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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

Transcription: Example Problems II

Hello, welcome back to

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

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