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 1 answerLast reply by: Professor HovasapianThu Nov 6, 2014 1:24 AMPost by xlr z on November 2, 2014Hello Raffi,sorry it is not related to this lecture but I did not see anything on Molecular structure such as wave function of H2, H2+ or H2- molecule or anything related to born oppenheimer approximation which is usually included in P-chem course. I was wondering if you will be making new lecture videos in the future or if you have notes on it and can post it online?thank you.

### The Hydrogen Atom VI

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 VI 0:07
• Last Lesson Review
• Spherical Component
• Normalization Condition
• Complete 1s Orbital Wave Function 4:08
• 1s Orbital Wave Function
• Normalization Condition
• Spherically Symmetric
• Average Value
• Example I: Calculate the Region of Highest Probability for Finding the Electron 21:19
• 2s Orbital Wave Function 25:32
• 2s Orbital Wave Function
• Average Value
• General Formula

### Transcription: The Hydrogen Atom VI

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

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

Let us go ahead and jump right on in.0007

Let me go ahead and do black today.0009

In the last lesson, we looked at the radial function R sub N L of R.0012

In particular, we look at the one that is orbital.0033

In particular, we looked at R 1, 0 of R which was 2/ A sub 0³/2 E ⁻R/ 2 α sub 0.0040

This is only the radial component of the wave function ψ.0065

There is also a spherical component or the angular component, the S 0, 0 of θ and φ.0069

S of 0, 0 happens to be 1/ radical 4 π.0102

Nice and simple, notice θ and φ going to show up.0116

This is just a constant function.0119

The normalization condition for the spherical component is as follows.0123

The normalization condition θ runs from 0 to π, φ runs from 0 to 2 π.0130

S00 conjugate × S00 and sin θ D φ D θ.0143

Once again, do not forget this because you are integrating over the surface of the sphere.0156

You are integrating spherical coordinates so you have to include this factor.0164

When you include R, the fact it was R².0168

The entire triple integral, this part becomes R² sin θ DR D φ D θ.0172

Do not forget this factor.0182

Again, integrating with respect to spherical coordinates.0188

This is just this, we end up with the normalization condition being 1/ 4 π, the integral from 0 to π,0193

the integral from 0 to 2 π sin θ D φ D θ.0210

Φ comes first because we are going φ first the integral and we will go to the outer integral.0219

There is nothing else going on here, that is just the standard normalization condition for the angular component.0225

This is a function of two variables so we end up having a double integral.0232

The normalization condition is always the integral of the conjugate of the function × the function itself.0237

The complete wave orbital, so the complete 1S orbital wave function is, this is ψ 1, 0, 0.0248

It is a function of R θ and φ, it is equal to the R1, 0 of R × the S0, 0 of θ and φ.0272

When I put this together with this and simplify, I end up with 1/ π A sub 0³¹/2 E ⁻R/ α sub 0.0284

We had the radial component S0, 0 happen to be a constant function so the entire wave function,0319

now that we have put together it turns out also to be only a function of R.0326

In the case of the S orbital, the 1S, 2S, 3S, 4S, it is always going to be this way.0331

All of the S orbital are only going to be functions of R only because the S sub 00 term is a constant term.0338

What this means is when you have three variables and it is only a function of one of the variables,0350

in this case the θ and φ, they sweep out a sphere that means the 1S orbital is spherical symmetric.0356

That is why we represent them as globes, bigger 3S, 4S, they are spheres.0362

A spherical symmetric, that is what this means.0368

This is where it comes from.0370

Let me go back to black here.0375

The normalization condition for ψ the entire wave function is the integral from 0 to infinity,0396

the integral from 0 to π, the integral from 0 to 2 π of ψ 1, 0, 0 conjugate × ψ 1, 0, 0.0414

Then, R² sin θ the inner integral is D φ D θ to the final integral is DR.0426

This is the normalization condition, that is all we are doing here.0436

We just wrote the following.0464

We just wrote the normalization condition as the integral from 0 to infinity, the integral from 0 to π,0467

the integral from 0 to 2 π ψ 1, 0,0* ψ 1, 0, 0 R² sin θ D φ D θ DR.0478

If it makes more sense to you, you can actually separate variables.0492

This is a triple integral, just integrating one at a time.0495

Working from the inside out, the inside integral is the D φ.0498

The outer integral with respect to θ and the next integral is with respect to R.0503

If it is more clear, you can separate the variables and write it like this.0508

You can write as the integral from 0 to infinity the R² DR, the integral from 0 to π sin θ D θ,0512

the integral from 0 to 2 π ψ 1, 0,0 ψ 1,0, 0 D φ.0524

In this particular case, you are just separating the variables.0541

You are setting up this integrand.0544

You are integrating with respect to φ first and then what you get is going to be your integrand including this0546

and you can integrate with respect to θ next.0553

What you get, you are going to put here integrand also including this and we are going to integrate with respect to R.0556

If this makes sense to you, more than this make sense to you, you are just working from right to left.0564

Here, you are working from inside out.0569

If you will put integrals before, back into calculus or multi variable calculus,0571

you were probably accustomed to seeing this one.0576

It does not really matter which one you use.0579

Since, ψ 1, 0, 0 = 1/ π A sub 0³ ^½ E ⁻R/ A sub 0 is a function only of R.0585

It is spherically symmetric sphere in space.0615

The electron can be found within that sphere but somewhere within that sphere.0634

Let us go ahead and work on some probabilities.0640

We have the integral from 0 to A, the integral from 0 to π,0643

the integral from 0 to 2 π of ψ 1, 0, 0 conjugate ψ 1, 0, 0 R² sin θ D φ D θ DR.0651

The ψ conjugate × the ψ 1, 0, 0, I have the function which is right there.0668

I’m just going to multiply it by itself and I end up with 1/ π A sub 0³ E⁻² R/ Α sub 0.0678

Let me go ahead and just call this Z so I do not have to keep writing it over and over again.0693

I’m just going to call this one big fat Z.0697

We are going to take this and we are just going to solve this when I put this Z into here and0701

we are actually going to integrate this one integral at a time, to see what we get when we do that.0714

Let us go ahead and we are going to integrate with respect to φ first.0721

This integral right here is going to end up being, or if you want this one.0725

We would do that one first.0737

We are actually doing the integration with respect to φ.0739

It is going to be 0 to 2 π of Z, whatever this thing is.0748

D φ, that is going to end up being 2 π × Z when we do that integral.0754

We are going to integrate with respect to θ next.0762

We are going to put what we get, the 2 π Z into here and we are going to solve this integral with respect to θ, the 2 π Z.0765

Z does not contain any θ so I could just pull it out.0780

0 to π sin θ D θ is going to give me 2 π Z × - cos θ from 0 to π.0786

When I end up solving this part, I'm going to get 4 π Z.0804

I got 4 π Z and that is going to equal, I have Z right here which is that thing.0814

Therefore, I will go ahead and write these again.0830

I will stick with red.0836

4 π Z is actually equal to 4 π and we said the Z is equal to 1/ π A sub 0³ E ⁻2R/ α sub 0.0842

The π cancel leaving me with 4/ A sub 0³ E ⁻2R/ Α sub 0.0859

The final integral gives us the probability of finding the electron within A of the radius.0877

The final integral is the integral from 0 to A of 4/ π A sub 0³ E ⁻2R/ α sub 0 R² DR.0887

This right here, remember this represents the probability that the electron is between R and DR.0913

When I integrate that probability from 0 to A, it gives me the probability that the electron is within A of the nucleus.0936

This expression, we have already seen it.0948

We saw it in the last lesson, it is the same expression for just the radial function.0954

Let me go back to black here.0961

We saw this already in the last lesson when dealing with only the R1, 0 of R.0964

It turned out to be the same probability for the entire wave function ψ 1, 0, 00994

precisely because θ and φ are not involved in the 1S orbital.1022

Because θ and φ are not part of the function, they are not part of ψ 1, 0, 0.1030

In other words, the S orbital are spherically symmetric.1042

Let us talk a little bit about average values.1059

Let me go ahead and put the line of demarcation here.1065

Let us move here, what about average values?1071

Recall that the average value for A is the integral of ψ conjugate × A × ψ × DV.1084

The average value R, the average value of where we are going to find the radius1102

is going to equal the integral from 0 to infinity, the integral from 0 to π.1108

This integral right here is just the radicals symbol for integral.1115

Since, we are dealing with function of three variables, we would have a triple integral.1118

0 to π, 0 to 2 π of ψ 1, 0, 0 conjugate × R × ψ 1, 0, 0 × R² sin θ D φ D θ DR.1125

When we actually end up taking the ψ × R × ψ conjugate is just multiplication.1144

It is ψ conjugate × ψ which we already know we will just multiply it by R.1151

We end up with the integral 0 to infinity when we take care of all of this, we get 4/ π A sub 0³ E ⁻2R/ Α sub 0 R,1156

which is this R, R² DR, which is going to end up equaling 4 / π A sub 0³.1179

The integral from 0 to infinity of E ⁻2R/ Α sub 0 R³ DR.1193

When we actually solve this integral, we get 3/2 A sub 0.1204

For this particular orbital, on average if I take a bunch of measurements of where the electron is going to be,1209

I'm going to find it is actually ½ bohr radius away from the nucleus.1218

Remember, we had a distribution that looks like this for the 1S orbital, something like that.1225

It is going to be between here and here, on average.1231

We are going to find it in a bunch of different places.1233

There is a place of highest probability, that on average for 100 measurement, 1000 measurement,1236

a million measurements, it is going to end up being ½ bohr radius is away from the nucleus.1243

That is all these means.1247

It is just an integral, it is just the function, the conjugate of the function × whatever it is that you are looking for.1249

In this case, we are looking for R radius how far away × the function and then just solve the integral1257

or let use mass software solve the integral.1264

Let us go ahead and look at an example here.1269

For the 1S orbital, calculate the region of the highest probability for finding the electron.1282

Highest probability means we want to maximize the probability.1287

We are looking for the peak, we are looking for the X value where the peak occurs.1298

The probability, the 1S orbital was founded already, it is 1/ π A sub 0³.1306

I will go ahead and put the R² E ⁻2R/ α sub 0 DR.1322

Remember, this part is the probability density.1332

The probability is the probability density × the differential element.1339

What we need to do is we need to maximize the probability density.1345

Remember the graph that we were looking at, the 1S orbital.1349

Probability density was on the Y axis, R was on this axis.1353

The probability density was on this axis.1357

We got something like that, we want to find that R value.1360

This is a probability density vs. R, that was the graph was.1368

In case we need to maximize that function.1371

We maximize something by doing exactly what you did in calculus.1376

We take the derivative, we set it equal to 0.1378

DDR of this function 4/ π A sub 0³ R² E ⁻2R/ α sub 0.1382

When you take the derivative of this function, you get 4/ π A sub 0³.1397

This is the product rule so it is going to be this × the derivative of that + that × the derivative of this.1404

We are going to get R² × E ⁻2R/ α sub 0 × -2/ α sub 0 + 2R × E ⁻2R/ Α sub 0.1410

This is going to equal 4/ π × A 0³ × E ⁻2R/ α sub 0 × - 2R²/ α sub 0 + 2R.1432

That is the derivative, we are going to set the derivative equal to 0.1455

This constant goes away, this E ⁻2R/ α sub 0 is never 0.1459

Therefore, this is the equation that we have to solve.1464

We get -2R²/ Α sub 0 + 2R.1469

We set it equal to 0, we end up with -2R × R/ Α sub 0 -1 is equal to 0.1476

We end up with R equal to 0 or we end up with R equal to α sub 0.1489

0 of course does not matter, that is out of the nucleus.1499

For the 1S orbital, calculate the region of highest probability for finding the electron.1503

The place we are most likely going to find electronic is 1 bohr radius away from the nucleus.1509

That all that is going on here.1516

We maximize the probability density because we were looking for the maximum region of highest probability.1518

Let us look at the 2S orbital.1531

We are looking at the 1S orbital, and the one that is complicated is the 2S orbital.1543

In other words, we are going to be looking for ψ 2, 0, 0.1549

Ψ 2, 0, 0 is equal to R 2, 0 × S 0, 0.1555

This S 0, 0 term shows up again and it is going to show up to all of the S orbital because the 3S orbital is going to be ψ 3, 0, 0.1562

4S orbital is going to be ψ 4, 0, 0, the S, 00 that term always shows up in the S orbital.1574

The S 0,0 term is part of the wave function.1590

In fact, for every S orbital the spherical harmonic will be S00 precisely because L = 0 and M = 0.1606

That is the S orbital.1644

The only thing that is going to change is the primary quantum number, 1, 2, 3, 4, 5, that is all that changes.1647

For every S orbital, the only variable for the wave function ψ is going to be R.1659

The total wave function, total hydrogen wave function for all S orbitals are only going to the functions of R.1678

They are only going to depend on the radius.1683

They are spherically symmetric.1685

Every S orbital is spherically symmetric.1691

For ψ 2, 0, 0 we end up with v1/ 32 π × 1/α sub 0³/2 × 2 - R/α sub 0 × E ⁻R/ 2 Α sub 0.1705

The average value of R for 2, 0, 0 is equal to exactly what you think.1734

It is the integral of the ψ conjugate of 2, 0, 0 × R × the ψ 2, 0, 0.1751

We solve that integral.1760

Ψ conjugate × ψ itself is just this thing multiplied by itself because this is a real function.1765

The conjugate is just ψ itself.1772

This becomes ψ² × R.1775

This thing × R.1778

We have ψ 2, 0, 0 conjugate × ψ 2, 0, 0 is equal to 1/ 32 π × 2 - R/ A sub 0² × R × E ⁻R/ α sub 0.1782

The 2’s cancel.1807

We are solving a triple integral.1811

We still have to solve a triple integral.1814

Even though it is a function of R, we still have to do the entire integral.1817

We already know that the integral from 0 to π, the integral from 0 to 2 π of the sin θ D φ D θ is equal to 4 π.1821

The inner integral, the angular part is going to equal 4 π.1834

We will just throw that in.1838

What we get is the average value of R for 2, 0, 0 is going to equal 4 π × this thing / 32 π A sub 0³.1840

Our final integral is the integration with respect to R.1859

I have done the triple integral, I just know that in this particular case because θ and φ are not involved in this function,1863

I will just go ahead and use this value in place of the inside two integrals.1871

It is going to be that 2 – R/ α sub 0² E ⁻R/ α sub 0.1879

Again, we have our R² DR, do not forget that term.1891

When I actually put this in a mathematical software, I'm going to get 6 Α sub 0.1894

For the 2S orbital, on average you will find the electron 6 bohr radius away from the nucleus.1901

Remember, for the 1S orbital it was ½ × the bohr radius so it was going to be ½.1909

For the 2S orbital, it is going to be 6 bohr radius away.1921

On average, it is going to be farther from the nucleus.1926

3S on average is going to be farther from the nucleus which is why1928

the pictures in your book represent the 1S as a small sphere, the 2S is a bigger sphere, the 3S is a bigger sphere.1932

On average, the electron is going to be further away.1938

That is why it looks the way that it does.1942

There is a general formula.1946

The average value of R for whatever S orbital 1S, 2S, 3S, 4S, 5S, is equal to 3/2 A sub 0 N².1955

Remember, the picture that we saw in the last lesson with the S orbitals, the 2S orbital has 1 node.1971

That means that if I take a sphere and I split it in half, the 1S orbital did not have a node,1980

which means that the electron can be anywhere in that sphere.1993

The 2S orbital had 1 node that means the electrons you are going to be in this region.1997

It is not going to be here or in this region.2009

I sliced it in half, you are looking at half a sphere that way.2011

The 3S orbital has 2 nodes.2019

There is going to be a place where the electron is not going to be.2023

There is going to be another place where the electron is not going to be.2025

The electron is either going to be this region or in this region or in this region.2030

That is what this means.2037

Do not forget that there are nodes.2037

When we talk about spherical symmetry, it does not mean that they can be anywhere in the sphere.2039

That only possible for the 1S orbital because it has no nodes.2044

But within the sphere itself, the 2S orbital has 1 node.2047

That means there is 1 sphere within that bigger sphere where you will never find the electron.2051

The 3S orbital has 2 nodes but if this is the main sphere, there are going to be 2 smaller spheres where you will not find the electron.2056

Again, pictorial representations of these electrons orbital are not very good because they are not giving you all the information.2065

Thank you for joining us here at www.educator.com.2078

We will see you next time, bye.2079