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Post by Van Anh Do on December 14, 2015

For example IV and V would it make sense to use only the radial part to calculate the radius of a sphere and <r> since those values are for R and only the radial part contains R?

The Hydrogen Atom Example Problems III

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
  • Example I: Show That ψ₂₁₁ is Normalized 0:07
  • Example II: Show That ψ₂₁₁ is Orthogonal to ψ₃₁₀ 11:48
  • Example III: Probability That a 1s Electron Will Be Found Within 1 Bohr Radius of The Nucleus 18:35
  • Example IV: Radius of a Sphere 26:06
  • Example V: Calculate <r> for the 2s Orbital of the Hydrogen-like Atom 36:33

Transcription: The Hydrogen Atom Example Problems III

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

We are going to continue our example problems for the hydrogen atom.0004

Let us get started.0006

The first example is show that the ψ 211 is normalized.0010

Now we are dealing with the entire wave function.0014

A function of R, a function of θ, and a function of φ.0019.2 We want to show that it is normalized.0017

Let us go ahead and write down what ψ 211.0023

Las time we are dealing with spherical harmonics, so now we are dealing with a full 3 quantum numbers 211, N, L, and M.0033

This particular wave function is equal to 1/ 64 π to ½ power Z/ α sub 0³/2 σ E ⁻σ/ 2 × sin θ E ⁺I φ, 0041

where σ is actually equal to ZR/ Α sub 0. 0070

Different books and different lists have some variation in how they actually list of functions,0079

depending on what they want to express as constants.0086

Whether they want to use exponents here or that I want to use √ signs, things like that.0088

If the wave functions in your book or in the particular list that you have to be looking at, 0094

do not look exactly like this, do not worry about it.0097

They are the same wave function, I promise you that.0100

When you integrate them, you are going to get the same answer.0105

I would not worry too much.0108

This is Z here, Z is just is the atomic number.0111

These are the orbitals, the wave functions for what they call hydrogen like atoms.0114

Basically, anything that something like helium, where why the electrons is been taken off.0120

It is a helium atom but it has been ionized so you are still talking about the 1 electron.0126

For our purposes, we are just going to take Z equal to 1 because the atomic number of hydrogen is 1.0131

We just wan to write the general equation for the hydrogen like orbital, 0137

that we are going to take Z equal to 1 and A sub 0 is the bohr radius.0142

We want to form the normalization integral which is the integral of ψ 211 conjugate × ψ 211.0148

We want this integral to equal 1 because we are trying to show that it is normalized.0167

We want this integral to equal 1.0174

Let us go ahead and write all this out.0184

Ψ 211 conjugate the integrand × ψ 211 is going to equal,0187

Let me see, the conjugate of this looks exactly the same except it is going to be E ⁻I φ instead of E ⁺I φ.0196

The 1/ 64 π to ½, 1/64 π gives us 1/ 64 π.0206

Again, we are taking Z equal to 1, 1/ A sub 0³/2 × 1/α sub 0³/2 is going to be 1/ Α sub 0³.0214

S σ is equal to Z R/α so this is just R/ A sub 0 E ⁻σ/ 2.0231

This is going to be -R/ 2 A0 sin of θ E ⁻I φ.0242

Yes, we still have some more, I think I have enough room here.0261

So × R/ A sub 0 E ⁻R/ 2 A sub 0 × sin θ × E ⁺I φ.0267

This, I multiply all the constants together.0282

This is for the conjugate and this is for the ψ 211.0286

When I put all this together, I get 1/ 64 φ A sub 0⁵ R² E ⁻R/ α sub 0.0294

R2 and R2 is going to be -2R.0318

Then, we have sin², sin θ sin θ is sin² θ E ⁺I φ E ⁻I φ × E ⁺I φ is just equal to 1.0323

This is our integrand.0334

The integral of ψ 211 conjugate × to ψ 211 is going to equal the integral from 0 to infinity.0339

Now, we are including R 0 to 2 π, 0 to π of this function which is 1/ 64 π A sub 0⁵ R² E ⁻R/Α sub 00351

sin² θ × the factor R² sin θ.0375

We are doing D θ D φ, we are going to do our last, D φ DR, inside to outside.0383

We have a function of R, a function of θ.0398

It does not look like we have a function of φ.0402

Let us go ahead and separate this out, = to 1/ 64 π A sub 0⁵ the integral from 0 to infinity, R² × R² is μ R⁴ E ⁻R A sub 0.0404

I’m going to separate out the variables, the R, θ, and φ, DR.0428

The integral from 0 to 2 π of D φ because there is no function of φ.0434

The integral from 0 to π if sin² θ sin θ, so we have our sin³ θ D θ.0439

We just put everything together.0450

All of the R’s we have put together with the r’s, under one integral.0452

The φ under one integral, the θ under one integral.0456

Now, we are going to solve these 3 integrals.0459

Integral number 1, our first integral which is equal to 0 to infinity of R⁴ E ⁻R/ Α sub 0 DR.0464

If you either let your software do it or if you look it up on a table, which is going to be on the inside cover back cover of your book 0484

or the inside front cover of your book, because this integral shows up a lot so it is definitely an integral that you are going to need.0491

I think you have actually seen it before, it is 4 !/ 1/α sub 0⁵.0497

It is going to end up being 24 Α sub 0⁵.0507

That takes care of the first integral.0512

The second integral, integral number 2 that is nothing more than 0 to 2 π of D φ and that is equal to 2 π.0514

That is the second integral.0526

The third integral, integral number 3 is equal to the integral from 0 to π of sin³ θ D θ.0530

I do not know if I have done it here manually but I think at this point it is best for you to just go ahead and put it into your software.0545

I do not necessarily know if I want to go through.0552

That is fine, I will go ahead and go through this process.0554

It is it is not a problem.0557

D θ =, I going to separate this out so I will go ahead and do the manual.0558

You do not need to, but this is what it looks like, in case you would do it.0564

0 to π sin² θ sin θ.0567

I have separated this out and I'm going to use a U substitution.0572

I'm going to let, sin² θ is actually equal to 1/ π is equal to 1 - cos² θ sin θ D θ.0577

I will use a U substitution on that.0589

I will let U equal cos θ.0592

Therefore, DU = - sin θ D θ.0594

Therefore, this integral is going to actually end up equaling - the integral from 0 to π 1 - U² DU, 0599

which is equal to - U – U³/ 3 from 0 to π, which is nothing more than U³/ 3 - U from 0 to π, 0612

which is equal to cos³ θ/ 3 – cos θ from 0 to π.0623

And this is going to equal -1/ 3, -and -1 I hope that I have done this correctly.0642

-1/3 -1 which is going to end up equaling 4/3, this is our third integral.0650

Now that we have the first integral, the second and third.0660

When we put this together, it is going to be this × that × that × the constant.0663

We have 1/ 64 π A sub 0⁵ × 24 A sub 0⁵ × 2 π × 4/3.0670

Everything comes up to 192/ 192 is equal to 1.0690

Yes, ψ 211 is normalized as written, good, very nice.0695

Example 2, we want to show that ψ 211 is orthogonal to ψ 310.0710

Let us write out what they are.0716

Ψ 211 is equal to 1 / √ 64 π × Z/ A sub 0³/2 ZR/ A sub 0 E ⁻Z R/ 2 A sub 0 sin θ E ⁺I φ.0719

Then, we have ψ of 310 that is going to equal 1/ 81 × 2/ π.0749

I think this is going to be to the ½ and then we have Z/ A sub 0³/2.0761

This time we have 6 ZR/ A sub 0 - Z² R²/ A sub 0² × E ⁻ZR/ A sub 0.0774

Actually, it is going to be 3 A sub 0 × cos of θ.0792

Those are the two wave functions ψ 211 and ψ 310.0800

We want to form the integral to show that they are orthogonal.0805

We want to take ψ 211 the conjugate multiply by ψ 310, we want to integrate that and we want to equal 0.0815

We want this to equal 0 because we are trying to demonstrate orthogonality.0830

Ψ 211 conjugate, I’m not going to go through in every step, 0842

at this point you should be reasonably familiar with taking conjugates when there is a complex there negated.0846

The ψ 211 × ψ 310 which is the integrand is going to equal,0853

I’m not going to write out all these constants over and over again.0860

I'm just going to call it the constant for 211 so let us call it ψ 211.0863

Again, we are taking Z equal to 1, the hydrogen atom.0867

R/ A0 E ⁻R/ 2 A sub 0 sin of θ E ⁻I φ because we are doing the conjugate × the constant for 310 × 6 R/ 0871

A 0 - R²/ A 0² × E ⁻R/ 3 A0 × the cos θ.0892

All of this is equal to ψ 211 × ψ 310 6.0909

I multiply the constants R, I'm going to get 6 R²/ A 0² – R³/ A 0³.0921

I just multiply everything together, E⁻⁵ R/ 6 A sub 0 sin θ cos θ E ⁻I φ.0934

That is my integrand.0961

Therefore, my integral of ψ 211 conjugate × ψ 310 is going to equal,0966

I'm going to go ahead, this is what I’m integrating in spherical coordinates.0977

I’m going to go ahead and just separate them at once, instead of doing it in 2 steps.0981

I'm going to have the integral from 0 to infinity of the R part.0985

6 R²/ A sub 0² – R³/ A sub 0³ E⁻⁵ R/ 6 A sub 0.0991

It is going to be R² DR 0 to 2 π of E ⁻I φ D φ × the integral from 0 to π of sin θ cos θ sin θ D θ.1001

Let us take a look at this.1038

I have this thing, this thing, and this thing.1040

The second integral is equal to 0, we have seen that before.1043

Integral number 2 is equal to 0.1047

Let me write what integral number 2 is, 0 to 2 π of E ⁻I φ D φ.1057

This is equal to 0, we know this from a previous example.1066

Therefore, our ψ 211 conjugate ψ 310 is equal to 0, which means they are orthogonal.1070

Exactly as we wanted.1087

You see that a lot of times you would not have to solve any integral.1090

The integral was very complicated but it is not complicated because a piece of it is really simple.1093

We already know that N = 0.1098

In one case it is this function.1100

In some other cases, you may have even and odd functions.1102

And if it ends up being the integral of an odd function/ a symmetric interval, that is going to equal 0.1105

Definitely makes life a lot easier.1112

Let us go ahead and go to example 3.1115

What is the probability that a 1S electron will be found within 1 bohr radius of the nucleus.1117

We want to evaluate this following integral.1123

The probability integral, we want to evaluate the integral from 0 to A sub 0.1129

Now, we are specifying an outer radius so it is no longer infinity of the ψ 1S conjugate ψ 1S.1140

This is the probability integral.1153

Normally, we just integrate it over the entire space and we want it equal to 1 to show normalization.1155

This is a normalization condition but now we are calculating the probability.1162

There is going to be some upper limit on this integral.1167

1S means that N is equal to 1 and S means that L is equal to 0.1172

Therefore, N is equal to 0.1182

What we are looking for is the ψ 100 function.1185

We need to form the ψ 100 conjugate × Ψ 100 the integrand.1191

Well, that is going to equal 1/ √ π × Z/ Α sub 0³/2 E ⁻ZR/ α sub 0 × 1/ √ π Z/ α sub 0³/2 E ⁻ZR / α sub 0.1200

Again, we are taking Z equal to 1.1230

Therefore, the Ψ 100 is going to equal,1233

This is 1 and this is 1, so it is going to equal 1/, π is taken care of, and A sub 0³ and 1245

it is going to be E ⁻R/α 0 –R/ Α 0 -2R/ Α 0.1255

The integral from 0 to Α sub 0 Ψ 100 conjugate ψ 100 is going to equal 1/ π Α 0³.1273

The integral from 0 to Α sub 0 of E ⁻2R / A sub 0 and this is R² DR.1286

That is the R version.1297

We have 0 to 2 π of our D φ and we have the integral from 0 to π of, we have sin θ D θ.1299

We already know what this is, this is 2 π.1318

We already know what this is, this is just 2.1322

We have 4 π over here, this integral right here.1325

Let us go ahead and do a little something with that particular integral.1329

Should I do it on the next page?1338

I’m just going to go ahead and take care of it here.1348

The first integral, integral number 1 is the integral from 0 to A sub 0 of R² E ⁻2R/ A sub 0 D R.1351

We would actually do a little bit of a U substitution here.1365

I'm going to call U, I'm going to call it R/ A sub 0.1369

Therefore, DU is equal to 1/ A sub 0 which means that A sub 0 × DU is equal to DR.1376

I’m going to put this into the DR and then when I do R², that is going to equal A sub 0² U².1391

I'm going to put that into there.1403

This integral, when I put 0 in for here into R, I get 0.1405

U = 0.1413

This lower limit of integration is going to end up becoming 0.1415

When I put A sub 0 into R, I'm going to get the upper integral of 1.1419

When I do the substitution, I’m end up getting the following.1424

I'm going to get the integral from 0 to 1 R² is A sub 0² U².1428

This is going to be E ⁻2U and then × DR, which is A sub 0 × DU, which ends up being A sub 0² × A sub 0.1441

We have A sub 0³ × the integral from 0 to 1 of U² DU.1459

I'm sorry, U² E ⁻2U DU.1467

When I go ahead and put this into my mathematical software, I'm going to end up getting 0.08083 × A sub 0³.1476

This is my first integral.1487

My integral number 2, which was the integral from 0 to 2 π of D φ, that is equal to 2 π.1489

And then my integral number 3, which is the integral from 0 to π of sin θ D θ, that is going to actually end up equaling 2.1502

When we put all of the 3 together, we have 1/ the constant π A sub 0³ × 0.08083 A sub 0³ × 2 π × 2.1515

That that, that that, what I end up getting is 0.3233.1540

There you go, that is my probability.1550

My probability is roughly 1/3 of finding the 1S electron within 1 bohr radius.1552

Let us try example number 4, what is the radius of the sphere that encloses a 60% probability of finding the hydrogen 1S electron?1565

Express your answer in terms of A sub 0, the bohr radius.1582

They are telling us that the probability is equal to 0.6.1588

The integral that we just solve, we are going to set that integral equal to 0.6 1591

and we are actually going to look for that upper limit of the integral on R.1596

We want to know what that is, what is the radius that encloses a 60% probability.1604

Let me go ahead and go to blue, change this up a little bit.1610

From the previous problem and if you want you can do the integral all over again, it is not a problem.1613

From the previous problem, which also concern the 1S orbital, we found that the probability 1624

was equal to 1/ π A sub 0³ 0 to some constant × A sub 0.1638

We are going to be looking for this constant ψ.1647

R² E ⁻2R/ Α sub 0 DR the integral from 0 to 2 π of D φ, the integral from 0 to π of sin θ D θ.1651

We found that all this is going to equal, this is 2, this is 2 π.1671

4 π 4 π, for on top the π cancel.1678

We are going to get 4/ A sub 0³ × the integral from 0 C A sub 0 R² E ⁻2R/ α sub 0 DR.1682

The probability integral is this, we are going to solve this integral.1699

We are going to get some equation in C and I’m going to set that C equal to our 60%, 1705

because we want the probability = 60%, I want to find out what C is 1710

because we want our answer expressed in terms of the A sub 0.1715

That is what we are going to do.1718

We will solve this integral then set it equal to 0.60 then solve for C.1720

Let us see what we have got.1752

Our probability integral is going to equal, I'm going to go ahead and do a little bit of U substitution here.1753

I’m going to go ahead and go to red.1766

Up here, I'm going to call U equal to R/ α sub 0 like I did before.1768

Which means that U² Α sub 0² is equal to R².1777

DU is going to equal 1/ α sub 0 DR means that DR is going to equal Α sub 0 DU.1783

When I put all of these into here, when I put DR, when I put this into DR.1793

When I put this in for R², I'm going to end up with 4 Α sub 0³/ Α sub 0³ × the integral from 0 to C.1796

When I put 0 into here, I'm going to get 0.1813

And then when I put C α sub 0 into here, I'm going to get C.1819

Now I have a new upper limit of integration and it is going to be U² E⁻² U DU.1826

This is by probability integral, I have converted this into this so I could deal with the C alone.1839

I'm going to go ahead and this cancels.1845

I’m going to go ahead and solve this integral by hand and so you can see it.1852

4 × the integral 0 to C U² E ⁻2U DU.1857

Just go ahead and do it with software but I would like you to go ahead and see it.1866

In this particular case, we have a function × a function so we are going to use something called integration by parts.1871

Integration by parts tends to get long, there is a short hand version of integration by parts called tabular integration.1876

Tabular integration, it works when your particular integration by parts integral is such that one of your functions,1886

I will go ahead and go back to blue.1893

When one of your functions differentiates down to 0 and the other one of your functions just integrates infinitely over and over again.1895

Integrals like this, where you have some polynomial and some exponential, they are perfect for tabular integration.1903

And what you actually do is this, you do U² over here, you do E ⁻TU, this is our U and this our DV.1910

Integration by parts notation, we differentiate this to TU.1920

We differentiate this down to 2 and we differentiate this down to 0.1926

We integrate this up to -1/2 E ⁻TU.1930

We integrate up to ¼ E ⁻2U and we integrate one more time up to -1/8 E⁻² U.1946

Now, we put them together like this.1952

We put that together with that one.1954

This together with that one, and this together with that one change in signs + - +.1956

This integral ends up actually equaling, it is going to be 4 × -1/2 U² E ⁻TU.1961

U is going to be this time this, this × this, this × this.1980

-U² E ⁻TU/ 2 - 2 U E⁻² U/ 4 -2 E ⁻TU/ 8.1986

This is going to equal – E⁻² U × 2U² + 2 U + 1, from 0 to C, by the way.2015

2U + 1 from 0 to C, it is going to equal - E⁻² C × 2 C² + 2 C + 1 - A -1 × 0 + 0 + 1.2034

We are going to get this integral equal to 1 - E⁻² C × 2 C² + 2 C + 1.2061

This is our probability, this is the solution to the integral.2076

We are going to set that equal to 0.6.2080

1 –E⁻² C × 2 C² + 2 C + 1 = 0.60.2084

When we go ahead and solve this equation, just put it in your software or just use a graphical calculator.2098

We end up with C equaling, E ⁻2C 2C² + 2 C + 1 = -0.40.2104

Move the 1over that side so I'm going to get E⁻² C × 2 C² + 2 C + 1.2128

These negatives cancel and I bring it back to the left.2138

It is going to be -0.40 is equal to 0.2141

This is the equation that I put my software or into my graphing calculator and I end up with C = 1.553.2146

For a 60% probability, our radius is equal to 1.553 α sub 0.2156

There you go, that is it.2174

Tedious math, but nothing that is straightforward math for the most part.2177

Of a 0.40, let us make that a little bit more clear.2183

What do we got next?2189

Calculate the average value of R for the 2S orbital of the hydrogen like atom.2202

Again, hydrogen like just means that you actually include the Z, the atomic number of the wave function.2208

You put in there for like hydrogen is just Z equal to 1.2213

This is what is going to leave the Z there.2216

Calculate the average value of R for the 2S orbital of the hydrogen like atom.2220

Let us see what we have got.2225

2S means that N is equal to 2.2227

It means that L is equal to 0, which means that M is equal to 0.2234

What we are looking at is our ψ 200.2241

The average value of R for 200 is going to equal the integral of ψ 200 conjugate × R operating on ψ 200.2249

This is the basic form for the average value of whatever it is we happen to be looking at.2265

Let us see what we have got.2273

Our ψ 200 is equal to 1/ √ 32 π × Z / α sub 0³/2 × 2 - σ × E ⁻σ/ 2.2275

Again, the σ is equal to ZR/ Α sub 0.2307

The ψ 200 conjugate is the same as this because there is nothing complex about this.2315

Therefore, this R operator, it just means multiply by R.2322

It does not matter where we put it.2330

I do not have to operate on this one first and then multiply on the left by this 1.2331

I could just multiply these 2 and then multiply by R.2335

R × ψ 200² is going to equal R × 1/ 32 π.2341

It is going to be Z³/ α 0³ × 2 - σ² E ⁻σ.2354

This × itself.2369

It is going to equal, let us pull the constants out.2373

Z³/ 32 π Α sub 0³, let us multiply this out for -4 σ + σ² E ⁻σ × R.2377

This integral here, it is going to be Z³/ 32 π A sub 0³ 0 to infinity 2408

0 to 2 π 0 to π of 4 -4 σ + σ² R × R² sin θ D θ DR.2431

It is actually going to be D φ DR.2453

This is going to equal.2466

The 0 to 2 π D φ and the 0 π of the sin θ D θ is going to equal of 4 π.2478

We end up with 4 π Z³/ 32 π A sub 0³.2488

The π go away and it is going to be integral from 0 to infinity of R³ × 4 - 4 σ + σ² E ⁻σ DR.2498

I'm going to go ahead and do a little substitution here.2526

Our σ is equal to ZR / A sub 0 which means that D σ is equal to Z/α sub 0 DR.2529

Therefore, DR is equal to Α sub 0 σ/ Z and R³ is going to equal Α sub 0³/ Z³ × σ³.2546

We just rearrange this and R³.2570

When we put all of these into this thing, we end up with a lot.2573

4 π Z³/ 32 π/ A sub 0³ × A sub 0³/ Z³ × A sub 0/ Z × the integral 0 2582

to infinity σ³ × 4 -4 σ + σ² E ⁻σ.2606

This × D σ =, all of these things cancel out.2620

That cancels that, that cancels that, + 8, we end up with A sub 0/ 8 Z, the integral from 0 2624

to infinity of σ⁵ and distribute this out and rearrange.2636

Σ⁵ -4 σ⁴ + 4 σ³ D σ.2643

We have that general integral formula, we have the integral formula, what we get from the table.2658

The integral from 0 to 8 of X ⁺N × E ⁻AX DX is equal to N!/ 8 ⁺M + 1.2669

When we do that for this one, this one, and this one.2685

I keep forgetting my exponential E ⁻σ D σ.2693

When I apply this formula to this integral, this × this, this × this, this × this, I end up with,2707

Let me go back to red.2717

I end up with A sub 0/ 8Z + 5! -4 × 4! + 4 × 3!.2719

I end up with 48 A sub 0/ 8 Z and I end up with 6 A sub 0/ Z.2733

There you go, that is what we wanted, our final answer .2744

The average value of R, on average the electron will be this far from the nucleus for the 2S electron.2752

We just went through the mathematics.2765

This goes here, this is that one.2769

Let us go ahead and give you some general formulas.2774

The average value of R sub NL is equal to N² A sub 0 / Z × 1 + ½ × 1 - L × L + 1/ this is that one and this is that one.2780

Now for hydrogen, Z is equal to 1 and this general formula reduces to A sub 0/ 2 × 3 N² - L × L + 1.2815

For example, if I wanted to find the average for the 3S orbital, N is equal to 3, L is equal to 0, M is equal to 0.2845

I will just put that in.2867

I will get A sub 0/ 2, N =3, 3² is 9, 3 × -27, 27 - L × L + 1 -0.2870

It is going to be 27/2 A sub 0.2881

If I were to do it for the 3P orbital, the 3P orbital here N =3 and P, we are talking about L is equal to 1.2887

We just plug these back into there.2898

The general formula for the hydrogen atom.2900

The general formula for any hydrogen like atom which includes Z.2902

Thank you so much for joining us here at www.educator.com.2910

We will see you next time, bye. 2912