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 VI

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
- The Hydrogen Atom VI 0:07
- Last Lesson Review
- Spherical Component
- Normalization Condition
- Complete 1s Orbital Wave Function 4:08
- 1s Orbital Wave Function
- Normalization Condition
- Spherically Symmetric
- Average Value
- Example I: Calculate the Region of Highest Probability for Finding the Electron 21:19
- 2s Orbital Wave Function 25:32
- 2s Orbital Wave Function
- Average Value
- General Formula

### Physical Chemistry Online Course

### Transcription: The Hydrogen Atom VI

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

*Today, we are going to continue our discussion of the hydrogen atom.*0004

*Let us go ahead and jump right on in.*0007

*Let me go ahead and do black today.*0009

*In the last lesson, we looked at the radial function R sub N L of R.*0012

*In particular, we look at the one that is orbital.*0033

*In particular, we looked at R 1, 0 of R which was 2/ A sub 0³/2 E ⁻R/ 2 α sub 0.*0040

*This is only the radial component of the wave function ψ.*0065

*There is also a spherical component or the angular component, the S 0, 0 of θ and φ.*0069

*S of 0, 0 happens to be 1/ radical 4 π.*0102

*Nice and simple, notice θ and φ going to show up.*0116

*This is just a constant function.*0119

*The normalization condition for the spherical component is as follows.*0123

*The normalization condition θ runs from 0 to π, φ runs from 0 to 2 π.*0130

*S00 conjugate × S00 and sin θ D φ D θ.*0143

*Once again, do not forget this because you are integrating over the surface of the sphere.*0156

*You are integrating spherical coordinates so you have to include this factor.*0164

*When you include R, the fact it was R².*0168

*The entire triple integral, this part becomes R² sin θ DR D φ D θ.*0172

*Do not forget this factor.*0182

*Again, integrating with respect to spherical coordinates.*0188

*This is just this, we end up with the normalization condition being 1/ 4 π, the integral from 0 to π,*0193

*the integral from 0 to 2 π sin θ D φ D θ.*0210

*Φ comes first because we are going φ first the integral and we will go to the outer integral.*0219

*There is nothing else going on here, that is just the standard normalization condition for the angular component.*0225

*This is a function of two variables so we end up having a double integral.*0232

*The normalization condition is always the integral of the conjugate of the function × the function itself.*0237

*The complete wave orbital, so the complete 1S orbital wave function is, this is ψ 1, 0, 0.*0248

*It is a function of R θ and φ, it is equal to the R1, 0 of R × the S0, 0 of θ and φ.*0272

*When I put this together with this and simplify, I end up with 1/ π A sub 0³¹/2 E ⁻R/ α sub 0.*0284

*We had the radial component S0, 0 happen to be a constant function so the entire wave function,*0319

*now that we have put together it turns out also to be only a function of R.*0326

*In the case of the S orbital, the 1S, 2S, 3S, 4S, it is always going to be this way.*0331

*All of the S orbital are only going to be functions of R only because the S sub 00 term is a constant term.*0338

*What this means is when you have three variables and it is only a function of one of the variables, *0350

*in this case the θ and φ, they sweep out a sphere that means the 1S orbital is spherical symmetric.*0356

*That is why we represent them as globes, bigger 3S, 4S, they are spheres.*0362

*A spherical symmetric, that is what this means.*0368

*This is where it comes from.*0370

*Let me go back to black here.*0375

*The normalization condition for ψ the entire wave function is the integral from 0 to infinity, *0396

*the integral from 0 to π, the integral from 0 to 2 π of ψ 1, 0, 0 conjugate × ψ 1, 0, 0.*0414

*Then, R² sin θ the inner integral is D φ D θ to the final integral is DR.*0426

*This is the normalization condition, that is all we are doing here.*0436

*We just wrote the following.*0464

*We just wrote the normalization condition as the integral from 0 to infinity, the integral from 0 to π, *0467

*the integral from 0 to 2 π ψ 1, 0,0* ψ 1, 0, 0 R² sin θ D φ D θ DR.*0478

*If it makes more sense to you, you can actually separate variables.*0492

*This is a triple integral, just integrating one at a time.*0495

*Working from the inside out, the inside integral is the D φ.*0498

*The outer integral with respect to θ and the next integral is with respect to R.*0503

*If it is more clear, you can separate the variables and write it like this.*0508

*You can write as the integral from 0 to infinity the R² DR, the integral from 0 to π sin θ D θ, *0512

*the integral from 0 to 2 π ψ 1, 0,0 ψ 1,0, 0 D φ.*0524

*In this particular case, you are just separating the variables.*0541

*You are setting up this integrand.*0544

*You are integrating with respect to φ first and then what you get is going to be your integrand including this *0546

*and you can integrate with respect to θ next.*0553

*What you get, you are going to put here integrand also including this and we are going to integrate with respect to R.*0556

*If this makes sense to you, more than this make sense to you, you are just working from right to left.*0564

*Here, you are working from inside out.*0569

*If you will put integrals before, back into calculus or multi variable calculus,*0571

*you were probably accustomed to seeing this one.*0576

*It does not really matter which one you use.*0579

*Since, ψ 1, 0, 0 = 1/ π A sub 0³ ^½ E ⁻R/ A sub 0 is a function only of R.*0585

*It is spherically symmetric sphere in space.*0615

*The electron can be found within that sphere but somewhere within that sphere.*0634

*Let us go ahead and work on some probabilities.*0640

*We have the integral from 0 to A, the integral from 0 to π, *0643

*the integral from 0 to 2 π of ψ 1, 0, 0 conjugate ψ 1, 0, 0 R² sin θ D φ D θ DR.*0651

*The ψ conjugate × the ψ 1, 0, 0, I have the function which is right there.*0668

*I’m just going to multiply it by itself and I end up with 1/ π A sub 0³ E⁻² R/ Α sub 0.*0678

*Let me go ahead and just call this Z so I do not have to keep writing it over and over again.*0693

*I’m just going to call this one big fat Z.*0697

*We are going to take this and we are just going to solve this when I put this Z into here and*0701

*we are actually going to integrate this one integral at a time, to see what we get when we do that.*0714

*Let us go ahead and we are going to integrate with respect to φ first.*0721

*This integral right here is going to end up being, or if you want this one.*0725

*We would do that one first.*0737

*We are actually doing the integration with respect to φ.*0739

*It is going to be 0 to 2 π of Z, whatever this thing is.*0748

*D φ, that is going to end up being 2 π × Z when we do that integral.*0754

*We are going to integrate with respect to θ next.*0762

*We are going to put what we get, the 2 π Z into here and we are going to solve this integral with respect to θ, the 2 π Z.*0765

*Z does not contain any θ so I could just pull it out.*0780

*0 to π sin θ D θ is going to give me 2 π Z × - cos θ from 0 to π.*0786

*When I end up solving this part, I'm going to get 4 π Z.*0804

*I got 4 π Z and that is going to equal, I have Z right here which is that thing.*0814

*Therefore, I will go ahead and write these again.*0830

*I will stick with red.*0836

*4 π Z is actually equal to 4 π and we said the Z is equal to 1/ π A sub 0³ E ⁻2R/ α sub 0.*0842

*The π cancel leaving me with 4/ A sub 0³ E ⁻2R/ Α sub 0.*0859

*The final integral gives us the probability of finding the electron within A of the radius.*0877

*The final integral is the integral from 0 to A of 4/ π A sub 0³ E ⁻2R/ α sub 0 R² DR.*0887

*This right here, remember this represents the probability that the electron is between R and DR.*0913

*When I integrate that probability from 0 to A, it gives me the probability that the electron is within A of the nucleus.*0936

*This expression, we have already seen it.*0948

*We saw it in the last lesson, it is the same expression for just the radial function.*0954

*Let me go back to black here.*0961

*We saw this already in the last lesson when dealing with only the R1, 0 of R.*0964

*It turned out to be the same probability for the entire wave function ψ 1, 0, 0 *0994

*precisely because θ and φ are not involved in the 1S orbital.*1022

*Because θ and φ are not part of the function, they are not part of ψ 1, 0, 0.*1030

*In other words, the S orbital are spherically symmetric.*1042

*Let us talk a little bit about average values.*1059

*Let me go ahead and put the line of demarcation here.*1065

*Let us move here, what about average values?*1071

*Recall that the average value for A is the integral of ψ conjugate × A × ψ × DV.*1084

*The average value R, the average value of where we are going to find the radius *1102

*is going to equal the integral from 0 to infinity, the integral from 0 to π.*1108

*This integral right here is just the radicals symbol for integral.*1115

*Since, we are dealing with function of three variables, we would have a triple integral.*1118

*0 to π, 0 to 2 π of ψ 1, 0, 0 conjugate × R × ψ 1, 0, 0 × R² sin θ D φ D θ DR.*1125

*When we actually end up taking the ψ × R × ψ conjugate is just multiplication.*1144

*It is ψ conjugate × ψ which we already know we will just multiply it by R.*1151

*We end up with the integral 0 to infinity when we take care of all of this, we get 4/ π A sub 0³ E ⁻2R/ Α sub 0 R,*1156

*which is this R, R² DR, which is going to end up equaling 4 / π A sub 0³.*1179

*The integral from 0 to infinity of E ⁻2R/ Α sub 0 R³ DR.*1193

*When we actually solve this integral, we get 3/2 A sub 0.*1204

*For this particular orbital, on average if I take a bunch of measurements of where the electron is going to be,*1209

*I'm going to find it is actually ½ bohr radius away from the nucleus.*1218

*Remember, we had a distribution that looks like this for the 1S orbital, something like that.*1225

*It is going to be between here and here, on average.*1231

*We are going to find it in a bunch of different places.*1233

*There is a place of highest probability, that on average for 100 measurement, 1000 measurement, *1236

*a million measurements, it is going to end up being ½ bohr radius is away from the nucleus.*1243

*That is all these means.*1247

*It is just an integral, it is just the function, the conjugate of the function × whatever it is that you are looking for.*1249

*In this case, we are looking for R radius how far away × the function and then just solve the integral *1257

*or let use mass software solve the integral.*1264

*Let us go ahead and look at an example here.*1269

*For the 1S orbital, calculate the region of the highest probability for finding the electron.*1282

*Highest probability means we want to maximize the probability.*1287

*We are looking for the peak, we are looking for the X value where the peak occurs.*1298

*The probability, the 1S orbital was founded already, it is 1/ π A sub 0³.*1306

*I will go ahead and put the R² E ⁻2R/ α sub 0 DR.*1322

*Remember, this part is the probability density.*1332

*The probability is the probability density × the differential element.*1339

*What we need to do is we need to maximize the probability density.*1345

*Remember the graph that we were looking at, the 1S orbital.*1349

*Probability density was on the Y axis, R was on this axis.*1353

*The probability density was on this axis.*1357

*We got something like that, we want to find that R value.*1360

*This is a probability density vs. R, that was the graph was.*1368

*In case we need to maximize that function.*1371

*We maximize something by doing exactly what you did in calculus.*1376

*We take the derivative, we set it equal to 0.*1378

*DDR of this function 4/ π A sub 0³ R² E ⁻2R/ α sub 0.*1382

*When you take the derivative of this function, you get 4/ π A sub 0³.*1397

*This is the product rule so it is going to be this × the derivative of that + that × the derivative of this.*1404

*We are going to get R² × E ⁻2R/ α sub 0 × -2/ α sub 0 + 2R × E ⁻2R/ Α sub 0.*1410

*This is going to equal 4/ π × A 0³ × E ⁻2R/ α sub 0 × - 2R²/ α sub 0 + 2R.*1432

*That is the derivative, we are going to set the derivative equal to 0.*1455

*This constant goes away, this E ⁻2R/ α sub 0 is never 0.*1459

*Therefore, this is the equation that we have to solve.*1464

*We get -2R²/ Α sub 0 + 2R.*1469

*We set it equal to 0, we end up with -2R × R/ Α sub 0 -1 is equal to 0.*1476

*We end up with R equal to 0 or we end up with R equal to α sub 0.*1489

*0 of course does not matter, that is out of the nucleus.*1499

*For the 1S orbital, calculate the region of highest probability for finding the electron.*1503

*The place we are most likely going to find electronic is 1 bohr radius away from the nucleus.*1509

*That all that is going on here.*1516

*We maximize the probability density because we were looking for the maximum region of highest probability.*1518

*Let us look at the 2S orbital.*1531

*We are looking at the 1S orbital, and the one that is complicated is the 2S orbital.*1543

*In other words, we are going to be looking for ψ 2, 0, 0.*1549

*Ψ 2, 0, 0 is equal to R 2, 0 × S 0, 0.*1555

*This S 0, 0 term shows up again and it is going to show up to all of the S orbital because the 3S orbital is going to be ψ 3, 0, 0.*1562

*4S orbital is going to be ψ 4, 0, 0, the S, 00 that term always shows up in the S orbital.*1574

*The S 0,0 term is part of the wave function.*1590

*In fact, for every S orbital the spherical harmonic will be S00 precisely because L = 0 and M = 0.*1606

*That is the S orbital.*1644

*The only thing that is going to change is the primary quantum number, 1, 2, 3, 4, 5, that is all that changes.*1647

*For every S orbital, the only variable for the wave function ψ is going to be R.*1659

*The total wave function, total hydrogen wave function for all S orbitals are only going to the functions of R.*1678

*They are only going to depend on the radius.*1683

*They are spherically symmetric.*1685

*Every S orbital is spherically symmetric.*1691

*For ψ 2, 0, 0 we end up with v1/ 32 π × 1/α sub 0³/2 × 2 - R/α sub 0 × E ⁻R/ 2 Α sub 0.*1705

*The average value of R for 2, 0, 0 is equal to exactly what you think.*1734

*It is the integral of the ψ conjugate of 2, 0, 0 × R × the ψ 2, 0, 0.*1751

*We solve that integral.*1760

*Ψ conjugate × ψ itself is just this thing multiplied by itself because this is a real function.*1765

*The conjugate is just ψ itself.*1772

*This becomes ψ² × R.*1775

*This thing × R.*1778

*We have ψ 2, 0, 0 conjugate × ψ 2, 0, 0 is equal to 1/ 32 π × 2 - R/ A sub 0² × R × E ⁻R/ α sub 0.*1782

*The 2’s cancel.*1807

*We are solving a triple integral.*1811

*We still have to solve a triple integral.*1814

*Even though it is a function of R, we still have to do the entire integral.*1817

*We already know that the integral from 0 to π, the integral from 0 to 2 π of the sin θ D φ D θ is equal to 4 π.*1821

*The inner integral, the angular part is going to equal 4 π.*1834

*We will just throw that in.*1838

*What we get is the average value of R for 2, 0, 0 is going to equal 4 π × this thing / 32 π A sub 0³.*1840

*Our final integral is the integration with respect to R.*1859

*I have done the triple integral, I just know that in this particular case because θ and φ are not involved in this function,*1863

*I will just go ahead and use this value in place of the inside two integrals.*1871

*It is going to be that 2 – R/ α sub 0² E ⁻R/ α sub 0.*1879

*Again, we have our R² DR, do not forget that term.*1891

*When I actually put this in a mathematical software, I'm going to get 6 Α sub 0.*1894

*For the 2S orbital, on average you will find the electron 6 bohr radius away from the nucleus.*1901

*Remember, for the 1S orbital it was ½ × the bohr radius so it was going to be ½.*1909

*For the 2S orbital, it is going to be 6 bohr radius away.*1921

*On average, it is going to be farther from the nucleus.*1926

*3S on average is going to be farther from the nucleus which is why *1928

*the pictures in your book represent the 1S as a small sphere, the 2S is a bigger sphere, the 3S is a bigger sphere.*1932

*On average, the electron is going to be further away.*1938

*That is why it looks the way that it does.*1942

*There is a general formula.*1946

*The average value of R for whatever S orbital 1S, 2S, 3S, 4S, 5S, is equal to 3/2 A sub 0 N².*1955

*Remember, the picture that we saw in the last lesson with the S orbitals, the 2S orbital has 1 node.*1971

*That means that if I take a sphere and I split it in half, the 1S orbital did not have a node,*1980

*which means that the electron can be anywhere in that sphere.*1993

*The 2S orbital had 1 node that means the electrons you are going to be in this region.*1997

*It is not going to be here or in this region.*2009

*I sliced it in half, you are looking at half a sphere that way.*2011

*The 3S orbital has 2 nodes.*2019

*There is going to be a place where the electron is not going to be.*2023

*There is going to be another place where the electron is not going to be.*2025

*The electron is either going to be this region or in this region or in this region.*2030

*That is what this means.*2037

*Do not forget that there are nodes.*2037

*When we talk about spherical symmetry, it does not mean that they can be anywhere in the sphere.*2039

*That only possible for the 1S orbital because it has no nodes.*2044

*But within the sphere itself, the 2S orbital has 1 node.*2047

*That means there is 1 sphere within that bigger sphere where you will never find the electron.*2051

*The 3S orbital has 2 nodes but if this is the main sphere, there are going to be 2 smaller spheres where you will not find the electron.*2056

*Again, pictorial representations of these electrons orbital are not very good because they are not giving you all the information.*2065

*Thank you for joining us here at www.educator.com.*2078

*We will see you next time, bye. *2079

1 answer

Last reply by: Professor Hovasapian

Thu Nov 6, 2014 1:24 AM

Post by xlr z on November 2, 2014

Hello Raffi,

sorry it is not related to this lecture but I did not see anything on Molecular structure such as wave function of H2, H2+ or H2- molecule or anything related to born oppenheimer approximation which is usually included in P-chem course. I was wondering if you will be making new lecture videos in the future or if you have notes on it and can post it online?

thank you.