WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, and welcome to www.educator.com, welcome back to Physical Chemistry.
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We are going to continue our example problems for the hydrogen atom.
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Let us get started.
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The first example is show that the ψ 211 is normalized.
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Now we are dealing with the entire wave function.
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A function of R, a function of θ, and a function of φ.0019.2 We want to show that it is normalized.
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Let us go ahead and write down what ψ 211.
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Las time we are dealing with spherical harmonics, so now we are dealing with a full 3 quantum numbers 211, N, L, and M.
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This particular wave function is equal to 1/ 64 π to ½ power Z/ α₀³/2 σ E ^-σ/ 2 × sin θ E ⁺I φ,
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where σ is actually equal to ZR/ Α₀.
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Different books and different lists have some variation in how they actually list of functions,
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depending on what they want to express as constants.
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Whether they want to use exponents here or that I want to use √ signs, things like that.
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If the wave functions in your book or in the particular list that you have to be looking at,
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do not look exactly like this, do not worry about it.
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They are the same wave function, I promise you that.
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When you integrate them, you are going to get the same answer.
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I would not worry too much.
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This is Z here, Z is just is the atomic number.
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These are the orbitals, the wave functions for what they call hydrogen like atoms.
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Basically, anything that something like helium, where why the electrons is been taken off.
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It is a helium atom but it has been ionized so you are still talking about the 1 electron.
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For our purposes, we are just going to take Z equal to 1 because the atomic number of hydrogen is 1.
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We just wan to write the general equation for the hydrogen like orbital,
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that we are going to take Z equal to 1 and A₀ is the bohr radius.
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We want to form the normalization integral which is the integral of ψ 211 conjugate × ψ 211.
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We want this integral to equal 1 because we are trying to show that it is normalized.
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We want this integral to equal 1.
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Let us go ahead and write all this out.
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Ψ 211 conjugate the integrand × ψ 211 is going to equal,
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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.
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The 1/ 64 π to ½, 1/64 π gives us 1/ 64 π.
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Again, we are taking Z equal to 1, 1/ A₀³/2 × 1/α₀³/2 is going to be 1/ Α₀³.
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S σ is equal to Z R/α so this is just R/ A₀ E ^-σ/ 2.
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This is going to be -R/ 2 A0 sin of θ E ⁻I φ.
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Yes, we still have some more, I think I have enough room here.
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So × R/ A₀ E ⁻R/ 2 A₀ × sin θ × E ⁺I φ.
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This, I multiply all the constants together.
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This is for the conjugate and this is for the ψ 211.
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When I put all this together, I get 1/ 64 φ A₀⁵ R² E ⁻R/ α₀.
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R2 and R2 is going to be -2R.
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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.
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This is our integrand.
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The integral of ψ 211 conjugate × to ψ 211 is going to equal the integral from 0 to infinity.
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Now, we are including R 0 to 2 π, 0 to π of this function which is 1/ 64 π A₀⁵ R² E ⁻R/Α₀
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sin² θ × the factor R² sin θ.
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We are doing D θ D φ, we are going to do our last, D φ DR, inside to outside.
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We have a function of R, a function of θ.
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It does not look like we have a function of φ.
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Let us go ahead and separate this out, = to 1/ 64 π A₀⁵ the integral from 0 to infinity, R² × R² is μ R⁴ E ⁻R A₀.
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I’m going to separate out the variables, the R, θ, and φ, DR.
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The integral from 0 to 2 π of D φ because there is no function of φ.
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The integral from 0 to π if sin² θ sin θ, so we have our sin³ θ D θ.
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We just put everything together.
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All of the R’s we have put together with the r’s, under one integral.
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The φ under one integral, the θ under one integral.
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Now, we are going to solve these 3 integrals.
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Integral number 1, our first integral which is equal to 0 to infinity of R⁴ E ⁻R/ Α₀ DR.
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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
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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.
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I think you have actually seen it before, it is 4 !/ 1/α₀⁵.
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It is going to end up being 24 Α₀⁵.
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That takes care of the first integral.
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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.
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That is the second integral.
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The third integral, integral number 3 is equal to the integral from 0 to π of sin³ θ D θ.
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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.
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I do not necessarily know if I want to go through.
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That is fine, I will go ahead and go through this process.
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It is it is not a problem.
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D θ =, I going to separate this out so I will go ahead and do the manual.
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You do not need to, but this is what it looks like, in case you would do it.
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0 to π sin² θ sin θ.
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I have separated this out and I'm going to use a U substitution.
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I'm going to let, sin² θ is actually equal to 1/ π is equal to 1 - cos² θ sin θ D θ.
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I will use a U substitution on that.
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I will let U equal cos θ.
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Therefore, DU = - sin θ D θ.
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Therefore, this integral is going to actually end up equaling - the integral from 0 to π 1 - U² DU,
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which is equal to - U – U³/ 3 from 0 to π, which is nothing more than U³/ 3 - U from 0 to π,
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which is equal to cos³ θ/ 3 – cos θ from 0 to π.
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And this is going to equal -1/ 3, -and -1 I hope that I have done this correctly.
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-1/3 -1 which is going to end up equaling 4/3, this is our third integral.
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Now that we have the first integral, the second and third.
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When we put this together, it is going to be this × that × that × the constant.
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We have 1/ 64 π A₀⁵ × 24 A₀⁵ × 2 π × 4/3.
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Everything comes up to 192/ 192 is equal to 1.
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Yes, ψ 211 is normalized as written, good, very nice.
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Example 2, we want to show that ψ 211 is orthogonal to ψ 310.
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Let us write out what they are.
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Ψ 211 is equal to 1 / √ 64 π × Z/ A₀³/2 ZR/ A₀ E ⁻Z R/ 2 A₀ sin θ E ⁺I φ.
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Then, we have ψ of 310 that is going to equal 1/ 81 × 2/ π.
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I think this is going to be to the ½ and then we have Z/ A₀³/2.
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This time we have 6 ZR/ A₀ - Z² R²/ A₀² × E ⁻ZR/ A₀.
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Actually, it is going to be 3 A₀ × cos of θ.
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Those are the two wave functions ψ 211 and ψ 310.
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We want to form the integral to show that they are orthogonal.
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We want to take ψ 211 the conjugate multiply by ψ 310, we want to integrate that and we want to equal 0.
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We want this to equal 0 because we are trying to demonstrate orthogonality.
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Ψ 211 conjugate, I’m not going to go through in every step,
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at this point you should be reasonably familiar with taking conjugates when there is a complex there negated.
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The ψ 211 × ψ 310 which is the integrand is going to equal,
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I’m not going to write out all these constants over and over again.
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I'm just going to call it the constant for 211 so let us call it ψ 211.
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Again, we are taking Z equal to 1, the hydrogen atom.
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R/ A0 E ⁻R/ 2 A₀ sin of θ E ⁻I φ because we are doing the conjugate × the constant for 310 × 6 R/
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A 0 - R²/ A 0² × E ⁻R/ 3 A0 × the cos θ.
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All of this is equal to ψ 211 × ψ 310 6.
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I multiply the constants R, I'm going to get 6 R²/ A 0² – R³/ A 0³.
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I just multiply everything together, E⁻⁵ R/ 6 A₀ sin θ cos θ E ⁻I φ.
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That is my integrand.
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Therefore, my integral of ψ 211 conjugate × ψ 310 is going to equal,
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I'm going to go ahead, this is what I’m integrating in spherical coordinates.
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I’m going to go ahead and just separate them at once, instead of doing it in 2 steps.
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I'm going to have the integral from 0 to infinity of the R part.
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6 R²/ A₀² – R³/ A₀³ E⁻⁵ R/ 6 A₀.
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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.
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Let us take a look at this.
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I have this thing, this thing, and this thing.
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The second integral is equal to 0, we have seen that before.
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Integral number 2 is equal to 0.
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Let me write what integral number 2 is, 0 to 2 π of E ⁻I φ D φ.
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This is equal to 0, we know this from a previous example.
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Therefore, our ψ 211 conjugate ψ 310 is equal to 0, which means they are orthogonal.
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Exactly as we wanted.
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You see that a lot of times you would not have to solve any integral.
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The integral was very complicated but it is not complicated because a piece of it is really simple.
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We already know that N = 0.
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In one case it is this function.
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In some other cases, you may have even and odd functions.
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And if it ends up being the integral of an odd function/ a symmetric interval, that is going to equal 0.
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Definitely makes life a lot easier.
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Let us go ahead and go to example 3.
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What is the probability that a 1S electron will be found within 1 bohr radius of the nucleus.
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We want to evaluate this following integral.
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The probability integral, we want to evaluate the integral from 0 to A₀.
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Now, we are specifying an outer radius so it is no longer infinity of the ψ 1S conjugate ψ 1S.
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This is the probability integral.
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Normally, we just integrate it over the entire space and we want it equal to 1 to show normalization.
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This is a normalization condition but now we are calculating the probability.
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There is going to be some upper limit on this integral.
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1S means that N is equal to 1 and S means that L is equal to 0.
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Therefore, N is equal to 0.
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What we are looking for is the ψ 100 function.
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We need to form the ψ 100 conjugate × Ψ 100 the integrand.
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Well, that is going to equal 1/ √ π × Z/ Α₀³/2 E ⁻ZR/ α₀ × 1/ √ π Z/ α₀³/2 E ⁻ZR / α₀.
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Again, we are taking Z equal to 1.
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Therefore, the Ψ 100 is going to equal,
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This is 1 and this is 1, so it is going to equal 1/, π is taken care of, and A₀³ and
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it is going to be E ⁻R/α 0 –R/ Α 0 -2R/ Α 0.
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The integral from 0 to Α₀ Ψ 100 conjugate ψ 100 is going to equal 1/ π Α 0³.
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The integral from 0 to Α₀ of E ⁻2R / A₀ and this is R² DR.
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That is the R version.
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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.
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We already know what this is, this is 2 π.
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We already know what this is, this is just 2.
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We have 4 π over here, this integral right here.
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Let us go ahead and do a little something with that particular integral.
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Should I do it on the next page?
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I’m just going to go ahead and take care of it here.
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The first integral, integral number 1 is the integral from 0 to A₀ of R² E ⁻2R/ A₀ D R.
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We would actually do a little bit of a U substitution here.
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I'm going to call U, I'm going to call it R/ A₀.
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Therefore, DU is equal to 1/ A₀ which means that A₀ × DU is equal to DR.
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I’m going to put this into the DR and then when I do R², that is going to equal A₀² U².
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I'm going to put that into there.
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This integral, when I put 0 in for here into R, I get 0.
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U = 0.
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This lower limit of integration is going to end up becoming 0.
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When I put A₀ into R, I'm going to get the upper integral of 1.
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When I do the substitution, I’m end up getting the following.
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I'm going to get the integral from 0 to 1 R² is A₀² U².
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This is going to be E ⁻2U and then × DR, which is A₀ × DU, which ends up being A₀² × A₀.
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We have A₀³ × the integral from 0 to 1 of U² DU.
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I'm sorry, U² E ⁻2U DU.
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When I go ahead and put this into my mathematical software, I'm going to end up getting 0.08083 × A₀³.
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This is my first integral.
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My integral number 2, which was the integral from 0 to 2 π of D φ, that is equal to 2 π.
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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.
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When we put all of the 3 together, we have 1/ the constant π A₀³ × 0.08083 A₀³ × 2 π × 2.
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That that, that that, what I end up getting is 0.3233.
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There you go, that is my probability.
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My probability is roughly 1/3 of finding the 1S electron within 1 bohr radius.
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Let us try example number 4, what is the radius of the sphere that encloses a 60% probability of finding the hydrogen 1S electron?
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Express your answer in terms of A₀, the bohr radius.
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They are telling us that the probability is equal to 0.6.
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The integral that we just solve, we are going to set that integral equal to 0.6
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and we are actually going to look for that upper limit of the integral on R.
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We want to know what that is, what is the radius that encloses a 60% probability.
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Let me go ahead and go to blue, change this up a little bit.
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From the previous problem and if you want you can do the integral all over again, it is not a problem.
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From the previous problem, which also concern the 1S orbital, we found that the probability
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was equal to 1/ π A₀³ 0 to some constant × A₀.
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We are going to be looking for this constant ψ.
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R² E ⁻2R/ Α₀ DR the integral from 0 to 2 π of D φ, the integral from 0 to π of sin θ D θ.
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We found that all this is going to equal, this is 2, this is 2 π.
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4 π 4 π, for on top the π cancel.
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We are going to get 4/ A₀³ × the integral from 0 C A₀ R² E ⁻2R/ α₀ DR.
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The probability integral is this, we are going to solve this integral.
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We are going to get some equation in C and I’m going to set that C equal to our 60%,
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because we want the probability = 60%, I want to find out what C is
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because we want our answer expressed in terms of the A₀.
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That is what we are going to do.
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We will solve this integral then set it equal to 0.60 then solve for C.
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Let us see what we have got.
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Our probability integral is going to equal, I'm going to go ahead and do a little bit of U substitution here.
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I’m going to go ahead and go to red.
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Up here, I'm going to call U equal to R/ α₀ like I did before.
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Which means that U² Α₀² is equal to R².
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DU is going to equal 1/ α₀ DR means that DR is going to equal Α₀ DU.
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When I put all of these into here, when I put DR, when I put this into DR.
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When I put this in for R², I'm going to end up with 4 Α₀³/ Α₀³ × the integral from 0 to C.
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When I put 0 into here, I'm going to get 0.
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And then when I put C α₀ into here, I'm going to get C.
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Now I have a new upper limit of integration and it is going to be U² E⁻² U DU.
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This is by probability integral, I have converted this into this so I could deal with the C alone.
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I'm going to go ahead and this cancels.
00:30:51.900 --> 00:30:57.300
I’m going to go ahead and solve this integral by hand and so you can see it.
00:30:57.300 --> 00:31:06.300
4 × the integral 0 to C U² E ⁻2U DU.
00:31:06.300 --> 00:31:11.400
Just go ahead and do it with software but I would like you to go ahead and see it.
00:31:11.400 --> 00:31:16.300
In this particular case, we have a function × a function so we are going to use something called integration by parts.
00:31:16.300 --> 00:31:25.700
Integration by parts tends to get long, there is a short hand version of integration by parts called tabular integration.
00:31:25.700 --> 00:31:33.600
Tabular integration, it works when your particular integration by parts integral is such that one of your functions,
00:31:33.600 --> 00:31:35.100
I will go ahead and go back to blue.
00:31:35.100 --> 00:31:42.900
When one of your functions differentiates down to 0 and the other one of your functions just integrates infinitely over and over again.
00:31:42.900 --> 00:31:50.200
Integrals like this, where you have some polynomial and some exponential, they are perfect for tabular integration.
00:31:50.200 --> 00:31:59.700
And what you actually do is this, you do U² over here, you do E ⁻TU, this is our U and this our DV.
00:31:59.700 --> 00:32:06.500
Integration by parts notation, we differentiate this to TU.
00:32:06.500 --> 00:32:10.500
We differentiate this down to 2 and we differentiate this down to 0.
00:32:10.500 --> 00:32:26.200
We integrate this up to -1/2 E ⁻TU.
00:32:26.200 --> 00:32:32.400
We integrate up to ¼ E ⁻2U and we integrate one more time up to -1/8 E⁻² U.
00:32:32.400 --> 00:32:34.000
Now, we put them together like this.
00:32:34.000 --> 00:32:35.900
We put that together with that one.
00:32:35.900 --> 00:32:41.300
This together with that one, and this together with that one change in signs + - +.
00:32:41.300 --> 00:32:59.900
This integral ends up actually equaling, it is going to be 4 × -1/2 U² E ⁻TU.
00:32:59.900 --> 00:33:05.800
U is going to be this time this, this × this, this × this.
00:33:05.800 --> 00:33:35.600
-U² E ⁻TU/ 2 - 2 U E⁻² U/ 4 -2 E ⁻TU/ 8.
00:33:35.600 --> 00:33:54.000
This is going to equal – E⁻² U × 2U² + 2 U + 1, from 0 to C, by the way.
00:33:54.000 --> 00:34:21.000
2U + 1 from 0 to C, it is going to equal - E⁻² C × 2 C² + 2 C + 1 - A -1 × 0 + 0 + 1.
00:34:21.000 --> 00:34:36.100
We are going to get this integral equal to 1 - E⁻² C × 2 C² + 2 C + 1.
00:34:36.100 --> 00:34:39.900
This is our probability, this is the solution to the integral.
00:34:39.900 --> 00:34:44.500
We are going to set that equal to 0.6.
00:34:44.500 --> 00:34:58.100
1 –E⁻² C × 2 C² + 2 C + 1 = 0.60.
00:34:58.100 --> 00:35:04.300
When we go ahead and solve this equation, just put it in your software or just use a graphical calculator.
00:35:04.300 --> 00:35:28.300
We end up with C equaling, E ⁻2C 2C² + 2 C + 1 = -0.40.
00:35:28.300 --> 00:35:38.400
Move the 1over that side so I'm going to get E⁻² C × 2 C² + 2 C + 1.
00:35:38.400 --> 00:35:41.100
These negatives cancel and I bring it back to the left.
00:35:41.100 --> 00:35:45.900
It is going to be -0.40 is equal to 0.
00:35:45.900 --> 00:35:56.200
This is the equation that I put my software or into my graphing calculator and I end up with C = 1.553.
00:35:56.200 --> 00:36:13.700
For a 60% probability, our radius is equal to 1.553 α₀.
00:36:13.700 --> 00:36:16.800
There you go, that is it.
00:36:16.800 --> 00:36:23.600
Tedious math, but nothing that is straightforward math for the most part.
00:36:23.600 --> 00:36:28.700
Of a 0.40, let us make that a little bit more clear.
00:36:28.700 --> 00:36:42.000
What do we got next?
00:36:42.000 --> 00:36:47.900
Calculate the average value of R for the 2S orbital of the hydrogen like atom.
00:36:47.900 --> 00:36:52.800
Again, hydrogen like just means that you actually include the Z, the atomic number of the wave function.
00:36:52.800 --> 00:36:56.300
You put in there for like hydrogen is just Z equal to 1.
00:36:56.300 --> 00:36:59.800
This is what is going to leave the Z there.
00:36:59.800 --> 00:37:05.200
Calculate the average value of R for the 2S orbital of the hydrogen like atom.
00:37:05.200 --> 00:37:06.900
Let us see what we have got.
00:37:06.900 --> 00:37:13.900
2S means that N is equal to 2.
00:37:13.900 --> 00:37:20.700
It means that L is equal to 0, which means that M is equal to 0.
00:37:20.700 --> 00:37:29.300
What we are looking at is our ψ 200.
00:37:29.300 --> 00:37:45.300
The average value of R for 200 is going to equal the integral of ψ 200 conjugate × R operating on ψ 200.
00:37:45.300 --> 00:37:53.400
This is the basic form for the average value of whatever it is we happen to be looking at.
00:37:53.400 --> 00:37:55.300
Let us see what we have got.
00:37:55.300 --> 00:38:27.200
Our ψ 200 is equal to 1/ √ 32 π × Z / α₀³/2 × 2 - σ × E ^-σ/ 2.
00:38:27.200 --> 00:38:35.600
Again, the σ is equal to ZR/ Α₀.
00:38:35.600 --> 00:38:42.100
The ψ 200 conjugate is the same as this because there is nothing complex about this.
00:38:42.100 --> 00:38:50.500
Therefore, this R operator, it just means multiply by R.
00:38:50.500 --> 00:38:51.400
It does not matter where we put it.
00:38:51.400 --> 00:38:55.400
I do not have to operate on this one first and then multiply on the left by this 1.
00:38:55.400 --> 00:39:01.400
I could just multiply these 2 and then multiply by R.
00:39:01.400 --> 00:39:14.100
R × ψ 200² is going to equal R × 1/ 32 π.
00:39:14.100 --> 00:39:28.900
It is going to be Z³/ α 0³ × 2 - σ² E ^-σ.
00:39:28.900 --> 00:39:32.700
This × itself.
00:39:32.700 --> 00:39:36.700
It is going to equal, let us pull the constants out.
00:39:36.700 --> 00:40:08.300
Z³/ 32 π Α₀³, let us multiply this out for -4 σ + σ² E ^-σ × R.
00:40:08.300 --> 00:40:31.200
This integral here, it is going to be Z³/ 32 π A₀³ 0 to infinity
00:40:31.200 --> 00:40:53.600
0 to 2 π 0 to π of 4 -4 σ + σ² R × R² sin θ D θ DR.
00:40:53.600 --> 00:41:06.000
It is actually going to be D φ DR.
00:41:06.000 --> 00:41:17.900
This is going to equal.
00:41:17.900 --> 00:41:28.500
The 0 to 2 π D φ and the 0 π of the sin θ D θ is going to equal of 4 π.
00:41:28.500 --> 00:41:38.500
We end up with 4 π Z³/ 32 π A₀³.
00:41:38.500 --> 00:42:06.100
The π go away and it is going to be integral from 0 to infinity of R³ × 4 - 4 σ + σ² E ^-σ DR.
00:42:06.100 --> 00:42:08.800
I'm going to go ahead and do a little substitution here.
00:42:08.800 --> 00:42:25.800
Our σ is equal to ZR / A₀ which means that D σ is equal to Z/α₀ DR.
00:42:25.800 --> 00:42:49.800
Therefore, DR is equal to Α₀ σ/ Z and R³ is going to equal Α₀³/ Z³ × σ³.
00:42:49.800 --> 00:42:53.300
We just rearrange this and R³.
00:42:53.300 --> 00:43:01.700
When we put all of these into this thing, we end up with a lot.
00:43:01.700 --> 00:43:26.200
4 π Z³/ 32 π/ A₀³ × A₀³/ Z³ × A₀/ Z × the integral 0
00:43:26.200 --> 00:43:39.800
to infinity σ³ × 4 -4 σ + σ² E ^-σ.
00:43:39.800 --> 00:43:44.000
This × D σ =, all of these things cancel out.
00:43:44.000 --> 00:43:56.200
That cancels that, that cancels that, + 8, we end up with A₀/ 8 Z, the integral from 0
00:43:56.200 --> 00:44:03.000
to infinity of σ⁵ and distribute this out and rearrange.
00:44:03.000 --> 00:44:18.100
Σ⁵ -4 σ⁴ + 4 σ³ D σ.
00:44:18.100 --> 00:44:29.500
We have that general integral formula, we have the integral formula, what we get from the table.
00:44:29.500 --> 00:44:44.900
The integral from 0 to 8 of X ⁺N × E ⁻AX DX is equal to N!/ 8 ⁺M + 1.
00:44:44.900 --> 00:44:52.700
When we do that for this one, this one, and this one.
00:44:52.700 --> 00:45:07.300
I keep forgetting my exponential E ^-σ D σ.
00:45:07.300 --> 00:45:17.100
When I apply this formula to this integral, this × this, this × this, this × this, I end up with,
00:45:17.100 --> 00:45:19.000
Let me go back to red.
00:45:19.000 --> 00:45:33.000
I end up with A₀/ 8Z + 5! -4 × 4! + 4 × 3!.
00:45:33.000 --> 00:45:44.400
I end up with 48 A₀/ 8 Z and I end up with 6 A₀/ Z.
00:45:44.400 --> 00:45:52.500
There you go, that is what we wanted, our final answer .
00:45:52.500 --> 00:46:05.200
The average value of R, on average the electron will be this far from the nucleus for the 2S electron.
00:46:05.200 --> 00:46:09.500
We just went through the mathematics.
00:46:09.500 --> 00:46:14.400
This goes here, this is that one.
00:46:14.400 --> 00:46:20.400
Let us go ahead and give you some general formulas.
00:46:20.400 --> 00:46:55.000
The average value of R sub NL is equal to N² A₀ / Z × 1 + ½ × 1 - L × L + 1/ this is that one and this is that one.
00:46:55.000 --> 00:47:25.500
Now for hydrogen, Z is equal to 1 and this general formula reduces to A₀/ 2 × 3 N² - L × L + 1.
00:47:25.500 --> 00:47:47.500
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.
00:47:47.500 --> 00:47:49.700
I will just put that in.
00:47:49.700 --> 00:48:01.600
I will get A₀/ 2, N =3, 3² is 9, 3 × -27, 27 - L × L + 1 -0.
00:48:01.600 --> 00:48:07.000
It is going to be 27/2 A₀.
00:48:07.000 --> 00:48:18.100
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.
00:48:18.100 --> 00:48:19.900
We just plug these back into there.
00:48:19.900 --> 00:48:22.100
The general formula for the hydrogen atom.
00:48:22.100 --> 00:48:30.000
The general formula for any hydrogen like atom which includes Z.
00:48:30.000 --> 00:48:31.900
Thank you so much for joining us here at www.educator.com.
00:48:31.900 --> 00:48:33.000
We will see you next time, bye.