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

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

### 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

### Physical Chemistry Online Course

### 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 0*0351

*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

0 answers

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?