For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### The Hydrogen Atom 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
- The Hydrogen Atom I
- Review of the Rigid Rotator
- Hydrogen Atom & the Coulomb Potential
- Using the Spherical Coordinates
- Applying This Last Expression to Equation 1
- Angular Component & Radial Component
- Angular Equation
- Solution for F(φ)
- Determine The Normalization Constant
- Differential Equation for T(a)
- Legendre Equation
- Legendre Polynomials
- The Legendre Polynomials are Mutually Orthogonal
- Limits
- Coefficients

- Intro 0:00
- The Hydrogen Atom I 1:31
- Review of the Rigid Rotator
- Hydrogen Atom & the Coulomb Potential
- Using the Spherical Coordinates
- Applying This Last Expression to Equation 1
- Angular Component & Radial Component
- Angular Equation
- Solution for F(φ)
- Determine The Normalization Constant
- Differential Equation for T(a)
- Legendre Equation
- Legendre Polynomials
- The Legendre Polynomials are Mutually Orthogonal
- Limits
- Coefficients

### Physical Chemistry Online Course

### Transcription: The Hydrogen Atom I

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

*Today, we are going to start our discussion of the hydrogen atom.*0004

*We are ready to put everything together and we are ready to talk about the orbitals, *0008

*the energy levels of the hydrogen atom.*0012

*One little bit of warning before we start here.*0014

*This is going to consist of multiple lessons and is going to be very heavily mathematical, *0018

*at least in the sense of all the symbols floating around on the page.*0023

*Ultimately what is important is, we want to be able to find the solutions for ψ,*0026

*the wave function for the electron that is moving around the proton in the hydrogen atom.*0034

*That is what is most important.*0039

*It is easy to lose our way, I'm going to present it in the normal fashion.*0041

*I’m going to present most of the mathematics, not all of it.*0047

*It is really important that you separate, do not get intimidated and just alienated from all the symbols on the page.*0050

*I will try to keep it clear as to what is important and what is just cultural information.*0060

*What is just part of your scientific literacy.*0066

*I could just present the solutions and say this is what it is and jump one to some problems *0068

*but we also want you to understand where does that these things come from because that is where real understanding takes place.*0072

*Ultimately, you are only responsible for the solutions *0078

*but we want to show you where the solutions come from and why they take the form it takes.*0081

*Just a little bit of warning, do not be intimidated, it is just symbols.*0085

*Let us get started.*0089

*In discussing the rigid rotator, we held R, the radius of the turn constant.*0092

*We allowed θ and φ to vary.*0120

*Something rotating this way, this way, is basically just something spinning around some fixed center.*0130

*You had the θ which is this way and the φ which is this way.*0138

*Those are the things that vary so it gave you all of the points on a sphere.*0141

*We are going to allow R to actually change.*0146

*We are going to allow not just move it about a sphere, we are going to allow it to be close.*0149

*We are going to allow it to be far, so all of space is going to be considered now.*0153

*We allow R to vary also.*0163

*For the hydrogen atom, we take as our model a fixed proton of the center, *0171

*a fixed proton at the origin so that will be our 0 position or origin position.*0197

*An electron of mass M sub E interacting with the proton via the Coulombic potential.*0205

*Basically, this is just a fancy way of saying we have a positive charge and a negative charge and *0231

*they are going to interact with each other based on the fact that positive and negative charge attract each other.*0234

*That is it, that is all we are saying.*0240

*The Coulomb’s, the potential is the potential energy is 1 charge of positive charge and negative charge,*0243

*the distance between them being R.*0250

*The potential energy is going to equal – E² / 4 π ε sub 0 R.*0253

*E here, it is the charge on 1 proton, the charge an electron in Coulomb’s.*0264

*It is going to be 1.602 × 10⁻¹⁹ C.*0271

*This ε sub 0, that is a permittivity of free space and its value is 8.854 × 10⁻¹² C² / J-m.*0278

*It is called the permittivity of free space.*0299

*R is the distance between the proton and the electron.*0315

*Basically, you just have this proton, center, and you already know it is already electron is moving around all over.*0329

*Not necessarily in circular orbits but now it is moving around this way, it is getting close, it is getting far, it can be anywhere.*0342

*Let us see what we have got.*0351

*Basically, we dealt with this rigid rotator which is a perfect sphere and all we have really done is allow the electron to move in and out.*0354

*Get close to the proton, get far from the proton.*0362

*We still want to work in spherical coordinates because the symmetry of the situation naturally lends itself to spherical coordinates.*0365

*That is why these various coordinates systems exist *0374

*because certain problems lend themselves to a particular coordinate system so the math actually becomes easier.*0377

*In spherical coordinates, let me go ahead and erase this.*0384

*Let me see, should I go ahead and put on this page.*0391

*A spherical symmetry of the model compels us to use spherical coordinates.*0394

*We are going to end up with something like this.*0426

*The Schrӧdinger equation is going to be H of ψ.*0428

*Ψ is a function of 3 variables, R, θ, and φ.*0431

*R, θ, and φ is equal to E × ψ of R, θ, and φ.*0436

*The Eigen function, Eigen value representation of that Schrӧdinger equation.*0447

*H is going to be, the Hamiltonian operator here is going to be – H ̅² / 2 × the mass of the electron Del² - the potential energy.*0451

*The potential energy was the Coulombic potential here, - E² / 4 π ε NR. *0466

*We know what del² looks like in Cartesian coordinates.*0483

*A spherical coordinates, it looks like this.*0492

*Let me go ahead and rewrite this.*0495

*H R is equal to - R² / 2 ME Del² -E².*0497

*Where the del² operator in spherical coordinates, when we actually make *0522

*the change of variable from cartesian to spherical, it looks like this.*0527

*It looks really complicated.*0531

*Do not let it scare you, it is just symbols.*0533

*We just want you to see it.*0534

*It is in your books, we want you to see it.*0536

*We want you to at least recognize it, that is all we ask.*0538

*Whatever R², DDR, R², DDR, so this is the del² operator in a spherical coordinate.*0542

*I'm sure that I’m forgetting something here, R sin θ DD θ sin θ DD θ +,*0561

*let me actually write it here, so I’m not squeezing it anywhere, + 1/ R² sin θ × E² D φ².*0578

*We see we have R, we have θ, and we have the φ.*0588

*We got the Laplacian operator in spherical coordinates.*0593

*That is all it is.*0598

*If we put this last expression into the equation 1, the original equation that we had *0601

*which was this ψ of R θ and φ is equal to energy × ψ R θ and φ.*0607

*In other words, if we apply this to this, we end up with something like this.*0620

*We apply this last expression, meaning this thing right, the whole thing, *0633

*that where this del square is equal to as, we apply this expression to, let us call this equation 1 and multiply by 2 M sub ER².*0642

*We get this thing.*0671

*Let us see if we can, that is getting crazy here.*0676

*We get –H ̅² DDR² D of ψ DR –H ̅² 1/ sin θ × DD θ sin θ D of ψ D θ + 1/ sin².*0680

*Here are the mistakes, I think there is an R² there and an R² there.*0726

*1/ sin² θ D² ψ.*0731

*Once again, I’m going to keep saying it this is not something that you have to know.*0740

*We just want you to see it because I think it is actually need to see it.*0744

*It is in your books but more than likely you are just sorting orders in your books, like the rest of us in school.*0748

*-2 M sub E R² E/ 4 π E sub 0 R + the energy × ψ = 0.*0754

*This is the equation that we end up solving.*0776

*It is really nice to see it because I think it is one intellectual achievement back *0779

*but we have managed to come up with this and would actually managed to solve it and be very successful in our solutions.*0783

*It is actually quite extraordinary.*0790

*It is really nice to know that you are in a position to appreciate this, see it, and understand what it means.*0792

*It is not necessarily the mechanics of the actual solution of it so you really should be very proud of where it is that you are.*0798

*This ψ, we said is actually a function of 3 variables.*0810

*It is a function of R, θ, and φ, so in order to solve this, we are going to separate variables.*0815

*We are going to write this ψ R θ and φ, we are going to write it in terms of some function just of R and some function of θ and φ.*0823

*The wave function for the hydrogen atom, we are going to separate them into two functions, a product of two functions.*0842

*One of them is just a function of the radius.*0849

*One of them is just a function of the angles θ and φ.*0851

*This one right here, this is called the radial component of the hydrogen wave function.*0858

*This is called the angular component.*0871

*We are going to deal these one at a time.*0876

*Since we have separated variables and written ψ as a product of two functions, we recovered two separate differential equations.*0887

*We are going to have a radial equation and we are going to have an angular equation.*0922

*We are going to deal with the angular one first.*0931

*We will deal with the angular equation first.*0945

*Let us go ahead and start working in blue.*0955

*When we do this, we are going to take these angular differential equation that we get *0958

*and we are going to subject it to further manipulation.*0964

*I’m not going to go through all of the manipulations but when we do that, *0966

*I will just say with further manipulation which is nothing more than simplifying it so we are going to deal with it better.*0971

*With further manipulation, we get sin of θ DD θ sin of θ DS D θ.*0978

*S is the function of θ and φ, the angular component that consist of the two variables.*0995

*Θ + D² S D φ² + β sin² θ × S is equal to 0.*1003

*This is for the angular component, the S θ of φ, this is the differential equation that we have solved in order to find S.*1020

*We are looking for S.*1032

*Notice that this equation is the same as the Schrӧdinger equation for the rigid rotator.*1034

*It is the same as the Schrӧdinger equation for the rigid rotator.*1058

*The angular portion of the hydrogen atom wave function is the same as the rigid rotator wave functions.*1071

*Let us see here, let us go ahead and do this.*1116

*We are looking for this S of θ and φ.*1121

*That is what we are looking for, the angular components.*1128

*We are dealing with this so we subject this to another separation of variables.*1131

*We are going to take this S of θ and φ because we are dealing only with that and deal with it by separating variables.*1138

*We are going to express this as a product of two functions, 1 just of θ, 1 just of φ.*1145

*I will call this T of θ and F of φ.*1153

*We are just making our life easier, this is what we do.*1159

*We start with 3 variables, we separated into 1 and 2 variables.*1161

*We deal with the 1 and 2 variables, we separate that into 1 and 1, that is what we are doing here.*1165

*The solutions for F of φ are F of φ is equal to this normalization constant A sub M E ⁺IM φ, *1173

*where M is another quantum number and it is equal to 0 + or -1, + or -2, and so on.*1204

*This is the normalization constant.*1216

*Let us determine what the normalization constant is so we can actually write a full function for this.*1228

*Let us determine the A sub M.*1238

*The normalization condition, we know that the normalization condition is that *1243

*the integral of the function conjugate × the function has to equal 1.*1256

*In this particular case, φ itself is greater than or equal to 0 and less than or equal to 2 π.*1265

*Those are our limits of integration, 0 to 2 π.*1271

*We have the integral from 0 to 2 π of S * sub M × FM = 1.*1276

*This is going to be the integral from 0 to 2 π.*1287

*We are going to take F sub M*, we are going to star this, we are going to conjugate this.*1294

*It is going to be A sub M, when we conjugate, we conjugate everything.*1303

*It is going to be the constant, conjugate of the constant E and – IM φ *1312

*because the conjugate of the IM φ is – IM φ × the function itself F sub M which is A sub M E ⁺IM φ D φ.*1318

*This is this, that ends up becoming this and this, ends up becoming the absolute value of A sub M².*1338

*The integral from 0 to 2 π, this and this may cancel out, - IN φ + IN φ is 0, which is equal to 1.*1352

*All you are left with is D φ, that is equal to 1.*1362

*Therefore, what we have is absolute value of A sub M² × 2 π.*1368

*This integral from 0 to 2 π of D φ is equal to 2 π is equal to 1.*1375

*That implies that A sub M is equal to 1 / 2 π M =0 + or -1, + or -2, and so on.*1382

*Our final function F of φ, this thing right here, what it looks like is the following.*1402

*F of φ is equal to 1/ 2 π, I will write it as, the exponent of ½ E ⁺IM φ.*1409

*Again, subject to these values for M.*1421

*This is the φ component of the angular component of the hydrogen wave function, fully normalized.*1425

*This is definitely one of the equations, one of the functions that we are interested in.*1436

*We are making good progress here.*1445

*We will see what we have got.*1448

*We had S of θ and φ.*1456

*When we separate variables for this, we end up with two differential equations.*1462

*We did mention the differential equation for the φ component, we just gave you the solutions.*1467

*For the other part, the T of θ.*1477

*This one I’m going to write down the differential equation.*1486

*We go to blue here.*1496

*We had this S of θ and φ, we separated into a function just of θ, and the function just of φ.*1499

*Let me go back to red.*1508

*We just found the F of φ, now we want to find the T of θ so we can multiply them together to get our full angular solution.*1508

*The differential equation for T of θ is the following.*1521

*It is going to be sin of θ T of θ DD θ sin of θ DT D θ + this β sin² θ = M².*1530

*The solution to this differential equation involves something called the change of variable.*1556

*We are going to change the variable and we are going to actually solve this equation.*1561

*The solutions will get our act going to be polynomials.*1567

*The variable inside those polynomials is actually going to be a cos θ.*1570

*It is actually is going to be a function of θ.*1574

*Solving this involves using a change of variable.*1582

*You do not have to know this, you do not have to reproduce this.*1596

*If you can follow it, that is great.*1602

*If not, this is just part of your general scientific literacy.*1604

*The change of variable we are going to use is going to be X = cos θ.*1607

*Θ is running from 0 to π so T of θ becomes a polynomial in X comes which will designate as P of X.*1615

*Under this change of variable, what ends up happening to the differential equation is the following.*1642

*I will go ahead and stay with red.*1648

*Under this change of variable, the differential equation becomes the following.*1651

*It becomes 1 -X² D² ψ DX² -2 X DP DX + β – M²/ 1 - X² × P is equal to 0.*1665

*Again, M is equal to 0 + or -1, + or -2, and so on.*1693

*This is a very important equation in physics, it is called the Legendre equation.*1702

*This is called the Legendre equation.*1710

*It shows up a lot.*1720

*Let me go ahead and go back to blue here.*1729

*What is the equation we solve?*1731

*When we solve this equation, in order for these polynomials, the solutions P of X to stay finite, we need it to stay finite.*1739

*This β, it must equal λ × L.*1755

*In order for the solutions to stay finite, this constant β has to equal this L × L + 1.*1768

*The value of L is another quantum number.*1778

*It is actually equal to 0, 1, 2, and so on but with a restriction that the absolute value of M has to be less than or equal to L.*1781

*There is a connection, M is actually dependent on L.*1798

*We write, we are just going to put this into β.*1805

*We write 1 - X² D² P DX² -2X DP DX + L × L + 1 - M²/ 1 - X² P = 0.*1814

*Where, L is going to take on the value 0, 1, 2, and so on.*1845

*M is going to take on the values + or -1, + or -2, all the way to + or – L.*1853

*The absolute value is less than or equal to L, which means that M is going to take on the values when we write them out explicitly 0, 1, -1, 2, -2, 3, -3.*1861

*If L is 4, then is going to be 4, -4 all the way up to that value.*1872

*Now, we have two quantum numbers that are going to arise here.*1876

*There are different sets of solutions for the various values of L.*1882

*Let us actually start writing this so we can read it.*1907

*For the various values of M, we are going to look first, let us first look at M = 0.*1911

*When M is equal to 0, the solutions which we symbolize as *1931

*P sub L of X are called appropriately Legendre polynomials because they are coming from Legendre equation.*1941

*The Legendre polynomials are very important.*1956

*I’m going to go ahead and list the first few Legendre polynomials.*1967

*We are taking M = 0.*1970

*The first few Legendre polynomials, P sub 0 of X is equal to 1.*1986

*P sub 1 of X now M is equal to 0, this subscript right here, that is the L value.*2000

*P sub 1 of X is going to equal X.*2006

*P sub 2 of X is going to equal ½ × 3 X² – 1.*2012

*P sub 3 of X is going to equal ½ 5X³ -3X.*2022

*We will go ahead and stop with P sub 4, that is going to equal 1/8 35 X⁴ - 30 X² + 3.*2032

*It is very important to remember.*2048

*Recall that X is not a variable here, they are the change of variable.*2054

*X is actually equal to cos of θ.*2061

*If I do P2 of X that is actually P2 of cos θ.*2071

*Wherever X is, I'm just going to stick in to cos θ.*2080

*In the case of the P2, I'm going to have ½ 3 cos² θ -1.*2083

*Θ is still the variable that we are concerned with.*2100

*Remember ψ is a function of R, θ, and φ.*2103

*All we have done is change the variable in order for us to actually solve this differential equation *2107

*and make it a little bit easier to deal with, that is all we have done here.*2111

*X is not the variable, XYZ is not a coordinate.*2114

*It is just a variable that we are using a hold the place of cos θ, the change of variable.*2117

*The function is really this thing right here because again we need a function of θ *2123

*but we are expressing it in terms of these polynomials.*2128

*This is very important to remember.*2131

*I will remind you of that several times because it is easy to lose ones way.*2132

*We all do, I do all the time.*2138

*Let us go back to blue here.*2140

*We said that L takes all the values from 0, 1, 2, 3, 4, and so on.*2145

*A couple of properties here of these polynomials.*2150

*When L is odd, the Legendre polynomial P of X is an odd function.*2153

*When L is even, the Legendre polynomial is an even function.*2169

*It is also true that the integral from -1 to 1 of P sub Q of X, P sub R of X, is equal to 0.*2181

*In other words, these Legendre polynomials are actually orthogonal.*2204

*They are mutually orthogonal.*2208

*That is the Legendre polynomials are mutually orthogonal.*2212

*If you are wondering where I get this -1 to 1, let me go to the next page and tell you.*2230

*The limits here are -1 to 1 because again X is equal to cos of θ.*2237

*Θ is running from 0 all the way to π.*2255

*X is cos θ so X goes from cos of 0 to cos of π.*2261

*Cos of 0 is 1 to -1.*2277

*It is θ is the variable that we are interested, θ runs from 0 to π.*2281

*When we take the cos of 0 like this one, we take the cos of π that gives us -1.*2285

*Our limit of integration, when we are integrating the polynomial is going to be -1 to 1 or 1 to -1.*2290

*The order does not matter, you are just going to switch signs.*2297

*When we are integrating with respect to θ, that is when we put 0 to π.*2301

*That is why -1 to 1, that is where the limits come from.*2304

*The last property, the coefficients in front of the polynomials, the ½, the 1/8, and so on, *2310

*the coefficients in front of the P sub L of X are chosen so that when we do the P sub L of 1 it always ends up equaling 1.*2320

*It is just for convenience, we have chosen it that way.*2341

*For example, P sub 2 of X is equal to ½ 3 X² -1.*2344

*P sub 2 of 1 is equal to ½ 3 × 1² - 1 is equal to ½.*2359

*3 × 3 – 2, we actually end up getting a value of 2.*2371

*We stick a ½ in front of it, in order for us to end up with a final value of 1.*2377

*This is the actual polynomial, we have adjusted it by putting the coefficient in front of it to make sure that P of 1 is equal to 1.*2382

*That is what gives us our final form of Legendre polynomials.*2391

*We will go ahead and stop this lesson here.*2396

*Thank you so much for joining us here at www.educator.com.*2397

*We will see you next time for a continuation.*2399

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