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Example Problems I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Example I: Calculate the Bond Length 0:10
  • Example II: Calculate the Rotational Constant 7:39
  • Example III: Calculate the Number of Rotations 10:54
  • Example IV: What is the Force Constant & Period of Vibration? 16:31
  • Example V: Part A - Calculate the Fundamental Vibration Frequency 21:42
  • Example V: Part B - Calculate the Energies of the First Three Vibrational Levels 24:12
  • Example VI: Calculate the Frequencies of the First 2 Lines of the R & P Branches of the Vib-Rot Spectrum of HBr 26:28

Transcription: Example Problems I

Hello and welcome back to, welcome back to Physical Chemistry.0000

Today, we are going to start our collection of example problems for molecular spectroscopy.0004

Let us jump right on in.0009

Example number 1, the rotational spectrum of hydrogen iodide shows a spacing between spectral lines of 3.82 × 10¹¹ Hz.0012

Calculate the bond length R sub E.0023

Let us see what we can do.0027

There are several ways we can do this under various units.0031

Let us go ahead and work in inverse cm because that is pretty much the standard unit of spectroscopic study and data.0034

Let us convert to the inverse cm.0042

Let me go ahead and work in blue here.0045

Let us convert to inverse cm.0050

3.82 × 10¹¹ Hz is equal to 12.852 inverse cm.0060

Please double check this for me because I think I may have 1 or 2 digits missing here.0093

Slightly different from the paper that I'm looking at while I’m writing this.0099

Please confirm this.0103

The numerical values are altogether not that important.0105

What is important is the process.0107

We went ahead and we converted to inverse cm, that is the unit we want to work in.0109

The spacing between the spectral lines is equal to 2B.0114

2B is equal to 12.852 inverse cm.0122

Now, here is where we have to be a little bit careful.0137

Sometimes you are going to work in inverse cm.0143

Sometimes you are going to work in inverse meters.0145

It all depends on what the actual units are of the other constants that you are using.0146

In this particular case, let us work in inverse meters.0152

Recall that 1 inverse cm is equal to 100 inverse meters.0160

This is the reverse of the normal 100 cm/ 1m.0165

Because it is inverse, it is 1 inverse cm is 100 inverse meters.0169

Be very careful.0174

If you are not quite sure why, go ahead and work it out by doing unit of cancellation.0175

2B is equal to 1285.2 inverse meters.0184

Therefore, B our rotational constant is equal to 642.6 inverse meters.0198

I will get back to that in just a second, sorry about that.0210

It is equal to planks constant/ 8 π² C × the reduced mass × R sub E².0212

The reduced mass is going to equal the product of the masses divided by the sum of the masses.0225

In the case of hydrogen, we are going to have 1.01 atomic mass units × 34.97 atomic mass units for iodine 127.0232

No, that is not correct. 0248

I think I have made a little bit of mistake in my calculations here.0255

34.97 that is the atomic mass for chlorine 35.0259

I'm going to go ahead and use 1 atomic mass unit for hydrogen.0274

I’m just going to go ahead and use my 127 divided by 1 + 127.0284

Whatever it is that ends up being, it is going to give you some value in the atomic mass units.0288

That number you are going to multiply by, let us go ahead and call it Z.0302

Z atomic mass units × 1.661 × 10⁻²⁷ kg per atomic mass unit.0307

Because we want the final value in kg.0318

Our final answer, this μ, our reduced mass whatever this 127 divided by 128, is going to be,0327

127 divided by 128, you are looking about 0.999.0337

It is essentially going to be 1.0343

I’m just going to go ahead and take my 1.66 × 10⁻²⁷ kg as my reduced mass.0345

That is the value that we actually end up putting here.0355

We have our B which is 642.6 inverse meters, that is equal to 6.626 × 10⁻³⁴ J-s0359

divided by 8 π² × 2.998 × 10⁸ m/ s.0381

Meters and meters, meters per second × 1. 66 × 10⁻²⁷ kg × R sub E².0391

I solved for the numbers.0411

And again, the numbers I’m getting might be slightly off.0413

The numerical values are not important.0416

What is important is the process, realizing that the spacing is 2B.0419

Put into the equation for B and solving for the RE².0422

I will just go ahead and use the numbers that I have written down here.0427

They may or may not be correct.0430

They are only going to be off by very little.0431

2.67 × 10⁻²⁰ m².0434

Therefore, the bond length is going to be 1.63 × 10⁻¹⁰ m or 163 pm.0443

Example number 2, the ground state equilibrium bond length for CO is 112.8 pm.0463

What is the value of B, the rotation constant?0470

That is the same problem except the other way around.0473

By the way, I should let you know before I start the problem, there is a wonderful app available for your phone 0477

that actually converts between frequency in Hertz, frequency in wave number, energy and wavelength.0504

I will give you all 4 values, it is really nice.0521

It is called spectral unit converter.0527

Let us go ahead and do this.0536

B is equal to/ 8 π² C μ R sub E².0539

Μ is equal to 12 × 16/ 12 + 16 × 1.661 × 10⁻²⁷ kg.0553

We end up with B equal to 6.626 × 10⁻³⁴.0570

This is the part I hate most in the quantum mechanics spectroscopy, is actually doing all the calculations.0578

The theoretical work is much nicer because it may be symbolic and drives you crazy 0584

but you have the tedium of doing the arithmetic.0588

8 π² 2.998 × 10⁸ m/ s and this reduced mass ends up being 1.139.0596

I hope you got the arithmetic is correct.0609

I never really know any more.0610

This is going to be 112.8 × 10⁻¹² m².0615

When I do all that, I end up with 193.2 inverse meters or 1.932 inverse cm.0627

That is the unit data that you will find in the spectroscopic data in your books or in a table.0644

Let us see what we have got.0653

Example number 3, if we assume that the kinetic energy of rotation for diatomic molecule 0656

can be approximated by the classical equation, the kinetic energy is ½ the rotational inertia × the angular velocity²,0661

the moment of inertia rotation are same thing.0669

Calculate how many rotations carbon monoxide in the J = 5 state makes in 1s?0672

Rotational constant for CO is B = 1.932.0677

Just happens to be what we just figured out.0682

The energy of the rotational state is equal to H ̅²/ 2 I × J × J + 1.0687

This is most of the problems where there is a bunch of ways to do this 0699

depending on what you want substitute into what, how you want to arrange it.0701

Do not feel that the way that I’m doing is actually the only way to do it.0706

It is just a question of putting equations together.0710

Setting this equation equal to this equation and solving for whatever it is you happen to solve.0713

In this case, the angular velocity.0720

That is the equation there.0726

K is equal to ½ I × ο².0729

Ο is angular velocity, it is radiance per second not revolutions per second.0734

We will multiply that by 2 π because there are 2 π radians in 1 revolution.0739

H ̅² is equal to H/ 2 π² which is equal to H²/ 4 π².0754

When I put that in there, I get that the energy is actually equal to H²/ 8 π² I × J × J + 1.0770

B is equal to H/ 8 π² I.0786

It is equal to the B ~ inverse cm.0804

This is in Hz, this is B = the rotational constant inverse cm × the speed of light.0812

I end up with is, this is B, when I put all of this in here I get E sub J is equal to H × B × C × J × J + 1.0821

This is the one that we want to set equal to ½ I ο².0840

When we solve for ο², it is equal to 2H B ~ C/ I × J × J + 1.0848

If B ~ is equal to H/ 8 π² CI, that means that 1/ I is equal to 8 π² C B ~ over that.0863

1/ I E, this is 1/ I, I will just multiply this by that and I end up with ο² = 16 B ~² π² C² J × J + 1.0886

I end up with ο² equal 16 × 1.932² × π² × 2.998 × 10¹⁰ cm/ s² inverse cm.0912

I just want to make sure my units match.0932

J = 5, 5 + 1 ο² ends up equaling 1.59 × 10²⁵.0937

Therefore, the ο is 4.0 × 10¹².0950

This is radians per second.0956

I’m going to multiply that by 1 revolution.0960

I revolution is 2 π radians.0963

I will divide by 2 π.0966

Sorry, I said earlier to multiply by 2 π.0967

Just make sure the units match.0969

It goes with radian and I end up with 6.34 × 10¹¹ revolutions per second.0972

In 1s, that molecule is spinning that many ×, it is very fast.0984

Potassium bromide shows an intense single peak at 213.0 inverse cm.0993

There is a force constant and a period of vibration.0999

The fundamental frequency of vibration is 1/ 2 π C × the force constant divided by the reduced mass ^½.1005

We need to find μ, the reduced mass of potassium bromide.1019

In this case I have not specified the isotope of potassium and bromide.1031

It is not specified, just take the most abundant isotopes.1036

In the case of potassium, we would have potassium 39 and we would have bromine 39.1040

We are using the potassium 39 and 79 for bromine.1053

You can actually use 39 and 79 just as is.1059

Or if you wan to be a little bit more exact, you can go ahead and look up 1063

the actual atomic mass unit weight for the individual isotopes.1066

Wikipedia, or any other source that you might have.1071

In this case, μ is going to equal 38.964 × 78.918 divided by 38.964 + 78.918.1075

I’m going to multiply all that by 1.661 × 10⁻²⁷ kg.1100

And I end up with a reduced mass of 4.33 × to 10⁻²⁶ kg.1107

Therefore, my fundamental vibrational frequency is 1/ 2 π C × K/ μ ^½, which gives us when I solve for K,1115

K is going to equal to 2 π C.1139

The fundamental vibration frequency, all of that² × μ.1147

K is equal to 2 π 2.998 × 10¹⁰ × the fundamental vibration frequency is 213.1156

That is a single peak.1173

The peak that you see is the fundamental vibration frequency of the ground vibrational state 213.0 inverse cm.1174

It is 10¹⁰ because we are centimeters per second, I’m going to go ahead and square all of that 1186

and multiplied by 4.33 × 10⁻²⁶ kg.1193

And what I end up with is a force constant equal to 69.71, that its unit wise is kg/ s².1200

Which is the same as kg/ s² × m/ m, which is the same as a N/ m.1213

Kg-m/ s²/ m, that is a Newton.1224

Our unit is N/ m.1230

Do not let this initial unit that you get throw you off.1234

A kg/ s² is equal to N/ m.1239

The period in any vibration is equal to 1/ the frequency.1246

The frequency is equal to, in terms of wave numbers just multiply the wave number in inverse cm by the speed of light.1254

You end up with 1 divided by 213.0 inverse cm × the speed of light 2.998 × 10⁸ N cm.1265

10¹⁰ cm/ s.1281

You will get a period vibration of 1.57 × 10⁻¹³ s.1285

We expected to be really small and really fast.1293

Example 5, chlorine 2 has a force constant of 319 N/ m.1304

We would like you to calculate the fundamental vibration frequency and E1, E2, the energies of the first 3 vibrational levels.1319

A is pretty straightforward, we have the equation E sub 0 = 1/ 2 π C, the force constant reduced mass ½.1329

CO₂, the reduced mass of chlorine 2 is going to equal 34.969² / 2 × 34.969 × 1.661 × 10⁻²⁷ kg.1345

What I end up with is a reduced mass of 2.90 × 10⁻²⁶ kg.1370

Therefore, our ν sub 0 = 1/ 2 π² × 2.998 × 10⁸.1380

I’m going to use the 10⁸ m/ s because the K is given in N/ m.1398

Meter is the unit that is there.1406

319.0 N/ m divided by 2.90 × 10⁻²⁶ kg ^½.1409

I end up with ν sub 0 = 55678, this is in inverse meters × 1 inverse cm is equal to 100 inverse meters, 1424

it is equal to 55678 inverse cm.1441

That is the fundamental vibration frequency.1450

Part B, the energy of a given a vibrational state is equal to this, × R + ½ inverse cm,1453

we have the vibrational term, that is equal to ν sub 0.1476

Once it is frequency, it is in inverse cm, it is that × R + ½.1482

Where R takes on the value 0, 1, 2, and so on.1500

Therefore, our G of 0 is equal to ½ N sub0 = 557/ 2 we get 278.5 inverse cm.1505

That is the energy for the ground state, that R = 0.1524

The R = 0 vibrational energy.1529

The vibrational energy of the first state is 3/2 and 0 is equal to 835.5 inverse cm.1535

G of 2 = 5/2 ν sub 0 = 1392.5 inverse cm.1551

If you would take the difference between this and this, the difference between this and this,1574

you will find the spacing is the same.1577

Which is the spacing that we expect, we expected it to be the same under the harmonic oscillator approximation.1580

For the molecule hydrogen bromide, our rotational constant is 253.771 GHz 1589

and our fundamental vibration frequency is 79.414 THz.1598

Under the rigid rotator harmonic oscillator approximation, calculate the frequencies of1604

the first two lines of the R and P branches for the vibrational rotation spectrum of HBR.1609

We would be using a rigid rotator harmonic oscillator approximation.1617

Let us see, where should we begin?1631

Let us go ahead and begin with the equations.1647

Let me go ahead and work in red.1649

For the R branch, let us go ahead and make some conversions to inverse cm, GHz, THz.1651

GHz we remember is 10⁹, THz is 10¹².1661

B, 253.771 GHz goes to 8.465 inverse cm.1667

That is equal to B~ and our ν sub 0 which is 79.414 THz that goes to 2648.966 inverse cm,1686

that is equal to our ν~.1707

The observed frequency for the R branch of the harmonic oscillator rigid rotator approximation1712

is ν sub 0 + 2B × J + 1 or J runs from 0, 1, 2, and so on.1721

The first line, our 1 line which represents the 0 to 1 rotational transition, not R1.1736

Let me go back.1754

The first line represents the 0 to 1 transition J = to J = 1.1757

This is the R 0 line.1770

That is going to equal, ν sub 0 which is 2648.966 + 2 × 8.465 × 0 + 1.1775

It is going to equal 2665.896 inverse cm.1803

The second line of the R branch, it represents the 1 to 2 transition.1817

The J value that we are going to be putting in are the initial J values.1825

The J value of the lower quantum state to where you are coming from.1829

Therefore, our R1 line is going to equal 2648.966 + 2 × 8.465 × 1 + 1.1834

Therefore, R1 is going to put us at 2682.826 inverse cm.1850

For the P branch, I will just go ahead and do the frequency of the P branch, that decreasing frequency.1866

The observed frequency we see is the fundamental vibration frequency – 2B.1878

In this case, J starts at 1, it is going from 1 to 0, 2 to 1, 3 to 2.1884

J = 1, 2, 3, and so on.1895

Therefore, our first P line is P1.1901

Let me show what this represents.1910

The first P line represents the transition from 1 to 0, J = 1 to J = 0.1911

That is the P1 line, that is going to be 2648.966.1919

That is that, -2 × 8.465.1927

Very simple because really all we are doing is putting numbers into equations but 1932

there are many equations and there are many numbers all over the place.1938

You have to be extra careful.1942

What we have been trying to do here with quantum mechanics, thermodynamics.1946

What is wonderful physical chemistry stuff.1951

We get a P1 of 2632.036 inverse cm.1954

The second P line which represents the transition from 2 to 1 J = 2 in the lower vibrational state to 1 in 1966

the upper vibrational state, that is P2 and that is going to equal 2648.966 - 2B J.1978

-2 × 8.465 × 2.1996

Therefore, our second line is going to be at 2615.106 inverse cm.2008

That is what we predict based on the data that we have been given.2016

Thank you so much for joining us here at

We will see you next time for more example problems.2024

Take care, bye.2027