Raffi Hovasapian

The Rigid Rotator III

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Table of Contents

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 The Rigid Rotator III

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Section 15: The Rigid Rotator: Lecture 3 | 35:19 min

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1 answer

Last reply by: Professor Hovasapian
Sat Apr 23, 2016 7:28 PM

Post by Tram T on April 21, 2016

Dear Prof. Hovasapian,

-I have question on the part when you mentioned having permanent dipole moment is a must so the electromagnetic radiation could 'grab' on the bond and 'rotate' the rotator faster:

-Is the same thing applied for vibrational transition where the dipole moment of the molecule must change either from already having a dipole moment to having another dipole or from not having a dipole moment to have one during vibration of the molecule?

-Is it why having a change in dipole moment also a must in vibrational transition in order for the electromagnetic radiation to 'grab' and 'pull' on to the 2 bonding atoms to make the 2 bonding atoms oscillate at higher frequency?

-If then why vibrational selection rule does not require permanent dipole moment but 'a change' in dipole moment instead?

-What does 'a change' in dipole moment during vibration really mean?

Thank you!!

The Rigid Rotator III

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

The Rigid Rotator III

Example I: Calculating Rotational Inertia & Bond Length

Intro 0:00
The Rigid Rotator III 0:11

When a Rotator is Subjected to Electromagnetic Radiation
Selection Rule
Frequencies at Which Absorption Transitions Occur
Energy Absorption & Transition
Energy of the Individual Levels Overview
Energy of the Individual Levels: Diagram
Frequency Required to Go from J to J + 1
Using Separation Between Lines on the Spectrum to Calculate Bond Length

Example I: Calculating Rotational Inertia & Bond Length 29:18

Example I: Calculating Rotational Inertia
Example I: Calculating Bond Length

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: The Rigid Rotator III

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

Today, we are going to close out our discussion of the rigid rotator which is a model for a rotating diatomic molecule.0004

Let us jump right on in.0011

I think I'm going to work with blue today.0014

I hope you guys can hear me.0017

In the last lesson, we saw that the allowed energies…0019

… of the quantum mechanical rigid rotator R A sub J0046

is equal to H ̅² / 2I × J × J + 1, where J is a quantum number that runs from 0, 1, 2, and so on.0056

And I here is equal to the reduced mass × the radius of the rotation, the distance between the two masses.0073

When a molecule rotator is subjected to electromagnetic radiation, it undergoes transitions from 1 energy state to the next.0083

For the rigid rotator, we are saying it is going to go from 1 to 2, 2 to 3, 3 to 4.0118

You may be asking yourself can I make the jump from, if I hit it with enough energy can I make the jump from 1 to 3, level 5 to 8?0141

In the case of the rigid rotator, just like in the case of a harmonic oscillator, the answer is no.0148

Any transition has to be made is going to be made between successive levels, 1 to 2, 2 to 3, 3 to 4, and so on.0153

4 to 3, 3 to 2, 2 to 1 and so on.0159

For the rigid rotator, the selection rule tells us about how the quantum number can change.0162

What levels we can jump to?0173

The rigid rotator, the selection rule is that δ J is equal to + or – 1.0176

Recall the selection rule for the harmonic oscillator, we call that R.0192

Δ R = + or -1, it is the same thing.0208

It can only make transitions between adjacent energy levels.0211

In addition to this δ J equaling + or -1, in order for a rotating molecule to actually absorb energy and0216

spin faster, move to the next energy level, it has to have a permanent dipole.0231

We will talk about this one later but it is something that you should now.0247

The molecule must possess a permanent dipole.0250

Basically, the idea is this.0259

You were hitting something with some electromagnetic energy.0262

The electric field of the energy that we actually hit with it, the electromagnetic energy0267

that we hit with it needs to be able to twist the molecule so they can make it spinning.0273

But it needs to make it spin faster.0283

Imagine I grab it and I just spin it faster, the electromagnetic energy needs to do that.0285

It needs to grab the molecule and actually spin it faster.0291

It is the force acting on the molecule.0295

If there is no dipole moment, if there is no positive and negative, as if it is just neutral, the electric field has nothing to grab onto.0299

The molecule does not experience the force of the electromagnetic field so we cannot actually spin it.0306

It cannot grab it and turn it faster.0313

It has to have a permanent dipole moment, a positive and a negative.0316

In order for that positive and negative end to feel the force of the electric field.0319

That is all that is going on.0324

That is why it have to have that.0326

In addition, the molecules must posses a permanent dipole in order to transition.0328

Something like the molecule it has a rotational spectrum because I know it has a permanent dipole.0344

Whereas something like nitrogen does not because there is no positive or negative end.0366

Therefore, there is nothing for the electric field to grab onto.0374

It just does not experience any force in the electric field.0377

Let us talk about some difference in energies.0383

So δ E is just going to be the E of J + 1 - E of the level right below it.0386

It is going to equal H ̅² all / 2I × J + 1 × J + 1 + 1 - J × J + 1.0399

Should I go through the multiplication?0420

That is fine, I will go ahead and go through it.0422

It is not a problem.0425

= H ̅² / 2I ×, we have J² + 3 J + J + 1.0426

This is going to end up being J + 2 - J × J + 1.0440

When you multiply these out and start canceling terms, you are going to get the J² + 3 J + 2 - J² – J = H ̅ / 2I × 2 × J + 1.0443

The 2 cancel leaving you with the δ E = H ̅² / I × J + 1.0474

In terms of the actual plank constant, H² / 4 π² I × J + 1.0493

This gives us the actual energies and again J is equal to 0, 1, 2, and so on.0505

These expressions right here, this and this, one in terms of the H ̅, one in terms of H,0519

they give us the actual change in energy from level to level.0525

So δ E is actually also equal to H ν.0532

H ν is equal to H² / 4 π² I × J + 1.0541

This H cancels this H, therefore, we have ν is equal to H / 4 π² I × J + 1.0551

Remember, I = MR² and K.0563

These are the frequencies, this equation right here.0570

These are the frequencies at which absorption transitions occur.0580

There are going to be very specific frequencies at which absorption transitions occur.0598

It is going to go from 0 to 1, 1 to 2, 2 to 3, and so on.0609

Rotational transitions they happen with frequencies in the microwave region or sometimes in the far infrared region.0617

It is a microwave region but also happened in the far infrared also.0646

In practice, let me go back to blue here.0655

In practice, we write the frequency is equal to 2B × J + 1 or B is equal to H / 8 π² I.0664

This B right here it is called the rotational constant.0689

The energies, the frequencies are actually expressed in terms of this rotational constant.0701

Also instead of frequency, we also use wave number which is actually more common.0711

We also use that ν tilde which is called wave number.0728

The wave number is anything with a tilde on it is just equal to the thing ÷ C, the speed of light.0737

We have ν tilde = ν/ C = 2B/ C × J + 1 and this 2B/ C that is the wave number, this B ̃.0744

B ̃ is just B ÷ the speed of light.0767

In this particular case, the B ̃ is going to equal to H ÷ 8 π² CI.0775

That is it, we have just taken B and we divide it by the speed of light.0786

Again, it is going to be expressed in terms of the wave number which is going to be0789

in units of inverse cm, that is the wave number or frequency which is going to be Hz.0793

It is just different ways of expressing the same thing.0799

Let us stop and think about this for a second.0802

What this thing gives us the J=0, 1, 2, and so on, what this gives is the energy needed.0811

Units of wave numbers, this inverse cm is actually a unit of energy.0833

What this gives is the energy needed.0838

Let us put it this way.0845

It is the energy that needs to be absorbed, in order to actually make the transition to the next energy level.0849

In order to make the transition from level J to the next level up J + 1.0880

When I put a J2, what this in effect?0898

This is going to be 2 + 1 is going to be 3, 2 × 3 is 6, 6 × this B ̃ that is the amount of energy0902

that needs to be absorbed in order to make the transition from 2 to 3.0912

Once again, what this ν gives us, what this equation gives us?0918

I’m sorry I forgot the J + 1.0926

What this equation actually gives us, it gives us the energy that needs to be absorbed in order to make the transition from J to J + 1.0932

This J is a level at which it is, in order make the transition from that to the next one up.0940

Let us go ahead and calculate a few.0947

You so ν sub 0 = 2B ̃ × 0 + 1 = 2B ̃.0953

At a wave number, 2B ̃, the energy transitions from level 0 to level 1.0976

Now ν sub 1 is equal to 2B ̃ × 1 + 1 that is equal to.0999

1 + 1 is 2, this is 4B.1008

This is the amount of energy for B, whatever this is energy.1012

We will worry about values in just a minute but this is the amount that is necessary1017

to be absorbed by something at level 1 in order to go to level 2 or B.1021

4B must be absorbed to go from level 1 to level 2.1038

Now, we are at level 2, let us calculate this one.1055

This is going to be 2B ̃ × 2 + 1 that is going to be 6B.1058

This is the amount of energy when you are at level 2.1067

When you are at level 2, this is the amount of energy that you need to absorb in order to make it to level 3.1070

6B must be absorbed to go from level 2 to level 3, that is what is happening here.1079

Notice the pattern 2B, 4B, 6B and it goes on.1098

In order to make the transition from level 3 to level 4, you need to absorb 8B ̃.1104

In order to make the transition from level 4 to level 5, you need to absorb 10B ̃.1109

In each case you need to absorb more energy to go to the next energy level up.1114

It is going to be important here and it tends to be a little bit confusing here for kids.1119

Now, when we actually see the spectrum, the spectrum itself, the lines are evenly spaced and the lines 2B apart.1124

You get a spectrum that looks like this.1153

Line, line, line, line, and this distance right here is always going to be 2B because you are going 2B, 4B, 6B, 8B.1156

This is very important.1166

I will do this in red.1167

Let me move over here.1169

I will write it down here.1178

On the spectrum, the lines are evenly spaced.1183

This does not mean that the energy levels are evenly spaced, they are not.1186

Each transition from 0 to 1 requires energy of 2B ̃, from 1 to 2 requires 4B ̃, from 2 to 3 requires 6B ̃,1217

that is how much energy that needs to be absorbed.1233

The spectrum itself, the lines are evenly spaced.1236

This is 2B, 4B, 6B, 8B.1240

This difference between them is the 2B.1242

However, that does not mean that the energy levels themselves are the same and we will demonstrate that in just a minute.1245

This does not mean that the energy levels are evenly spaced.1249

They are only evenly space on the spectrum itself.1252

The energies are as follows.1256

The levels of the individual energy levels are E sub J is equal to H ̅² / 2I × J × J + 1.1271

The energy level of 0 is equal to, when I put 0 in for this, I end up with 0.1291

Energy level of 1, I put 1 in for here, 1 + 1 is 2, 2 × 1 is 2.1300

2 H ̅²/ 2I = 2 H ̅²/ 2I.1307

The 2 cancels and I’m left with H ̅² / I.1314

My energy level for 2, I'm going to end up with 6 H ̅² / 2I which is just going to be 3 H ̅² / I.1320

When I calculate the third energy level, I'm going to end up with 12 H ̅² / 2I which is going to be 6 H ̅² / I.1332

Let me just go ahead and do one more.1344

Energy of a4th level is going to be 20 H ̅² / 2I which is going to equal 10 H ̅² / I.1346

Notice that the spacing between the energy levels is not constant.1359

From here to here, I have 2 H ̅² / I.1363

From here to here, I have 3 H ̅² / I.1367

From here to here, I have 4 H ̅² / I.1370

The energy levels themselves are different.1374

It is the spectrum spacing that is the same.1377

Once again, the spacing between the lines of the spectrum are evenly spaced.1386

The space between them is always going to be 2B ̃.1406

Let us go ahead and draw what this looks like.1410

Let us see if I should go ahead and do it on this page.1415

I’m going to go ahead and do it on the next page.1419

Let me go ahead and start down here.1423

Let me start with the 0 level and I will go to 1 level and I will go to the 2 level and I will go to the 3 level.1426

In case I got 0, 1, 2, 3.1445

I’m going from the 0 level to the 1 level, I'm going to have to absorb 2B amount of energy.1457

In order to go from 1 to 2, I'm going to have to absorb a total of 4B energy.1464

In order to go from 2 to 3, I’m going to have to absorb the total of 6B.1473

The amount of energy in order to go from 3 to 4, sure enough I'm going to have to absorb 8B.1479

The spectrum is just going to tell me how much energy I’m actually absorbing.1486

Here is going to be to 0 level, I'm going to get a line at that.1496

I’m going to get a line here and here.1502

This is 2B, this tells me 4B, this is 6B, and this is 8B.1510

The spacing in between is constant on the spectrum.1521

This is the spectrum, the amount of energy that takes to go from 1 level to the next1526

is not constant because the levels, the energies themselves are not constant.1532

This definitely distinguish between the energies and the amount of energy that we need to absorb in order to make the transition.1538

That is what is going on here, it is the spectrum spacing.1545

This 2B that is constant on the spectrum not here.1548

Let us see what we have got.1556

We have frequency = 2B × J + 1 or wave number = 2B ̃ × J + 1.1561

In either case, we have J goes 0, 1, 2…1575

We have B equal 2H / 8 π² I.1583

We also have the B ̃ = H / 8 π I² C × I.1590

These expressions give the frequency and or wave number required to transition, required to go from level J to level J + 1.1600

Of course, the energy of each level is given by H ̅² / 2I × J × J + 1.1633

These are the actual energies of the levels starting with J = 0 point energy.1643

Since the rotational inertia is equal to the reduced mass × the radius², notice the I shows up in those expressions.1685

We can use the separation between lines on the spectrum which we know is 2B to find R, the bond length of the molecule.1697

Profound, this is how we find bond lengths for most molecules, always the ones that have permanent dipole moment.1735

Let us go ahead and do an example.1751

Transitions between the rotational energy levels occur in a microwave for infrared region of the electromagnetic spectrum as we said.1760

One such spectrum for hydrogen bromide HBR79 shows a separation between adjacent lines of 16.7 inverse cm.1772

Use this data to calculate the rotational inertia and the bond length R for the molecule.1785

We know the spacing is 2B.1792

2B ̃ is equal to this 16.70 inverse cm.1805

B ̃ is equal to H / 8 π² CI.1815

2B which is 2 H / 8 π² C × I is equal to 16.70.1829

Let us go ahead and find I.1842

We want calculate I and we want to calculate R.1844

The first thing that we need to do is we need to find I.1846

We rearrange and we get I is equal to 2H ÷ 8 π² × 16.7.1849

We have 2 × 6 point this is planks constant, 626 × 10⁻³⁴ J/ s ÷ 8 π² × C1859

which is going to be 3.0 × 10¹⁰ cm/ s and then × our 16.7 inverse cm.1895

And then when we end up actually solving for this, calculating it, which is not my favorite part.1910

I still do not like all these numbers floating around, they make crazy.1917

But in any case, 5.594 × 10⁻⁴⁶ and we end up with a unit of kg m².1920

We went ahead and we found I, that is the rotational inertia of this particular system which is the HBR molecule.1932

Let me see, do I have another page?1943

We know that I is equal to the reduced mass × R².1954

We have I, we have the reduced mass, we can find that.1960

We need to find R, the bond length.1965

The first thing we need to do now is go ahead and find the reduced mass.1967

Let us go ahead and do that on the next page.1972

The reduced mass is equal to mass 1 × mass 2 ÷ mass 1 + mass 2.1978

The mass 1, the bromine is going to bromide, it is going to be 79 atomic mass units × hydrogen1990

which is 1 atomic mass unit ÷ the 79 + 1, which is the 80 atomic mass units.2002

We are going to need to change that to kg.2011

It is 1.661 × 10⁻²⁷.2013

There are that many kg in 1 atomic mass unit and we end up with a reduced mass equal to 1.64 × 10⁻²⁷ kg.2020

We are ready to solve.2037

I is equal to reduced mass R² which means that R is going to equal the rotational inertia ÷ the reduced mass the square root of that2039

And that is going to equal 5.594 × 10⁻⁴⁶ kg m² ÷ the reduced mass which is 1.64 × 10⁻²⁷.2051

It is going to be kg, kg cancels kg.2073

This is the square root and the square m² gives us a bond length of 5.8 × 10⁻¹⁰ m or 5.8 angstroms.2077

There we go, we use our microwave spectra data and we know the distance between the absorption bonds is going to be 2B ̃.2096

We use that information to find the rotational inertia and we use the rotational inertia to find the bond length.2108

That is all.2115

Thank you so much for joining us here at www.educator.com.2116

We will see you next time, bye.2118

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