Raffi Hovasapian

The Harmonic Oscillator II

Slide Duration:

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 Harmonic Oscillator II

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Section 14: The Harmonic Oscillator: Lecture 2 | 43:04 min

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1 answer

Last reply by: Professor Hovasapian
Tue Apr 12, 2016 3:42 PM

Post by Tammy T on April 12, 2016

Dear Prof. Hovasapian,

-Watching from 14-20 minute mark, the Energy of Harmonic Oscillator or the Energy of vibrating diatomic molecule E=hv(r+1/2) is the Potential aka. the Potential Energy of the vibrating molecule? Does Kinetic Energy counted in this Energy of Harmonic Oscillator?

-Is "r" in Energy E formula the Energy level? It is mentioned that it is 'a quantum number'. Is r the n quantum number?

Thank you!

2 answers

Last reply by: Professor Hovasapian
Wed Dec 30, 2015 1:18 AM

Post by bohdan schatschneider on December 29, 2015

Dr. Hovasapian, why no mathematical explanation regarding Taylor series expansion and the selection of the quadratic term for the potential? I came here looking for why the linear term in the expansion vanishes...

The Harmonic Oscillator 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

The Harmonic Oscillator II

Intro 0:00
The Harmonic Oscillator II 0:08

Diatomic Molecule
Notion of Reduced Mass
Harmonic Oscillator Potential & The Intermolecular Potential of a Vibrating Molecule
The Schrӧdinger Equation for the 1-dimensional Quantum Mechanic Oscillator
Quantized Values for the Energy Level
Ground State & the Zero-Point Energy
Vibrational Energy Levels
Transition from One Energy Level to the Next
Fundamental Vibrational Frequency for Diatomic Molecule
Example: Calculate k

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 Harmonic Oscillator II

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

Today, we are going to continue our discussion of the quantum mechanical harmonic oscillator.0004

Let us dive right on in.0008

A vibrating molecule can be modeled by the harmonic oscillator.0011

This is all that is happening, this is going back and forth like this.0017

A vibrating molecule can be modeled by the harmonic oscillator.0028

A diatomic molecule has two masses that are moving, not one fix end and the other one moving back and forth like that.0053

A diatomic molecule has two masses that oscillate and no fixed end.0067

In general, if you have like a really huge atom and a tiny atom, for all practical purposes, the huge atom is not going to move very much.0088

You can consider that fixed end and the tiny atoms itself would vibrate back and forth like that.0098

We cannot always guarantee you that so we need something a little bit more sophisticated than just a fixed wall and one mass moving.0103

The situation we have is this.0111

Something like that where you have mass one and mass two.0116

There are some modifications that we make actually to the equations of motion.0122

Instead of one equation of motion, we have two equations of motion, one for each mass.0127

We combine them and we come up with a new differential equation.0131

When I can to go through that process but will say this much.0134

We will not discuss the modifications due to the equations of motion but we end up with this.0141

But we end up with a differential equation that we end up with is μ D² X DT² + KX equal 0.0169

This is exactly the same equation that we came up with for the classical harmonic oscillator0181

with one fixed end of the wall and one body, one mass moving back and fourth.0185

The only difference is instead of the mass of the single ball moving back and forth, this is μ something called the reduced mass.0191

I will tell you what it is.0198

Where μ is called the reduced mass and is given by the μ is equal to 1/ M1 + 1/ M2.0202

It is also equal to M1 +,0232

I'm sorry to not μ, this is 1/ μ.0235

M1 + M2/ M1 M2.0240

If you like μ directly is M1 M2/ M1 + M2.0245

Basically, what we have done when we modify the equations of motion,0253

we have come up with this way of combining the masses into something called reduce mass.0256

By combining I do not just mean adding, it is this, this is the relationship.0261

The reduced mass is 1/ M = 1/ M1 + 1/ M2.0265

We have combined them and we have basically taken this two body problem and we have turned it into a one body problem.0270

This is a single equation, this is a single number.0277

This happens a lot in classical mechanics and classical physics.0280

In fact, we can do this, we can take a two body problem or three body problem and turn it into a one body problem.0285

It is actually quite extraordinary that we can do this.0292

It is all we have done.0296

Everything is the same, the only difference is this thing called the reduced mass which is a combination of the masses of the two objects.0297

In this case, the two atoms or maybe the two blocks of cement, whatever it is you have to be dealing with.0305

This is possible to do but you will end up with the same equation.0310

You end up with the same general solution, that is what makes this beautiful.0314

Let me go to blue actually.0324

This notion of reduced mass has allowed us to treat two body problem as a one body problem.0328

The same as before.0357

The equation, μ D² X DT² + KX is equal to 0, is exactly the same as before.0362

The general solution is the same, everything is the same.0391

Everything that we did in the previous lesson applies here.0396

The general solution is the same.0400

We have X of T equals C1 × the cos of ω T + C2 × the sin of ω T.0407

Where ω is the angular velocity.0419

Ω is K/ M, the only difference is ω is K.0422

This time instead of /M it is /μ.0428

Everything else is the same, nothing is changed.0430

Wherever we see mass before, we are replacing it with a reduced mass.0433

Let us take a look, we are talking about molecules.0440

We are talking about atoms and atoms that we come together back and fourth, they oscillate like this.0445

Let us look at how good an approximation of the harmonic oscillator potential is to the actual intra molecular potential.0454

This intermolecular should be inter atomic potential but that is fine.0503

I will just leave this as intermolecular.0513

They are potential of the vibrating molecule.0520

These are one of those terms that are stuck.0524

We are talking about potential that exist between the two atoms of the molecule.0527

Intermolecular means between two molecules.0533

Basically, you have something that can look like this.0539

I will go ahead and draw something like that.0542

In this axis, we are going to have the energy.0545

On this axis, we are going to have the distance that one atom is to the other atom.0550

I will go ahead and mark S of 0 here.0557

And this is just S, this is the distance that the two atoms are away from each other.0560

We are end up getting something like this.0564

Let me go ahead and put a minimum mark here.0567

I will go ahead in red.0580

I will go ahead and draw something like that, sort of magnified.0582

This right here, we will go back to blue.0599

This curve right here is the potential energy, that is the ½ KX².0601

That is the parabola, that is the potential energy of our harmonic oscillator.0608

As we pull something further apart the potential energy increases.0614

They want to be pulled back together.0619

As we squeeze them together, the potential energy increases.0622

They want to push themselves apart.0625

There are someplace where the equilibrium position is actually perfect where the energy is minimized.0626

That is this point right here.0632

This is the potential energy, this is the actual potential energy curve of0634

what happens when you take two atoms and you squish from together.0647

Over here, there are an infinite distance apart from each other.0651

As you start bring them together, bring them together,0654

as we are bringing the atoms from an infinite distance apart from each other, the closer and closer and closer.0657

The attraction that they feel toward each other is actually going to drop the potential energy.0662

The potential energy is going to drop.0667

It is going to hit a point where they are as close as they are going to be to each other where energy is minimized.0670

If you push them any closer together, the energy rises again.0675

This point right here, this little minimum that is the actual bond length.0680

That is the length at which the atoms are comfortable.0685

They are as close as they are going to be to each other.0689

It pull them apart, the potential energy gets bigger.0691

You push them together, the potential energy gets bigger.0694

This is an approximation based on the harmonic oscillator.0698

Again, harmonic oscillator you pull them apart, the potential energy increases down to 0.0702

It push them away from equilibrium position goes to 0.0707

For very small displacements which is pretty much what happens when a molecule vibrates and a molecule is not going like this.0713

There are not huge displacements away from the equilibrium position.0721

They are very small and for very small displacements this parabolic approximation to the actual potential energy curve is very good.0724

In fact, you can see them overlap beautifully and that is what we are doing.0736

We are a approximating this inter nuclear potential.0741

It is not intermolecular, it is inter nuclear potential.0746

Sorry, earlier in the day I was thinking about intermolecular forces and I think I’m getting them mixed up.0759

The inter nuclear potential, this is what actually happens when you take atoms and bring them closer together.0764

This pair of parabola is an approximation, this is the harmonic oscillator.0770

For very small displacements, they are right on top of each other.0774

It was a perfect match.0778

It is only when we get it to huge oscillations that they start to deviate from each other0779

or our parabolic model and the potential energy does not fit with the data anymore.0787

But it is never going to be that way.0793

For normal vibration of frequencies the displacement is very small.0795

We are good, let us go ahead and write this down.0802

For small vibrations about the equilibrium position it is an excellent approximation.0808

Of course that place which is the minimum, that distance, that distance they are from each other, that is the bond length.0846

It makes sense.0853

Now the Schroeder equation and we want talk about a quantum mechanical harmonic oscillator.0856

The Schrӧdinger equation for the 1 dimensional quantum mechanical harmonic oscillator looks like this.0865

It is going to be –H ̅²/ 2 and again we use the reduced mass.0888

D² ψ/ DX² + the potential energy × ψ equals the energy × ψ.0896

We know what the potential energy is, it is ½ KX².0905

When we put B equals ½ KX².0909

When we put that into here, rearrange this equation, we get D² ψ DX² + 2 μ/ H ̅² × energy - ½ KX² × ψ equals 0.0914

This is the equation that we end up solving.0939

This is the Schrӧdinger equation for the harmonic oscillator.0941

When we solve this, when we solve this equation which is actually not an easy thing to do0944

because the coefficients are no longer constant.0954

We will start with the energies first.0959

We get the following quantized values for the energy, that right there.0963

For the energy E of the system, we get that E is equal.0984

Let me do this in blue.0997

The energy of the system was equal to H ̅ × K/ μ ^½ × R + ½,1000

Where R is equal to 0, 1, 2, and so on.1013

R is another quantum number.1019

It is another quantum number so are 0, 1, 2, it can only take on those values.1026

The energy of the system is quantized, it cannot be any energy.1032

It is going to be one energy and then it is going to be another energy.1036

It does not make a nice smooth transition jumps.1041

It is quantized.1044

K/ M ^½ is just ω.1047

H ω R + ½ those are the energies for the quantum mechanical harmonic oscillator.1051

Since frequency equals ω/ 2 π and E is also equal to planks constant × frequency × R + ½.1061

Again, ω and K/ μ ^½ power and frequency equals ω/ 2 π,1084

Which is going to equal 1/ 2 π K μ ^½.1107

In most books you will see that the energy is equal to H ̅ ω × something that looks like a μ + ½.1119

Μ= 0, 1, 2, and so on.1142

These V are modified μ, what ever it is that they call this thing in the books.1148

Let me go ahead and write it out.1157

This and this, look too much alike in a book.1161

I reserve that symbol for frequency and R for the quantum number.1175

When we plot the energies, we end up doing something like this.1201

We have our parabolic.1205

This is R = 0, R =1, R=2, R=3.1212

Again, as we are getting further and further away, our potential is increasing.1224

This one, our energy sub 0, our 0 state energy is equal to ½ H ̅ ω.1230

Our energy for level 1 is equal to 3/2 H ̅ ω.1239

Our energy 2, is equal to 5/2 H ̅ ω.1247

Our energy 3, and so on, equals 7/2 H ̅ ω.1252

Again, we are just plugging them into these values.1257

Notice that the successive energy levels ½ H ̅ W, 3/2 H ̅ W, 5/2 H ̅ W, 7/2 H ̅ W.1264

They are the same, the jump is the same.1275

It is H ̅ ω.1278

Notice that the successive energy levels have equal spacing.1284

That equal spacing is H ̅ ω.1308

Notice, specially that the ground state energy is E0 ½ H ̅ ω.1313

It is not equal to 0.1338

This is very different from the classical harmonic oscillator.1342

The classical harmonic oscillator, when the mass is sitting at its equilibrium position, it is not moving, it is not stretched or compressed.1345

There is no potential energy, there is no kinetic energy, it is 0.1351

The ground state of the classical harmonic oscillator is 0.1354

The ground state of the quantum mechanical harmonic oscillator is not 0.1357

There is always some vibration going on, that is what is happening here.1361

The ground state, the one with a quantum number R=0 is called the 0 point energy.1367

The 0 point energy does not mean it is 0.1389

Sometimes it will be, sometimes it will not be.1393

In the case of a quantum mechanical harmonic oscillator, it is not.1397

The fact that it is not 0 it actually comes from the uncertainty principle.1404

It does not come from as a result of the uncertainty principle.1418

It is a result of the uncertainty principle and we will show you how.1422

The energy of the system is equal to its kinetic energy + its potential energy.1436

The kinetic energy can be written as P²/ 2 × the mass.1442

½ mass × velocity² is the same as the momentum² / twice the mass + ½ KX².1448

Let me go to red here.1459

We have momentum and we have position.1463

In order for the energy to be 0, I have to be able to make that position arbitrarily 0 and the momentum arbitrarily 0.1466

And we know from previous work that we cannot do that.1474

These two operators do not commute.1476

You cannot specify to an arbitrary degree of precision or accuracy both the position and the momentum simultaneously.1479

As you make one better, the other one gets worse.1488

But you cannot arbitrarily make them both.1491

You can bring the error of both to 0, which would make the 0.1494

As a result of that, because the energy is kinetic + potential, you have the momentum and1498

the position showing up in the same expression, you are never going to get something which is 0.1503

That is where it comes from.1509

Let us go back to black here.1520

If we take the quantum mechanical harmonic oscillator as the model for an oscillating diatomic molecule or a vibrating diatomic molecule,1526

then the vibration energy levels of the molecule are given by what we said before E sub R = H ̅ × ω × R + ½.1561

It is giving me the different energy values of the different vibrational states of this vibrating molecule.1590

R goes from 0, 1, 2, and so on.1597

Molecule can transition from one energy level to the next if it absorbs or emits radiation energy of frequency μ.1604

It is vibrating, its molecule is vibrating.1649

If I hit it with some radiation, the frequency of the radiation matches the energy change, the H ̅ ω,1651

the system is going to start vibrating at the next level up.1660

Or if it is vibrating to the next level up, it releases enough energy that happens to match certain frequency which is given by,1664

if the energy of that particular frequency happens to be H ̅ ω, then it will go from a higher energy state to the next lower energy state down.1673

It is just making transitions between energy states.1681

A molecule can transition from one energy level to the next, if it absorbs or emits radiation of frequency μ.1685

The change in energy is equal to planks constant × μ.1692

The radiation of a given frequency has energy equal to planks constant × that frequency.1698

Later in the course, we will demonstrate that the quantum mechanical harmonic oscillator1704

only allows transitions between successive energy levels, that is the δ R =+ or -1.1727

In other words, if I'm at the energy level 1 and if I want to get to the energy 5, I cannot just go directly from 1 to 5.1755

I have to go to 2, to 3, to 4, to 5.1762

In can only jump go up or down in individual stages.1765

Successive energy levels, I cannot make a huge leap like an electronic transition or something.1769

In this particular case, I have to pass through the successive stages.1773

This right here, this DR is equal to + or -1.1777

In other words, R 12345 54321 is called the selection rule.1782

Let us write here it is called the selection rule.1789

+ or - 1 is called the selection rule and you are going to see quite of them.1793

The change in energy from one state to another is the energy of R + 1 - the energy of R, the one stage above,1805

- the one stage that you are coming from.1814

Let us do it this way, H ̅ × K of μ ^½ × R + 1 + ½ - R × K/ μ ^½ × R + ½.1819

This is going to equal H ̅ × K/ μ ^½ × R + 3/2 - R - ½.1841

R and R, you end up with H ̅ K/ μ ^½ 3/2 - 1/2 is 2/2 = 1.1856

It equals that, and that equals H ̅ ω.1874

The change in energy, the transition from one energy level to the next one, either up or down is going to be H ̅ ω.1881

The change in energy is equal to planks constant × the frequency.1892

You see that the change in energy between successive energy levels of the quantum mechanical harmonic oscillator is H ̅ ω or H ̅ × K/ μ ^½.1896

Let us set them equal to each other.1911

We have H μ is equal to, this H ̅ is equal to H/ 2 π.1915

I’m just going to write it that way.1924

It is equal to H/ 2 π × K/ μ ^½.1925

The H's cancel and you are left with μ is equal to 1/ 2 π × K/ μ ^½.1935

In order for it to actually make the transition from one energy level to the other,1949

the quantum mechanical harmonic oscillator it has to absorb or emit energy of frequency that is given by this.1953

Nothing more than the spring constant and the actual reduced mass.1960

We are going to express this in terms of something called a wave number,1965

which is something that is what you actually the scales that you see in spectroscopic data are actually expressed in wave numbers,1968

not necessarily in frequencies, more often than not.1976

Anything with a little tilde sign over it is a wave number.1981

It is nothing more than the actual thing divided by the speed of light.1986

In this particular case, it is going to be 1/ 2 π × the speed of light K / μ ^½.1990

That is it, just take the frequency and divide by ψ.2000

That is all you are doing.2002

Where this μ tilde is the wave number.2007

It is called the wave number in units of inverse centimeter.2016

Since successive energy states are separated by the same energy which is H ̅ ω δ E is the same for every transition.2027

Δ E is the same for every transition so when we actually irradiate something or when we measure the radiation that actually being radiated.2069

We measure the radiation that actually gives off when it is dropping back down to is ground state.2074

It is all the same energy level from 5 to 4, 4 to 3, 3 to 2, 2 to 1.2080

All we measure is one value.2088

The spectrum for this only gives us one line at one frequency or one particular wave number.2090

Spectrum predicted by μ or μ tilde consists of a single line.2098

We see the spectrum predicted by that, predicts that it should be represented by a single line.2122

The prediction is good and fits well with the actual data that we correct.2134

The prediction is good and this single line is called the fundamental vibration of frequency.2146

For diatomic molecules, this line appears at around 10⁻³ inverse cm.2178

Or somewhere in the range of 10⁻⁴ 10⁻⁵ Hz, if you want to express it in terms of frequency.2209

This falls into the infrared range.2225

Now we can use IR spectra data and this thing.2231

We can use that to actually find force constant defined values of K for individual diatomic molecules.2251

How stiff is the bond.2265

That is what we are doing, we are finding K.2268

How stiff is the bond when it vibrates, is the atom going a lot or it is going a little really fast, what is it?2270

Let us go ahead and do an example.2279

Let me go ahead and do this in blue.2283

Our example is going to be, the IR spectrum of H 35 CO has a single line at 2886 inverse cm.2288

Calculate K, the force constant, for this molecule.2327

Let us go ahead and do it.2335

Let us see, we start with our basic equation which is μ tilde = 1/ 2 π C × K / μ ½.2338

We are going to get, 2 π C that² is equal to K/ μ.2364

Therefore, our force constant is going to be μ × 2 π C that².2378

We just have to work out all the values.2389

Let me write it again, K = μ × 2 π C μ tilde².2395

Let us go ahead and calculate μ.2403

Μ we said it is going to be M1 M2/ M1 + M2.2408

Hydrogen is 1 atomic mass unit or dealing with HCL 35, the chlorine is 35.2420

The atomic mass units/ 1 + 35 atomic mass units × 1.661 × 10⁻²⁷ kg/ atomic mass unit.2431

And we end up with μ = 1.615 × 10⁻²⁷ kg.2453

Therefore, K is equal to 1.615 × 10⁻²⁷ kg × 2 π × 3.0 C, which is 3.0 × 10⁸.2464

This is going to be m/ S × 2886 inverse cm.2495

It is going to be ×, I’m dealing with meters and cm.2508

I have to go 100 cm/ 1m and all of that is going to be².2511

This whole thing, that is going to be².2524

When I actually do this, I get K is equal to 478 kg/ s².2529

A Newton is a kg m/ s².2541

Therefore, a N/ m is equal to kg m/s² m.2549

The m cancels leaving me kg/ s² which is exactly the unit that we got.2557

We got N/m.2563

K =478 N/ m that is the force constant of the bond between Hydrogen and Chlorine 35.2566

Thank you so much for joining us here at www.educator.com.2582

We will see you next time, bye.2583

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