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 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
- Example I: Calculate the Frequencies of the Transitions
- Example II: Specify Which Transitions are Allowed & Calculate the Frequencies of These Transitions
- Example III: Calculate the Vibrational State & Equilibrium Bond Length
- Example IV: Frequencies of the Overtones
- Example V: Vib-Rot Interaction, Centrifugal Distortion, & Anharmonicity

- Intro 0:00
- Example I: Calculate the Frequencies of the Transitions 0:09
- Example II: Specify Which Transitions are Allowed & Calculate the Frequencies of These Transitions 22:07
- Example III: Calculate the Vibrational State & Equilibrium Bond Length 34:31
- Example IV: Frequencies of the Overtones 49:28
- Example V: Vib-Rot Interaction, Centrifugal Distortion, & Anharmonicity 54:47

### Physical Chemistry Online Course

### Transcription: Example Problems II

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

*Today, we are going to continue our example problems in molecular spectroscopy.*0004

*Let us jump right on in.*0008

*Our first example is the force constant for carbon monoxide is 1857 N/m.*0010

*The equilibrium bond length = 11.83 pm.*0019

*Using the rigid rotator harmonic oscillator approximation,*0025

*construct a table of energies for the first 5 rotational states for the vibrational levels R = 0 and R = 1.*0029

*The 5 rotational levels for R = 0, 5 rotational levels for R = 1.*0039

*Specify which transitions are allowed and calculate the frequencies of each transition,*0043

*associating each with its appropriate branch R or P.*0049

*Let us see what we can do.*0054

*Let me go ahead and work in blue today.*0056

*Under the rigid rotator harmonic oscillator approximation,*0061

*our energy equation is going to be energy RJ is equal to ν sub 0 R + ½ + this rotational constant × J × J + 1.*0065

*That is it, it is the vibrational energy + the rotational energy.*0083

*We have to find ν sub 0 that, and this, before we can actually start running J from 0 to 5.*0088

*Let us see what we can do.*0107

*Ν sub 0, the definition is 1/2 π C × the force constant divided by the reduced mass ^½.*0109

*We have to find the reduced mass.*0124

*The reduced mass is equal to the product of the weights divided by the sum of the weights.*0128

*12 × 16, 12 for carbon 16 for oxygen, divided by 12 + 16.*0135

*This is in atomic mass units, I’m going to multiply that by 1.661 × 10⁻²⁷ kg per atomic mass unit.*0142

*My atomic mass units cancel and I'm left with a reduced mass in kg.*0153

*When I do this calculation, I should get,*0159

*I hope that you are confirming my arithmetic, I am notorious for arithmetic mistakes.*0161

*The arithmetic is less important, the process is important.*0167

*The equations that you choose, that is what is important.*0171

*I may clear my throat a few more times than usual during these few lessons today.*0176

*× 10⁻²⁶ kg.*0181

*I want to keep writing these numbers over and over again.*0189

*When I go ahead and put this value of K into here, when I put this value ν into here and run this calculation,*0192

*I'm going to get a ν sub 0 equal to 214354.49.*0201

*The answer is actually going to be in inverse meters because this is N/ m.*0212

*When I convert that, I will just move that decimal over a couple of times,*0218

*I'm going to get 2143.54 inverse cm.*0221

*The problem is, the issues with spectroscopy are like many issues throughout quantum mechanics and thermodynamics.*0228

*They are just for conversion issues, just watch your units.*0234

*That is really all you have to watch out for.*0239

*We have that number.*0242

*Let us go ahead and find the rotational constant.*0248

*Let me go ahead and do that on the next page here.*0251

*Our rotational constant is equal to, the definition H/ 8 π² C × the rotational inertia.*0255

*We need I to put it into here.*0267

*I = that, that².*0273

*Therefore, I = R 1.139 × 10⁻²⁶ kg × RE is a 112.83 × 10⁻¹² m.*0277

*What we get is I = 1.45 × 10⁻⁴⁶ kg/ m², the unit of rotational inertia.*0299

*We put this value in here with all of the other values.*0314

*I will go ahead and write this one out, it is not a problem.*0323

*E = 6.626 × 10⁻³⁴ J-s divided by 8 π² × 2.998*0328

*× 10⁸ m/ s × 1.45 × 10⁻⁴⁶, that is the I.*0344

*When I do that, I will get 193.04 but this is going to be in inverse meters because a joule has meters.*0355

*That is going to give us, when I convert this, move the decimal over twice 1.9304 inverse cm.*0367

*Remember, inverse cm and inverse meters, it is the other way around.*0376

*It is going to be 100 inverse meters in 1 inverse cm, which is why this one goes to the left.*0381

*We have that and we have that.*0390

*We said that the energy was, let me go ahead and rewrite the equation.*0392

*The energy was E= sub RJ equal to ν sub 0 × R + ½.*0397

*This is the rigid rotator harmonic oscillator approximation + B ̃ J × J + 1.*0406

*Let me make my J a little more clear here.*0417

*For the R = 0 vibrational state, that is going to be E⁰ and J is going to run through these quantum numbers.*0420

*That is going to equal, when I put R and 0 into this equation, it becomes ½ ν sub 0 + B ̃ × J × J + 1.*0435

*And for the R = 1 vibrational state, I just put 1 into this equation and I get the E sub 1 J is equal to 3/2 ν sub 0 + B ̃ J × J + 1.*0448

*Let me double check that.*0464

*Therefore, our E0 J, when I actually put the ν 0 that I got and the B that I got in here,*0469

*I'm going to get equation that says 1071.77 + 1.9304, which is the rotational constant, × J × J + 1.*0478

*Now, I just set up a table of values.*0490

*I have J over here and I have my energy 0 J over here.*0492

*I’m going to take 0, 1, 2, 3, and 4.*0498

*I get 1071.8, I get 1075.6, I get 1083.4, I get 1094.9, I get 1110.4.*0502

*These are energies not frequencies of absorption or emission.*0524

*These are not spectral lines, these are not frequencies.*0528

*These are the energies of the actual rotational states in the R = 0 vibrational state.*0530

*Let us go ahead and do the energies for 1 vibrational state.*0539

*When I put the values in, I get 3, 2, 1, 5.*0546

*In other words 3/ 2, the ν sub 0.*0549

*3215.31 + 1.9304 which is the rotational constant, × J × J + 1.*0551

*Let us go ahead and set up my table of values again.*0570

*This is going to be E1 J.*0573

*I have 0, 1, 2, 3, and 4.*0576

*I have 3215.3, I have 3219.2, 3226.9, 3238.5, and 3253.9.*0583

*Let us go ahead and let me do it on one page here.*0607

*Let me see if I want to just go ahead and do this one in red.*0614

*I got J and I got E0 J.*0617

*Over here, I have J and I have E1 J.*0623

*I have 01234, 01234.*0630

*These are the first to 5 table of values.*0636

*I had 1071.8, 1075.6, 1083.4, 1094.9, and 1110.4.*0639

*Over here, I have 3215.3, 3219.2, 3226.9, 3238.5.*0657

*Again, I want them on the same page so I can see them together.*0673

*3253.9, the energies of the 0 vibrational state, the first 5 rotational for the 1 vibrational state.*0677

*The allowed transitions, they also asked about that.*0687

*The allowed transitions, δ J is + or -1.*0690

*The allowed transitions, the selection rule is δ J = + or -1.*0699

*Therefore, for the 0 state to the 1 state, the 0 vibrational state to the 1 vibarational state, I can go 0, 1, 2, 3, 4.*0709

*I have 0, 1, 2, 3, 4.*0724

*I can go from 0 to 1 that is + 1.*0727

*1 to 2, 2 to 3, and 3 to 4.*0731

*Let me do this one in black.*0736

*I can go from the lower 0, 1 to 0, to the 1, 3 to 2, 4 to 3.*0738

*Those are my allowed transitions.*0746

*The 0 state to 1 state, I can go 0 to 1, 1 to 2, 2 to 3, 3 to 4.*0749

*Or the 0 to 1 vibrational state, I can go rotational 1 to 0, 2 to 1, 3 to 2, and 4 to 3.*0753

*Those are my allowed transitions, the frequencies of those particular transitions.*0762

*The frequencies of these transitions, in other words the frequencies of the spectral lines of the R and P branches,*0773

*we have ν sub R branch is equal to ν sub 0 + 2B × J + 1, where J takes on the values 0, 1, 2, and so forth.*0817

*The P branch is going to be ν sub 0 -2B J.*0835

*Here, J takes on the values 1, 2, 3, and so on.*0844

*My ν sub R is going to be 2143.54 + 2 × 1.9304 × J + 1.*0852

*My ν sub P, those branches are going to be 2143.54 -2 × 1.9304 × J.*0876

*The R is 0 branch that represents the 0 to 1 transition, that was going to take place at 2147.4.*0894

*I will just put the J values in here.*0904

*J = 0, J = 1, J = 2, J = 3, J = 4.*0905

*That is the R0, let me be label it with the particular J value.*0912

*We put the R branch for J value 0 and it is going from 0 to 1.*0919

*When we do the P branch, it is actually going to be a P1 because it is going to be going from 1 to 0.*0924

*The R1, that represents the 1 to 2 transition.*0932

*The R2 represents the 2 to 3 transition.*0937

*The R3 represents the 3 to 4 transition.*0941

*These are the frequencies that we have.*0946

*We have 2151.3, we have 2155.1, and we have 2159.*0948

*We have the P1 transition, the P1 line, the P2 line, the P3 line, and the P4 line.*0962

*This one represents the 1 to 0 transition.*0971

*This one represents the 2 to 1 transition.*0973

*These are subscripts or the beginning energy level.*0976

*The energy level of the lower vibrational state.*0979

*For the lower rotational state, on the lower vibrational state.*0984

*In other words, the departure state not the arrival state.*0991

*I like that better, departure and arrival.*0995

*I think the mathematical terms are actually a lot better.*0997

*This one represents the 3 to 2 transition and this one represents the 4 to 3 transition.*1003

*These are 21392139.7, 2135.8.*1012

*These are going down, these are going up.*1020

*R branch increases, P branch decreases.*1022

*2132 and 2128.*1027

*There you have it, we found the energies, we found the allowed transitions.*1035

*These are the frequencies of the actual transitions that are allowed.*1040

*These are the labels for them representing this particular transition.*1043

*Let us go back to red here.*1048

*Be very careful to distinguish, especially on test.*1054

*On a test, you are going to be stressed out and your mind is going to be racing.*1061

*You need to really slow down when dealing with spectroscopy because they can be asking about energy,*1065

*they could be asking about absorption, emission, frequency.*1070

*Those are not the same.*1072

*Be careful to distinguish between the energy of a given level and the energy of transition,*1074

*the energy of transition between two energy levels.*1113

*The transition between two energy levels that is what we see in spectra are the transition energies from one level to another.*1129

*One final thing to notice, we found the energy levels for the E sub 0 J and E sub 1 J.*1161

*We used equations, the ν sub R, ν sub P to actually find those frequencies.*1173

*What you could have just done is take the difference between energies in the table.*1179

*Notice that the frequencies of transition,*1184

*instead of using equations for ν R branch and ν of the P branch,*1203

*we could have just taken δ E between the two vibrational states*1235

*for the appropriate J values.*1262

*For example, if I wanted the R1 line,*1271

*Let us go ahead and start with R0 line.*1279

*The R0 line that represents the transition from 0 to 1.*1280

*We have the energies already in the table.*1289

*We could have just taken the energy of 1 1 - the energy of 0 0.*1290

*This is the vibrational number, that is the vibration number.*1304

*Upper lower, the transition going from the 0 to 1 transition.*1308

*Upper lower, we could have just done that, just subtract the two values.*1316

*I will do that to the problem a little bit later on in this lesson.*1319

*Let us see what is next.*1325

*Using the table of parameters below and equations which correct for anharmonicity and vibration rotation interaction,*1330

*construct a table of energies for the first 5 rotational states of hydrogen iodide, for vibrational levels R = 0 and R = 1.*1344

*Specify which transitions are allowed and calculate the frequencies of these transitions,*1353

*associating each to its appropriate peak in the R and P branches.*1356

*The same sort of thing except now we need to make adjustment to the rigid rotator harmonic oscillator*1361

*and we are correcting for anharmonicity and vibration rotation interaction.*1367

*Notice, we are not correcting for centrifugal distortion.*1371

*We can always correct for all three, for two of those things, or for one of those things.*1374

*It just depends on what the problem is asking for.*1378

*We have our table of parameters, our rotational constant, our Α sub E,*1381

*our fundamental frequency and our correction for anharmonicity.*1389

*We said that the energy for the rigid rotator harmonic oscillator approximation*1399

*that was E sub RJ = ν sub 0 × R + ½ + B × J × J + 1.*1405

*This is the rigid rotator harmonic oscillator approximation.*1417

*It does not really matter.*1422

*The dependence of B on R, B sub R is equal to B sub E - Α sub E × R + ½.*1424

*This thing goes in where B is, that is the adjustment for vibration rotation interaction.*1438

*The adjustment for harmonicity, the energy is ν sub E × R + ½ - this X sub E ν sub E × R + ½².*1446

*This is the adjustment for anharmonicity.*1465

*Now when we put this together, when we put both of these into this,*1469

*we get a new equation for the energy which is RJ is equal to,*1474

*It gets a little complicated, a little long, no doubt about that.*1482

*R + ½ - X sub E ν sub E × R + ½² and the rotational term which is going to be B sub E - Α sub E × R + ½ × J × J + 1.*1487

*This is the ν for the two corrections that we have to make.*1510

*Notice again, this problem does not ask us to account for centrifugal distortion.*1516

*If it did, we have one more – that D thing, D × J² J + 1², but that does not matter here.*1520

*For R = 0, we get E of 0 J is equal to ½ ν sub E -1/4 X sub E ν sub E + this B sub E - Α sub E.*1530

*Again, these are just a bunch of parameters that we just have to account for.*1551

*It is not terribly a big deal.*1554

*R is 0 here so we can just go ahead and put the -1/2 Α sub E × J × J + 1.*1557

*This ends up being E sub 0 J is equal to 1154.*1569

*I will just put the numbers in.*1577

*In other words, I will put the ν sub E in, we put the X sub ν E in, the BE, the Α E, we just put it in.*1579

*That is it, nothing strange.*1586

*9.911 + 6.4275 × J × J + 1.*1588

*For the R = 1 vibrational state ,we get E of 1 J that is going to equal 3/2 ν sub E - 9/4*1601

*X sub E ν sub E + B sub E – 3/2 Α sub E × J × J + 1.*1614

*And we end up as far as numerically is concerned, we end up with 3463.521 - 8189.199.*1629

*This is nothing more than just a bunch of tedious arithmetic, that really is L comes down to.*1645

*+ 6.2585, which is why I personally love theory as opposed to the tedium.*1650

*That is why we work with symbols, you do not want to do those arithmetic.*1662

*× J × J + 1.*1665

*This equation right here, let me go to blue, we use this equation and we use this equation.*1669

*We create a table of values.*1676

*We have J, J is going to take on the values 0, 1, 2, 3, 4, that is the first 5 states.*1680

*E sub 0 J, all the values are in inverse cm.*1689

*E1 J we have 1144.6, 1157.5, we have 1183.2, we have 1221.7,*1695

*we have 1273.1, we have 3374.3, we have 3386.8, we have 3411.9, 3449.*1713

*My head is already spinning, believe me.*1733

*I completely understand, this is not something that you actually get used to.*1735

*After a long time doing this, you lose your mind from what are all these numbers.*1739

*Your head starts to spin.*1748

*3449.4 and this is going to be 3499.5.*1751

*The allowed transitions are δ J = + or -1.*1760

*We are going to have the 0 to 1, 1 to 2, 2 to 3, 3 to 4.*1767

*Or we are going to have the 1 to 0, 2 to 1.*1773

*Basically, all you have to do when you need to find the actual frequencies of the absorption,*1777

*the frequencies of the spectral lines, you are just going to take for example the 0 to 1 transition,*1783

*you are just going to take this number - this number.*1788

*The 1 to 2 transition, this number - this number.*1793

*You do not actually need the equations, which is why the energy equation is really the most important.*1796

*If you have that, you can get everything else.*1801

*The δ J = + or -1, when we do that, let us go ahead and create a new table of values here.*1805

*Once again, let us say δ J = + or -1.*1815

*The R0, R1, R2, R3, that represents the 0 to 1 transition, this is the 1 to 2 transition,*1823

*this is the 2 to 3 transition, this is the 3 to 4 transition.*1834

*This is going to be at 2242.2.*1838

*It is just the second number the second column - the first number the first column.*1842

*The energy of the 1 - the energy of the 0.*1848

*The energy of the 2 - the energy of the 1.*1852

*For the two vibrational states.*1854

*We do not necessarily need equations for Ν sub ρ and ν sub P.*1858

*2254.4, 2266.2, 2277.8, and we have the P branch.*1863

*This is going to be P1, P2, P3, P4.*1875

*This represents the 1 to 0 transition.*1880

*This represents the 2 to 1 transition.*1883

*This represents the 3 to 2 transition.*1886

*This represents the 4 to 3 transition.*1888

*This is going to be 2216.8.*1893

*This is going to be 2203.6.*1897

*This is going to be 2198.2.*1901

*And this is going to be 2176.3.*1905

*I really like the black.*1918

*If we wanted specific equations for ν sub ρ and ν sub P, basically the observed frequencies.*1920

*These numbers, we have to do some algebra, that is the problem.*1934

*We would have to do a lot of algebra actually.*1945

*We have to do a lot of algebra.*1952

*Essentially, what you have been doing is the following.*1954

*For ν sub R, you take the energy of the upper states and subtract the energy of the lower state.*1959

*That is what you are doing.*1963

*You are going to end up with, it is going to be the energy E R + 1 J + 1 - ERJ.*1964

*You have to do all that and come up with some final equation.*1975

*For the P branch, that is going to end up being E R + 1 J - 1 ERJ.*1978

*Based on the equation above, that long equation that we had for the energies,*1990

*you have to take one for the upper state then subtract that long equation, one for the lower state.*1994

*Go through all the algebra and come up with some equations, which is essentially what you see in your books.*1999

*Most of what you see in your books, the equations, the derivations of the spectroscopy section are just that.*2004

*They are just taking one energy and subtracting other energy.*2010

*It tends to look complicated but it is just a lot of algebra.*2013

*In order to come up with equations for the spectroscopic lines that we see,*2016

*based on whatever correction we are making it.*2021

*It is either going to be the harmonic oscillator rigid rotator approximation for the energy.*2023

*You can either make a correction for vibration rotation interaction.*2029

*You can make a correction for anharmonicity.*2033

*You are going to make a correction for centrifugal distortion.*2035

*Either 1, 2, or all 3 of those.*2038

*The equation is longer for every correction that you make.*2041

*You remember all this from the first lesson, the initial lesson that we did in spectroscopy*2046

*where all these things are coming from.*2050

*That it is not this ocean of equations that you are dealing with.*2053

*It is just the energies of the upper state - the energies of lower state that give you the frequency of the spectral line.*2057

*Let us do some more examples here.*2068

*I think we are on example number 3 now.*2070

*Wait, I think we are missing some data here.*2078

*We do not have this data but that is not a problem.*2084

*I actually have it right here with me, I will go ahead and write it.*2086

*For some odd reason, it did not end up on this page, my apologies.*2088

*Calculate B sub 0, B sub 1 of the equilibrium bond length for the R = 0 vibrational state and*2091

*the equilibrium bond length for R = 1 vibrational state, the B sub E and Α sub E for carbon monoxide.*2100

*I apologize, it did not show up on this slide.*2118

*Ν observed, we observed an R 0 line of 2173.81.*2122

*We observed R1 line at 2177.58.*2132

*We observed a P1 line at 2166.15.*2140

*And we observed a P2 line at 2162.27.*2146

*Given this data for two R lines and the first two lines of the P branch, how can I calculate all of these?*2153

*We see B0 and B1, that equations I want to deal with are going to only account for the vibration rotation interaction.*2162

*I do not see anything else here.*2184

*I do not see any X sub E, ν sub E.*2185

*I do not have to worry about the anharmonicity.*2187

*I do not see a D here so I do not have to worry about centrifugal distortion.*2190

*I see B sub 0 B sub 1, the equations that I’m going to be looking at,*2193

*I'm going to be looking at the rigid rotator harmonic oscillator corrected for vibration rotation interaction.*2197

*That is how you choose your equation.*2203

*You take a look at what constants are at your disposal, what they want, and I will tell you what to choose.*2205

*We see B0 and B1, we consider only the vibration rotation correction.*2211

*I always like to begin with the basic equation and then add to that.*2234

*It is just a nice way to keep sort of refreshing yourself and see that the equation over and over and over again.*2238

*Our harmonic oscillator rigid rotator equation E sub RJ, the energy is ν R + ½ + B × J × J + 1.*2245

*I do not need that parenthesis at J × J + 1.*2263

*The vibration rotation correction, that one says that this B, the rotational constant depends on R.*2267

*B sub R is equal to B sub E - Α sub E × R + ½.*2279

*This we put into here and get a new equation.*2289

*The corrected equation for the vibration rotation interaction is equal to,*2297

*Sometimes, I will put a tilde over the E, sometimes not.*2302

*It is in wave numbers, do not worry about it.*2305

*That is equal to, it is going to be ν R + ½*2308

*+ this thing B sub E - Α sub E R + ½ × J × J + 1.*2323

*Unless specifically stated otherwise,*2338

*always take the R = 0 to the R = 1 vibration transition.*2357

*That is the one that you want to take.*2365

*At normal temperatures, room temperatures, most molecules are actually going to be in their ground vibrational state.*2368

*They are also going to be in their ground electronic state.*2374

*However, at normal temperatures, it is the rotational states that molecules tend to be the higher states,*2378

*they cannot be the ground state.*2384

*J = 0, they can be anywhere from like J = 2 or 3, above that.*2386

*At normal temperatures, most molecules are in the vibrational ground state and electronic ground state.*2393

*It is only the rotational state that there are in excited states at normal temperatures.*2401

*We will talk more about that when we do statistical thermodynamics next.*2405

*When we look at this, our new observe, what we are looking for is this.*2419

*We need an equation that is going to be the energy of the upper state - the energy of lower state.*2423

*The energy being the equation that we just had.*2433

*Our R branch is going to be E of 1 J + 1 - E of 0 J.*2435

*Our ν sub P is going to be E of 1 J-1 E0 J.*2448

*When I do the algebra for these and I put the energy equation that I just had in the previous page,*2459

*one for the upper state and one for the lower state.*2464

*When I work out that algebra, here is what I get.*2466

*When we work out these differences, we get a rather daunting looking equation actually.*2475

*Ν sub 0 + B1 - B0 J² + 3 B1 - B0 J + 2 B1.*2491

*The J takes on the value starting from 0, 1, 2, and so on.*2512

*For the P branch, I get ν sub 0 + this same term B1 - B0 J².*2518

*The third term is different, this is going to be B1 - B0 × J.*2529

*Here, the J takes on the values of 1, 2, 3.*2540

*And again, for the R branch 0, so on.*2543

*The P branch is 1 and so on.*2545

*Let me go to blue here.*2561

*The R is 0 line that represents the J = 0, that is going to be ν sub 0 + 2 B1.*2565

*In other words, I put J = 0 into this equation and I see what I get.*2577

*I get ν sub 0 + 2 B1.*2584

*The data gave us the frequency of absorption, the ν sub 0, the R sub 0 line is at 2173.81.*2588

*The R1 line that is where J is equal to 1, that is ν sub 0 + 6 B1 – 2 B0 is equal to 2177.58.*2602

*The P sub 1 line that represents that J = 1.*2622

*I’m going to use this equation, the one for the ν sub P.*2627

*I get ν sub 0 -2 B0 = 2166.15 inverse cm.*2631

*For the P2 line that is the J = 2, I get ν sub 0.*2642

*I get + 2 B1 -6 B0.*2647

*And that one was observed at 2162.27.*2654

*These are the equations that I’m going to work with now.*2660

*I’m going to take the R1 equation - the P1 equation.*2668

*When I do that, I get 6 B1 = 11.43.*2674

*Therefore, B1 is equal to 1.905 inverse cm.*2684

*I found B1.*2694

*Now, I take the R0 line and I subtract from it the P2 line.*2696

*When I do that, I get 6 B0 = 11.54.*2701

*Therefore, B0 is equal to 1.923 inverse cm.*2711

*We found B1, we found B0.*2722

*BR, this B sub R that we just found is equal to H/ 8 π² C × I.*2726

*I is the reduced mass × this, whatever this is².*2739

*Now, we need to find the equilibrium bond length for each of these.*2744

*When I put the values in, I will get the following.*2751

*When I rearranged this for RE², R sub E vibrational state 1² is equal to 6.626 × 10⁻³⁴.*2754

*I’m not going to put the units in and I’m just going to put the numbers in.*2771

*8 π² 2.998 × 10¹⁰.*2774

*I went ahead and use cm directly.*2782

*1.139 × 10⁻²⁶ that was the reduced mass and then the 1.905 that is this number right here.*2786

*This goes down here, this comes up here.*2797

*When I solve for this, I get that it is equal to 113.6 pm.*2800

*The same for R0, when I do the same for R0, R sub E for the 0 state, I end up with a value of 113 pm.*2812

*We found RE 1, RE, we found B0, we found B1.*2832

*Let us go ahead and find for B sub E and Α sub E.*2836

*We said that b sub R is equal to B sub E – α sub E × R + ½.*2850

*We know what B0 is.*2862

*B0 is equal to B sub E – ½ Α sub E.*2866

*We know what B1 is, let us put that 0 into here, 1 into here for R.*2874

*We get B sub E - 3/2 Α sub E.*2881

*We know what B0 is, we already found it 1.923.*2888

*1.923 = B sub E – ½ Α sub E.*2892

*And we know that 1.905 which is B1, that = B sub E - 3/2 Α sub E.*2903

*Two equations and 2 unknowns.*2914

*Let us go ahead and take, this is equation 1 and this equation 2.*2916

*Let us go ahead and take equation 1 - equation 2.*2921

*When you do that, you end up with Α sub E is equal to 0.018 inverse cm.*2928

*Put this into one of these other equations and you end up with B sub E is equal to 1.932 inverse cm.*2944

*There you go, I hope that make sense.*2957

*Let us see what is next.*2962

*The IR spectrum of ML shows a fundamental line at 1877.62 inverse cm and the first overtone at 3728.66 inverse cm.*2970

*Find the values of ν sub E and X sub E ν sub E for ML.*2983

*Let us go ahead and work in black here.*2993

*The frequencies of the overtones for the anharmonic oscillator*2998

*are given by G of R – G of 0, just looking at vibrational transitions right now.*3027

*G of R to G of 0, the anharmonic oscillator.*3037

*Under normal temperatures, most molecules are in the ground vibrational states.*3043

*They are all going to start at the 0 vibrational state.*3046

*But δ R is no longer + or -1.*3050

*It can be + or -1, + or -2, + or -3.*3053

*The + or -1 is the fundamental.*3056

*The + or -2, those are the first overtone.*3059

*The first overtone, second overtone, third overtone.*3062

*The G of R equation that is equal to ν sub E × R + ½.*3066

*The anharmonic oscillator – X sub E ν sub E R + ½².*3076

*When we take the G sub R – G of 0, when we do the algebra,*3078

*what we get is the observe line equal to ν sub E × R – X sub E ν sub E × R × R + 1.*3093

*R1 runs from 1, 2, 3, and so on.*3109

*The fundamental transition is the transition from 0 to 1, that is the one that we see.*3117

*That is the brightest one that we see.*3131

*That is equal to, I put into this equation, fundamental 0 to 1.*3135

*I put R = 1, it is going to be ν sub E × 1 - X sub E Ν sub E × 1 × 1 + 1.*3142

*We end up with ν sub E - 2 X sub E ν sub E.*3157

*It tells us what that is already.*3178

*The fundamental is 1877.622.*3179

*1877.622 that is one of the equations that we are going to use.*3184

*The first overtone represents the transition from 0 to 2.*3191

*It is a weaker line.*3198

*That is going to be ν sub E × 2.*3202

*Just using this equation right here, taking care of the R = 1.*3208

*Now the first overtone R = 2 - X sub E ν sub E × 2 × 2 + 1.*3214

*What we end up with is 2 ν sub E – 6 X sub E ν sub E, it tells us what the first overtone is 3728.*3227

*3728.66 that is the second equation that we are going to use.*3240

*When you solve simultaneously, I will not go through the process.*3247

*I will let you go ahead and do that.*3249

*This equation and this equation, when you solve simultaneously,*3250

*you are going to get a value of ν sub E is equal to 1904.20 inverse cm and X sub E ν sub E is equal to 13.29 inverse cm.*3253

*That is it, nice and easy.*3274

*I know it is not nice and easy, believe me I do.*3279

*I have been there, the stuff still gives me grief too.*3282

*Let us take a look at our final example.*3289

*I’m actually wondering whether I should leave this example off because the first two lessons took care of it.*3291

*Of course, we had an example from the previous lesson that did this but we will see.*3297

*In examples 6 of the previous lesson, we asked you to calculate the frequencies of the first two lines of the R and P branches*3304

*for the vibration rotation spectrum of HBR under the harmonic oscillator rigid rotator approximation.*3312

*Here we ask you to repeat the problem.*3320

*But instead of the harmonic oscillator rigid rotator approximation,*3323

*we asked you to do so by accounting for all three corrections we made to the HIR equations.*3326

*We accounted for vibration rotation interaction for centrifugal distortion and for harmonicity.*3335

*Use the following spectroscopic parameters as necessary.*3340

*This is a great example because we get a chance to put everything together,*3343

*the HRR approximation and all of the corrections that we made.*3350

*I’m not going to write all the equations here.*3367

*I can do so in the last couple of lessons, basically what we are going to be doing is,*3368

*in this one you are going to take the harmonic oscillator approximation.*3372

*You are going to make the corrections for all three of these things.*3378

*The vibration rotation interaction, the centrifugal distortion, and the anharmonicity.*3381

*You got to find an equation for the total energy.*3385

*You are going to take the upper level - the lower level.*3388

*In other words, you are going to find δ E.*3391

*That is going to give you your equation.*3395

*You do not necessary have to go through the algebra, the equations are actually have been done for you.*3398

*They are in your books, I think we actually did in our lessons too.*3402

*When you do it, the equations that you come up with are the following.*3404

*Ν sub R, that is going to be ν sub 0 × 1 – 1.*3411

*2648, we are going to have ν sub E × 1 - 2 XE + 2 × BE × J + 1 - Α sub E × J + 1 × J + 3 - 4 D × J + 1³.*3434

*The J values are going to run from 0, 1, 2, and so on.*3465

*This is the equation that you get when you take the total energy,*3471

*the correction for all three things and the energy of the upper state – the energy of the lower state.*3475

*It is a lot of algebra but this is the equation you come up with.*3482

*This is the frequency of the spectral line that we should see.*3484

*The ν sub P branch that is going to be ν sub E × 1 - 2X sub E -2 B sub E × J - Α sub E × J × J - 2 + 4D J³.*3492

*Here J = 1, 2, and so on.*3525

*These are the two equations that we actually get.*3529

*Notice the equation for the spectral line changes depending on what corrections we are making.*3532

*In this case, we corrected for all three.*3537

*They give us ν sub E, we have that one.*3542

*We are going to need that parameter.*3544

*We are going to need this parameter.*3546

*We are going to need this parameter.*3549

*We do not need those others.*3554

*We also need the X sub E but they gave us N sub E ν sub E.*3555

*Actually, you have to take this.*3559

*We would need that.*3561

*We would have to take this parameter divided by that, in order to get just the X sub E.*3561

*Let us go ahead and do that first.*3567

*The data gives ν sub E and X sub E ν sub E.*3570

*Let us find X sub E alone.*3586

*The X sub E is equal to X sub E ν sub E divided by ν sub E.*3591

*We end up with 45.217 divided by 2648.975 and we end up with X sub E = 0.01707 inverse cm.*3599

*We put all these parameters into the equations.*3617

*What we end up with, ν of the R sub 0 line.*3621

*That is going to equal 2648.975 × 1 - 2 × 0.01707 + 2*3636

*× 8. 465 × 0 + 1 - 0.2333 × 0 + 1 × 0 + 3 - 4 × 3.457 × 10⁻⁴ × 0 + 1³.*3653

*When I calculate that, I end up with 2574.768 inverse cm.*3683

*This is my R0 line.*3693

*I should see it there.*3696

*Notice, we expected something lower than what we get with the rigid rotator harmonic oscillator approximation.*3699

*Exactly what we got.*3706

*Now, when I do the same for the R1 line, I will write all this out.*3708

*What I end up with is 2590.522 inverse cm.*3714

*Notice the spacing between the lines.*3724

*The spacing between the spectral lines is this line - this line.*3728

*This line - this line, the absolute value there.*3729

*The spacing is ν of R1 - ν of R0, that is equal to 15.755 inverse cm.*3735

*2B E = 2 × 8.465 that is equal to 16.93 inverse cm.*3750

*You see that the spacing, the corrections that we have made.*3763

*On the actual spectrum, the spacing is less than 2B.*3766

*2B is 16.93.*3769

*The corrections that we made give us the spacing which is about less than 2B, which is exactly what we expect.*3771

*For the R branch, the spacing is smaller.*3780

*For the P branch, the spacing will actually be bigger than 2B.*3783

*The 2B is the rigid rotator harmonic oscillator approximation.*3787

*Let us go ahead and do the P branch.*3792

*I think I have one more page, If I’m not mistaken, yes I do.*3794

*I got ν of P1 and when I put into the equation for the P1, I end up with 2541.8.*3801

*Let me write this one, at least.*3822

*It is going to be 2648.975 × 1 - 2 × 0.01707 - 2 × 8.465*3824

*× 1 - 0.2333 × 1 × 1 - 2 + 4 × 3.457 × 10⁻⁴ × 1³.*3844

*This will give me 2541.843 inverse cm.*3863

*And when I do the same for the P2 line, I end up with 2524.690.*3870

*Now, the difference here, the δ ν.*3885

*In other words, one of them - the other, that is going to equal 17.153, which is greater than 2B,*3891

*which is exactly what we expect for the P branch.*3903

*That is it, thank you so much for joining us here at www.educator.com.*3910

*We will see you next time.*3913

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