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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### 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
- Example I: Derive a Formula for the Degeneracy of a Given Level n
- Example II: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
- Example III: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
- Example IV: Orbital Functions

- 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

### Physical Chemistry Online Course

### 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 and*0679

*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 of*1939

*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 variable*1967

*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

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