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### Example Problems 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 0:00
• Example I: Three Dimensional Box & Eigenfunction of The Laplacian Operator 0:37
• Example II: Positions of a Particle in a 1-dimensional Box 15:46
• Example III: Transition State & Frequency 29:29
• Example IV: Finding a Particle in a 1-dimensional Box 35:03
• Example V: Degeneracy & Energy Levels of a Particle in a Box 44:59

### Transcription: Example Problems I

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

Today, we are going to start doing some example problems on the Schrӧdinger equation, particle in a box,0004

particle in a 2 dimensional box, 3 dimensional box, things like that.0012

We are going to be doing a lot of problems.0016

The only real way to wrap your mind around any of this is doing a ton of problems.0019

The only warning that I have is as you already figured out quantum mechanics can be notationally intensive.0023

The best advice I can give is keep calm, cool, and collect it, and work slowly.0030

Let us jump right on in.0035

Our first example problem is let ψ be a wave function for a particle in a 3 dimensional box.0039

Find del² of the ψ and show that the ψ is an Eigen function of Laplacian operator.0046

Let us see what we have got.0056

First of all, let us start off with our equations.0057

Again, I think is always a good idea to just write down the equations that you know simply0060

because repetition locks it in your mind, makes you feel more comfortable with it.0060

Let us go ahead and write down our equations.0070

Ψ for a 3 dimensional box, N sub X N sub Y N sub Z = 8/ ABC ^½ × sin of the N sub X π/ AX × sin of the N sub Y π/0073

B × Y × sin of N sub Z × π/ C × Z.0096

That is our wave function.0107

The del² that was that D² / DX² + D² DY² + D² DZ².0116

What we have to do is we have to do is to apply the Laplace operator to this function.0133

I know what you are thinking, I'm thinking the same thing.0143

Our del² ψ = del² DX² + del² DY² + del² DZ² of ψ.0148

Again, operators they can distribute so we are going to do this, we are going to do this, and we are going to add.0166

Let us just jump right on in.0176

Let us go ahead and do the second partial derivative of the X with respect to X,0180

Which means we are going to hold this constant and this constant.0191

The only thing we are going to differentiate is this.0195

Let me actually do this in red.0198

This first term, we are just going to differentiate this function because we are holding all the other variables constant.0200

Let me go ahead and write this out.0210

When I take the first derivative of this, I’m going to write it this way.0215

N sub X π/ A × X, when I take the first derivative, just that D means take the derivatives.0219

I end up with the following.0227

I end up with NX π/ A × cos of NX π/ A × X that is the first derivative.0228

If I take the second derivative, if I differentiate again, I end up with N sub X² π²/ A² and the derivative of the cos is –sin,0238

I get the negative sin there and I get N sub X π/ A × X.0253

This is my second derivative.0262

Since, I’m holding everything else constant, this is the only thing that I need to concern myself with.0265

My D² DX² of ψ is going to equal to -8/ ABC ^½.0271

I will take this thing , -sin N sub X² π²/ A².0290

I have this × the rest of this.0301

I have included this, that is here.0305

The negative sin from here, this thing the N sub X² π²/ A² that is here.0308

Now, I have this × this and this.0314

What I get is a sin of N sub X π/ A × X × sin of N sub Y π/ A.0317

BY × sin of N sub Z × π/ C × Z.0329

This is the first thing that I want.0342

Let me go ahead and circle it, that is the first thing that I want.0346

That is my first distribution.0351

I have done 1/3 of my operation so far.0353

It is going to be same, these functions are the same.0358

It is the exact same thing except this time with respect to Y and with respect to Z.0360

When I operate on ψ with the second partial with respect to Y and operate with respect to Z,0368

I end up with the following.0375

I'm just working very carefully.0379

Let me go to red.0382

I get D² DY² that is going to equal the same thing as before.0385

Everything is the same except now the variable is different.0391

You get -H/ ABC ^½ and this time you have N sub Y² π²/ B² and everything else is the same.0395

Sin of N sub X π/ A × X, you will get sin of N sub Y π / B × Y × sin N sub Z π/ C × Z.0409

The last one, we get the del²/ del Z² it looks exactly the same.0425

8/ ABC¹/2 except now we differentiate it with respect to Z.0433

We have N sub Z², let me make this a little bit more clear here.0441

N sub Z² × π/ C² and everything else is the same.0452

Sin of N sub X π/ A × sin of N sub Y π/ B.0459

There is an X here, there is a Y here.0468

Sin of N sub Z π/ C × Z.0472

You can see that it is really easy for a problem in quantum mechanics to go south on U for no other reason than for arithmetic or missing some letter.0479

It is going to make you crazy. It is going to make you want pull your hair out but that is the nature of the game.0488

We have our 3 partial derivatives.0493

We have operated on the wave function.0495

We have this and this and the previous one, so now we just add them up.0499

That was the last part of the operator, we have to add them.0504

When we add them up, adding these 3 expressions together, notice the sins are all the same.0507

The -8/ ABC¹/2 that is the same, you have the π² that is the same.0529

When you put all the terms together, what you end up with is the following.0533

Let me write out the operator here.0541

Del² of ψ N sub E= - 8/ ABC ^½ × we have π² that is going to be × N sub X²/ A² + N sub Y²/0548

B² + N sub Z²/ C² × this thing.0574

I have the sin of N sub X π/ A × X.0581

I should have made a little bit more room here.0588

Sin of N sub Y π/ B × Y × sin of N sub Z π/ A × Z.0590

This is my final solution, the first part of the problem, that is it right there.0608

I wanted to apply the del operator ⁺to the wave function or particle in a 3 dimensional box that it is right there.0616

On the second part I have to show that it is actually an Eigen function of the operator.0625

For a function to be an Eigen function of an operator.0632

Let me write this down.0636

For a function F, speak in general terms, for a function F to be an Eigen function of an operator A, remember operators have a little hat on top of them.0639

The following must hold, this is the definition.0664

The following must hold applying the operator to F gives me some constant × F.0671

In other words, when I apply an operator to a function, what I end up getting is just some constant × that function.0682

If that holds, then the function is an Eigen function of the operator.0690

In other words, an F is an Eigen function of the operator A and L happens to be the Eigen value associated with that particular Eigen function.0697

That is what is happening.0707

Notice what we have, our wave function.0710

I will do this in black here.0714

Our original wave function ψ, in other words, what we want is this.0719

We want Del² ψ to equal sum constant.0723

I will just use λ again.0730

To be some λ × ψ.0732

I want to get ψ back so I operated on it with the del² operator and I got this.0734

My function, my original ψ is this one right here.0743

It includes this, that, and this.0745

That right there is ψ.0749

I operated on it, I got this thing to actually to equal.0763

I can write it this way, that is my function.0768

I end up with del² of ψ = I'm left with - π² × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C² × ψ.0772

Thing is just the actual function ψ.0804

Sure enough, when I operated on it, I got this thing.0808

This thing is nothing more than a constant which is that × the original function.0812

Therefore, ψ my wave function for a particle in a 3 dimensional box is an Eigen function of the Laplacian operator.0819

This thing happens to be the associated Eigen value for the associated function.0830

Let me write that down.0837

Do I have an extra page here?0838

I do, so let me go ahead and use it.0841

Our ψ N sub X N sub Y N sub Z is an Eigen function of the Laplacian operator.0846

The corresponding Eigen value is - π² × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².0867

That is it so again, I have an operator and I operate on some function F.0897

If what I end up getting when I do that is just the constant × the function back, that means that F is an Eigen function of this operator.0902

That means it is very intimately connected and λ is the corresponding Eigen value for this particular Eigen function.0913

There are going to be many Eigen functions for a given operator.0920

It one of those Eigen functions has an Eigen value.0923

What you are doing is this, it is saying when you take the derivative, when you operate on a function you are just getting that function back.0926

All you are doing is multiplying it by some constant.0933

That is very extraordinary and we will say more about it as we go on in the course.0936

Let us go on to another problem here.0945

Example 2, what are the most likely positions of a particle in a 1 dimensional box of length A when it is in the N = 2 state?0950

What is the most likely position of a particle?0959

In other words, what position in a 1 dimensional box?0962

This is 0 and this is A, where along this interval is the highest probability that0967

I’m going to find the particle when the wave function is in the 2 state?0979

That is all the question is asking.0986

Let us go ahead and write down our equations.0989

We know that the equation for a 1 dimensional box ψ sub N = 2/ A.0991

I hope you do not mind.1002

I’m going to make one slight change here.1004

It is a capital here, I am accustomed to using the small a so I’m going to change this, of length A.1007

I’m just accustomed to writing the small a.1015

We have 2/ A ^½ × sin of N π/ A × X.1018

This are actual wave function for a particle in a 1 dimensional box.1029

We said that it is in the N = 2 state.1034

Ψ of 2 = 2/ A ^½ × sin of 2 π/ AX.1037

We just actually pick the value.1047

We are talking about probabilities here.1051

We said that ψ conjugate × ψ or in the case of a real function just ψ² is the probability density or in other words, the probability.1055

When we speak of probability density, we can go ahead and say probability.1081

We said that when you multiply the conjugate of ψ × ψ or in the case of a real function just ψ² is a probability density.1085

This is the function that we have to maximize.1095

This is the function we have to maximize.1098

Let me write a little bit better.1104

Ψ² is the function we must maximize because we wanted the most likely positions.1107

Once we have the function which we do, ψ² for the probability, we want to maximize the probability.1124

That is the function we have to maximize.1130

You remember from calculus, maximize means that the DDX of this ψ² function = 0.1131

We are going to form ψ², we are going to differentiate it with respect to X.1149

We are going to set it equal to 0 and we are going to find the X values that actually maximize it.1152

What you are going to get are some of that maximize and minimize it.1159

We are going to have to pick the ones that maximize it.1161

That is what we are doing.1162

This is a straight maximum problem from single variable calculus.1164

Let us go ahead and write down,1169

The ψ * × ψ which is the same as ψ² because we are dealing with a real function.1171

That is just going to equal this function × itself.1178

We get 2/ A × the sin² of 2 π/ A × X.1182

This is our ψ².1196

We need to go ahead and differentiate that function.1198

When we do DDX of our ψ², I will leave the 2/ A.1203

I’m going to differentiate this, this is going to be × 2 × sin of 2 π/ A × X.1213

Chain rule so I differentiate this, I get the cos of 2 π/ A × X and I differentiate what is inside.1228

It is going to be the 2 π/ A.1238

That is the derivative with respect to X of this function.1242

When I simplify it, let me go ahead and put some things together here.1246

I end up with 4 π/ A² × 2 × sin of 2 π/ A × X × cos of 2 π/ A × X.1253

I’m going to use the identity, 2 sin θ cos θ = sin of 2 θ cos θ.1270

I'm going to rewrite this using that identity and I end up with the following.1284

I end up with the DDX of our ψ² is going to equal 4 π/ A² × sin of 4 π/ A × X.1291

That is our derivative.1308

That is the derivative that we now need to set equal to 0.1310

That means that sin of these 4 π/ A × X = 0 because this is just a constant.1315

Now sin of angle = 0 whenever this 4 π/ X = 0 π 2 π 3 π 4 π 5 π 6 π 7 π.1325

Therefore, this is true.1342

Whenever 4 π/ A × X = some integer × π, where K = 0, 1, 2 and so one.1344

That is what makes this possible.1359

Rearrange for X, go ahead and cancel the π, and I get X = K × A/ 4.1360

For K= 0, 1, 2, 3, 4.1372

Try different values of K.1379

Let me go ahead and rewrite my function ψ² just so I have it.1385

My ψ² which we have actually maximized.1391

You have to go back to the original function not the derivative.1396

Ψ² = 2/ A × the sin² of 2 π/ A × X this is the function that we have maximized.1399

We differentiated it, set it equal to 0, these are the values.1411

When K =0, what you will end up with is.1416

Let me put K and 0 into here, and we put this value of X that we get into here.1424

What you will end up with is the following.1434

You end up with 2/ A × the sin² of 2 π/ A × K 0.1436

This is 0 that means X is 0 × 0 = 0.1447

Let us try K = 1.1454

When I do K= 2 in here, I get 2A = 4, X = 2A/ 4.1458

I put that into this X so I end up with 2/ A × the sin² of 2 π/ A × 2A/ 4.1470

The sin of π is 0, so again this = 0.1492

Let us try K=4.1497

4 is as far as we can go, I will tell you why in just a minute.1509

When we use K =4, we get 4A/ 4 that is X.1512

We put that into this and we end up with,1518

Again, we are doing the original function sin² of 2 π/ A × 4A / 4, that cancels.1521

A and A what you get is a sin of 2 π², sin² of 2 π.1534

The sin of 2 π is 0 so we end up with 0.1540

These are the minimum values.1544

4 is as far as we go for K and here is why.1549

When K = 4, 4/ 4 will give us A.1552

We got as far as we are going to go.1556

Again, we are going from 0 to A that is our box.1558

We cannot go farther than that.1564

We cannot go for example 2, 6A/ 4 which is 3/ 2.1566

That is out here.1571

We are sticking here is the case, from 0 to 4.1573

If K is equal 0, 2 and 4 give us minimum values that means that the 1 and the 3, when K = 1 and K = 3,1576

that is going to give us the maximum values.1584

When K = 1 and K = 3 give us the Max values of this function.1592

With K = 1, X = ¼ and when K = 3 X = 1 A/ 4.1614

It is A/ 4.1632

This is going to be 3A/ 4.1633

Therefore, let us do an extra page.1640

A/ 4 and 3A/ 4 are the most likely positions of finding a particle in a 1 dimensional box for the N = 2 state.1646

What is it that we have done?1681

We have the wave function, let me go back to black.1683

We have the wave function ψ.1690

We know the probability or the probability density.1692

In this particular case, because it is a real function, under normal circumstances is this one.1694

This, because it is a real function it is just ψ².1700

We form the ψ² and we differentiated it.1704

We took the derivative of it.1708

We set the derivative equal to 0.1709

We found the values of X that make that true and we took those values of X for different values of K 0, 1, 2, 3, 4.1714

We plug them back in the original equation to see which one give us a minimum value,1725

which in this case was 0 and which one give us a maximum value which was in this case, you end up with a sin =1.1729

That is what we did.1738

We found out that when K = 1 and K = 3, that gives us the maximum values.1739

Therefore, when we put that back into the X = KA/ 4 which gives us our max and min values,1746

we end up with A/ 4 and 3/ 4 as the most likely positions to find a particle.1755

I hope that makes sense.1762

Let us try our next example here.1766

For an electron in a 3 dimensional box which sides are 1 nm × 2 nm × 4 nm and X, Y, Z respectively,1772

X is 1, Y is 2, Z is 4, calculate the frequency of the radiation required to stimulate a transition from the state 123 to the state 234.1781

In other words, N sub X is 1, N sub Y is 2, N sub Z is 3.1793

For a particle in a box, we have 3 quantum numbers.1798

In this case, 123 state.1800

I want to stimulate it to the 234 state.1802

How much energy do we need?1808

Or what is the frequency of the radiation that I need?1810

Let us go ahead and see if we can work this one out.1813

Let us start off with our equations.1818

I know the energy N sub X, N sub Y, N sub Z, is equal to planks constant/ 8M × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².1820

The change in energy from the state is going to be the energy of the 234 state - the energy of the 123 state.1845

Energy of the 123 is the difference between them.1857

Let us find the energy of the 234 and the energy of the 123.1865

The energy of the 234 state, we just plug the values in.1868

That is fine, let us go ahead and do this way.1874

I will write down what these values are.1886

A²/ 8 × the mass of the electron ×,1888

Well, NX² NX is 2 so this is going to be 2²/ A is 1 nm.1894

1 × 10⁻⁹².1902

You want to work in meters, you want to work in kg, you have to watch the units.1907

Time is going to be in seconds, mass is going to be in kg.1912

Length is going to be in meters as energy is going to be in Joules.1916

+ N sub Y,1922

N sub Y is 3²/ B².1924

B is going to be 2 × 10⁻⁹² + 4²/ 4 × 10⁻⁹² because that is 4 nm long.1928

Planks constant H = 6.626 × 10⁻³⁴ J/ s and the mass of the electron, the rest mass of the electron is 9.109 × 10⁻³¹ kg.1944

When I put these values in to here and run the calculation on my calculator, I get an energy for state 234 = 4.37 × 10⁻¹⁹ J.1963

Pretty standard range.1979

Let us go ahead and calculate the energy of 123.1982

The energy of the 123 state is again, the same thing A²/ A × the mass of the electron.1985

This time, we have 1²/ 1 × 10⁻⁹² + 2²/ 2 × 10⁻⁹² + 3²/ 4 × 10⁻⁹².1993

And when I go ahead and calculate this, I end up with 1.54 × 10⁻¹⁹.2014

Again, I hope that you are going to be checking my arithmetic.2021

I am notorious for arithmetic mistakes.2024

Therefore, the change in energy is equal to, like we said the energy of the 234 state –2029

the energy of the 123 state is going to end up equaling 2.83 × 10⁻¹⁹ J.2036

We know that the change in energy = planks constant × the frequency of the radiation2047

which implies that the frequency of the radiation = the change in energy ÷ planks constant.2053

We get 2.83 × 10⁻¹⁹ J ÷ 6.626 × 10⁻³⁴ J/ s.2061

Joules cancel and we end up with a frequency of 4.27 × 10¹⁴ inverse seconds which is as you now is a Hertz.2074

There you go, nice and simple.2088

You just plug the values in, that is all you got to do.2091

Let us see, what do we got next?2097

For a 1 dimensional box of length A, calculate the probability of finding a particle between A/ 2 and 3A/ 4.2106

Let us do this one in blue.2117

We have 0 and we have A, they are saying calculate the probability of finding a particle between A/ 2 and 3A/ 4.2120

This is 3/4 of A.2134

What is the probability?2136

What are the chances of me actually finding a particle right there?2139

Let us do it.2143

This is a 1 dimensional box so let us start off with our 1 dimensional wave equation2144

which is equal to 2/ A ^½ power × sin of N π/ A × X, that is our equation.2150

We also know that the probability of finding a particle between, when X is between some left and right point,2160

that is our L, this is our R, the probability of that is equal to the integral from L to R of ψ × ψ DX.2173

This is the probability density when we integrate over the entire interval that we are interested in,2185

we end up getting the probability of finding a particle there.2192

This is the definition of probability.2196

You just take the complex conjugate × the function, or in this case it is a real function.2199

Ψ² and you integrate it, that is all you are doing.2203

We know what ψ² is, ψ² is just equal to 2/ A × sin² of N π/ A × X.2208

It is just this thing squared.2219

Our probability is actually equal to, in this particular case it is going to be the integral from A/ 2 to 3A/ 4 of 2/ A sin² N π/ A × X DX.2223

Again, you can use mathematical software for this.2245

You can use a table of integrals for this.2247

It just depends on what it is that I happen to be doing.2250

In this particular case, I’m going to go ahead and use a table of integrals to look up this integral,2252

to see what it is, put the values in so I did this manually.2257

There are going to be other times when you just want some good straight numerical value and you just plug it into your software.2260

It does not really matter.2264

From a table of integrals I found the following.2268

Let us go ahead and do this in red.2271

The integral of sin² B X DX = X/ 2 - sin of 2 BX/ 4B.2276

In this particular case, our B is this N π/ A.2293

I'm going to just plug in N π/ A whenever and wherever I have a B and I’m going to evaluate this.2301

We end up with the following.2309

We end up with this integral, this constant comes out of course.2311

I have the integral of sin² N π/ X DX and again this is going to be our B here.2321

This is what I’m going to use.2329

I end up with the following.2331

I end up with the probability of being equal to 2/ A, that came out.2333

And when I integrated this thing, it is going to be X/ 2 - sin 2 N π/ A × X ÷ 4 N π/A.2338

We are evaluating it from A/ 2 to 3 A/ 4.2360

When I put these values in, here is what I get.2368

I’m going to write all this out.2371

2/ A × 3A/ 4 /2.2374

I get 3A/ 8 - sin of 2 N π/ A × X which is 3 A/ 4 all of that / 4 N π/ A -, now we will do A/ 2 in here.2380

I get A/ 4 - sin of 2 N π/ A × A/ 2 all over 4 N π/ A.2410

I just put this in here and this in here.2431

I’m just evaluating the integral when I start combining things 3A/ 8 -2A/ 8, I get A/ 8.2433

Here I get some cancellations.2444

The A and A cancel, the 2 and 4 cancel here, the 2 and 2 cancel, the A and A cancel.2446

I end up with the following.2455

Let us see here.2457

This one, I end up with sin of N π.2458

The sin of N π, regardless of when N is going to be 0, this term actually goes to 0.2463

It drops out.2468

When I take 3 A/ 8 – 2 A/ 8, I get A/ 8.2470

What I’m left with here, 2 and 3 is 6.2475

Here is what I end up getting.2491

I end up with the following.2495

I end up with 2/ A × A/ 8 - A × sin of 3 N π/ 2 all over 4 N π.2497

When I multiply, when I distribute this I end up with the following.2515

I end up with the probability equaling.2519

The A cancels so I end up with ¼ - 1/ 2 N π × sin of 3 N π/ 2.2526

This is my probability.2542

Whatever N happens to be, I will put it in and this is going to be a probability that we are seeking.2544

This is the probability that I will find a particle between A/ 2 and 3 A/ 4.2551

Notice something really interesting here.2563

As N goes to infinity, 1, 2, 3, 10, 20, 30, 40, 50, 100,2566

The term 1/2 N π goes to 0.2574

The probability ends up equaling ¼.2589

Here is how you calculate the probability.2601

It is going to be ¼ - something, ¼ + something depends.2603

But as you take N higher and higher and higher, as quantum numbers become larger and larger and larger,2607

the probability just comes down to ¼.2614

In other words, this particular area accounts for ¼ of the interval.2621

Therefore, my chances of actually finding the particle there is ¼ or 25%.2627

This is classical behavior and here is the correspondence principle in action.2635

The probability is going to change based on what N is.2641

It is not going to be ¼.2644

For N even it is going to be 1/4 but for N odd it is going to be different values ¼ - something, ¼ + something.2646

But as N gets higher and higher and higher, this term goes to 0 because this term goes to 0,2652

which means that the probability is going to be fixed at ¼.2659

If I do a million experiments, I'm going to find 1/4 of the time that the particle is going to be2665

in this particular area because it accounts for 1/4 of the interval.2670

Again, the correspondence principle says that as quantum numbers become larger and larger and larger,2674

quantum mechanical results which is this thing, they end up approaching classical mechanical results which is just the ¼ in general.2679

Let us see our last problem here.2693

Let A = B = 2C for the sides of a 3 dimensional box.2702

What are the degeneracy of the first five energy levels of a particle in this box?2707

The first five energy levels are the lowest, the next highest, next highest, the next highest, different numbers.2713

Let A = B = 2C.2719

When we talk about degeneracy, we set A = B =C, a perfect cubed.2722

In this case, two of the sides are equal but one of the sides is not.2726

What is going to happen?2729

Are there going to be degeneracy?2730

Are there not going to be degeneracy?2732

Let us find out.2733

Let us go ahead and do this in blue.2736

Here we have A= B and we have A = 2C.2738

Therefore, C = A/ 2.2744

When I take C² I'm going to get A²/ 4.2750

The energy in a 3 dimensional box, the energy of N sub X, N sub Y, N sub Z = planks constant²/ 8 M ×2755

N sub X²/ A² + N sub Y² / B² + N sub Z²/ C².2768

But in this particular case, A = B and C happens to equal A/ 2.2780

I can plug in, I can fix this so the energy actually ends up becoming H²/ 8M × N sub X²/ A² +2786

N sub Y²/ A² because B = A + 4 N sub Z²/ A²,2804

Because C² = A²/ 4.2817

When I flip it, the 4 comes on top and I’m left with A².2821

Now A² is everywhere in the denominator, I can pull it out as a constant.2824

I'm left with H²/ 8 M A² × N sub X² + N sub Y² + N sub Z².2829

This right here, this is a constant.2843

This is what is going to change.2846

N is going to change, it is going to vary independently.2848

111, 121, 131, 657 whatever.2850

This is what we are going to analyze right here.2854

This term is what we will analyze and which will give us the different values of the energy, what we will analyze.2859

Let us go ahead and start.2871

I think I would actually do this on the next page.2873

Let me rewrite the expression.2878

Our energy = H²/ 8 × M × A² × N sub X² + N sub Y² + 4 N sub Z².2881

We will go ahead and take the 111 state.2900

For 111, I’m just going to plug the 1 here, the 1 here, the 1 here, I end up with,2903

1² + 1² + 4 × 1² that is going to equal 1 + 1 + 4 that is going to equal 6.2913

6 × H²/ 8 MA² that is the energy of the 111 state.2924

I hope that makes sense.2932

I’m going to write it this way.2937

I will put E of 111 is equal to, this is what we are analyzing.2943

It is going to be 1 + 1 + 4 = 6 that is level 1.2950

We are going to try different values in here.2959

We are going to try 121, 211, 212, and you end up with different numbers.2960

You can try it at any random order but you are going to arrange the energy levels increasingly.2966

Here is what you come up with when you do this.2971

The energy of the 211 state, that is going to be 4 + 1 + 4 that is going to equal to 9.2974

For this particular state, the energy is going to be 9 H²/ 8 MA².2989

This is the constant so I’m just working out the numbers in the parentheses.2998

Now the energy of the 121 state that equals, if I put the 121 into this expression, I end up with 1 + 4 + 4 = 9.3002

Sure enough, it ends up having the same energy.3018

Let us try 131, energy of 131.3022

It is going to be 1 + 9 + 4 is going to equal 14.3025

Let us try the energy of the 311 state.3035

That is going to be 9 + 1 + 4.3040

It is going to equal 14 so we see this.3044

When we do the energy of the 321 state, I will put the 321 into here and I get 3² is 9, 2² is 4 + 4, that is going to equal 17.3053

The energy of the 231 state is going to be 4 + 9 + 4 so I end up with 17.3073

Let us try the 112 state, energy of 112 = 1 + 1 + 16 = 18.3087

This is level 1, it has 1 degeneracy.3104

This is level 2, it is level 2 because it is the next highest energy based on whatever N can be.3109

It is the next highest but it is the lowest of the bunch.3116

I can put in any N I want, I’m going to get a bunch of numbers.3120

I'm going to end up with a bunch of energies.3124

It is going to be 6 × this thing, 9 × this thing, 9 × this thing, 14 × this thing.3127

What I'm looking for are the number of degeneracy for that particular energy level.3131

For the energy level of 9 H²/ 8 MA², 2 states have that same energy.3136

The 211 state, the 121 state.3143

This particular energy level is 2 fold degenerate.3146

It has a degeneracy of order 2.3150

That is what I'm doing.3151

The question asked, find the degeneracy of the first five energy levels for a particle in a 3 dimensional box.3153

Here is level 1, here is level 2, here is level 3, I see it has 2 degeneracy.3163

Here is level 4, it has 2 degeneracies.3171

Here is level 5, it has no degeneracy.3174

It is just one thing.3178

That is what we have done.3180

The degeneracy of order 1, its next level up has degeneracy of order 2, the third level is 2 fold degenerate,3182

the fourth level is 2 fold degenerate, and the fifth level is not degenerate at all.3190

It has 1 energy level.3197

It might be, I continue on to see, that is the whole point.3200

I have to try each different state.3203

I hope that actually makes sense.3207

We calculated the formula for the energy and we just tried different values of N by varying the NX, NY, NZ separately3210

to end up with some increasing energies of the particles of different states.3219

These are 2 different states, 2 entirely different wave functions that they have the same energy 9 H²/ 8 MA².3225

For this particular setup, where A = B = 2C, the 131 and the 311 state also happen to have the same energy.3233

The 321 and 231 state happen to have the same energy.3241

That is all we are doing here, that is all we have done.3245

I hope that makes sense.3248

Do not worry about it, this is just the first set of example problems.3251

We are going to be doing a lot of example problems here.3254

I’m going to be doing them in the next several lessons.3257

In the meantime, thank you so much for joining us here at www.educator.com.3260

We will see you next time, bye.3264