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 0:00
- Example I: Harmonic Oscillator 0:12
- Example II: Harmonic Oscillator 23:26
- Example III: Calculate the RMS Displacement of the Molecules 38:12

### Physical Chemistry Online Course

### Transcription: Example Problems II

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

*Welcome back to Physical Chemistry.*0002

*Today, we are just going to continue doing our example problems for the harmonic oscillator and the rigid rotator.*0003

*Let us jump right on in.*0010

*The first example for the harmonic oscillator in the ψ sub 3 state.*0014

*I will forgot to subscript that, ψ sub 3 state showed that the average value of X² is equal to 7 H ̅/ 2 × KM¹/2.*0020

*This example is going to be very long.*0038

*This is going to take several pages and the reason that it is, is because I decided at least for this example set,*0041

*for example 1 and 2, I decided to actually do them explicitly.*0047

*Rather than relying on mathematical software, I want to go through at least 1 or 2 times*0052

*how to actually put the integral together and how to actually solve the integral by hand.*0058

*It is just going to be a long example.*0066

*It is going to be a lot of symbolism on the page but there is nothing here that is actually difficult.*0069

*It is just keeping all of the algebra straight, that is the only difficult part.*0072

*Let us jump right on in and see what we can do.*0076

*For the ψ sub 3 state, we want to show that this is the case.*0080

*You remember that the average value or recall, I should say,*0085

*that the average value of some quantity is equal to the integral of the wave function operator.*0089

*This is the integral that we have to form.*0100

*We have to take the function, operate on it, whatever operator is that we are dealing with and*0103

*multiply by the conjugate of that particular wave function and then integrate that.*0107

*So setting up the integral should be not a problem.*0111

*It is actually solving it that is going to end up being the most problematic, simply because it is tedious.*0113

*You do not have to do this because you have math software.*0120

*On an exam, you do not have to be given anything difficult because you have to do a reasonably quick integration.*0122

*Let us see what we have got.*0129

*Let us go ahead and erase this.*0131

*We have X², this thing is going to equal - infinity to infinity of ψ sub 3, that X² ψ sub 3 DX.*0136

*This is the integral that we actually have to solve.*0154

*Ψ sub 3 is equal to, let me write it over here.*0159

*Ψ sub 3 is actually equal to Α Q³/ 9 π × 2 Α X³ -3X × E ^- α X²/ 2.*0166

*Where recall, Α is shorthand for K × μ/ H ̅²¹/2.*0185

*Our ψ*, the integrand X² ψ, that is equal to this is just multiplication by X².*0204

*Multiplication is commutative so I do not have to apply to this one and then multiply it.*0214

*I just pull the X² out.*0219

*This is a 3, this is a 3, I forget the state, sorry about that.*0223

*This is a real function so this is just X² ψ of 3 × ψ of 3.*0228

*We are just going to end up squaring it.*0235

*I think I forgot something.*0242

*This is 1¹/4, the Α³/ 9 Π is going to be the biggest issue.*0243

*It is just remembering to write down all the little symbols.*0250

*Let us just take this nice and slow.*0255

*This is ψ sub 3, so when I multiply this by itself, I'm going to end up with the following.*0257

*I'm going to end up with Α³/ 9 π¹/2.*0275

*I have the X² and I have the 2 Α X² - 3X.*0287

*This is² and this multiplied by itself becomes - α X².*0295

*This = Α³/ 9 π¹/2 × X² ×, I’m going to expand this out.*0303

*It is going to be 4 Α³ X.*0316

*We see that we are already starting to have some issues here.*0324

*Hopefully, we can take care of these issues before it gets crazy.*0328

*This is going to be X⁶ -12 Α X⁴ + 9 X² × E ⁻Α X².*0333

*I’m going to go ahead and distribute the X so we had the Α³/ 9 π¹/2 × 4 Α² X⁸ -12 Α X⁶ + 9 X⁴ E ^- Α X².*0348

*That is what the integrand is.*0372

*The integrand which is this is equal to this thing.*0374

*We will go ahead and do this in red.*0392

*This X² thing is going to equal 2 × Α³/ 9 π¹/2 the integral from 0 to infinity.*0395

*The 2 of -infinity to infinity, I just split it up into 2 + 0 + infinity, so we have this thing.*0414

*4 Α² a X⁸ -12 Α X⁶ + 9 X⁴ × E ^- α X² DX.*0421

*This is what we actually have to solve.*0436

*You are going to be using a table or you could be using math software to do this.*0441

*On your exams, they only have to give you a very easy function to integrate.*0445

*Or they are going to give you from a table of integrals to actually work with so*0455

*you can do the integration reasonably quickly because you cannot sit there for 30 minutes doing one problem.*0459

*But in this particular case, we are going to go through this by hand individually.*0464

*Let us jump right on in and see what we can do.*0469

*I’m going to write this integral on the next page and let us take it from there.*0473

*Let me go back to black and let me write what we have.*0479

*We have the X ^² the average value is going to equal 2 × α³/ 9 π¹/2 × the integral from 0 to infinity of 4 α² X⁸ -12 α X⁶.*0486

*Α is just a constant, + 9 X⁴ × E ^- α X² DX.*0511

*You are going to be seeing this integral a lot in quantum mechanics for the harmonic oscillator.*0521

*We are going to break this up into 3 integrals.*0527

*Essentially, we are going to break it up into this × this, this × this, this × this.*0531

*Our first integral is going to be the integral from 0 to infinity.*0540

*We will worry about this factor at the end.*0548

*4 Α² X⁸ E ⁻Α X² = 4 α² the integral from 0 to infinity of X⁸ E ⁻α X² DX.*0553

*That is going to be our first integral.*0573

*Our second integral is going to be the integral from 0 to infinity of -12 Α X⁶ E ^- Α X Squared DX,*0578

*which is going to equal -12 Α the integral from 0 to infinity of X⁶ × E ^- Α X² DX.*0592

*That is the second integral.*0608

*Our third integral is going to be the integral from 0 to infinity of 9 X⁴ × E ^- α X² DX,*0612

*which is going to equal 9 integral of 0 to infinity of X⁴ E ⁻α X² DX.*0628

*The integral form of that we are actually going to use is going to be this one.*0640

*We will use the following general integral formula that we get from our table.*0655

*The integral of the form 0 to infinity X ⁺2N E ^- Α X² is equal to 1 × 3 × 5 ×… all the way to 2N -1/ 2 ⁺N + 1 A ⁺N × π/ A¹/2.*0672

*This is the general formula, X ⁺2N E ⁻α X².*0713

*Notice, we have X⁴ X X⁶ X⁸ E ⁻α X².*0717

*Here, the A is the Α.*0725

*Let us go ahead and do integral number 1 first.*0731

*Integral number 1, 4 α² 0 to infinity of X⁸ × E ^- Α X² DX.*0737

*In this particular case, X⁸ is equal to X² × 4.*0763

*Therefore, our N is going to equal 4.*0770

*Since N = 4, the 2 N -1 is equal to 2 × 4 is 8, that is equal to 7.*0774

*The value of this integral is equal to 1 × 3 × 5 × 7/ 2 ⁺N + 1, which is going to be to the 5th.*0784

*It is going to be Α⁴ × π/ Α¹/2.*0801

*Let us make this α a little clear, there we go.*0811

*That is the first integral.*0813

*Therefore, our 4 Α² × the integral from 0 to infinity of X⁸ E ^- α X² DX is going to equal 4 Α² × this thing,*0823

*which ends up being 105/32 Α⁴ × π/ Α¹/2.*0843

*That is our first integral.*0857

*Let us do integral number 2.*0864

*Our integral number 2 is -12 Α the integral from 0 to infinity, this time we have X⁶ E ^- Α X² DX.*0870

*X⁶ is equal to X² × 3 so N is equal to 3, 2 × 3 -1, 2 -1 is equal to 5.*0883

*The integral portion, we have 1 × 3 × 5/ 2 ⁺N + 1 = 4.*0896

*It is going to be Α³ × π/ Α¹/2.*0909

*The total integral -12 Α the integral from 0 to infinity of X⁶ E ⁻Α X² DX is going to equal -12 Α ×, then we will multiply this.*0917

*1× 3 × 5 and do some cancellations.*0932

*We are going to end up with 15/ 16 α³ × π/ Α¹/2, that is our second integral.*0935

*We are getting there, slowly but surely.*0951

*It is not a problem.*0952

*We have the third integral.*0955

*The integral number 3 that was equal to 9 × the integral from 0 to infinity, this time X⁴ E ⁻α X² DX.*0959

*X⁴ is equal to X² which means that N = 2 and 2 × 2 -1 is equal to 3.*0971

*Therefore, we have 1 × 3/ 2³ Α².*0987

*We are just using the general formula π/ Α¹/2.*0994

*This integral 9 0 infinity of X⁴ E ^- α X² DX is equal to 9 ×, this thing is 3/ 8 Α² × π/ Α¹/2.*1000

*That is our third integral.*1024

*Let us not forget our normalization constant and our factor of 2.*1030

*Let us remember our normalization constant and our factor of 2.*1038

*Our normalization constant was Α³/ 9 π¹/2 and 2.*1061

*We have the following.*1072

*We have our X² average value is equal to 2 and we will have our Α³/ 9 π¹/2.*1074

*We have the first integral, the first integral was 4 Α² × 105/32 Α⁴.*1088

*The second integral was -12 Α × 15/16 Α³ + 9 × the 3/8 Α² × the π/ Α¹/2.*1110

*We are going to have a whole bunch of cancellations.*1139

*This is the answer, we just need to simplify this thing algebraically.*1142

*Let us go ahead and cancel, we got the π¹/2 π ^½.*1149

*We can take care of the π, we can cancel this α with one of these α to give us an Α².*1153

*You do have all kinds of cancellations.*1161

*You are going to have a 4, this is going to be 8, this is Α².*1164

*Here, this is going to leave me with an Α².*1168

*This Α and this Α is going to leave me with an Α².*1172

*An Α² here, when I do the cancellations, when I multiply, when I simplify, I should get the following.*1177

*I should get 2 Α / 3 × 105/ 8 Α² -45/ 4 Α² + 27/ 8 Α².*1188

*Let me see what I have got here.*1222

*This Α cancels these squares, when I multiply everything out, 3 cancels with the 105.*1225

*I get 35 up here and I get 15 here, I get 9 here.*1236

*I can go ahead and multiply those out.*1264

*I think I lost my way.*1267

*Let me change something here.*1271

*Let me start again.*1280

*I will just write down what is that I have written.*1284

*I think that is going to probably be the best thing to do.*1286

*I have got 2 Α/ 3 × 105/ 8 Α² -45/ 4 α² + 27/ 8 Α² 2 Α/ 3 × 105 -90 + 27/ 8 α².*1290

*Α goes with Α, 3 goes that becomes 35.*1324

*This becomes 30, this becomes + 924.*1330

*We are going to end up with is going to be 7/ 2 Α, that is the final answer.*1339

*However, but α is equal to √ K μ/ H ̅.*1349

*That is going to equal 7 H ̅/2 × √ K μ, which is exactly what we wanted to show.*1364

*That takes care of that example, manual integration.*1377

*Quantum mechanics is not particularly difficult.*1381

*It is the calculations that are very very tedious.*1383

*You hove software at your disposal, please use the software or table of integrals if you prefer.*1385

*Example 2, let us see what we have got.*1393

*It is going to be about the same thing, it is not a problem.*1399

*For the ψ sub 3 state, let us go ahead and make this a subscript here.*1409

*For the ψ sub 3 state of the harmonic oscillator show that the average value of*1413

*the momentum² is equal to 7 × H ̅ × √ K × μ all divided by 2.*1418

*We have this is equal to -infinity to infinity.*1430

*It is going to be ψ sub 3 DX, this is the integral that we have to solve.*1440

*In this particular case, we have to integrate this twice and multiply by that.*1454

*We cannot just pull this out, multiply these 2, and then apply the operator.*1458

*We have to apply the operator once, we have to apply the operator twice.*1464

*Let us go ahead and write down.*1469

*We have ψ sub 3, it is equal to Α³/ 9 π¹/4 × 2 Α X³ -3 X E ⁻α X² / 2.*1470

*We have our momentum operator, it is equal to – I H ̅, I will go ahead and just use the partial notation.*1492

*That is fine, I will use the one variable so it does not really matter.*1502

*We will just do DDX which implies that P², it is just this thing applied to itself.*1507

*You end up with – H ̅² D² DX².*1519

*For now, I'm going to go ahead and leave aside this normalization constant and this constant right here.*1528

*I’m just going to go ahead and differentiate and deal with the part, and I will bring the constant back.*1538

*Let us go ahead and we would have to differentiate ψ sub 3 so we have to differentiate this function.*1544

*Let us do it this way, I’m going to write it 2 Α X³ -3 X × E ⁻Α X²/ 2.*1553

*I’m going to do this.*1568

*The symbol that I’m going to use, I’m going to write the function and this D means differentiate.*1571

*When I differentiate it, this is the product.*1576

*It is going to be this × the derivative of that + that × the derivative of this.*1578

*What I end up with is the following.*1582

*I end up with 2 Α X³ -3 X, this × the derivative of that.*1584

*The derivative of that is E ⁻Α X²/ 2 × the derivative of this.*1593

*This is going to end up being × -Α X + this × the derivative of that, + E ^- α X²/ 2 × 6 Α X² -3.*1602

*This is going to equal E ^- α X²/ 2 ×, I’m going to multiply this and this,*1621

*I end up with -2 Α² X⁴ + 3 Α X² + 6 Α X² -3.*1630

*Clearly, the biggest problem here, which is the biggest problem in most mathematics is not the mathematics, it is the algebra.*1646

*I hope to God that we have not missed a symbol.*1654

*If we have, let us find out.*1656

*It = E ^- α X²/ 2 and we will goo ahead and combine some common terms here.*1659

*We have 6 α X² and 3 α X², we have 9 α X².*1667

*I’m going to put something positive first instead of this.*1672

*-2 Α² X⁴ -3, so that is the first derivative.*1674

*Now, we are going to differentiate again because we are taking the derivative twice.*1681

*When we differentiate again, we end up with the following.*1686

*It is a product functions so we have E ⁻α X²/ 2 × 18 Α X -8 Α² X³.*1694

*This × the derivative of that, - that × the derivative of this.*1718

*+ 9 Α X² – 2 α² X⁴ – 3 × E ⁻α X²/ 2 × – α X.*1722

*My final result is going to be E ⁻α X²/ 2 ×,*1746

*I’m going to have 18 Α X when I combine all my terms.*1754

*-8 Α² X³ -9 Α² X³ + 2 Α³ X⁵ + 3 α X.*1757

*This right here is that.*1778

*We have to multiply this on the left by ψ sub 3 because that is the integrand.*1789

*Now, ψ sub 3 × ψ sub 3 is going to be 2 Α X³ -3 X × 8 ^- α X²/ 2 ×, what is that we got right there.*1797

*E ^- Α X²/ 2 and I will left this expanded.*1823

*I’m going to combine some of these terms like this one and this one.*1828

*We have 21 Α X - 17 Α² X³ + 2 Α³ X⁵.*1831

*=, this ends up being E ⁻α X².*1854

*I’m going to multiply this by that, I'm going to have 123456 terms.*1859

*I’m going to end up with 42 α² X⁴ - 34 α³ X⁶ + 4 α⁴ X⁸ – 63 α X² + 51 α² X⁴.*1865

*I’m glad I have my software, it is the greatest invention in the world.*1893

*6 Α³ X⁶ which ends up equaling E ⁻α X² × 93 Α² X⁴ – 40 α³ X⁶ + 4 Α⁴ X⁸ -63 Α X.*1898

*We can finally put in everything.*1929

*What you should get is, our P² is equal to.*1935

*Let us bring back the -H ̅².*1941

*Let us bring back the Α³/ 9 π¹/2.*1944

*This is going to be the integral of 0 to infinity.*1951

*I’m going to actually factor a 2 in here, so let me go ahead and put that 2 here.*1954

*-2 × that, it is going to be the E ⁻α X² × what it is that I just wrote.*1961

*93 Α² X⁴ - 40 Α³ X⁶ + 4 α⁴ X⁸ – 63 α X² DX.*1970

*This is the one that we actually end up wanting to solve.*1992

*We are going to end up breaking this up into 4 integrals.*1996

*This one, this one, this × this one, and this × this one.*2000

*I'm not going to do that explicitly, I just can write out what is that we actually get when you go through all of the math.*2005

*That is fine, I'm will just go ahead and write down all of the arithmetic.*2015

*-2 H ̅² Α³/ 9 π¹/2 ×, it is going to end up being 93 Α² × 3/ 8 Α² -40 α³ × 15/ 16 Α³ +*2020

*4 Α⁴ × 105/ 32 Α⁴ -63 α × 1/ 4 Α × that π/ Α¹/2.*2056

*When I multiply, simplify, cancel, and do all that stuff, I'm going to end up getting –H ̅² Α -14/ 4*2077

*which is equal to 7/2 H ̅² Α.*2092

*Since the Α B what it is, when we put in 7 H ̅², Α is going to be √ K μ/ H ̅.*2101

*We get a cancellation there and you end up with your final answer.*2117

*You end up with 7 H ̅ √ K × μ out of the two, which is exactly what we wanted.*2120

*A little tedious but I thought it might be nice to solve a couple of problems to go through this manually.*2130

*Let us see here, do I have another page?*2139

*I can use one, in fact I do.*2143

*Let us go back.*2146

*In general, we generalized all of this.*2148

*It is nice to go through the problem but in general you have the following.*2153

*In general, for the harmonic oscillator, for ψ sub R, the average value of X² is going to equal 1/ α × R + ½.*2160

*It is always going to equal that, which is equal to H ̅/ K × μ¹/2 × R + ½.*2190

*The average value of the square of the momentum is going to equal H ̅² Α × R + ½ which is equal to H × K × μ¹/2 × R + ½.*2206

*In this particular case, this particular problem that we just did R = 3. So we just put it in.*2238

*3 + ½ is 7/2, you end up with the answer.*2243

*If you work ψ sub 6, it would be 13/2, that is it.*2246

*This is the general expressions and if you remember from a previous discussion that*2255

*the average value of X for the harmonic oscillator is actually equal to 0.*2259

*The average value of the momentum is actually equal to 0.*2264

*Now, we have the average value of X and its momentum.*2268

*The average value of X², the average value of the momentum².*2271

*These are all very important values.*2275

*You remember the relationship between this and this, and this and this, gives you the uncertainties.*2277

*Let us go ahead and finish off with a reasonably simple problem.*2288

*It seems to be a lot is actually quite simple.*2295

*For this example, of the root mean square displacement is just a square root of what it is that we just found.*2297

*We found this X² term, the average value.*2302

*The root means square displacement is just the square root of this number.*2306

*It is just this raised to the ½ power.*2310

*Given the following table of fundamental vibration frequencies and equilibrium bond lengths,*2316

*calculate the root mean square displacement of the molecules for the R = 0 state for the harmonic oscillator.*2322

*Then, compare them to their respective equilibrium bond lengths.*2330

*We have a molecule, we have N14 and this 16.*2335

*I can erase it.*2345

*You should actually be superscript it, sorry about that.*2348

*Basically, this is just the inter molecule, these two should be down below.*2353

*This is N2 and this is O2, there is nothing more than N2 and O2 molecules.*2360

*To the inter molecules, the fundamental vibration frequency is 23, 30 and this is in wave numbers.*2366

*It looks like I forgot about the 7 superscript of the numbers.*2374

*The equilibrium bond length of the bond is 109.4.*2380

*You remember the N2 is a triple bond, something like that, remember from general chemistry.*2383

*For the oxygen, it is not 14.*2391

*I’m just having a great time here, this is oxygen 16.*2393

*The oxygen molecule, its fundamental vibration frequency is 1556 inverse cm.*2397

*Its equilibrium bond length is 120.7.*2407

*These are actually in pm.*2412

*And if you remember, a pm is actually a 10⁻¹² m.*2417

*Milli, micro, nano, and pico.*2423

*Let us see what we have got.*2428

*They give us the fundamental frequency.*2431

*They want us to find the root mean square displacement of molecules.*2435

*I guess the first thing we probably need to do is find K.*2442

*Because we want us to find this, they want us to find that.*2453

*We said in the previous example, when we generalize that, that is just equal to H ̅/ K μ¹/2 × R + 1/2.*2457

*We need to go ahead and find what this K is and we are also going to be probably end up finding μ also.*2481

*Let us go ahead and do that.*2489

*We are going to go ahead and do the nitrogen molecule first.*2491

*Let us take care of that.*2496

*We will go ahead and do nitrogen first.*2497

*The fundamental vibration frequency is equal to 1/ 2 π C × K/ μ¹/2.*2503

*When I rearrange this I, end up with K, a force constant is actually going to be equal to*2515

*2 × π × C × this fundamental vibration frequency² × μ.*2525

*Let us see what we have got.*2537

*We said from the previous problem, we said that X² is going to equal 1/ Α R + ½.*2539

*For R = 0, we are just going to get 1/ 2 Α.*2551

*This is going to actually end up equaling H ̅/ 2 × K μ¹/2.*2561

*Let us go ahead and find some things here.*2574

*We just said that K is equal to this.*2578

*We need to find this and I would end up taking the square root of it.*2583

*Μu for this was just, it was mass 1 × mass 2/ mass 1 + mass 2.*2588

*14 × 14/ 14 + 14, you are going to end up with actually 7 atomic mass units.*2596

*Let me go ahead and do explicitly here.*2604

*It is going to be 14 × 14 atomic mass units/ 14 + 14.*2607

*I will multiply that by 1.661 × 10⁻²⁷ kg/ atomic mass unit.*2614

*We end up of μ equaling 1.163 × 10⁻²⁶ kg.*2624

*I hope that the arithmetic is correct, I’m not exactly sure if I did.*2632

*Our K is going to equal 1.163 × 10⁻²⁶ kg × 2 × π × 3 × 10¹⁰ cm/ s × 2330 inverse cm².*2638

*When I do that, I get a K is equal to 2243.*2671

*Our X² is going to equal H ̅/ 2 K μ¹/2.*2679

*That is going to equal, H ̅ is 1.055 × 10⁻³⁴ J-s divided by 2 × 2243 × 1.163 × 10⁻²⁶¹/2.*2692

*I hope that I do my arithmetic correctly.*2722

*It is going to be 1.033 × 10⁻²³.*2726

*When I take the square root of that, I'm going to end up with 3.214 × 10⁻¹² m or 3.214 pm.*2733

*There you go, from the previous problem we had a formula for that.*2757

*Let us say it is R = whatever.*2763

*We found that, and we realized that we have to find K and we have to find μ.*2766

*We used the fundamental frequency that they gave us to actually find a value of K.*2771

*We put it in, we found the average value of X² and we took the square root in order to get the root mean square displacement.*2775

*We found that its 3.214 pm for the nitrogen molecule.*2786

*As far as the oxygen molecule is concern, I'm going to go ahead and let you do that one yourself.*2792

*It is the exact same process.*2797

*Thank you so much for joining us here at www.educator.com.*2800

*Take care, I will see you next time, bye.*2802

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