Enter your Sign on user name and password.

Forgot password?
Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Physical Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Fri Nov 6, 2015 3:05 AM

Post by Tyler Westover on November 5, 2015

There is a missing square root in the radial part of the wave function. (6:34)

The Hydrogen Atom IV

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
  • The Hydrogen Atom IV 0:09
    • The Equation to Find R( r )
    • Relation Between n & l
    • The Solutions for the Radial Functions
    • Associated Laguerre Polynomials
    • 1st Few Associated Laguerre Polynomials
    • Complete Wave Function for the Atomic Orbitals of the Hydrogen Atom
    • The Normalization Condition
    • In Cartesian Coordinates
    • Working in Polar Coordinates
    • Principal Quantum Number
    • Angular Momentum Quantum Number
    • Magnetic Quantum Number
    • Zeeman Effect

Transcription: The Hydrogen Atom IV

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

Today, we are going to again continue our discussion of the hydrogen atom.0004

Let us jump right on in.0008

We have been dealing with this hydrogen atom wave function, this ψ which is a function of R θ and φ,0013

which we separated into a radial component R of R and a spherical component S of θ and φ.0025

In the last three lessons, we have been dealing with S of θ and φ.0033

We are going to go ahead and deal with R.0036

We find this R of R.0040

The equation that we need to solve, you will see this, know this, or understand it completely but it is nice to see it.0046

The equation to solve here to find the radial component which is a function of R single variable is -H ̅² / 2 ×0056

the mass of the electron R² DDR R² D of R DR + H ̅² L × L + 1/ 2 mass of the electron R² - E² / 4 π E sub 0 × R - E R = 0.0071

When we solve this equation, we find that the energy for the state N,0109

there is a yet another quantum number M E sub N = - M sub E,0124

this is a subscript E so the mass of the electron E × E⁴ / 32 π² ε² H ̅² N².0132

Or N take some of values 0, 1, 2, 3, and so fourth.0152

If we introduce a bore radius which is symbolized A sub 0 which is equal to 4 π ε H ̅² /0159

the mass of the electron E², we get for the energy.0182

This is a perfectly, it is not a problem or you can use this one, E sub N = - E²/ 8 π E sub 0 A sub 0 N².0191

Again, N take some of values 0, 1, 2, and so forth.0209

This gives us the energy of that particle for that with respect to the radial function.0214

We also found the following.0229

Notice, we have a L in that equation, we also discover and we have introduced this new quantum number N.0233

There is a relationship between N and L.0241

We also find that the relation between N and L is N is greater than or equal to L + 1.0244

But we write it more familiarly this way, this definitely remember.0268

It is written as L which is going to be greater than or equal to 0, less than or equal to N – 1.0285

There you go.0296

This is the important relation between L and N -1.0297

The solutions for the radial equation.0303

We have discussed the energies for the radial equation, now the solutions for the radial equation.0305

We go ahead and do this in blue.0311

The solutions for the radial component, the radial function they depend on two quantum numbers.0316

Let me write this our.0338

On two quantum numbers, N and L.0341

The symbolism is going to be the following.0353

We have R and L which is a function of R is equal to -.0356

Is there a minus in there?0364

I have to double check that.0368

We get N – L – 1!/ 2 N × N + L!³ × 2 / N A sub 0 ⁺L + 3/ 2.0370

Do not worry about it, this is just the general solution, what it looks like.0406

We want you to see what it looks like.0411

The actual functions themselves are a lot easier to deal with, when you put in the L, the N,0413

and all of the number sort of condensing become something.0418

Do not get me wrong, I’m not saying this is not complicated, but it is not as bad as it looks.0421

R ⁺L E ⁻R/ N A sub 0,0430

I would make a little bit more room here so let me make these a little bit smaller.0441

This is going to be × R ⁺L × E ⁻R/ A N sub 0 × L sub N + L super 2 O + 1 the function of the 2R/ N A sub 0.0446

I have these things right here, do it in red.0477

These are called, these L, N + L, super 20 + 1, they are functions of 2R/ N A sub 0,0485

these are called the associated Laguerre polynomials.0503

We had the associated Legendre polynomials.0524

We have the Legendre polynomials.0527

We have the associated Legendre functions0529

These are the associated Laguerre polynomial or associated Laguerre functions.0532

The first few are as follows.0537

The first few associated Laguerre polynomials.0544

Again, notice it depends on L and it depends on N, L and N.0569

For N = 1, I’m sorry I think in the previous slide it is N starts at 0.0576

It does not start at 0, N starts at 1.0589

For N = 1, L = 0.0593

We get L of 1, 1 of X is equal to 1.0601

For N = 2, we have the possibility L = 0 and we have a possibility L = 1, because L runs from 0 all the way to N – 1.0610

If N is 2, starts at 0 and move all the way to N -1, 2 -1 is 1.0624

For N2 L0, we get L2 sub 2 super 1 of X is equal to – 2! × 2 – X.0631

Here we have L, again N + L 2 + 1 this is 3.0646

The superscript is 2L + 1, 2 × 1 is 2 + 1 is 3, so L3 3 of X is going to equal -3!.0653

Let us do N = 3, for N = 3 we have L equal 0, we have L = 1, and we have L = 2.0666

For the N3 L0, we have L3 1 of X is equal to -3! × 3 -3 X + ½ X².0676

For N = 3 L = 1, we would end up with L4 3.0697

The 4 comes from N + L 3 + 1 is 4.0703

The 3 comes from 2L + 1, 2L + 1 is 3.0707

This is going to equal -4! × 4 – X.0713

Here, N + L the subscript is 5, the superscript is 2L + 1.0720

2 × 2 + 1 this is going to be 5.0726

This as a function of X is going to equal -5!.0729

Let us go back to blue.0741

The complete wave function ψ which is a function of R θ and φ for the atomic orbitals of the hydrogen atom R,0746

ψ the subscript depends on 3 quantum numbers N, L, and M = R sub NL0786

which is a function of R × S sub L super M which is a function of θ and φ.0799

N runs from 1, 2, 3, and so on.0809

L goes from 0, 1, 2, all the way to N -1 and M takes on the value 0, + or -1, + or – 2, all the way to + or –L.0819

N is the quantum number.0845

L depends on N, M depends on L, this is the symbolism for it.0848

The energy of the particle = - E/ 8 π ε R radius N².0855

Notice that the energy only depends on the N quantum number.0876

The normalization condition is the integral from 0 to infinity, the integral from 0 to π.0902

The integral from 0 to 2 π of ψ conjugate N LM × ψ N LM R² sin θ D φ D θ DR D φ, because this inside when it is φ.0923

D θ, the next one is θ, the final one DR it is that.0958

Make sure you place in order.0962

These functions the ψ N LM, they form an orthonormal set.0971

The integral of 1 × the other = 0.0976

They form an orthonormal set.0982

This is what we have come to, this is what the last 3 lessons a bit about.0987

We have been working up to this, this ψ N LM which is a function of R θ φ.0989

This is why we have had all of this machinery.0995

It is truly extraordinary when you sort of step back from this and you realize that0999

the human mind has elucidated all of this and done so successfully.1002

It is absolutely staggering.1008

I'm still amazed by every single day.1010

For the orthonormal set, what else can we say about this?1013

I would like to notice something here.1019

Let me go back to blue.1022

Notice that the R² sin θ D φ D θ DR is the differential volume element.1033

In other words, the equivalent of DX DY DZ but you notice it is not just D φ D θ DR, that is DX DY DZ.1060

There is this extra factor that we have to multiply by, in order for it to actually work,1070

when we take integrals in spherical coordinates.1076

We have three variables to triple integral.1080

For a single variable function, it is a single integral.1083

Two variable function, double integral.1085

Three variable function, triple integral.1087

We integrate one at the time.1088

Now in Cartesian coordinates, we have this DX, DY, and DZ.1091

Under a change of variable for spherical coordinates, the relationship is as follows.1111

The relationship between the X, Y, and Z, and R θ of φ goes like this.1134

X = R sin θ cos φ, Y is equal to R sin θ sin φ, and Z is equal to R × the cos of θ.1139

Given this change of variable, the DX DY DZ is actually equal to R² sin θ D φ D θ DR.1161

There has to be this extra thing that we have to multiply by when we do a change of variable.1178

When we are doing these integrals in polar coordinates, you cannot just write D φ D θ DR,1183

you have to write R² θ D φ D θ DR at the end.1189

That is the differential volume element.1193

I want to actually show you where does that this actually comes from.1203

I’m going to show you where this comes from.1208

It comes from the following.1214

This comes from the following.1221

Is it really all that necessary?1225

No, it is not necessary.1228

If you want, you are more than welcome to get in touch with me and I'm happy to send you a brief little half page,1233

one page showing you where this actually comes from.1239

But it is actually not that important for our purposes.1242

What is important is that, that actually has to be there.1245

The moral of the story is when working in polar coordinates,1251

in other words when doing integrals in polar coordinates,1263

please make sure to use the R² sin θ D φ D θ DR as your differential volume element.1266

It is very important, it must be there.1286

Let us see, the hydrogen wave functions ψ N LM depend on three quantum numbers.1292

The first one N, it is called the principle quantum number.1319

It takes all on the values 1, 2, 3, 0, on and on.1334

The energies depend only on this principle quantum number.1343

E sub N depends only on an M.1347

The next one we have is L, this is called the angular momentum quantum number.1356

We have come to the point where this is the important stuff right here.1370

This is what we want to know.1374

This is what we are going to use to actually solve our problems.1375

We have the wave function and we want to be able to understand what the relationship is1378

among these quantum numbers and how it relates to a particular energy, orbital, and things like that.1382

This is called the angular momentum quantum number and values that take on are 0, 1, 2, all the way to M -1.1387

L depends on M.1399

Now, some things to know about this one.1401

The angular momentum L of the electron is completely determined by L quantum number.1404

It is completely determined by L, according to the following.1427

The magnitude of the angular momentum is equal to H ̅ × √ L × L + 1.1435

That is the relation, when you have L you have the magnitude of the angular momentum of the electron.1445

The direction of the angular momentum but you had the magnitude of the angular momentum.1454

I will start on the next page here.1462

Note that the function itself, the radial function NL which is the function of R, the radial component of ψ depends on both N and L.1470

We assign letter values to various values of L.1516

We have seen this already.1533

We have L = 0, that is the S orbital.1536

L = 1 that is the P orbital.1537

L = 2 that is the D orbital.1540

L = 3 that is the F orbital.1542

L = 4 that is the G orbital.1546

L = 5 that is the H orbital, and just so on.1549

Our final quantum number, we have M.1556

We have M, that is our final quantum number, that is called the magnetic quantum number.1560

M takes on the values 0, + or -1, + or – 2, all the way to + or – L.1566

M depends on L.1578

M has 2L + 1 values.1585

M completely determines the Z component of the angular momentum.1595

The way that L completely determines the magnitude of the angular momentum,1621

M determines the magnitude of the Z component of the angular momentum.1630

Z is equal to M × H ̅ so with some constant, some integral multiple of H ̅.1641

M is called the magnetic quantum number because in the presence of a magnetic field,1653

the energy E sub N ends up depending not only on N but it ends up depending on the value of M also.1686

The energy E sub N depends on both N and M.1698

In the absence of the magnetic field, the E sub N has a 2 O + 1 degeneracy.1709

If L is equal to 1, 2 × 1 is 2 + 3, all of those 3 orbitals, they have the same energy E sub N.1735

That is what that means, that is what degeneracy means.1743

If let us say, L = 2, then 2 × 2 is 4, 4 + 1 is 5.1746

That means that those 5 D orbitals for N = 3, they all have the same energy.1755

They are all degenerate.1760

When you put that in a magnetic field, all of those 5 levels, those 5 orbitals, they no longer have the same energy.1762

The magnetic field actually ends up splitting up the energies, it separates them.1769

When you have taken out the magnetic field, all those orbitals collapse so they become the same energy.1773

We will show you what it looks like in just a second.1779

In the absence of the magnetic field, the energy has a 2 0 + 1 degeneracy.1781

The splitting of the energy of the E sub N into various levels, it is not the E sub N that we are splitting into various levels.1788

It is the splitting of the energies into various levels instead of just the one level that was based on N.1810

The various level is called the Zeeman effect.1824

Let us go ahead and take a look at with this looks like.1837

Do I have one more page here?1838

Yes, I do.1840

I think I will go ahead and do it on the next page.1841

For N = 2 and L = 1, M is going to equal 0, -1 + 1.1846

When we have no field over here, when we turn on the magnetic field over here,1863

when there is no field this is the N = 2, this is the L = 1, N = 0 -1 + 1.1876

These 3 orbitals, they all have the same energy level.1891

It just depends on N = 2 because there is no field.1895

The I places the magnetic field, this field actually ends up splitting the energies.1899

Now all of these orbitals, they end up having different energies.1904

Here is what it looks like.1907

I have got N = +1, N = 0, N = -1.1920

Now, the energies no longer just depend on N, they also depend on what N is.1928

The absence of the field, you get a 2 0 + 1 degeneracy.1934

These are all orbitals, that is an orbital, that is an orbital.1939

They all have the same energy to degenerate.1943

You put in a magnetic field then all of the sudden they have different energies.1945

PX PY PZ that is what is going on here.1950

There are 2 × 1 + 1 = 3 levels with the same energy, that is what degeneracy means.1957

Same energy that just depends on N.1972

Each has a different energy.1980

The energy depends on N and it depends on M.1991

In this case, we said N = 2, L = 1.1996

L = 1 is the P orbital so these 3 are going to end up being our PX PY PZ.2002

In the absence of the magnetic field, the P orbitals all have the same energy.2012

If they are there, the PX PY PZ based on X =0, 1, and -1.2017

Put a magnetic field, the energy split.2024

The PX PY PZ has different energies.2027

That is all that is happening here.2030

Thank you so much for joining us here at www.educator.com.2033

We will see you next time.2034