For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### 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

### Physical Chemistry Online Course

### 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 simply *0060

*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 Laplacian operator, del² was this thing.*0110

*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 that *0967

*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 radiation *2047

*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 equation *2144

*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 be *2665

*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 separately *3210

*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

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