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The Hydrogen Atom Example Problems V

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: Derive a Formula for the Degeneracy of a Given Level n 0:11
  • Example II: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ 8:30
  • Example III: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ 23:01
  • Example IV: Orbital Functions 31:51

Transcription: The Hydrogen Atom Example Problems V

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

Welcome back to Physical Chemistry.0002

Today, we are going to continue with our example problems for the hydrogen atom.0004

Let us get started.0009

Our first problem is the following, the energy of a given orbital is a function of the primary quantum number N.0013

The relation is as follows, E sub N = - E²/ 8 π A sub 0 A sub 0 N².0022

Again, this is the permittivity of free space, the A sub 0 A sub 0 is the Bohr radius and N is the primary quantum number.0030

With the exception of N = 1, each primary energy level is degenerate and for N = 2, N = 3, N = 4, so on.0039

In other words, they are difference states with the same energy.0048

Derive a formula for the degeneracy of a given level N.0053

That is, for each value of N, how many states have energy E sub M.0057

We want you to derive the formula for that.0065

Let us see what we have got.0069

We know that 4 each N, the L value 0,1, 2, and so on, all the way to N -1.0073

For each L, we know that M is equal to 0, + or -1, + or -2, all the way to + or – L.0095

I will go to the next slide here.0112

For each L value, there are 2 L + 1 states.0119

Therefore, we form the following sum.0137

The sum as L goes from 0 to N -1 of 2 L + 1. 0154

For every L value, we have 0, + or -1, + or -2, + or -, whatever all the way to + or – L.0164

For every L, there is 2 L + 1 state.0176

L is going to run from 0 all the way to N -1.0183

Whatever N we choose, we are going to add 2 L + 1 state for L0, 2 L + 1 states for 1, 0187

2 L + 1 state for 2, 2 L + 1 all the way to N -1.0194

This is the sum that we need.0199

This is the number of states that have the particular energy E sub M.0201

The sum forms an arithmetic sequence.0209

If you remember what an arithmetic sequence is, it is a sequence of numbers 0230

where each number differs from the one before it by a fixed number.0235

+ 5, 13579 is an arithmetic sequence and the difference is 2.0240

1 to 3 is 2, 3 to 5 is 2, 5 to 7 is 2, and you are adding which is why we call it arithmetic.0246

When you are multiplying like for example 2, 4, 8, 16, in the case you are multiplying by 2, 0252

each term was called a geometric sequence.0259

This is an arithmetic sequence.0261

The sum is actually equal to, when L = 0, we get 1.0264

When L = 1, we get 3.0270

When L = 2, we get 5.0273

We are going to add them all the way to, N -1 we put N -1 in here.0276

It is going to be 2 × N -1 + 1.0283

L is equal to 1 + 3 + 5 +… + 2N - 2 + 1.0290

The final term of the sequence is going to be 2 N -1.0299

We have a closed form formula for the sum of an arithmetic sequence.0304

The sum of an arithmetic sequence with N terms is equal to the first term + the last term × 0312

the total number of terms divided by 2.0336

In this particular sequence, the first term is 1.0340

The last term is 2 N -1.0348

The number of terms we have is N terms, from 0 to N -1 is N terms.0352

Our sum is equal to 1 which is the first term, + 2 N -1 × N/ 2.0363

That is equal to N²/ 2 is equal to N².0373

There you go.0388

For each N, there are N² states having energy E sub M, when this particular molecule is not in a magnetic field.0391

Let us go ahead and say it now.0426

When you place something in a magnetic field, the magnetic field actually ends up splitting the energy.0432

Now, the energy levels are going to degenerate.0438

You end up actually separating the energy levels which is why the spectrum often shows, 0441

when you put in a magnetic field several lines instead of just one line.0446

For each N, there are N² states having energy E sub N when not in a magnetic field.0450

I will go ahead and write what I just said.0462

Recall that when an atom is placed in a magnetic field, the field splits E sub N into 2 L + 1 levels.0466

Let us see.0511

For the angular portion of the ψ 321 32 -1 orbital, use the following linear combinations 0513

to represent the spherical harmonics S21 and S2 -1 as functions of the real variables θ and φ.0522

Let us talk about what is going on.0530

For the angular portion of the ψ orbital 321, use the following linear combinations 0532

to represent the spherical harmonics as functions of the real variables θ and φ.0539

Let us just recall it a little bit of our theory here.0543

I will use ψ 321 as the example.0550

321, that is made up of a radial function 32, which is a function of R.0553

Our spherical harmonic 21, which is a function of θ and φ.0566

The problem is, when M is not equal to 0, the spherical harmonics they are complex.0575

They are real in θ but they are complex in φ, E ⁺I φ E ⁻I φ.0587

Mathematically, this is does represent a problem but what chemist decided to do is, 0594

they decided to see if there are ways to actually change these complex functions and just represent them as real functions.0599

What they decided to do, I think we did not talk about this in the previous lesson, they took the spherical harmonic.0605

When we say angular portion, this is the radial portion of the wave function.0611

This is the angular portion of the wave function.0618

Radial, angular, or spherical harmonic, basically when we say angular portion.0621

They decided to see if they can represent these as real functions, real functions of θ and φ as opposed to complex functions.0627

They did that by taking linear combinations of them so that the complex part would cancel out.0636

They tried a whole bunch of linear combinations until they got one that is actually ended up working.0642

The reason they are able to do this was because we just said that these things are degenerate.0647

For functions that have the same energy, if I take a linear combination of them, they have the same energy.0651

In other words, they still satisfied the Schrӧdinger equation.0658

Basically, what we did is we ended up creating, taking S21 and then taking S2 -1,0663

adding them together or subtracting them depending and coming up with a new orbital that actually is a real function.0671

Instead of just dealing with S21 directly, we decided to fiddle around with it and0679

come up with a new orbital that is just a function of real variables θ and φ.0684

Not a function of complex variables.0692

That is all that is happening here, they still satisfy the Schrӧdinger equation.0694

They still have the same energy.0699

They are still Eigen functions of the Schrӧdinger operator, of the angular momentum² the angular momentum operator.0701

For all practical purposes, they are kind of the same function but they behave better.0708

That is all we have done, we have taken a function, we have created a new function 0713

from a pair that happen to have the same energy to create a function that behaves better.0716

Once again, use the following linear combinations to represent a spherical harmonics as functions of the real variables θ and φ.0724

This is one thing we are going to do, this is the other thing we are going to do.0731

We are going to take S21, we are going to add it to S2 -1.0735

We are going to multiply it by 1/ √ 2.0739

We are going to see what function we come up with.0741

That is going to be the DXZ orbital.0744

Let us go ahead and see if we can do this.0748

Let us go ahead and talk about what S2 and S1 are so we can look them up in our book.0750

S21 is equal to 15/ 8 π¹/2 sin θ cos θ E ⁺I φ.0756

We see this is a complex function and S2 -1 is equal to 15/ 8 π ^½.0772

This part is the same, sin θ cos θ except this E ⁻I φ.0786

Again, a complex function.0791

Chemists, for one reason or another decided they want one or more complex functions so they decide to add these,0793

subtract them, subject them to some linear combination of these 2,0800

to see if they can come up with functions that behave better.0803

These are the linear combinations that they came up with.0806

Let us go ahead and see what we can do.0809

DXZ , we are going to do this one first.0813

I think I will go back to black.0827

Let us go ahead and do blue.0831

DXZ = this thing, 1/ √ 2 × S1 S2.0834

That is equal to 1/ √ 2 ×, I’m going to pullout this and this are the same.0842

It was 15/ 8 π ^½ , the sin θ and the cos θ are the same.0854

I’m going to write E ⁺I φ + E ⁺I φ.0865

I'm going to write this as 1/ √ 2 15/ 8 π ½ sin θ cos θ ×, I’m going to express these in there.0876

Polar form, I’m going to write this as cos φ + I sin φ, Euler’s formula as an expression E ⁺I something.0894

E ⁺I θ = cos θ + I sin θ and then + E ⁻θ is going to end up equaling cos θ - I sin θ + cos φ - I sin φ.0907

This and this cancel out.0924

What I'm left with is, 1/4 ^½ 15/ 4 π ^½.0929

I have taken a π and I have written it to make a little bit easier to deal with.0955

Sin θ cos θ × 2 cos of φ.0959

1/4 is ½, the 1/2 and 2 cancel and what I end up with is 15/ 4 π ^½ sin θ cos θ cos of φ.0966

This is DXZ, this is the orbital that they came up with.0986

Notice, S21 S2 -1 had E ⁺I φ and E ⁻I φ, the complex went away.0990

I have this orbital which is sin θ cos θ of φ.1000

Real variable and real variable.1006

I took orbitals that are complex, I combine them to create an orbital which is real.1009

That is all I have done.1013

I personally, do not particularly like that they have done this.1016

What this does is when you actually plot this out using a polar plot, 1020

this is what gives you the famous dumbbell sheet orbitals and all the orbital shapes that you see in your general chemistry book.1026

They are also in your physical chemistry book.1035

The lobe is this way, that is what these are.1037

When they plot these, it gives a certain directional character in space along the X axis, along the ZY axis, 1042

and between the X and Y axis, and between the Y and Z axis.1048

As we get these lobes, they actually give you some information.1053

I did not know, some× they do, some× they do not.1055

But that is what is happening here.1059

They want to turn complex orbital into real orbitals.1060

OK so let us go ahead and do DYZ.1064

DYZ we said is equal to 1/ π √ 2 × S21, this time - S2-1.1068

This is going to equal 1/ I √ 2, do the exact same thing as I did before.1080

15/8 π ^½, this is sin θ cos θ.1086

It is actually nice to go through these which is why I included them in example problem.1094

I think it is nice to actually go through the mathematics and see where these functions come from.1098

The complex version and also the real version of the end up plotting.1105

Some books it just feels as though there is a sort of throwing these equations out like which one do I use.1108

You can use the complex but chemists tend to use the real version, the real variable versions.1114

This is θ and this is going to be E ⁻I φ.1121

This is going to end up equaling 1/ I √ 2 15/8 π ^½ sin θ cos θ.1134

Now, we have cos φ + I sin φ, then we have - cos φ - I sin φ.1149

- +, we end up with 1/ I √ 2 × 15/8 π ^½ sin of θ cos of θ.1165

Cos of φ - cos φ they go away, I sin φ.1181

You end up with 2 I sin of φ.1185

You end up with 1/2 I × 15/4 π ^½ sin θ cos θ × 2 I sin φ.1192

The 2 I cancel and you are left with 15/ 4 π ^½ sin θ cos θ sin φ.1210

This is equal to the DYZ orbital.1228

There you go.1236

Let me see if I have another page here that I can use.1238

I do, let me go ahead and do that.1243

Let us talk about what is actually happened here.1249

Sorry, I want to make sure you understand what is happening globally.1262

Ψ 321, I think that is what we had right and ψ 32-1.1267

We took the 21 and 2 -1 portions.1279

We took the spherical harmonic S21 and S2 -1, both of these were complex.1283

What we did is we found a linear combination, 1/ √ 2 adding these.1291

Basically, what I want to do is I want to find a way to turn this into a real function, a function of real variables.1296

I want to turn this into a function of real variables, instead of complex and complex.1303

By taking 1/ √ 2, this + this, I end up converting this one into the version that has the sin θ cos θ cos φ.1310

I ended up by taking 1/ √ 2 I × this – this.1325

I end up converting it into the version that has the sin θ cos θ sin φ.1331

I have 2 spherical harmonics.1341

I ended up converting them to 2.1344

Now, I call this one DXZ and I call this one DYZ.1350

Start off with 2, I ended up with 2, but now they are functions of the real.1356

Θ θ and φ, there is no complex number to be found.1360

That is all that is that I have done.1365

I sort of change the way they look but it is still the same function.1367

I still end up getting the same numbers out, whatever calculations with them.1370

That is all I have done, I have dressed them up.1375

Repeat the procedure for example 2, except this time with S22 and S2 – 2.1383

Usind the following linear combinations, which you should notice are the same as for example number 2.1389

Now, using the spherical harmonics S22 and S2 -2, can I find the linear combination of them which I have right here.1394

We are actually giving it to them explicitly.1403

1/ √ 2 × the sum of them and 1/ √ 2 × the difference of them.1405

We got a little bit of an issue here.1410

This is 2, this is – 2, this is 2, and this is -2, sorry about that.1420

It is the same linear combination, I'm just taking 1/ √ 2 × the sum of I/ √ 2 × the difference.1426

Again, converting this through this into a function of real variables.1432

That is all that is happening here.1439

Let us go ahead and take care of these.1443

We have got a DXZ and it is not DXZ actually, this one and this one, what happened here?1447

This one was actually going to be the DX² - Y² orbital and this one is actually going to end up being the DXY orbital.1463

There we go, now we are good.1475

We can actually do the problems.1479

Let us start off with the DX² - Y².1481

DX² - Y² = 1/ √ 2.1486

I will rewrite this, S22 + S2 -2 that is going to equal 1/ √ 2.1495

Let us actually write out what S22 S2 -2 are.1506

S22 is equal to 15/ 32 π ^½, this is going to be sin² θ E ⁺I φ.1510

S2 -2 = 15/ 32 π ^½ sin² θ E ⁺I – 2 φ.1528

This is S22, this is S2 -2.1544

It is going to equal , the 15/ 32 π we have to include that.1548

The sin² θ is the same so I pulled that one out as a factor.1559

And I'm left with E ⁺I 2 φ + E ⁻I 2 φ.1563

That is going to equal, DX² - Y² is going to equal, 1576

I’m going to write that as 1/ 2.1583

I pulled out, this is now going to be 16 π.1585

I pulled out a 2 out of there.1589

½ sin² θ, exactly the same as before.1591

Now, it is cos 2 φ + I sin 2 φ + cos 2 φ - I sin 2 φ.1599

That cancels with that and I'm going to end up getting 15/16 π¹/2 sin² θ × the cos of 2 φ.1611

Actually, 2 × the cos of 2 φ.1634

This and this cancel and I'm left with my final DX² - Y² is equal to 15/ 16 π ^½ sin² θ cos of 2 φ.1636

There we go, that takes care of that one. I'm going to do D of XY.1658

This was going to be 1/ I √ 2 × S22 - S2 -2.1667

And this is going to equal 1/ I √ 2 15/ 32 π ^½.1676

The sin² θ comes out, we have E ⁺I2 φ – E ⁻I 2 φ.1692

Let us go ahead and pull out a 2 here.1704

√ 2 cancels √ 2, so I get 1/2 I × 15/ 16 π ^½ sin² θ.1706

The same thing, this time we have cos 2 φ + I sin of 2 φ - cos of 2 φ - I sin of 2 φ.1720

Cos of 2 φ - cos of 2 φ, this and this they add together.1736

I end up with 1/ 2 I × 15/ 16 π ^½ sin² θ × 2 I sin 2 φ.1742

2 I cancels the 2 I, and I'm left with the D of XY orbital is equal to 15/ 16 π ^½ × sin² θ × sin of 2 φ.1763

There we go.1789

Basically, I have 2 complex spherical harmonics S22 and S2-2,1790

I need functions of real variables not functions of real and complex variables.1810

I have taken this one and the combination with this one, 1815

I have transformed it into the one that involves the sin² θ cos 2 φ.1819

I have taken this one and by combining it with this one, in a linear combination that ended up working out really nicely,1826

I ended up with the one that gave us the sin² θ sin 2 φ.1833

Of course, there are the constants, the 15/ 16 π in front, that is all I have done.1838

I have taken complex spherical harmonics and I turned them into real functions and end up giving me the same values.1844

I have dressed up the function, I have made it look better.1853

I make it easier to handle, as far as real variable is concerned.1856

Mathematically, the fact of the matter is, it is easier to handle than this one.1859

It is not any more difficult to handle a complex variable function as opposed to a real variable function.1864

In fact, in my personal opinion, complex variable functions are easier to handle in a lot of ways.1871

Again, for one reason or another, historically, chemists prefer to deal with just functions of real variables.1877

These have the advantage of you can plot them.1882

The plots that you see, those bell shaped plots, all of the polar plots for the densities of the electrons, 1888

the DC², the XY, the YZ, the lobes here and there, you are actually seeing plots of these polar plots of those.1895

Let us move on to the next one.1909

A lot of information here.1913

It is not difficult, it is just extra information.1916

Here is when you actually have to look back on your book to find the functions that you need.1920

The following are the change of variable relations between the Cartesian and spherical coordinates.1926

X is equal to R sin θ cos φ.1931

Y is equal to R sin θ sin φ.1933

Z is equal to R cos θ.1936

The following are the real variable representations of the angular portions of1939

the spherical harmonics of the hydrogen atom wave functions for L = 2.1944

In other words, the D orbitals.1949

Pretty much what we just did.1951

We did this one, this one, this one, and this one.1953

The one for DZ³², the one for ψ 320.1960

In other words, the S20 that one did not need to be converted into a functional real variable1967

because it is already a function of a real variable as it stands.1974

I do not have to subject it to any linear combination.1977

It is the others, the 21 2 -1, the 22 2-2, spherical harmonics that I had to subject 1981

to linear combinations to convert them to these equations.1988

The ones in real variables.1992

This on, I can just take as is.1994

These are the real variable representations of the D orbitals.1996

Ψ 320, we call it DZ².2002

Ψ 321, we call it DXZ.2005

Ψ 32 -1, we call it DYZ.2008

322, re call it DX ⁻Y².2010

32 -2, we call it DXY.2015

What you are charged with doing is the following.2021

Use the information given to explain why these orbital functions are actually given the labels DZ², DXZ, DYZ, DXY², and DXY.2025

Why we call it ψ 320, 321, 32-1, 322, 32 -2.2037

We do but this is how we refer to them.2044

Why do we refer to them like this?2047

Let us see what we have got.2053

Let us go ahead and take the first one.2056

Let us go ahead and take the first one which is the ψ 320, that function is 3 cos² -1.2060

I actually need to write my things down here.2077

X = R sin θ cos φ.2089

Y = R sin Θ sin φ.2096

Z = R cos θ.2101

I'm going to change this and I'm going to express it in terms of X, Y, and Z.2105

3 cos I² θ -1.2112

Cos² Θ is nothing more than Z²/ R².2118

I’m going to write this as 3 Z²/ R² -1.2123

There you go, there is my Z² because the spherical coordinate version, 2130

once I convert it using the conversions between spherical and Cartesian version which is 3 Z²/ R², 2142

because it is something proportional to Z², I end up calling this DZ².2151

That is where the DZ² comes from.2158

Let us do the next one.2162

Let us go ahead and do the sin θ cos θ cos φ.2164

We can consider that one to be the ψ 321.2176

Sin θ, I’m going to convert this to XYZ form.2181

Let us see what it turns out to be.2188

Sin θ, I'm going to use this one.2191

It is going to be X/ R cos φ.2195

The cos θ is going to be Z/ R.2205

The cos φ, I’m going to take this one cos φ is equal to X/ R sin θ.2212

This is equal to XZ X/ R² × R sin θ cos φ.2224

R sin θ cos φ is nothing more than X.2246

This is X φ X/ R² X, those cancel and I'm left with X Z/ R².2250

There is my XZ, something proportional to 1/ R².2262

This orbital, that is why this is called D sub XZ.2266

Let us do our next one.2275

This one we can consider the 32 -1 orbital.2277

That one has sin θ, I’m not using the constants in front of 15/ 16 π with those orbital, R sin θ cos θ sin of φ.2281

Let me write it here again for reference.2301

X = R sin θ cos φ.2304

Y = R sin Θ sin φ.2310

Z = R cos θ.2314

Θ, φ, and R all over the place.2319

Sin θ cos θ sin φ.2322

Sin θ is equal to Y/ R × the sin of φ.2324

Cos θ is equal to Z/ R.2339

I’m converting it back to the Cartesian form.2343

Sin of φ is equal to Y/ R sin Θ.2350

This is equal to YZY/ R² × R sin θ sin φ which is equal to YZY/ R².2357

R sin θ sin φ is just Y so that goes away, leaving me YZ/ R².2376

There is my YZ, this is my DYZ orbital.2387

That is why I call it DYZ because ψ 321 orbital, when I express those, the sin and cos, 2393

and things like that, in terms of XYZ, I end up getting something which is 1/ R² × YZ.2401

We call it DYZ.2409

What else have we got?2415

Let us say the ψ 322, and that one the trigonometric portion is sin² θ × the cos of 2 φ.2432

For the cos of 2 φ, let us use the identity cos² φ - sin² φ.2449

We can write this as sin² θ × cos² φ - sin² φ.2473

Again, using these conversions I end up with the following.2491

This ends up being equal to X²/ R² cos² φ.2494

The cos² φ is equal to X²/ R² sin² θ.2504

The sin² of φ is going to be Y²/ R² sin² θ.2517

This is going to equal X² × X² - Y²/ R² × R² sin² θ cos² φ.2531

This is equal to X² × X² - Y²/ R² × X².2556

The X² cancel and I'm left with X² - Y²/ R².2572

It is 1/ R² × X² - Y².2581

This is where we get our DX² - Y² label.2586

This is what we call it the X²- Y² orbital.2592

Let us do our final one.2599

I think it was ψ 32 -2 and that one was sin² θ sin of 2 φ.2601

Let us go ahead and use an identity here.2616

We will use the identity.2619

Sin² θ × 2 sin φ cos φ, pretty standard double angle identity first.2621

Let us go ahead and write again.2636

We had on this page, X = R sin θ cos φ.2638

Y = R sin θ sin φ.2646

Z = R cos of θ.2651

When we put these into these, we end up getting X² / R² cos² φ × 2 × the sin of φ 2654

which is Y/ R sin θ × cos of φ, which is X × R sin θ which is equal to X² × 2 XY/ R² × R² sin² θ cos² φ, 2673

which is equal to X² × 2 XY/ R² × X².2707

The X² cancel and I'm left with 2 XY/ R².2717

XY proportional to the 2/ R².2726

Therefore, this orbital is why we actually call it the DXY orbital.2731

I hope that makes sense.2739

Thank you so much for joining us here at www.educator.com.2740

We will see you next time, bye.2743