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### Example Problems I

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

• Intro 0:00
• Equations Review 0:11
• Energy of the Harmonic Oscillator
• Selection Rule
• Harmonic Oscillator Wave Functions
• Rigid Rotator
• Selection Rule for Rigid Rotator
• Frequency of Absorption
• Wave Numbers
• Example I: Calculate the Reduced Mass of the Hydrogen Atom 11:44
• Example II: Calculate the Fundamental Vibration Frequency & the Zero-Point Energy of This Molecule 13:37
• Example III: Show That the Product of Two Even Functions is even 19:35
• Example IV: Harmonic Oscillator 24:56

### Transcription: Example Problems I

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

We have talked about the harmonic oscillator.0004

We talked about the rigid rotator and now we are going to do some example problems.0007

Let us jump right on in.0010

Before we jump on in, I always like to recall what it is that we learned.0012

Take a look at the major equations so we have a certain set of tools so that we know what we are going to be dealing with.0016

Let us recall what we have.0021

Let us recall what we have.0026

I think I’m going to work in blue this time.0034

The energy of the harmonic oscillator.0037

The energy was E sub R is the state, R is a quantum number.0039

It is equal to H ̅ × K/ ν¹/2 × R + ½, where R is equal to 0, 1, 2, and so on.0057

Other versions of this are H ̅ O R + ½ or this ω here is going to be this thing,0075

this is the angular velocity, if you want think about it that way.0087

That is that.0089

Another version of that is H × the frequency R + ½.0091

We have the ω is equal to this K / ν¹/2.0099

This ν is equal to 1/2 π K/ μ¹/2 is equal to ω/ 2 π.0109

Just so we have everything on the table here.0123

K is the force constant of the bond of the harmonic oscillator.0126

The μ is the reduced mass.0136

The definition was 1/ μ is equal to 1/ mass 1/ mass 1 + 1/ mass 2.0148

Or if you prefer, μ = M1 × M2/ M1 + M2.0163

I will go ahead and do on the next page.0178

The selection rule for the harmonic oscillator was that δ R = + or -1.0183

The selection rule for the harmonic oscillator, when the harmonic oscillator0194

is moving from one energy state to another, it is not going to jump from the first state to the fifth state.0198

It is going to go one at a time.0202

It is going to go up by 1 or down by 1, that is all that means.0204

The observe frequency of radiation, we hit it.0209

When we hit, some system was rotating, some quantum system is rotating with a certain amount of energy.0236

It is going to absorb radiation as certain frequency.0242

What frequency is that?0247

That is what this is.0248

The observed is equal to 1/ 2 π K/ μ¹/2.0252

We often represent this actually in terms of wave numbers, which the unit is inverse centimeters.0264

And that is going to equal 1/ 2 π C K/ μ¹/2.0270

In spectroscopy, we tend to use these units, this wave numbers.0281

The unit is inverse cm.0288

This thing is just that divided by C.0292

Anytime you see a tilde over something, just find the corresponding variable and just divide by the speed of light.0297

Adjacent states are the same, δ E apart.0306

State 1, state 2, state 3, state 4, are the same δ E apart.0321

When we take a look at spectrum for the harmonic oscillator, it is going to be a single line.0328

The spectrum is a single line at this.0335

Let us take a look at the harmonic oscillator wave functions.0354

We are just collecting our little bits of information, the most important bits of information that0359

we know to solve these problems.0364

The harmonic oscillator wave functions.0366

We have ψ sub R is equal to N sub R H sub R, the argument is Α¹/2 X E ^- Α X²/ 2.0373

Where Α is equal to K × μ/ H ̅²¹/2.0392

The normalization constant, it actually changes with R.0405

It is equal to 1/ 2 ⁺R × R!¹/2 × Α/ π¹/4.0408

These are the Hermite polynomials.0426

This H sub R are the Hermite polynomials for our rigid rotator.0432

The harmonic oscillator, now rigid rotator.0455

The energy of the rigid rotator, we have energy sub J is equal to H ̅/ 2I J × J + 1 K,0458

or J is going to equal to 0, 1, 2, 3, and so on.0488

I here is the rotational inertia, it happens to equal the reduced mass × the radius²,0495

where R is the distance between the masses.0504

Of course, this μ is the reduced mass which we saw earlier.0523

We also have D sub J is equal to 2 J + 1.0526

This the degeneracy, this is the number of states that actually have this energy 2 J + 1.0531

If you are at the J = 2 state, you are going to have 2 × 2 + 4, 2 × 2 is 4 + 1 is equal to 5.0541

You have 5 different states that have the energy.0550

The selection rule for the rigid rotator is δ J = + or -1.0558

You can only move to different states, passing through adjacent states.0568

The frequency of absorption transition is the frequency is equal to H/ 4 π² I × J + 1.0576

Which is more commonly written as frequency is equal to 2 B × J + 1, where this B is equal to this H / 8 π² I.0606

B is called the rotational constant, very important.0649

In terms of wave numbers, we have that it is equal to 2 B J + 1.0659

The only difference is like we said, any time you see a tilde over, you just divide everything by the speed of light.0681

We have H/ 8 π² CI.0687

Let us go ahead and start doing our example problems.0693

The first one is going to be very simple.0697

Just computation of a reduced mass.0700

Calculate the reduced mass of the hydrogen atom.0706

Use the full decimal values listed in your book for the masses of the proton and the electron.0708

We said that the reduced mass is going to equal mass 1 × mass 2/ mass 1 + mass 2.0714

Very simple.0727

We have 9.109390 × 10⁻³¹ kg × 1.672623 × 10⁻²⁷ kg and divided by this + that.0729

9.109390 × 10⁻³¹ + 1.672623 × 10⁻²⁷, all of that is kilograms.0762

When we do the math, we end up with 9. 104432 × 10⁻³¹ kg.0782

Kg × kg is kg² on top.0795

When we add the numbers, this many kilograms × this many kilograms, the units are multiplied so it will be squared.0799

When we are adding, it is just 1 unit of kilogram, that cancels that is why we end up with kg here.0807

Nice and simple.0814

Example 2, the force constant K of the nitrogen 14 nitrogen 14 bond is 2243 N/ m.0819

Calculate the fundamental vibration frequency and the 0 point energy of this molecule.0831

The fundamental vibration frequency is the frequency that we actually see at that single spike,0838

that absorption spike on the spectrum.0845

That is called the fundamental vibration frequency.0848

We are also going to be finding the 0 point energy.0852

The energy where we are at the 0 state, where R = 0.0856

Let us see, we have ν ̃ is equal to 1/ 2 π C × K/ μ¹/2 and this is in units of inverse CM.0862

We are going to be using wave numbers.0877

Let us go ahead and calculate the reduced mass.0879

The reduced mass is going to be, we need that first.0883

We have K, we have reduced mass, that is the only one that we have to find here.0889

Like we said, it is going to be M1 M2/ M1 + M2.0894

It is going to equal, this is nitrogen 14 14 so we have 14 atomic mass units × 14 atomic mass units/ 140902

+ 14 atomic mass units × 1.66 × 10⁻²⁷ kilograms per atomic mass unit.0915

That and that, they all cancel.0933

What I’m left with is 1.162 × 10⁻²⁶.0935

When we are given the molecule this way, these are not g or kg, they are atomic mass units.0942

To work in mass units then you have to convert.0948

1 atomic mass unit is equal to that many kilograms.0951

We want the mass in kg.0955

This is equal to 1/ 2 π × 3.0 × 10¹⁰.0962

We have to use the speed of light in cm/ s, not m/ s.0974

K is 22 43 N/ m, 1.162 × 10⁻²⁶ kg¹/2.0979

You end up with 2331 inverse cm.1001

The spectrum would look, I would see a single spike at 2331.1011

Let us go ahead and do our 0 point energy.1018

Do I have another page here? Yes I do.1023

I think I can go ahead and do the 0 point energy.1025

We said that the energy is going to equal H μ × R + ½.1039

We want it to be for R = 0, the 0 point energy.1049

Our energy sub 0 is going to equal H μ/ 2.1054

Let us see what we have got here.1063

We know that μ/ C is equal to ν ̃ which means, therefore, that ν itself is equal to ν ̃ × C1066

because ν ̃ is actually the value that they gave us, or the value that we just found because the formula is actually ν.1079

Let us go ahead and calculate ν.1089

Ν is equal to this 2331 inverse cm × the speed of light 3.0 × 10¹⁰ cm / s.1091

The centimeter and centimeter go away.1104

And I'm left with a frequency of 6.993 × 10¹³ inverse seconds which is Hz.1106

Therefore, our 0 point energy is going to equal ½ × planks constant 6.626 × 10⁻³⁴ J-s × 6.993 × 10¹³.1118

/ s, that goes away and we will do our calculation.1145

If I have done my arithmetic correctly which I probably have not, so please check it for me.1149

2.33 × 10⁻²⁰ J, I really dislike arithmetic.1152

that the problem of quantum mechanics is not the difficulty.1166

The problem of quantum mechanics is the tedium of the calculations.1169

Let us do example 3, show that the product of 2 even functions is even but1176

the product of 2 odd functions is even, and that the product of an odd and even function is odd.1181

What this has to do with quantum mechanics?1187

You will see in a minute in our next problem, when we start actually doing integrals.1189

We are going to be integrating odd functions, even functions, odd functions × odd, odd functions × even functions.1192

It makes our integration easier, I just want to do this real quickly.1200

Let us recall what even and odd mean.1205

An even function means that if I do F of – X, that I end up just getting my original X back.1207

In other words, if I have some function F of X, if every place there is an X, I stick a –X then and I simplify it,1219

if I end up getting the original function back, that is an even function.1225

Something like X² is an even function, something like cos of X is an even function.1229

An odd function, it means that if I stick in a - X wherever I see an X in my function,1235

I end up actually getting the negative of the original function back – F of X.1243

X³ is a perfect example of an odd function.1247

Sin X is a perfect example of an odd function.1250

An even function is symmetric around the Y axis.1253

An odd function is symmetric about the origin.1259

X³ is symmetric about the origin, that is the difference between the two.1266

That is the geometric version, this is the algebraic version.1270

We want to show that the product of 2 even functions is even.1275

The first one is both are even.1280

The product function is going to be F of X × G of X.1286

The P of – X, because we want to show that the product is odd or even, so we put in a - X wherever there is an X.1299

The P of - X is just going to equal F of -X × G of –X, but they are both even functions.1307

If they are even, F of – X is F of X.1318

G of –X is G of X, so this is just F of X × G of X.1321

That just = P of X.1328

P of - X = P of X so it is even.1333

P of -X = P of X which means we have an even function.1344

The product of 2 even functions is an even function.1351

Let us take a look at B.1359

B says that if they are both odd, if you multiply an odd function and an odd function, you should get an even function.1364

In this case, both odd.1371

We need to find P of – X.1375

P of - X is equal to F of - X × G of –X.1378

F of X and G of X are both odd which means that F of -X is –F of X and G of -X is -G of X.1386

When I multiply them together, the negative cancel so I still end up with F of X and G of X, which is equal to P of X.1397

Therefore, P of -X = P of X.1405

When both functions are odd, their product is even.1411

C, we have one odd and we have one even function.1424

We do P of –X, P of -X is equal to F of -X × G of –X.1432

If one of them is odd, let us say the F is odd, that is going to be –F of X.1441

G of –X, that one is even, that is just going to be G of X.1451

We are going to end up with -F of X G of X, that is going to be –P of X.1455

Therefore, P of -X is equal to –P of X, that is the definition of an odd function.1464

If you have odd function and even function, their product is odd.1473

Let us see what we have got.1490

We can go ahead and apply this to a particular problem.1492

For the harmonic oscillator, demonstrate that ψ 1 and ψ 2 are orthogonal, and that ψ 1 and ψ 3 are orthogonal.1498

Remember what orthogonal means.1508

Let us just go ahead and write the wave functions first.1511

I think I would go back to black.1512

Let us write out, we have to do this explicitly.1515

Let us just write off the wave functions first.1518

We have ψ 1, it is going to equal 4 α³/ π¹/4 × X E ⁻α X²/ 2, that is the wave function for ψ 1.1521

Ψ 2 is equal to Α/ 4 π¹/4 × 2 Α X² -1 × E ⁻Α X²/ 2, that is ψ sub 2 for the harmonic oscillator.1541

The wave function ψ 3 is equal to α³/ 9 π ¼ 2 Α X³ -3 X E ⁻Α X²/ 2.1561

These are the 3 wave functions.1582

Let us recall what orthogonal means.1584

Orthogonal for the harmonic oscillator means, the harmonic oscillator moving along the X axis.1588

Theoretically, we have to integrate over all of space so it can go to infinity, positive infinity and negative infinity.1605

Ψ * ψ MN, we need that integral to equal 0.1616

If the integral = 0, it demonstrates that they are orthogonal.1623

I need to demonstrate that ψ 1 ψ 2, the integral of the product of those two is equal to 0 and1626

the integral of ψ 1 ψ 3 is equal to 0.1632

I need to do these integrals explicitly.1635

Notice that, this E ^- Α X²/ 2, this is an even function.1641

Even × you have this X here, this is an odd function here.1654

Ψ sub 1 is odd, ψ sub 2 is even because of that X², and this is odd.1663

The ψ sub 1 is odd, ψ sub 2 is even, ψ sub 3 is odd.1674

Let us see what we have got.1683

In general, ψ sub R is even when R is even.1687

One is an odd number, it is an odd function.1704

2 is an even number, it is an even function.1707

3 is an odd number, it is an odd function.1709

That is the general case for the harmonic oscillator.1712

In general, ψ sub R is even when R is even and ψ sub R is odd when R is odd.1714

Let me go back to black here.1727

Ψ * 1 ψ * 2, is equal to ψ 1 ψ 2 because it is a real function.1733

The complex conjugate of a real function is the function itself because ψ is real.1742

Basically, what you have now is an odd function × an even function.1754

You have an odd function × an even function.1758

What do we say?1762

An odd function × an even function is going to be an odd function.1763

It is going to end up being odd.1769

This ψ 1 ψ 2 is odd.1773

Recall, when we integrate -A to A, when we integrate along a symmetric interval,1779

some odd function the integral actually = 0.1793

The integral from - infinity to infinity of ψ 1 ψ 2 is the integral of an odd function about a symmetric interval that is equal to 0.1800

You do not have to actually have to solve the integral.1812

Therefore, ψ 1 and ψ 2 are orthogonal.1816

Let us do ψ 1 and ψ 3.1827

Ψ 1 is odd and ψ 3 is odd.1831

Ψ 1 * ψ 3 is equal to ψ 1 ψ 3, odd × odd, we are going to end up with an even function.1843

In this case, ψ 1 ψ 3 is even.1859

We do not have things simple like that so we would actually have to do this integral explicitly.1862

We must integrate explicitly.1869

Let us go ahead and move on to that.1881

Our integral is going to look like this.1883

The integral from -infinity to infinity of ψ 1 ψ 3 DX is going to equal,1885

I’m going to pull out the normalization constants out of the integral.1896

It is going to be 4 Α/ π¹/4 × Α/ 9 π.1900

My numbers are getting crazy.1918

To the ¼ × the integral from 0 to infinity, because it is a symmetric interval, I can just multiply the integral by 2.1921

Instead, of going from –infinity to 0 is the same as 0 to infinity.1934

Therefore, I will just go 2 × 0 to infinity of X × E ^- Α X²/ 2 × 2 Α X³ - 3X × E ⁻α X²/ 2 DX.1938

This is the integral, this is what I have to solve.1961

Do not worry, I know it looks really scary but we have software.1964

I’m just going to go ahead and call this as Z.1969

I do not want to keep writing the normalization constant.1970

This is actually going to equal to Z × the integral from 0 to infinity.1974

I’m going to multiply the X × this thing and I’m going to multiply this and this.1982

When I multiply this and this, the two goes away so I’m left with the following.1986

I'm left with 2 Α X⁴ - 3X² × E ⁻Α X² DX.1990

You can use a table of integrals or you can just use your math software.2002

When you do this integral, when you solve it, it is equal to 0.2007

I’m not going to go ahead and solve it manually.2012

I actually do so in the next lesson, but in this case, one of the nice things2014

about having math software is you can do complicated integrals like this.2020

Thank you so much for joining us here at www.educator.com.2025

We will see you next time, bye.2027