spin Quantum Number: Term Symbols III	38:03
	Term Symbols & Atomic Spectra	57:49
Section 20: Term Symbols Example Problems
	Example Problems I	1:01:20
	Example Problems II	56:34
Section 21: Equation Review for Quantum Mechanics
	Quantum Mechanics: All the Equations in One Place	18:24
Section 22: Molecular Spectroscopy
	Spectroscopic Overview: Which Equation Do I Use & Why	50:02
	Vibration-Rotation	59:47
	Vibration-Rotation Interaction	46:22
	The Non-Rigid Rotator	29:24
	The Anharmonic Oscillator	30:53
	Electronic Transitions	1:01:33
Section 23: Molecular Spectroscopy Example Problems
	Example Problems I	33:47
	Example Problems II	1:01:05
	Example Problems III	33:31
Section 24: Statistical Thermodynamics
	Statistical Thermodynamics: The Big Picture	1:01:15
	Statistical Thermodynamics: The Various Partition Functions I	47:23
	Statistical Thermodynamics: The Various Partition Functions II	54:09
Section 25: Statistical Thermodynamics Example Problems
	Example Problems I	48:32
	Example Problems II	57:30

Transcription: Vibration-Rotation Interaction

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

In the last lesson, we started our discussion of molecular spectroscopy.0004

In this lesson, we are just going to go ahead and continue our discussion.0008

Let us get started.0011

This is the vibration rotation spectrum for HCL.0014

Here, we have the R branch, sometimes you going to see it like this0022

and sometimes you are going to see it the other way when the peaks are actually pointing up.0029

It does not really matter, the thing you will remember those, the R branches with the frequencies are increasing,0033

the P branch is the one with the frequencies are decreasing.0040

R is left and right, in this case the frequencies increasing this way, the R branch is over here.0044

This is a decreasing frequencies of P branch over here.0053

This is not unnecessary bit of information.0056

Some of the things that we want to notice here.0060

I will stick with black.0063

Some things to notice, the first thing you want to notice is that the distance between the lines in the 2 branches are not equal.0067

In other words, the distance from here to here is not the same as the distance from here to here.0079

We would expect 2B, 2B, 2B, 2B, 2B in the P branch, in the R branch but that is not the case.0084

These are narrower and as we actually increase, go further along the R branch, the gap tends to get smaller.0092

And as we go further along the P branch, the gap actually gets bigger.0099

We will say 1 the distance between lines in the 2 branches are not equal.0108

The distances are not equal.0129

For the R branch, the gaps get smaller as J increases this way.0139

For the P branch, the gap get larger, the gaps get wider as J increases.0163

We have the following for the harmonic oscillator rigid rotator approximation.0182

We had that the energy RJ, R is the vibrational quantum number.0187

J is the rotational quantum number.0195

The total energy is equal to the vibration term + the rotational term and we had prime R + ½ + this constant × J × J + 1.0198

R goes from 0, 1, 2.0223

J = 0, 1, 2, and so on.0228

In this particular case, B the rotational constant is equal to planks constant over 8 π² ψ × the rotational inertia.0234

The rotational inertia happens to be the reduced mass × the equilibrium bond length².0246

This is the expression for the energy under the harmonic oscillator rigid rotator approximation.0255

Let us take a look at this next image here.0262

This green line represents the potential curve for the harmonic oscillator, that is what we based it on.0268

You remember the potential energy was equal to ½ KX², a parabola.0278

The actual potential curve is not parabolic, that is the blue line.0286

This is the actual potential curve, that is 2 molecules come together.0289

From an infinite distance, the energy decreases and may reach a certain length between the two.0295

If you push them any closer together, the potential energy rises.0300

If you pull them apart, the potential energy rises.0304

This down here, that point of lowest energy is what we call the bond length.0307

R sub E equilibrium bond length far as the distance between them.0311

You will notice that the harmonic oscillator approximation is good for the lower vibration states.0315

By the way, these ν right here, these are the quantum numbers or what I call R.0324

For our purposes, the reason I use R for the vibrational quantum number instead of the ν,0332

you have probably seen in your books is I want to confuse it with the actual frequency absorption0339

or transmission or other frequencies that we happen to talk about.0345

These are the vibrational states R = 0, R = 1, R =2, R =3, 4 and 5.0348

For the lower vibrational states, the parabolic, the harmonic oscillator0355

actually is a pretty good approximation for the non harmonic oscillator.0360

The actual, the real potential energy curve.0365

Very very good approximation.0368

Under normal temperatures, most of the molecules are going to be in the R = 0 vibrational state0371

and we will show you why that is later when we discuss statistical thermodynamics.0378

You know that that is the case, for that vibrational state R = 0, the harmonic oscillator is a great approximation.0383

As the vibrational state increases, in other words, as R increases, the value of R sub E, the equilibrium bond length,0391

I use a capital here, I just happen to use a small letter.0425

The value of RE, the bond length also increases.0429

You can see this on the potential energy curve.0443

Notice, if I stay just for the harmonic oscillator, if I just go straight up,0446

I'm right down the middle of in between the potential energy curve along the energy level.0451

But as R starts to increase, the vibrational states go up.0459

You go higher and higher levels.0466

The halfway mark on the blue line tends to start to shift further and further to the right.0469

What happens is that the equilibrium bond length tends to go further and further to the right.0477

In other words, the equilibrium bond length have to get wider and wider and wider.0483

The bond length increases as we move up the vibrational states.0486

As the bond length increases, the rotational constant B~ decreases precisely because the definition of that is equal to,0496

Remember in the previous page, 8 π² ψ, that RE², this is the rotational inertia.0515

As something in the denominator increases, the value itself is defined by this ratio decreases.0524

We express the independence of the rotational constant on R, the vibration quantum number as B sub R.0537

We put that little subscript there to let us know that this B actually depends on which vibration we happen to be in.0566

Is it 0, 1, 2, 3, 4, things like that.0572

We can rewrite the energy equation E sub RJ equal to + ½ + B sub R × J × J + 1.0576

Nothing is different, all that we have done is actually put this little BR here.0600

There is an equation that actually relates to the BR to something.0604

We will get to that.0613

We call this dependence of the rotational constant B on R, the vibration rotation interaction.0616

The vibration rotation interaction is going to be one of the corrections that0644

we are going to have to make to our harmonic oscillator rigid rotator approximation equation.0653

There would be 3 corrections that we are going to make.0659

The first one is this, the vibration rotation interaction.0661

Later, we are going to correct for anharmonicity of the anharmonic oscillator and we are going to correct for the non rigid rotator.0665

This is the first of the corrections, the vibration rotation interaction.0676

What we are going to do is we are going to examine the transition from R0 to R1,0682

that is the equation that we are going to look at.0687

Let us examine the R = 0 to the R = 1 transition.0693

Therefore, the δ R is equal to + 1, 0 to 1.0708

The first case we have, we said that the δ J can equal + or -1.0715

Δ R = that and δ J is equal to + or -1.0732

The first case that we are going to look at, case 1, we are going to take a look at the δ J = + 1.0736

We would be making the transition from the 0 vibrational state to the 1 vibrational state.0746

Within those vibrational states, you have a bunch of rotational levels.0750

We are going to see the transitions that take you from 0 to 1, 1 to 2, 2 to 3, things like that,0754

between the rotational states.0761

E sub R, the frequency that we see on the spectrum or that we expect to see on the spectrum,0766

is going to be for the R branch because the δ J + 1 represents the R branch.0774

It is equal to upper energy - the lower energy.0782

It is going to be E1 J + 1 - E0 J, this is the upper energy level, this is the lower energy level.0787

I will go through the algebra, at least for the first one here.0800

This is going to equal, R is 1, 1 + ½ + E sub 1.0806

I’m specifying my R, my quantum number, × J + 1 × J + 2 - the lower energy.0822

This is going to be 0 + ½ + B0 × J × J + 1.0834

All I'm doing is I'm actually putting these quantum numbers 1 and J + 1, 0 and J into our equation for the energy.0846

The upper energy level, the lower energy level, and subtracting the lower from the upper.0853

That gives me the frequency of the absorption or the transmission that I see.0858

This is going to equal, after a little bit of algebra which we will actually do right here.0867

+ B1 × J² + 3J + 2.0873

And I hope that you are confirming my algebra because I am notorious for algebraic errors.0880

- the - distributes over everything.0887

-B0 J² – B sub 0 J = ν + B1 J² + 3 B1 J + 2 B1 – B sub 0 J² - B sub 0 × J.0890

Our final equation that we have is this going, the frequency that we observe for the R branch0929

is going to be the fundamental frequency + B1 - B0 J² + 3 B1 - B0 × J + 2 B0.0939

This is one very important equation.0963

Under this dependence that the rotational constant has on the quantum number or the vibration quantum number R,0966

we have come up with an equation from the 0 to 1 vibration level for the R branch.0975

This actually gives me the frequencies of absorption that I should see.0983

Something very very important to remember.0990

In this case, J is the initial rotational state for lower.0992

It is the lower the initial, we are talking about absorption here.0999

We went from a lower to a higher, J is the lower rotational state.1003

It is a quantum number of the lower rotational state.1010

When we do one for P branch, that is going to equal, the upper energy level1017

which is going to be vibrational level 1 and this is δ J = -1.1027

We are still going from vibrational level 0 to 1.1045

The rotational levels are going to go from 1 to 0, 2 to 1.1048

The J values are going to go down by 1.1055

What we have is the following.1058

We have the P branch is going to equal the upper energy level which is 1 J-1 - the lower energy level which is 0 J.1060

We can go ahead and go through the algebra.1073

I’m not going to go ahead and go through it.1076

I will just go ahead and write out the equation that you get.1077

P is equal to ν + B1 - B0.1081

I should we stop using these tilde because it is really a lot of excess of symbolism that makes you crazy.1088

But I guess that is the nature of quantum mechanics, symbols.1092

-B1 + B0 × J, this equation.1097

J is the lower rotational state, the quantum number of the lower rotational state, those two equations.1111

Now, as R increases like we said, B sub R decreases.1127

This implies that B1 is less than B0.1147

B1 being less than B0, if we look at these equations that we got, these two equations.1155

The one for the P branch, and 1 for the R branch.1160

As J increases in the two equations above, the R branch gaps decrease and the P branch gaps increase.1162

In other words, once I put J = 0, 1, 2, 3, 4, into the R branch J = 1, 2, 3, 4, 5 in the P branch,1208

when I take the differences between 1 rotational state and another, the R branches, the difference is going to get smaller.1220

When we started this lesson, we notice that the R branch actually tend to decrease.1228

The P branch, the lines of the spectrum are going to widen as J increases.1233

These two equations explain that behavior.1239

Vibration rotation interaction explains the appearance of a lines or spectra.1245

The equation, the dependence of B on R is given by an explicit relation B sub R1272

is equal to something called B sub E - Α sub E × R + ½.1293

B sub E and α sub E are 2 part spectroscopic parameters that are tabulated for different diatomic molecules.1303

The values for B sub E and Α sub E are available in tables of spectroscopic data.1316

And we will see some those tables in the example problems later on.1334

Let us go ahead and do a quick example here.1350

Basically, in the problems you are going to get two source.1355

You are going to be given some spectroscopic data and you are going to be asked to actually find the spectroscopic parameters.1357

Or you are going to be given the spectroscopic parameters and you are going to be as to find the lines spectroscopic data.1365

It goes both ways.1371

We will see examples of that when we did example problems.1373

Let us go ahead now and take a look at an example.1376

Example 1, this is going to be only example for this particular lesson.1382

I'm collecting all of the example problems for afterward when we do them all at once after the other.1387

I want you to see at least one here. The numerical values spectral lines are put into tables.1395

The lines of the P and R branches are labeled with a branch they belong to,1400

+ the initial rotational state of the transition line it represents.1404

For example, for the R branch, the first line is labeled R sub 0.1408

The reason is because it belongs to the R branch, it represents the transition from the 0 to 1 rotational state.1412

The second line is R1.1418

It is in the R branch and repressed transition from 1 to 2.1421

These subscript are the lower rotational states, the initial rotational state.1425

The third line R2, be very careful to keep the indices very clear and distinct.1429

For the P branch, the first line is P1.1436

The P branch because the transition is going from the initial state rotational state of 11439

to the upper rotational state which is actually the 0.1445

This represents the 1 to 0 transition, δ J = -1.1449

The second line is P2, it goes from 2 to 1.1453

I hope that makes sense.1458

Using the equations above, like what we just did for ν sub R and ν sub P,1460

and the data table of frequencies for the HCL vibration rotation spectrum below,1464

or the numerical values from the spectrum directly, your choice what you want to use the spectrum,1469

with numbers or from the table that we provide.1475

Find the rotational constant for the 0 vibrational state, the rotational constant for the 1 vibrational state,1478

the equilibrium bond length for the 0 vibrational state, and equilibrium bond length for the 1 vibrational state.1487

Use 1.63 × 10⁻²⁷ kg for the reduced mass of HCL.1495

Here, in this particular case, we have given you both.1503

We have given you the spectrum and here is the table.1506

The table looks like this.1509

Basically, just read and write off the spectrum.1510

The R is 0 branch that is this one right here, it is 2906.1514

The R1 branch was 2926.1523

When I go to the P branch over here, this is P1, this is R0.1528

This is R1 P1, 2865.1535

I’m sorry, it is actually not that clear here.1540

The P2 branch 2844.1542

Again, these are in inverse cm.1549

We would be working almost exclusively in inverse cm.1550

Some of the problems might give you data in Hertz, megahertz, gigahertz, things like that.1553

We will always be converting to inverse cm because for spectroscopy, inverse cm is the standard unit.1559

Sorry about that, this is what I just have at the table.1570

It was later that I decided to put the spectra in.1577

In general, sometimes you need a table, sometimes you are going to be given a spectrum.1580

You have to be prepared to work from either one.1584

Let us go ahead and start this problem.1589

Let us go ahead and rewrite our equations.1591

The R branch spectral line that we see is going to be the ν,1594

the fundamental vibration frequency + B1 - B0 J² + 3 B1 – B 0 J + 2 B1.1603

That is the equation for the frequency for the R branch.1625

For the P branch, what we are going to see is ν + B1 - B0 × J² - B1 + B0 × J.1629

These equations and we are going to be working with.1647

Let us go ahead and do the ν for R 0 and R sub 0.1650

R branch, the first line is 0 to 1 transition.1672

Our R value = 0, this small R value = 0.1678

That is going to be, I will just put them in.1683

I'm sorry, let me start again here.1701

And so we have ν, this is going to be R sub 0.1706

The 0 is the J value.1715

We are going from R branch, we are going from the 0.1720

Let us do it this way.1726

R0 represents the 0 to 1 transition for the J.1729

That is a little bit better.1733

This is going to be ν + this thing initial J0, that is going to be 0.1735

J is 0 that is going to be 0 + 2 B1, that one is equal to 2906.1744

We have the R1 transition, that represents the transition from J = 1 to J = 2.1760

I just put 1 in for J in here, I get this + B1 - B0 × 1² which is 1 + 3 B1 - B0 + 2 B1.1766

I combine all like terms and I get ν + 6 B1 – 2 B0 and that is equal to 2926.1797

The two equations we have are this one and we have this one.1815

Let me go ahead and work in red.1825

I got B0 + 2 B1 = 2906 and I have ν + 6 B1 -2 B0 = 2926.1828

This represents the R branch.1850

Let us go back to black and go ahead and do the P branch.1855

Our P1 that represents the transition from 1 to 0.1857

In this particular case, the initial J value is 1.1862

I’m going to put 1 into this equation, the 1 for ν sub P.1865

What we end up with is ν + B1 - B0 × 1 - B1 - B0.1873

And that we say is equal to 2865.1892

The P2 line that represents the transition from J = 2 to J = 1, that is going to be that.1902

It is going to be + 2 into this equation, 2² is 4 so it is going to be +4.1917

B1 -4 B0 -2 B1 -2 B0 and that is going to equal 2844.1924

What I have here, the other equations that I have are ν B1, that goes away.1945

- 2 B0 = 2865, and the second equation here I have that, + 2 B1 – 6 B0 = 2844.1957

This represents the P branch.1980

This is the R branch, let me go ahead and rewrite.1993

I will just make sure my numbers are correct here.2010

What I'm going to do, I’m going to rewrite these equations on the next page so that I actually have them.2018

My R branch, my R is 0 is ν + 2 B1 = 2906.2026

My R1 = ν + 6 B1 – 2 B0 = 2926.2042

My P1 =, I have ν – 2 B0 = 2865.2054

And I have my P2 branch is ν + 2 B1 -6 B0 = 2844.2066

We want B0 and B1, the rotational constant for the 0 vibration level and for the 1 vibration level.2081

Let us go ahead and take, I'm going to take these 2 equations.2097

Let me actually do this one in blue.2102

I'm going to take these two and I’m going to form R0 - P2.2106

I’m going to take the first equation.2115

I’m going to subtract this second equation.2118

What you end up with is when you take this equation - this equation, this and this go away.2121

2 B1 – 2 B1 go away.2135

This – this, what you end up with is 6 B1 = 2906 – 2844.2138

You should get 62.2150

Therefore, the value of the one that we end up getting is 10.33 inverse cm.2154

I'm going to go ahead and take R1 - P1.2164

I’m going to take R1 - P1, this - that cancels.2171

Did I make a mistake here?2184

Let me go back, these indices floating around is enough to make you crazy.2194

I have something written on my piece of paper but I just want to make sure that my indices are actually correct here.2201

Ν + 2B1 = 2906.2208

Ν + 6 B1 – 2 B0 = 2924.2210

Again, this is 2926.2222

I have my P1 which is ν 0 -2 B0, that is correct, 2065.2235

I have ν + 2 B1 -6 B0 = that.2244

I think everything should be okay.2250

Let us try this again.2251

Let us take R0 - P2.2252

That was my mistake, sorry, a little bit of a notational error.2263

We just have to be very careful.2267

We are taking R0 – P2.2267

The ν cancel, the 2 B1 – 2 B1 that cancels 6 B0, my apologies for that.2272

This is 6 B0 is equal to 62, 2906 -2044.2282

Again, I hope that I have done my arithmetic correctly.2293

B0 not B1, is what should be 10.33 inverse cm.2296

There you go, sorry about that.2302

We will go ahead and take the R1 - P1.2307

R1 - P1, the ν they cancel.2310

The B0 cancel and we are left with 6 B1 = 2926 - 2865 =59.2319

I think again, please check my arithmetic.2334

Again, the arithmetic is unimportant.2337

It is the process that is important.2339

B1, other value of 9.83 inverse cm.2341

Note that B1 is less than B0, 9.83 is less than 10.33.2349

We need to find the equilibrium bond lengths.2362

We need to find the R, E, for the R = 0 and find the RE for the R = 1.2364

B, we know that B is equal to planks constant over 8 π² × ψ × the reduced mass RE².2380

I wonder if I should go ahead and do this in terms of,2407

B1 is equal to 6.626 × 10⁻³⁴ J/ s.2411

Here is what we have to watch our units.2424

We are working in inverse cm but joules, kilograms, and meters, and stuff like that,2427

you have to make sure the your units match.2433

We have 8 π², I suppose we can go ahead.2436

2.998 × 10¹⁰ centimeters per second.2448

The reduced mass is 1.63 × 10⁻²⁷.2458

We are looking for R sub 0², this is for B1 R1.2464

We said that one was actually equal to 9.83.2473

We found that value when we solve this equation for R sub E.2478

We get R sub E for the R = 1 value = 1.705 × 10⁻²⁰ m.2485

I will stick with what it is that I have written here.2499

This is meters, therefore equilibrium bond length is going to equal 1.305 × 10⁻¹⁰ m which is 138.5 picometers.2511

It tends to be expressed in terms of picometers.2531

B0, that is equal to 6.626 × 10⁻³⁴ divided by, we have 8 π² × 2.998 × 10¹⁰ cm/ s.2533

This is joule seconds × 1.63 × 10⁻²⁷ kg × R sub E for 0².2564

And we said that one was equal to 10.33.2577

We will get this equal to 1.662 × 10⁻²⁰ m which means the equilibrium bond length is equal to 1.289.2581

1.289 × 10⁻¹⁰ meters which means you are looking at 128.9 picometers.2598

Notice that the one for the vibration state R = 0 is 128.2613

When we moved up a vibrational state to R = 1, the equilibrium bond length increased.2620

Let us do one last thing here.2637

We said that the relationship B sub R is equal to its equilibrium bond length - this Α sub E × R + ½.2639

From the example, we got a B1 value = 9.83 and we got a B sub 0 value = 10.33.2657

I think we can use these values in this equation for that different R.2682

We can find the actual parameters that are tabulated in the data.2689

We can now use B0 and B1 to find B sub E and Α sub E.2696

B sub 1 = B sub E - Α sub E × 1 + ½.2717

We said that that one is equal to 9.83 and the P sub 0 that is equal to be B sub E - Α sub E × 0 + ½.2729

We said that that = 10.33.2743

There you go, you have two equations and two unknowns.2747

The two unknowns are B sub E and Α sub E.2752

This is just 3/2, this is ½, you have these.2755

I will leave that to you.2761

Two equations and two unknowns.2765

The two unknowns are B sub E and Α sub E, the spectroscopic parameters.2769

Thank you so much for joining us here at www.educator.com.2779

We will see you next time, bye.2781

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