For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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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
- Example I: Calculate the Bond Length
- Example II: Calculate the Rotational Constant
- Example III: Calculate the Number of Rotations
- Example IV: What is the Force Constant & Period of Vibration?
- Example V: Part A - Calculate the Fundamental Vibration Frequency
- Example V: Part B - Calculate the Energies of the First Three Vibrational Levels
- Example VI: Calculate the Frequencies of the First 2 Lines of the R & P Branches of the Vib-Rot Spectrum of HBr

- 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

### Physical Chemistry Online Course

### Transcription: Example Problems I

*Hello and welcome back to www.educator.com, 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-s*0359

*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 of*1604

*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 approximation*1712

*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 www.educator.com.*2022

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

*Take care, bye.*2027

1 answer

Last reply by: Professor Hovasapian

Wed Dec 14, 2016 1:19 AM

Post by Ashley Robbins on December 13, 2016

How do I find the fundamental vibrational frequency and bond dissociation constant if I am only given the energy states and their transition energies?