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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Wed Mar 25, 2015 4:40 PM

Post by matt kruk on March 25, 2015

hi professor,

so the reason all of the solutions to these wave equations are series solutions because they are a probability distribution since there is no definite means that the atom/wave is anywhere but a probability problem correct?

1 answer

Last reply by: Professor Hovasapian
Sun Nov 2, 2014 3:31 AM

Post by xlr z on November 1, 2014

when calculating for the radial part of the wave function for n=2, l=1, you multiplying bohr radius in the denominator by 2, could you explain why because at 5:10 you wrote the general form of the function and bohr radius was not multiply by n.
so is it (2/a)^l+3/2) or (2/na)^(l+3/2)?

Thank you

The Hydrogen Atom V: Where We Are

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 V: Where We Are 0:13
    • Review
    • Let's Write Out ψ₂₁₁
    • Angular Momentum of the Electron
    • Representation of the Wave Function
    • Radial Component
    • Example: 1s Orbital
    • Probability for Radial Function
    • 1s Orbital: Plotting Probability Densities vs. r
    • 2s Orbital: Plotting Probability Densities vs. r
    • 3s Orbital: Plotting Probability Densities vs. r
    • 4s Orbital: Plotting Probability Densities vs. r
    • 2p Orbital: Plotting Probability Densities vs. r
    • 3p Orbital: Plotting Probability Densities vs. r
    • 4p Orbital: Plotting Probability Densities vs. r
    • 3d Orbital: Plotting Probability Densities vs. r
    • 4d Orbital: Plotting Probability Densities vs. r
  • Example I: Probability of Finding an Electron in the 2s Orbital of the Hydrogen 45:40

Transcription: The Hydrogen Atom V: Where We Are

Hello, welcome back to, and welcome back to Physical Chemistry.0000

today, we are going to continue our discussion of the hydrogen atom.0003

we are going to stop and just take a quick look at where we are, what we have done, just so we do not lose our way.0006

We have done a lot of quantum mechanics in the last four lessons.0016

I want to take this lesson that just to stop, take a look at what is that we have done because there is a lot of mathematics, a lot of symbolism, where we are, what we had at our disposal.0019

just so we do not lose sight of the big picture.0029

That is the biggest problem with quantum mechanics is there so much mathematics going on.0031

There is so much symbolism going on that you tend to lose your way.0035

Ultimately, we are concerned with the wave function.0038

once we have that wave function, we do certain things to it.0042

we operate on that wave function.0045

as pretty much what quantum mechanics comes down to.0047

you find the wave function and then you operate on that wave function, by means of a bunch of integrals.0050

let us see where we are.0056

Let me go ahead and work with blue here.0059

we wanted to find this ψ of R θ φ, the wave function for the electron in the hydrogen atom.0065

just going back over what we did in the last 4 lessons.0102

we wrote the ψ function of R θ φ, the three variables we wrote it as a product function.0105

as a product of a function just of R called the radial component and as a product of the R the radial component and S which is a function of the θ and φ is called the angular component.0112

we also wrote and we broke this up into a function of 2 variables.0126

we wrote S of θ φ as T of θ and F of φ.0139

We just broke up into a bunch of functions.0148

we found that ψ depends on the three quantum numbers.0152

for example, we might have ψ of N LM, we express that as R and L as a function of R.0171

I'm going to have to stop saying that, I hope by now that we realize that R is just a function of R, the radius and the S is a spherical harmonic is a function of the θ and the φ.0183

R and L, and we have S LM, this is the symbolism for it.0197

N ran from 1, 2, 3, and so on.0204

L depends on N and it ran from 0, 1, 2, all the way to N -1.0209

M depends on L, and it was 0, + or -1, + or – 2, all the way to + or – L.0218

the expressions for the R and S were as follows.0237

this is what we found.0241

we found that R and L, I’m going to go ahead and put the R = -, we have got N – L -1! ÷ 2N × N + L!.0243

this one is going to be cubed and this one is going to be raised to ½.0273

here we have 2/ A sub 0 ⁺L + 3/2.0278

It is going to be multiplied by R ⁺L.0284

there is going to be an exponential term E ⁻R/ N A sub 0 and × this L + L 2L + 1.0287

its argument is 2R/ N × A sub 0, where this L N + L, 2L + 1, 2R/ N A sub 0 are called the associated Laguerre polynomials.0300

this what we be found when we solved one of those differential equations that we found.0333

the partial differential equations that we found for the hydrogen atom.0337

this was one expression.0340

this was the expression for R.0342

this is what we found right here.0345

it looks complicated, it is not when you actually put in the different values of N, L, and M.0348

it actually condenses and becomes reasonably straightforward to deal with.0353

we will see that in just a minute.0357

Do not let the general form intimidate you.0359

it is just a bunch of numbers that you are putting in.0362

that is the R part that we found.0366

Let me go back to blue here.0368

we found the S L sub M which is a function of θ and φ.0372

we found that to be equal to 2L + 1 × L - the absolute value of M!/ 4 π × L + the absolute value of M! ^½.0378

P sub L absolute value of M, its argument is cos θ and its exponential E ⁺IM φ, where this P sub L superscript absolute value of M cos θ, these are the associated Legendre functions or Legendre polynomials.0404

let us go ahead and write out, stop, and just write out the wave function for a particular collection of N, L, and M.0444

let us write ψ 2, 1, 1.0452

in this case, N is the first quantum number.0464

N is 2, L is 1, and M = 1, this particular wave function.0467

let us see what we have got.0477

It is going to get a little bit messy but again it is just a bunch of numbers that is in there.0478

do not worry about it, we will condense it into something really nice.0481

let us go ahead and do this in red.0486

We are going to do R of 2, 1, R sub NL.0491

R of 2, 1 which is a function of R is going to be - 2 -1 -1!/ 2 × 2 × 2 + 1 !³ ^½.0495

we have got 2/ 2 A0¹ + 3/2 R¹ E ⁻R 2 α sub 0.0522

I’m just putting in the N and the L, that is all I'm doing.0537

this is going to be L of 2 + 1 and this is going to be 2 × 1 + 1.0543

Write N + L, 2 + 1, 2L + 1, 2 × 1 + 1, this is going to be 2R/ N A sub 0/ N is 2 A sub 0.0551

this part right here, when we look up the particular Laguerre polynomial for this one, it just ends up being -3!.0569

Always look it up, it ends up being -3!.0579

that means that we get R of 2, 1 is equal to -3!.0585

That is a minus so it actually becomes positive.0597

Then, we put some things together, we get 1/ 4 × 3!³, this is going to be the ½ power.0601

I’m hoping that I’m not forgetting anything here.0611

It is a little crazy, tedious, just keeping track of every little number that is really the biggest problem of quantum mechanics I find.0614

3/ 2, let me make that a little clearer.0622

this is 1/ A sub 0, this is going to be to the 5/2 RE ^- R/ 2 × A sub 0.0624

there we go, we have the radial portion of the wave function for the 2, 1, 1 state.0638

Let us go ahead and do the spherical portion.0647

we have S1, 1 θ φ that is going to equal L is 1, M is 1, N is 2.0652

we have 2 × 1 + 1/ 4 π × 1 -1 !.0664

0! is 1, 1 + 1!, all of that raised to the ½ power × 1, 1, 1 of cos θ E ⁺I φ.0676

in this particular case, P1, 1 when we look it is equal to 1 – X² ^½, this is the argument for the X so it becomes 1 - cos² θ¹/2.0696

1 - cos² that, remember your Pythagorean identity from trigonometry, it is equal to sin².0716

and √ sin² is equal to sin θ.0721

P1, 1 is just equal to sin θ.0726

Therefore, our S1, 1 of θ φ ends up equaling 3 × 1/ 4 π 2! ^½.0730

there are different ways that you simplify these numbers.0750

I’m just putting in so we actually see everything.0753

sin of θ E ⁺I φ, there you go, that is your spherical harmonic.0756

that is the angular component of the hydrogen atomic wave function.0764

when we put them together, therefore, we have 4 ψ of 2, 1, 1 which is equal to R of 2, 1 × S1, 1.0768

we end up with ψ2, 1, 1 is equal to 3! × 1/ 4 π × 3!³ ^½ × 1/ A sub 0⁵/2 × 3/ 4 π × 2!.0784

This is not π, this is 4 × 3! because it was from the radial component.0821

You see that is very easy to lose your way here.0835

R of E - R/ 2 Α sub .0839

Let me make the minus sign a little bit more clear.0845

× sin θ × E ⁺I φ.0850

that is a version of the ψ 2, 1, 1.0859

just plugging in the numbers, plugging in the N, the L, and the M, to actually get the particular wave function that we are talking about.0862

we know that we worked out the wave functions but we also know some things about the angular momentum.0872

it is very important.0879

Let me go back to blue here.0881

we also know some things about the angular momentum of the electron, we know that the magnitude of the total angular momentum is equal to H ̅ × √ L × L + 1.0883

you notice the magnitude of the angular momentum depends only on the quantum number L.0930

the Z component of the angular momentum that is equal to M × H ̅.0936

we call N, the principle quantum number.0948

we call L, the angular momentum quantum number.0955

we call M, the magnetic quantum number.0965

The last bit of information that we have, I’m going to go ahead and write this over here.0973

the particular energy which depends only on N, as long as it is not a magnetic field is going to be E²/ 8 × π × the permittivity constant × the Böhr radius × N².0979

this is the particular energy associated with a given orbital and it depends only on N, provided the electron, the atom is not in a magnetic field.1000

I should go ahead and that is not a problem.1018

when the wave functions are listed in books, you are definitely going to find several lists of these hydrogen wave functions in your book in the chapter on the hydrogen atom.1024

when the wave functions are listed in books, they are simplified to various degrees.1044

what I mean by that is, they are simplified to various degrees.1056

and what I mean is that different authors and they tend to have combined certain things into one value but say they take the R/ α sub 0 and they call it Z, or they call it B, or sigma, or something.1062

they simplify it to various degrees, in order for it not to look so complicated.1080

when you see them in your book, they might actually look different than what it is that I'm writing here.1088

that is just simply because different authors write them in different ways to simplify them to various degrees.1093

Do not worry about it, it is the same wave function.1097

The wave functions that are listed in books, they are simplified to various degrees.1101

do not be concerned that what I have written or will write differs from what you see in your book.1107

they are the same wave function.1154

They are just written differently with different parameters that combine different things.1156

Let me give you an example of a couple of these representations that you might see.1172

Let me do this in red.1175

one representation, I will just write a couple of the wave functions.1178

I will write too many of them.1183

one such representation is as follows.1185

you might see this in your book.1191

you might see where N is equal to 1, where L is equal to 0, and M is equal to 0.1194

you will see ψ 1, 0, 0, written as 1/ radical π × Z/ A sub 0³/2 E ⁻B.1201

in this particular case, Z is the atomic number and the reason we put the atomic number here, I know we found them for the hydrogen atom but we actually assume that these orbitals are also valid for all the other atoms of the periodic table.1223

you just plug in Z, Z is the atomic number, that is all it is.1242

B is that factor so B is going to be equal to Z × R/ Α sub 0.1247

Or the α sub 0 is the bohr radius and the numerical value Α sub 0 is 5.292 × 10⁻¹¹ N.1257

of course, in the case of the hydrogen Z is just 1, it is the atomic number.1276

For the wave function, for N is equal to 2, L is equal to 0, M is equal to 0, you will see this one, you might see this one in your book.1282

Ψ 2, 0, 0, is equal to 1/ 32 π under the radical Z/ Α sub 0³/2 and it is going to be 2 - B E ⁻B/ 2, or again, B is that.1293

We are just writing it in a way that you can use it, that is all it is.1323

Let us go ahead and do one more for good measure.1328

for N = 2, for the L = 1, and M = 0, this is going to be ψ 2, 1, 0, that is equal to 1/ 32 π Z/ α sub 0³/2.1332

this time there is going to be a B × E ⁻B/ 2 and is going to be multiplied by cos θ.1352

You might see this one, this one, this one in your book or you might see it differently.1360

maybe they chose a different variable for B.1370

maybe they did not put the Z in there at all.1373

maybe they have a different constant.1375

Maybe they combine the constants in different ways.1377

I just want to let you know what you see in your book is not necessary what you see here, but they are the same functions.1380

when you start integrating units, you are going to get the same number.1386

Let us see what we have got.1391

let me go back to blue, let me actually go back to black.1396

pictorial descriptions of the wave functions.1401

these wave functions they are orbitals, that is what they are.1406

They are orbitals, the 1S, 2S, 2P, the 3S, the 3P, the PX, PY, PZ.1414

That is what these different wave function represent.1420

They represent the orbitals.1421

They represent places where you may find the electron.1423

you remember from the general chemistry that you have already seen pictures of what these orbitals look like.1427

I will show you some pictures here, that is not a problem.1432

However, one of the things that I want to emphasize is that the pictures that you see are there to help you wrap your mind around where the electron might be in space.1435

They are actually not very good representations of that truth of what is going on, where the electron is.1444

I would really like if you actually start moving away from pictorial versions, geometric notions of orbitals and start moving towards the mathematical description of these orbitals because that is what it is.1452

an orbital is a mathematical description of where a particle might be, how it might be moving.1463

we want to actually think in terms of the mathematics.1469

you want to use the geometry to help you understand but you do not want to rely on it because it is going to lead you astray, if you rely too much on the geometry.1472

it is the algebra, the calculus, the mathematics that is actually describing what is going on.1480

pictorial descriptions of the wave functions of the orbitals are difficult and misleading because ψ is a function of three variables.1486

what that means is that we actually need a 4 dimensional space in order to represent, in order to graph a wave function.1524

we would need a 4 dimensional space, 4 dimensional graph, R, θ, φ, and ψ, the value that you get.1535

The pictorial descriptions of wave functions are difficult because ψ is a function of 3 variables.1561

In order to graph it, you need a 4 dimensional space.1565

We do not have a 4 dimensional space, we have no way of representing the 4 dimensional space on a sheet of paper.1568

that is a problem with these things.1574

for example, if you have Y = F of X, you have the X and you have the Y variable.1577

it was a function of 1 variable but in order to graph it, you need a 2 dimensional plane.1583

That is what the graph is.1587

if you have a function of 2 variables, X and Y, you have Z = some function of X and Y.1590

now you have 3 variables that you need a graph, you need to graph the X, you need to graph the Y, and then you need to graph what Z is when you actually solve the equation Z = XY.1596

you need a 3 dimensional.1604

in this case, we have 3 independent variables.1607

we have 1 dependent variable ψ, therefore we need a 4 dimensional space to graph it.1610

we do not have a 4 dimensional space.1615

the truth is, we have no real way of graphing this, an actual wave function.1617

we can consider them separately.1624

Let me go back to blue.1628

Therefore, consider the radial and spherical and angular components separately.1629

the radial component is easy to visualize because it is just a function of 1 variable R.1675

all we have to do is graph in the plane really nice.1680

the radial component which is R as a function of R, the single variable is easy to visualize because it depends only on R.1683

let us look at the one as orbital, the easiest orbital 1S ψ 1, 0, 0.1716

let us look at the 1S orbital.1724

that is going to be the R 1, 0 of R.1737

1, 0 N is 1, L is 0.1741

the function R 1, 0 as a function of R is 2/ α sub 0³/2 E ⁻R/ α sub 0.1746

This is one of the representations of the radial function for 1, 0.1762

the normalization condition is the integral from 0 to infinity of R conjugate × R and then R² DR.1769

remember, we are integrating this over a surface of the sphere, so we have to have that DV component.1800

Remember, R² sin θ D θ DR D φ, do not forget this R² term.1806

Let me mark that down in red.1815

Do not forget this R² term when doing your integrations, when integrating in spherical coordinates.1821

this part right here, we have R conjugate × R R² DR.1847

0 to infinity of R conjugate of 1, 0 × R of 1, 0, R² DR is equal to, here is my R.1860

R is a real function so R conjugate is the same as R, so we are just going to have R².1878

when I Square that, I get the following.1884

I will get 4 / A sub 0³ × the integral 0 to infinity of E⁻² R/ A sub 0 R² DR.1886

this is the normalization condition for the R 1, 0 function.1911

we are just taking a look at the simplest orbital.1916

this right here, this integrand including the DR, this is the probability.1918

this is the normalization condition.1934

this represents the probability.1937

this is the probability that the electron in this orbital lies between R and DR.1941

this part right here, without the DR that is a probability density.1960

we are going back several lessons to when we initially started this.1970

this is the probability density × its differential element.1973

in this particular case, R gives you the probability that it is between, that it is in that little differential length element.1983

we have another page here.1995

the probability of the 1S orbital is equal to 4/ A sub 0³ E⁻² R/ A sub 0 R² DR.1997

in general, the probability is equal to, we are talking just about the radial function here.2027

we are not talking about the entire wave function, just the radial function here.2038

we have broken it down, = R and L conjugate × R and L R² DR.2042

we will leave it like that.2055

this is the probability to take the radial, again we are just talking about the radial function here.2064

we want to be able to graph some radial dependents, as we move a certain distance R away from the nucleus, where is the electron?2071

what are the chances that I'm going to find it here or there?2083

this easy to do because it is only a function of one variable R so we can plot it in a 2 dimensional plane.2086

this is the probability right here.2092

when we plot these, when we plot the probability densities which is just this part.2096

when we plot the probability densities vs. R the radius for various values of N and L, here is what we get.2115

this is a plot of the 1S orbital, the probability density that is this thing right here.2149

this is R² R², the radial functions².2156

this right here is just R conjugate × R N sub L, you have that R² term.2160

this is the probability density on the Y axis.2172

On this X axis, we have R, that is all that is going on here.2175

The fact that is ÷ the bohr radius is just the way of standardizing it.2178

Do not to worry about that.2183

this is the probability density vs. R.2185

notice what you have got.2189

as you move away from the nucleus, as you get further and further away from it, this measures probability that you will actually find the electron within that far from the nucleus.2190

we see about here is what is actually the highest probability.2207

The electron is going to be spending most of its time between here and here, that is what this means.2213

notice a couple of things, notice we have 0 nodes, and notice when the probability density goes to 0, where you absolutely will not find it.2220

and 0 does not count for probability.2235

the initial, the nucleus is 00 point, it does not count for a node.2237

there is no nodes.2240

this is N = 1 and L = 0.2243

this is the 1S orbital.2246

the 1S orbital as you get further from the nucleus, the electron is going to spend, it is going to be there and then as you get past this point, you are probably not going to find the electron anywhere passed 4 units from the nucleus.2251

Let us go to 2S orbital.2268

this is a 2S orbital, this is N = 2 and L = 0.2274

Notice, you have 1 node right there.2282

You have 1 node, what this means is that as you get farther from the nucleus, you will probably find the electron somewhere between here and here or between here and here.2290

what this node tells you is that you will never ever find the electron there.2300

that is all the node means.2307

you are never going to find the electron there.2308

It will be closer to the nucleus or farther from the nucleus.2310

these high points represent the maximum probability that they will probably be there.2314

that is what this represents.2320

notice that the 00 mark does not count as a node, that is just the origin.2324

For the 3S orbital, we are graphing for radial function, that is all we are doing.2331

the 3S orbital here, N = 3, L = 0.2339

L = 0 is the S orbital.2344

Notice, you have 2 nodes.2348

3S orbital has 2 nodes.2354

the 1S orbital had no nodes.2357

The 2S orbital have 1 node.2359

the 3S orbital has 2 nodes, you see a pattern.2362

it tells me that the 3S orbital, I can find the electron here or here or here.2364

at this point and this point, I'm not going to find electron ever.2369

Let us go to the 4S orbital, N is equal 4, L is equal to 0, that is the S orbital.2375

we have 1, 2, 3, nodes.2385

Let us do some P orbitals.2398

we set for various values of N and L.2401

here, we took N from 1, 2, 3, 4, we left L at 0.2404

we are just looking at different orbitals.2408

we have the 2P orbital and again this is just the radial component that we are looking at.2413

we are not looking at ψ yet.2424

it is R and L and S LM, all we are looking at is the radial component because as a function of 1 variable, we can actually see what happens graphically.2427

what we are doing is we are plotting the probability density of the y axis and the radius on the X axis.2437

the further we get from the nucleus, what is the most likely place that the electron is going to be.2444

that is what this graph represents.2449

the 2P orbital, the N is 2 and the L is equal to 1.2452

L is 1, that is P orbital.2457

Notice, 0 nodes.2460

Let us do the 3P orbital.2464

Basically, what this says is that in the P orbital, this right here underneath about the orb, that is the most likely place that you are going to find the electron in a P orbital.2470

You are not probably going to find it out here, you are not going to find it here.2480

You are going to find it somewhere between here and here.2484

The 3P orbital, the N is equal to 3 and L is equal to 1.2489

Notice, we have 1 node.2495

the 3P orbital, you are going to find the electron here or you are going to find the electron somewhere around here.2500

you will never find it here, that is what this says.2505

you will never find it there.2507

Let us look at the 4P orbital.2512

4P orbital, N is equal 4, L is equal to 1.2519

How many nodes do you have?2527

You have 1,2, there are some patterns that we are going to elucidate here.2528

I’m going to reiterate this over and over again.2536

sorry to keep repeating myself, this was very important.2538

there is a lot of mathematics here, a lot of graphs that you are going to be seeing, do not lose your way.2540

Do not lose the 4 from the 3.2546

Do not forget, we are looking only at the radial function of ψ right now.2548

ψ which is a function of R, θ, φ, R of R, and S of θ and φ.2575

We are only looking at this one, that is all.2589

That is what we are graphing.2591

Let us move on to D orbital.2596

this is going to be a D orbital.2603

this is going to be the 3D orbital.2607

N is equal to 3 and L is equal to 2.2609

Notice, we have 0 nodes.2613

In a D orbital, you are most likely to find the electron between here and here, with the highest probability about right there.2616

that is all we are doing, we are graphing probability density vs. distance from the nucleus.2623

Let us look at the 4D orbital.2629

the 4D, N is equal to 4 and L is equal to 2.2631

Notice, we have 1 node.2636

Do I have another page here?2641

I do not.2643

when we examine all of the values of N and all of the values of L to see how many nodes depending on what N is, depending on what L is, whether it is S, P, D orbital, we come up with the following.2647

I will write it over here.2661

When we examine all the N and L values, and analyze the number of nodes, we find the following.2664

we find that the number of nodes is equal to N - L-1.2695

For the 4D orbital, N is equal to 4, 4 -2 is 2, 2 -1 is 1.2712

The D orbital has 1 node.2719

There is going to be one place in the D orbital where the electron will never be found.2721

that is all that means.2726

these are graφcal representations of probability density vs. distance from the nucleus with graphing the radial function only.2728

Let us go ahead and do a quick example here.2737

calculate the probability of finding electron in the 2S orbital, the hydrogen atom within 1 bohr radius of the nucleus.2741

once again, calculate the probability of finding an electron, the 2S orbital of the hydrogen within 1 bohr radius of the nucleus.2751

2S means that N is equal to 2 and it means that L is equal to 0.2764

Therefore, we are going to be looking at the R 2, 0 function.2776

that is the function that we are going to be working with.2782

when we put the 2 and 0 into our particular value of R.2786

I will just go ahead and write it out.2792

R of 2, 0 of R is going to equal -2 -0 -1!/ 2 × 2 × 2 + 0 !.2797

I think there is a cubed there also and a 1/2.2829

2/ 2 × sub 0 × R⁰ E ⁻R/ 2 Α sub 0 × 2 - R/ α sub 0 × – 2!.2835

we end up with R of 2, 0 is going to end up equaling 1/ 8 × A 0³ ^ ½ × 4 -2 R/ Α sub 0 × E ⁻R/ 2 Α sub 0.2859

this is our value of R 2, 0.2885

this is our wave function, it is the radial wave function.2888

we want to find the probability of finding electron in this orbital within 1 Böhr radius of the nucleus.2890

the probability is equal to the integral of 0, 2 A sub 0 within 1 Böhr radius R 2, 0 conjugate × R 2, 0 R² DR.2900

that is the φ integral that we have to solve.2921

we have our R, we multiply R by itself.2924

in this particular case, R conjugate is the same as R because R 0 is a real function.2928

It is not a complex function.2935

R conjugate of 2, 0 × R 2, 0 is going to be nothing more than R 2, 0².2943

when I square this, when I multiply it all by itself, I get that the probability is equal to,2952

this thing is going to come out of the integral.2961

It is going to be 1/ 8 Α sub 0³ the integral 0 to A sub 0.2962

when I square this part, I'm going to get 4 -2 R/ α sub 0².2973

and this is going to be E ⁻R/ 2 α sub 0 × E⁻² R/ 2 α sub 0.2981

it becomes E⁻² R/ 2 α sub 0.2994

the 2 is cancel so you are just left with α sub 0, R² DR.2996

that is my answer right there.3004

the probability is very simple.3010

the probability is always the same.3012

anytime it ask you for probability, it is always going to be some integral of what ever function you are dealing with.3014

the conjugate × the function itself.3020

in this case, the function is a real so it is just the function² × some volume element.3023

that is all you are doing, with appropriate limits of integration A to B, whatever that happens to be.3028

in this case, we want to find it within 1 böhr radius of the nucleus.3035

the nucleus is 0, 1 böhr radius is A sub 0 so this is the probability.3038

I have the function, I² it.3044

I put it into here and now all I have to do is go ahead and put in a mathematical software3047

and let the software solve the integral and give you some number.3052

I did not bother to putting this into my software and solving it, because I would leave that to you.3055

you know when I teach this course, I actually have the kids just leave it in this form,.3063

the software will do the work for you.3071

I do not need for them to actually do the integration.3073

That is calculus work, that is not altogether that important.3075

at this point, it is important to be able to construct the integral.3078

software will do the rest.3082

so it is up to you if you want to put in your software and see what answer you will get,3084

I think if you really interesting for you to take a look at it.3086

but other than that, that is your answer.3089

Thank you so much for joining us here at

We will see you next time, bye.3094