Physical Chemistry The Hydrogen Atom Example Problems IV

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

So today, we are going to continue our example problems for the hydrogen atom.0004

Let us get started.0008

The first one says, plot the probability density vs. the radius for the 2S radial function as seen below.0012

Find the values of the radius such that the probability density achieves a maximum.0020

We have seen this plot before in a previous lesson.0025

This is a plot of the probability density for the radial function.0028

That is this thing right here.0036

We basically take the radial function, the radial portion of the wave function.0037

We square it and of course we have that R² factor because we are dealing in spherical coordinate, that is the Y axis.0041

This of course is the radius itself, that is the X axis and it is in increments of the bohr radius A sub 0.0048

For example, A sub 0, 2A sub 0, 3A sub 0, and so on.0057

We want to find the values of the radius such that the probability density achieves a maximum.0062

In other words, where the most likely places to actually find the electron is far as distance from the nucleus.0068

We would be looking for the X value that gives this point and the X value that gives this point.0075

We want to find this X value and this X value analytically.0079

Let us go ahead and get started.0085

Let us go ahead and start over here.0093

I will go ahead and stick with black.0095

2S means that the N value is equal to 2 and L = 0.0098

The radial function is going to be R20, that is the function that we are going to use.0111

The integral is the integral of R20 * R 20 R² DR.0118

This is the probability integral.0137

The normalization integral, if you will.0139

It is just a function itself, the conjugate multiplied by the function R² DR.0141

This is what gives us the probability.0146

This R 20 * × R20 × R², in other words the integrand, it is the probability density.0152

We remember this from very early on.0170

The function, the conjugate of the function × the function,0174

that actually is in this particular case because a spherical coordinates, we have this extra term right here.0177

The integrand without the DR that gives us the probability density.0183

When we multiply by the DR, that gives us the probability when we integrate it.0187

It is the probability density of the radial function.0193

This is the one that we need to maximize.0201

We need to find this, take the derivative, set the derivative equal to 0 and solve the equation for R.0203

The probability density of the radial function.0209

Let us go ahead and see.0211

R20 is equal to, based on the fact that N = 2 and L = 0.0213

Again, we have a formula for the radial function.0220

I’m going to do it explicitly here.0223

It is going to be - 2 -0 -1!/ 2 × 2 + 0!³ ^½ × 2/ 2 ×0225

A sub 0⁰ + 3/2 R⁰ E ⁻R/ 2 A sub 0.0256

I’m just plugging in all the values for N and L into that equation that we have for the radial function,0268

the general equation, you can find it in the previous lesson or you will find it in your book.0274

Of course, we have the L, this is going to be 2 + 0 2 × 0 + 1.0280

The argument of this particular polynomial is going to be 2R/ 2 A sub 0.0289

When we go ahead and work all that out, in this particular case the L 21, R, A sub 0,0302

is going to equal -2! × 2 - R/ α sub 0.0314

When we actually work all of this out, we get R of 20 is equal to 1/32¹/2 × 1/ A sub 0³/20323

E ⁻R/ 2 A sub 0 × 2 - R/ A sub 0 × 2.0342

When we take the conjugate of this, multiply it by this, and then multiply by R²,0360

what we are looking for is this, * × R 2 sub 0 R².0366

We are going to end up with 1/ 8 A sub 0³ E ⁻R/ A sub 0 × 2 - R/ A sub 0² × R².0374

This is the function that we wanted.0395

This is the function that we are going to take the derivative of and set equal to 0.0397

This is the probability density.0404

This is what was graphed in the previous page.0406

Maximize means take the derivative with respect to R of this thing, this R20 * R20 R²,0411

take the derivative and set it equal to 0.0431

We are going to take the derivative of this function.0434

Let me see, should I go ahead and do it on this page or should I do it on the next page?0446

I can go ahead and just start on the next page here.0453

This one I will do in red here.0456

The derivative is going to equal 1/8 A sub 0³.0463

It is going be a little long, of course, it is quantum mechanics so it is long.0470

/ A sub 0 × 2 - R/ A sub 0² R² + E ⁻R/ A sub 0 × - 2/ A sub 0 × 2 - R/ A sub 0 ×0478

R² + E ⁻R/ A sub 0 × 2 - R/ A sub 0² × 2 R.0509

This is just the standard derivative just tends to be a little bit longer, that is all.0525

What are we going to do?0536

Hopefully, I have not forgotten a + or – somewhere, that is always the case.0539

8/ A sub 0³ E ⁻R/ A sub 0 × R × 2 - R/ A sub 0 × - R/ A sub 0 × 2 - R/ A sub 0 –0546

2 R/ A sub 0 + 2 × 2 - R/ A sub 0.0581

All of that is the derivative and all of that is going to be equal to 0.0593

We have something × something × something × something = 0.0599

We have to set each of these factors equal to 0.0607

Over here, we have the R equal to 0, that is one possibility.0610

We have 2 - R/ A sub 0 equal to 0 which means that R = 2 A sub 0, that is another possibility.0618

Of course, we have this last factor which we said equal to 0.0628

I’m going to go ahead and multiply this out.0632

It is going to be - 2 R/ A sub 0 + R² / A sub 0² -2 R/ A sub 0 + 4 - 2 R/ A sub 0 and that is going to equal 0.0639

When I put this together, I'm going to get -2 R A sub 0 + R².0662

I will multiply all of these by A sub 0² -2 R A sub 0 + 4 A sub 0² -2 R A sub 0 = 0.0670

I'm going to end up with R² -6 R A sub 0 + 4 A sub 0.0686

That is my quadratic equation.0696

This is AR² + BR + C =0, basic quadratic equation.0698

Let me go ahead and use the quadratic equation here.0711

Here, the coefficients are, you have 1, 6 A sub 0 is the B, and of course you have this.0718

That is fine, I will go ahead and actually work this out.0732

R is equal to - B so we have 6 A0 + or - √ B² -4 is C so we have 36.0734

A sub 0² -4 × A × C so it is going to be -16.0745

A sub 0²/ 2 A which is going to be 2.0752

We are going to end up with 6 A sub 0 + or – 8 sub 0 × 2 √ 5/ 2.0757

This of course, this is going to give us 3 A sub 0 + or – A sub 0 × √ 5.0774

Our final answer is going to be 3 + or - √ 5 × A sub 0.0788

That is actually going to equal, one of the answers I will do over here.0797

One of the answers is going to be 0.76 A sub 0, that is the 3 – √ 5.0802

The other answer is going to be 5.24 A sub 0.0809

These are our two answers and this is the exact answer, if you do not want to do it in decimal form.0814

Notice, one of them is R = 0.0821

That is just the radius equal 0, that is just use to graph it.0824

The graph would like this and like that.0828

That is this point.0834

R = 2 A sub 0 is going to end up being the minimum.0836

That is where it is going to the probability density actually goes to 0.0838

The two maxima, here and here, that is what those are.0842

That takes care of that problem.0848

Let us see what the next problem is.0856

Our model for the hydrogen atom has the electron interacting with a proton via the Coulombic potential.0859

The Coulombic potential is basically just the one that you remember from your study of electricity and magnetism.0865

Just charges attracting each other, the potential exists between them.0873

The Coulombic potential, the potential energy is -E² / 4 π ε sub 0 R.0876

Ε sub 0 is the permittivity of free space and E is the charge in Coulombs.0884

The operator V it just means multiplied by V sub R.0893

For the 2P Z electron, we want you to show that the average potential energy = twice0898

the over all energy and that the kinetic energy = – the overall energy.0907

Let us see what we have.0919

We are dealing with the 2P Z electron.0921

I think I will go to black.0923

For the 2 PZ electron, we have N is equal to 2, we have L is equal to 1, that is the P,0931

and we have M is equal to 0.0941

The Z subscript always means that M is equal to 0.0944

What we are looking at here is the wave function for ψ 210.0949

Ψ 210 that is the 2 PZ electron.0956

Let us see what that is.0959

When we look that up, 210, we get the following.0961

We get 1/ √ 32 π × 1/ A sub 0³ × σ E ⁻Σ/ 3 - σ/ 2.0965

I think I have written this incorrectly but that is okay.0994

Cos θ, and here our σ is equal to R/ α sub 0.0998

Let us see here.1007

If I’m not mistaken, I think this is going to be 3/2.1010

We are looking for the average kinetic energy, average potential energy, average kinetic energy,1019

and we are trying to establish this relationship.1029

We know what this is so we need to find the average value, the expectation value.1032

The average value expectation value depends on what your teacher calls it, I call it both.1040

We know already that the expectation value or the average value,1045

the integral is given by the integral of the wave function itself conjugate × the operator operated on the function itself.1048

This is the integral that we need to form ψ * V ψ.1062

This is a real function so ψ * is just ψ itself.1072

All we have to do is multiply this function by itself and multiply by the Coulombic potential.1077

When we do that, we get the following.1085

We get that the average value of the potential energy is going to equal -E² / 128 π².1088

I think it should be to be the 4th π² A sub 0 A sub 0³ × the integral.1105

The integral was actually a triple integral because we have 3 variables.1118

It is going to be the integral from 0 to 2 π of D φ, the integral from 0 to π of cos² θ sin θ D θ.1124

It is going to be the integral from 0 to infinity of 1/ R Σ E ⁻σ R² DR.1145

We are dealing with a triple integral.1158

We are dealing with something in spherical coordinates.1159

And it is so this is what the total integral looks like.1161

The integral from the previous page where we just used ψ, VC.1164

When I multiply everything out, this is what I get.1168

Again, we have we have seen this several times already.1173

This is not a big deal.1175

This integral right here is equal to 2 π.1177

This integral right here is equal to 2/3.1182

I’m not going to go ahead and go through each individual integral.1185

When we put that together, we end up with this being equal to, when we take this 2/3 and 2 π and put it all here,1190

we end up with -E²/ 96 π E sub 0 A sub 0³ × the integral from 0 to infinity of R σ E ⁻Σ DR.1200

Now, we need to solve this integral.1224

We know that σ is equal to R/ α sub 0 which means that R is equal to σ α sub 0 D Σ = 1/ α sub 0 DR,1227

which means that DR is equal to D σ α sub 0.1245

When I substitute all of these back into this integral, I end up getting -E² / 96 π E sub 01252

A sub 0³ × the integral of σ ⁻σ D σ.1266

This integral, I have seen it several × before.1278

It is just equal to 3!, it is just equal to 6.1281

I'm certain I forgot, there is actually A sub 0² here.1285

When I substitute all of these back in, some of these A sub 0 actually show up here.1300

They are pulled out as constants.1304

Whet I end up getting is the following.1306

I end up getting -6 E² A sub 0²/ 96 π E sub 0 A sub 0³.1308

This goes with that, leaving 16.1325

This knocks that out and I’m left with -E²/ 16 π E0 A sub 0.1329

The average potential energy is this thing right here.1347

Let us go ahead and jump to the next page here.1355

Let me write that.1360

I have my average, it is -E² / 16 π E sub 0 A sub 0.1364

Let me go to blue.1381

The energy sub N is equal to –E sub²/ 8t π E sub 0 A sub 0 N².1385

N is equal to 2 so this is going to end up being 32.1402

2 × 2 is 4, 4 × 8 is 32, so the energy of level 2 is going to end up being -E²/ 32 π E sub 0 A sub 0.1407

The average potential energy is this value and the total energy is that value.1424

Everything is the same, the only difference is the 16 and the 32.1433

What we have is, we have ½ × the average potential energy ½ × this is equal to the energy itself.1437

Or V is equal to 2 E, which is one of the things that we wanted to prove.1452

Just like actually finding the expectation value for the potential energy.1460

The average value of the potential energy using the integral definition of it.1465

We just worked out on integral which is really what we do most of time.1471

Now, the total energy is equal to the kinetic energy + the potential energy.1475

It is equal to the kinetic energy + the potential.1487

We just solved the potential, it is twice the energy.1491

Now, we have this equation, I’m just going to move this over the other side.1498

I end up with is - the energy is equal to average of the kinetic energy.1504

This was the other thing that we wanted to prove.1512

The relationship ends up being the average potential energy is equal to twice the average energy.1520

It is equal to - × the kinetic energy.1532

That is the fundamental relationship that exists for this particular orbital.1537

Let us see what the next problem says.1546

We are taking from example 2, except for the wave function ψ sub 311, we did ψ 210.1549

We want to do it for 311, that is find the relation among the average kinetic,1557

the average potential, and the average total energy.1562

Let us do the same thing, let us solve the integral.1566

Let us start off with what ψ 311 is, that is equal to 1/ 81 √ π 1/ A sub 0³/2.1570

Just a more complicated function, that is all, not a big deal.1584

6 σ - σ² E ⁻σ / 3 × sin of θ × E ⁺I φ.1588

This was complex, not a problem though.1601

Its conjugate is going to be E ⁻I φ.1604

When we multiply ψ conjugate × ψ, the complex parts can actually go away.1606

Again, it is always a good idea to write down what it is that you want.1613

We want to form the following integral.1617

The integral of ψ 311 conjugate × the operator V operating on ψ 311.1621

That is the integral that we want to form.1632

Therefore, this is the average potential energy.1635

Therefore, the average potential energy is going to equal -E² / 4 π ε 0 × 1/ 6561 π × 1/ A sub 0³ ×1642

the integral from 0 to 2 π of the D φ part, the integral from 0 to π of the sin² θ sin θ D θ part.1668

We will bring it here because the integral is long.1687

× the integral from 0 to infinity of 1/ R × 6 σ.1692

This 1/ R came from the Coulombic potential.1700

Let us put R into under the integral for the R variable.1703

6 σ - σ²² E⁻² σ/ 3 R² DR.1707

Let us go to red here.1723

This integral right here was equal to 2 π.1726

This integral right here, when I solve that one, that integral is equal to 4/3.1735

When I put everything together, the 4/3, the 2 π, multiply all this out, I end up getting the following.1746

Let me now go to red.1753

It is going to equal -4 E² × 2 π/ 3 × 4 π E0 × 6561 π A sub 0³.1756

I left everything there, I had not canceled anything yet.1779

0 to infinity of² R, this is going to be R ×, when I square this 36 σ² - 12 σ³ + σ⁴ E ^- 2/3 σ DR.1782

Once again, for this integral I have σ is equal to R/ A 0, which means that R is equal to σ A 0.1809

D Σ = 1/ A 0 DR which means that DR is equal to A 0 D σ.1832

When I substitute all of these into this integral, what I end up getting is the A 0 come out.1842

What I end up getting is the following.1851

I should go ahead and write it all out this one or do the cancellation.1866

I think I will go ahead and write it all to actually see the cancellation.1873

When I plug all of these in to this integral, I’m going to leave this alone.1876

What I would end up with is the following.1881

I end up with -4 E² 2 π A sub 0 A sub 0², that came from all the work that I just did with the σ.1883

3 × 4 × π × ε sub 0 × 6561× π A sub 0³, the integral from 0 to infinity of σ × 36 σ² -12 σ³ + σ⁴ E⁻²/3 σ D σ.1898

4 and 4, π and π, A sub 0 and A sub 0, that cancels that.1930

What I end up with is the following.1945

My average potential energy is going to equal - 2 E²/ 3 × 6561 × π ε sub 0 A sub 0 ×1953

the integral of 36 σ³ -12 σ⁴ + σ⁵ E⁻²/3 σ D σ.1969

When I'd taken this integral, can I break it up into 1 integral, 2 integral, 3 integral.1986

I have seen this integral several times.1991

I end up the following.1993

I end up with, 36 × 3!/ 2/3⁴ - 12 × 4!/ 2/3⁵ + 5!/ 2/3⁶.1995

All of this ends up equaling to 16 × 81/ 16 -288.2030

I hope you do not mind that I’m actually going through all of the arithmetic here.2044

I figured that at least do it for this one integral.2047

+ 120 × 729.2051

I know that you guys are more than capable of doing the arithmetic.2055

I can just go ahead and give the answer but figured out that just go through with it here.2058

= 17,496/ 16 - 34,992/ 16 + 21,870/ 16.2062

That integral = 4374/ 16.2081

K so we have 4374/ 16 × -2 E².2089

Remember, the constant 3 × 6561 π A sub 0.2101

A sub 0 actually is going to equal - E²/ 36 π E sub 0 A sub 0.2107

Our average potential energy is equal to -E² / 36 π E sub 0 A sub 0.2119

A3, let me do it in blue.2132

Energy level 3 is equal to -E² / 8 π E sub 0 A sub 0 3² is equal to - E² / 72 π E sub 0 A sub 0.2142

Once again, we have ½ × this is equal to this.2163

Therefore, the average value of V is equal to twice the value of the energy.2174

We ended up with the same result.2184

Once again, the energy is equal to the kinetic energy + the potential energy is equal to the kinetic energy.2187

The potential is plus twice the full energy.2198

Once again, we have this equation that we solve.2204

We bring that over there, we end up with - the energy = the average kinetic energy.2207

Again, we have that the average potential energy is equal to twice the total energy is equal to - twice the kinetic energy.2221

It is the same relation that we had for the 210 orbital, 311 orbital.2241

This result is true in general.2249

This is where we are going to begin a little bit of general discussion.2253

This result is true in general.2257

This result is true in general, when the potential energy is the Coulombic potential.2292

When V of R is the Coulombic potential.2306

If you happen to be dealing with the Coulombic potential, the -E²/ 4 π E sub 0 R,2313

you are always going to get the relationship that the average potential energy = twice the total energy2321

= –twice the kinetic energy.2328

When V is any potential whatsoever, ½ KX² 3 R³ cos, whatever potential,2335

Let me go back to black here.2347

When V is any potential whatsoever, then the general expression for the relationship2352

between the kinetic energy and the potential energy, is as follows.2373

Where V is equal to the function of X, Y, and Z, now we are dealing with all three space.2416

This is something called the Virial theorem, the quantum mechanical Virial theorem.2426

The earlier result is also called of Virial theorem for the Coulombic potential.2431

This is the general expression for the Virial theorem.2436

This is called very important, this is called the Virial theorem.2439

Let us take a look at what will we get.2455

We are dealing with Coulombic potential.2457

Potential energy was this -E²/ 4 π E sub 0 A sub 0 R.2460

When that is the case, the relationship among the average potential energy, the energy, and the average kinetic energy is this.2467

In general, for any kind of potential whatsoever, not the Coulombic potential but any potential,2475

the relationship is the average value of given some potential energy as a function of X, Y, Z, or R θ φ, whatever.2481

If I take X × the partial with respect to X, Y × the partial with respect to Y, Z × the partial with respect to Z.2493

If I take the average value of that, in other word solve the integral.2501

I'm going to end up getting is going to equal twice the average of the kinetic energy.2506

This is the quantum mechanical Virial theorem.2511

Another version of the Virial theorem is this one.2514

Another version, which might be the version that you actually see in your book.2521

Another version of the Virial theorem reads as follows.2525

When the potential energy of the particle has the form V is equal to some constant, I’m going to use K.2545

Some constant KX⁸, then the average values of kinetic and potential energies are related2580

by twice the average kinetic is equal to this constant A × the potential.2616

This is that, so this is another version of the Virial theorem.2627

This might be the one that you actually see in your book.2632

This is the general expression of the Virial theorem or the quantum mechanical Virial theorem for any potential, whatsoever.2634

This is specifically if the potential is in this form.2643

For the problems that we did for the 210 or the 311 orbital, these are the Coulombic potential.2647

In that particular case, this was the relationship that we got.2653

This is the general expression right here.2658

Whatever the form of the potential is, if you do this and then take the average value of that,2665

the integration, you end up twice the average value of the kinetic energy.2671

This is a profoundly deep and important theorem.2675

Not just in quantum mechanics, in classical mechanics as well.2679

Let us a little bit more about this.2685

Let me actually go to black here.2686

The Coulombic potential is -E² / 4 π ε sub 0 A sub 0 × R, which is the same as2698

–E sub 0²/ 4 π ε sub 0 A sub 0⁻¹.2723

This thing is just a constant.2734

Basically, what you end up with is a constant × R⁻¹.2736

Based on what we just wrote, that all conversion of the Virial theorem,2744

twice the kinetic energy is equal to - 1 × the potential energy, which is exactly what we got before except we want a – here.2749

We went ahead and to confirm what it is that we already did.2767

If we use the larger expression, the one with the partial derivatives, the Coulombic potential is written as this.2777

The Coulombic potential can be written as, here we have R.2801

When we are given 3 space, R is just equal to vx² + Y² + Z².2812

This is the spherical coordinate designation for something ends in Cartesian coordinates.2819

If you have a point in 3 space, there are some vector that goes from the origin to that point.2827

That distance is X² + Y² + Z².2832

If I put this into here, my potential expressed as X, Y, Z is going to be -E²/ 4 π E sub 0 A sub 0 X² + Y² + Z².2836

When I form X DB DX + Y, the derivative of this with respect to Y,2858

and Z the derivative of this with respect to Z, when I do that and I take the average value of it,2874

I end up actually getting that -V = 2K.2887

All of these are the same thing, the quantum mechanical Virial theorem.2893

Let us go ahead and see what our next example brings.2905

Use the quantum mechanical Virial theorem in its full partial derivative form to show2914

that the average potential = the average kinetic = 1/2 of the energy for the harmonic oscillator.2918

Interesting, once again, the full partial derivative form is this X DDX, Y DDY, Z DDZ = 2K.2927

We need to show that for the harmonic oscillator, this is the relationship that exists2936

between average potential and the average kinetic and the energy.2941

For the harmonic oscillator in 3 dimensions, the potential,2947

is equal to the potential X, Y, Z is equal to ½ K1 X² + ½ K2 Y².2966

Earlier, we just do one harmonic oscillator, something that is sliding back and forth like this.2984

But now in 3 dimensions, it can be oscillating this way or this way.2989

We have to include X, Y, Z.2992

It is just the potential, we know the potential of the harmonic oscillator is ½ KX².2995

We just add the Y potential and the Z potential which happens to be the same.3003

½ K sub 3 Z².3007

Let us just go ahead and find DV DX.3014

DV DX, that is nothing, that is nothing because we are holding these, we are doing partial derivatives.3021

We are holding these constant, we end up with K 1X.3025

The partial with respect to Y is going to equal to K 2Y.3033

The partial with respect to Z is going to equal K 3Z.3038

I’m going to multiply this X DV DX, I’m just forming this thing.3045

That is all I’m doing, I’m forming that thing.3052

X DVDX + Y DVD Y + Z DVDZ.3055

That is going to equal K1 X² + K2 Y², X × K1 X is K1 X².3067

Y × K2 Y is K2 Y².3078

Z × DV DZ, Z × K3 Z is K3 Z², + K3 Z².3082

This is equal to twice the ½ K1 X².3092

Twice of this is equal to this, I’m just rewriting it.3101

+ 1/2 K2 Y² + ½ K3 Z².3106

Well ½ K1 X Squared + 1/2 K2 Y² + 1/2 K3 Z², that is this thing.3114

That is V already.3122

What we have is this thing that we just did, this X DVDX + Y DVDY + Z DVDZ is equal to twice the potential energy.3125

That means this is equal to this.3152

We also know this from what we just saw that the Virial theorem, = this 2V = 2K.3157

When you to remove the 2, you get V = K, which is the first part of what it is that we have to prove.3175

The total energy is equal to the kinetic energy + the potential energy.3186

The total energy =, the kinetic energy = the potential energy so this is just twice that.3196

Of course, we have the final result.3206

We just divide by 2, it does not really matter.3208

½ of the energy, this was the other thing that we want to prove.3210

That is it, all based on the quantum mechanical Virial theorem.3216

Whenever you are given the potential V, if you form X DVDX + Y DVDY + Z DVDZ,3222

the average value of that is going to always equal twice the kinetic energy.3236

That is the quantum mechanical Virial theorem.3241

Let us see what we have got.3246

Let us see what is next.3252

In problem 5 of the previous lesson, we found that the average value of R is equal to 6 A sub 0/ Z3255

for the 2S orbital, for the hydrogen atom.3264

If it is just hydrogen, the Z = 1 that is just 6 A0.3267

Find that σ sub R², find the variance for the 2S orbital.3272

Recall that σ sub R² is equal to the average value of R² - the average value of R quantity².3279

The average value of R 6A/ Z, we already have this one.3289

Let me go to red.3295

What we need to find is the average value of R².3296

We must find the average value of R².3308

The 2S orbital means that N is equal to 2, L is equal to 0, and M is equal to 0.3313

We are looking for the wave function 200.3323

Well , the average value of R² is going to equal the integral of ψ 200 conjugate × R² ψ 200.3328

The operator R² just means multiply by R², multiply by R and multiply by R again.3345

What we end up getting, I’m not going to go ahead and write the wave function ψ 200,3352

you can go ahead and look that up.3356

What we end up actually getting is the following.3359

We end up with the 1/ 32 π so it is just going to be ψ is real.3361

It is just going to ψ 200² × R² 1/ A sub 0³ × the integral from 0 to 2 π D φ.3366

Again we are always working in spherical coordinates here, D θ.3379

0 to π of sin θ D θ 0 to infinity of R² × 2 - σ² E ⁻Σ R² DR.3385

This integral is equal to 2 π.3408

This integral is also going to end up equaling 2 π.3411

We end up with 1/ 8 A sub 0³ × the integral from 0 to infinity of R⁴ × 2 -σ² E ⁻Σ DR.3414

We have the same way, σ = R/ α sub 0, which means that R = σ × α sub 0.3432

D σ = 1/ α sub 0 R, that is the same α.3443

DR, which means that DR is equal to A sub 0 D σ.3454

When we put all of these in here, under the integrand, we end up with the following.3460

We end up with E sub 0⁴ × A sub 0/ 8 A sub 0³ the integral from 0 to infinity of Σ⁴ × 4 - 4 Σ + σ² × E ⁻σ D σ.3465

All of that is going to equal, when I cancel and work some things out, I'm going to end up with A sub 0²/ 8 ×,3495

I’m going to separate the integrals out.3509

That is fine, I will just go ahead and write it here.3513

From 0 to infinity of 4 σ⁴ E ⁻σ D σ - the integral from 0 to infinity of 4 σ⁵ E ⁻σ D σ +3516

the integral from 0 to infinity.3535

I want to work this entire one myself.3539

-σ D σ.3542

It is going to end up equaling A sub 0²/ 8 × 4 × 4! -4 × 5! + 6!.3548

When I worked all this out, I'm going to end up with 42 E sub 0².3562

Our σ R² is equal to the average value R² - the average value of R².3571

That is going to equal what we just got.3581

This is this one, so it is going to be 42 A sub 0² -6 A sub 0².3583

That is going to equal 6 A sub 0².3594

That is our variance.3599

Do I have another page here?3601

I do, let me go ahead and go to the next page.3605

The general formula, we just found for one particular orbital.3612

The general formula for the average value R² is as follows.3620

The average value of R² is equal to N⁴ A sub 0²/ Z × 1 + 3/2 × 1 – L × L + 1 -1/3 / N².3631

There you go, that is the general expression for R².3663

In the previous lesson, we also have a general expression for the average value of R.3669

We have those based on just the quantum numbers N and L.3675

Everything else is very easily taken care of.3684

Thank you so much for joining us here at www.educator.com.3687

We will see you next time, bye.3689