WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com.
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Welcome back to Physical Chemistry.
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Today, we are just going to continue doing our example problems for the harmonic oscillator and the rigid rotator.
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Let us jump right on in.
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The first example for the harmonic oscillator in the ψ₃ state.
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I will forgot to subscript that, ψ₃ state showed that the average value of X² is equal to 7 H ̅/ 2 × KM¹/2.
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This example is going to be very long.
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This is going to take several pages and the reason that it is, is because I decided at least for this example set,
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for example 1 and 2, I decided to actually do them explicitly.
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Rather than relying on mathematical software, I want to go through at least 1 or 2 times
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how to actually put the integral together and how to actually solve the integral by hand.
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It is just going to be a long example.
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It is going to be a lot of symbolism on the page but there is nothing here that is actually difficult.
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It is just keeping all of the algebra straight, that is the only difficult part.
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Let us jump right on in and see what we can do.
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For the ψ₃ state, we want to show that this is the case.
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You remember that the average value or recall, I should say,
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that the average value of some quantity is equal to the integral of the wave function operator.
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This is the integral that we have to form.
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We have to take the function, operate on it, whatever operator is that we are dealing with and
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multiply by the conjugate of that particular wave function and then integrate that.
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So setting up the integral should be not a problem.
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It is actually solving it that is going to end up being the most problematic, simply because it is tedious.
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You do not have to do this because you have math software.
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On an exam, you do not have to be given anything difficult because you have to do a reasonably quick integration.
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Let us see what we have got.
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Let us go ahead and erase this.
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We have X², this thing is going to equal - infinity to infinity of ψ₃, that X² ψ₃ DX.
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This is the integral that we actually have to solve.
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Ψ₃ is equal to, let me write it over here.
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Ψ₃ is actually equal to Α Q³/ 9 π × 2 Α X³ -3X × E ^- α X²/ 2.
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Where recall, Α is shorthand for K × μ/ H ̅²¹/2.
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Our ψ*, the integrand X² ψ, that is equal to this is just multiplication by X².
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Multiplication is commutative so I do not have to apply to this one and then multiply it.
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I just pull the X² out.
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This is a 3, this is a 3, I forget the state, sorry about that.
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This is a real function so this is just X² ψ of 3 × ψ of 3.
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We are just going to end up squaring it.
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I think I forgot something.
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This is 1¹/4, the Α³/ 9 Π is going to be the biggest issue.
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It is just remembering to write down all the little symbols.
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Let us just take this nice and slow.
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This is ψ₃, so when I multiply this by itself, I'm going to end up with the following.
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I'm going to end up with Α³/ 9 π¹/2.
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I have the X² and I have the 2 Α X² - 3X.
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This is² and this multiplied by itself becomes - α X².
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This = Α³/ 9 π¹/2 × X² ×, I’m going to expand this out.
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It is going to be 4 Α³ X.
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We see that we are already starting to have some issues here.
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Hopefully, we can take care of these issues before it gets crazy.
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This is going to be X⁶ -12 Α X⁴ + 9 X² × E ^-Α X².
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I’m going to go ahead and distribute the X so we had the Α³/ 9 π¹/2 × 4 Α² X⁸ -12 Α X⁶ + 9 X⁴ E ^- Α X².
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That is what the integrand is.
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The integrand which is this is equal to this thing.
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We will go ahead and do this in red.
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This X² thing is going to equal 2 × Α³/ 9 π¹/2 the integral from 0 to infinity.
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The 2 of -infinity to infinity, I just split it up into 2 + 0 + infinity, so we have this thing.
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4 Α² a X⁸ -12 Α X⁶ + 9 X⁴ × E ^- α X² DX.
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This is what we actually have to solve.
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You are going to be using a table or you could be using math software to do this.
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On your exams, they only have to give you a very easy function to integrate.
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Or they are going to give you from a table of integrals to actually work with so
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you can do the integration reasonably quickly because you cannot sit there for 30 minutes doing one problem.
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But in this particular case, we are going to go through this by hand individually.
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Let us jump right on in and see what we can do.
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I’m going to write this integral on the next page and let us take it from there.
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Let me go back to black and let me write what we have.
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We have the X ^² the average value is going to equal 2 × α³/ 9 π¹/2 × the integral from 0 to infinity of 4 α² X⁸ -12 α X⁶.
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Α is just a constant, + 9 X⁴ × E ^- α X² DX.
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You are going to be seeing this integral a lot in quantum mechanics for the harmonic oscillator.
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We are going to break this up into 3 integrals.
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Essentially, we are going to break it up into this × this, this × this, this × this.
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Our first integral is going to be the integral from 0 to infinity.
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We will worry about this factor at the end.
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4 Α² X⁸ E ^-Α X² = 4 α² the integral from 0 to infinity of X⁸ E ^-α X² DX.
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That is going to be our first integral.
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Our second integral is going to be the integral from 0 to infinity of -12 Α X⁶ E ^- Α X Squared DX,
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which is going to equal -12 Α the integral from 0 to infinity of X⁶ × E ^- Α X² DX.
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That is the second integral.
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Our third integral is going to be the integral from 0 to infinity of 9 X⁴ × E ^- α X² DX,
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which is going to equal 9 integral of 0 to infinity of X⁴ E ^-α X² DX.
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The integral form of that we are actually going to use is going to be this one.
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We will use the following general integral formula that we get from our table.
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The integral of the form 0 to infinity X ⁺2N E ^- Α X² is equal to 1 × 3 × 5 ×… all the way to 2N -1/ 2 ⁺N + 1 A ⁺N × π/ A¹/2.
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This is the general formula, X ⁺2N E ^-α X².
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Notice, we have X⁴ X X⁶ X⁸ E ^-α X².
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Here, the A is the Α.
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Let us go ahead and do integral number 1 first.
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Integral number 1, 4 α² 0 to infinity of X⁸ × E ^- Α X² DX.
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In this particular case, X⁸ is equal to X² × 4.
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Therefore, our N is going to equal 4.
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Since N = 4, the 2 N -1 is equal to 2 × 4 is 8, that is equal to 7.
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The value of this integral is equal to 1 × 3 × 5 × 7/ 2 ⁺N + 1, which is going to be to the 5th.
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It is going to be Α⁴ × π/ Α¹/2.
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Let us make this α a little clear, there we go.
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That is the first integral.
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Therefore, our 4 Α² × the integral from 0 to infinity of X⁸ E ^- α X² DX is going to equal 4 Α² × this thing,
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which ends up being 105/32 Α⁴ × π/ Α¹/2.
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That is our first integral.
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Let us do integral number 2.
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Our integral number 2 is -12 Α the integral from 0 to infinity, this time we have X⁶ E ^- Α X² DX.
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X⁶ is equal to X² × 3 so N is equal to 3, 2 × 3 -1, 2 -1 is equal to 5.
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The integral portion, we have 1 × 3 × 5/ 2 ⁺N + 1 = 4.
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It is going to be Α³ × π/ Α¹/2.
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The total integral -12 Α the integral from 0 to infinity of X⁶ E ^-Α X² DX is going to equal -12 Α ×, then we will multiply this.
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1× 3 × 5 and do some cancellations.
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We are going to end up with 15/ 16 α³ × π/ Α¹/2, that is our second integral.
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We are getting there, slowly but surely.
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It is not a problem.
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We have the third integral.
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The integral number 3 that was equal to 9 × the integral from 0 to infinity, this time X⁴ E ^-α X² DX.
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X⁴ is equal to X² which means that N = 2 and 2 × 2 -1 is equal to 3.
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Therefore, we have 1 × 3/ 2³ Α².
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We are just using the general formula π/ Α¹/2.
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This integral 9 0 infinity of X⁴ E ^- α X² DX is equal to 9 ×, this thing is 3/ 8 Α² × π/ Α¹/2.
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That is our third integral.
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Let us not forget our normalization constant and our factor of 2.
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Let us remember our normalization constant and our factor of 2.
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Our normalization constant was Α³/ 9 π¹/2 and 2.
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We have the following.
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We have our X² average value is equal to 2 and we will have our Α³/ 9 π¹/2.
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We have the first integral, the first integral was 4 Α² × 105/32 Α⁴.
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The second integral was -12 Α × 15/16 Α³ + 9 × the 3/8 Α² × the π/ Α¹/2.
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We are going to have a whole bunch of cancellations.
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This is the answer, we just need to simplify this thing algebraically.
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Let us go ahead and cancel, we got the π¹/2 π ^½.
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We can take care of the π, we can cancel this α with one of these α to give us an Α².
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You do have all kinds of cancellations.
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You are going to have a 4, this is going to be 8, this is Α².
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Here, this is going to leave me with an Α².
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This Α and this Α is going to leave me with an Α².
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An Α² here, when I do the cancellations, when I multiply, when I simplify, I should get the following.
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I should get 2 Α / 3 × 105/ 8 Α² -45/ 4 Α² + 27/ 8 Α².
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Let me see what I have got here.
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This Α cancels these squares, when I multiply everything out, 3 cancels with the 105.
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I get 35 up here and I get 15 here, I get 9 here.
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I can go ahead and multiply those out.
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I think I lost my way.
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Let me change something here.
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Let me start again.
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I will just write down what is that I have written.
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I think that is going to probably be the best thing to do.
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I have got 2 Α/ 3 × 105/ 8 Α² -45/ 4 α² + 27/ 8 Α² 2 Α/ 3 × 105 -90 + 27/ 8 α².
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Α goes with Α, 3 goes that becomes 35.
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This becomes 30, this becomes + 924.
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We are going to end up with is going to be 7/ 2 Α, that is the final answer.
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However, but α is equal to √ K μ/ H ̅.
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That is going to equal 7 H ̅/2 × √ K μ, which is exactly what we wanted to show.
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That takes care of that example, manual integration.
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Quantum mechanics is not particularly difficult.
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It is the calculations that are very very tedious.
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You hove software at your disposal, please use the software or table of integrals if you prefer.
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Example 2, let us see what we have got.
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It is going to be about the same thing, it is not a problem.
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For the ψ₃ state, let us go ahead and make this a subscript here.
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For the ψ₃ state of the harmonic oscillator show that the average value of
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the momentum² is equal to 7 × H ̅ × √ K × μ all divided by 2.
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We have this is equal to -infinity to infinity.
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It is going to be ψ₃ DX, this is the integral that we have to solve.
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In this particular case, we have to integrate this twice and multiply by that.
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We cannot just pull this out, multiply these 2, and then apply the operator.
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We have to apply the operator once, we have to apply the operator twice.
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Let us go ahead and write down.
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We have ψ₃, it is equal to Α³/ 9 π¹/4 × 2 Α X³ -3 X E ^-α X² / 2.
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We have our momentum operator, it is equal to – I H ̅, I will go ahead and just use the partial notation.
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That is fine, I will use the one variable so it does not really matter.
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We will just do DDX which implies that P², it is just this thing applied to itself.
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You end up with – H ̅² D² DX².
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For now, I'm going to go ahead and leave aside this normalization constant and this constant right here.
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I’m just going to go ahead and differentiate and deal with the part, and I will bring the constant back.
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Let us go ahead and we would have to differentiate ψ₃ so we have to differentiate this function.
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Let us do it this way, I’m going to write it 2 Α X³ -3 X × E ^-Α X²/ 2.
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I’m going to do this.
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The symbol that I’m going to use, I’m going to write the function and this D means differentiate.
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When I differentiate it, this is the product.
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It is going to be this × the derivative of that + that × the derivative of this.
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What I end up with is the following.
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I end up with 2 Α X³ -3 X, this × the derivative of that.
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The derivative of that is E ^-Α X²/ 2 × the derivative of this.
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This is going to end up being × -Α X + this × the derivative of that, + E ^- α X²/ 2 × 6 Α X² -3.
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This is going to equal E ^- α X²/ 2 ×, I’m going to multiply this and this,
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I end up with -2 Α² X⁴ + 3 Α X² + 6 Α X² -3.
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Clearly, the biggest problem here, which is the biggest problem in most mathematics is not the mathematics, it is the algebra.
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I hope to God that we have not missed a symbol.
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If we have, let us find out.
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It = E ^- α X²/ 2 and we will goo ahead and combine some common terms here.
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We have 6 α X² and 3 α X², we have 9 α X².
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I’m going to put something positive first instead of this.
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-2 Α² X⁴ -3, so that is the first derivative.
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Now, we are going to differentiate again because we are taking the derivative twice.
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When we differentiate again, we end up with the following.
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It is a product functions so we have E ^-α X²/ 2 × 18 Α X -8 Α² X³.
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This × the derivative of that, - that × the derivative of this.
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+ 9 Α X² – 2 α² X⁴ – 3 × E ^-α X²/ 2 × – α X.
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My final result is going to be E ^-α X²/ 2 ×,
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I’m going to have 18 Α X when I combine all my terms.
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-8 Α² X³ -9 Α² X³ + 2 Α³ X⁵ + 3 α X.
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This right here is that.
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We have to multiply this on the left by ψ₃ because that is the integrand.
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Now, ψ₃ × ψ₃ is going to be 2 Α X³ -3 X × 8 ^- α X²/ 2 ×, what is that we got right there.
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E ^- Α X²/ 2 and I will left this expanded.
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I’m going to combine some of these terms like this one and this one.
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We have 21 Α X - 17 Α² X³ + 2 Α³ X⁵.
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=, this ends up being E ^-α X².
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I’m going to multiply this by that, I'm going to have 123456 terms.
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I’m going to end up with 42 α² X⁴ - 34 α³ X⁶ + 4 α⁴ X⁸ – 63 α X² + 51 α² X⁴.
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I’m glad I have my software, it is the greatest invention in the world.
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6 Α³ X⁶ which ends up equaling E ^-α X² × 93 Α² X⁴ – 40 α³ X⁶ + 4 Α⁴ X⁸ -63 Α X.
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We can finally put in everything.
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What you should get is, our P² is equal to.
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Let us bring back the -H ̅².
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Let us bring back the Α³/ 9 π¹/2.
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This is going to be the integral of 0 to infinity.
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I’m going to actually factor a 2 in here, so let me go ahead and put that 2 here.
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-2 × that, it is going to be the E ^-α X² × what it is that I just wrote.
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93 Α² X⁴ - 40 Α³ X⁶ + 4 α⁴ X⁸ – 63 α X² DX.
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This is the one that we actually end up wanting to solve.
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We are going to end up breaking this up into 4 integrals.
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This one, this one, this × this one, and this × this one.
00:33:25.600 --> 00:33:35.300
I'm not going to do that explicitly, I just can write out what is that we actually get when you go through all of the math.
00:33:35.300 --> 00:33:40.600
That is fine, I'm will just go ahead and write down all of the arithmetic.
00:33:40.600 --> 00:34:16.400
-2 H ̅² Α³/ 9 π¹/2 ×, it is going to end up being 93 Α² × 3/ 8 Α² -40 α³ × 15/ 16 Α³ +
00:34:16.400 --> 00:34:37.600
4 Α⁴ × 105/ 32 Α⁴ -63 α × 1/ 4 Α × that π/ Α¹/2.
00:34:37.600 --> 00:34:52.400
When I multiply, simplify, cancel, and do all that stuff, I'm going to end up getting –H ̅² Α -14/ 4
00:34:52.400 --> 00:35:01.400
which is equal to 7/2 H ̅² Α.
00:35:01.400 --> 00:35:16.900
Since the Α B what it is, when we put in 7 H ̅², Α is going to be √ K μ/ H ̅.
00:35:16.900 --> 00:35:19.900
We get a cancellation there and you end up with your final answer.
00:35:19.900 --> 00:35:29.900
You end up with 7 H ̅ √ K × μ out of the two, which is exactly what we wanted.
00:35:29.900 --> 00:35:39.400
A little tedious but I thought it might be nice to solve a couple of problems to go through this manually.
00:35:39.400 --> 00:35:43.200
Let us see here, do I have another page?
00:35:43.200 --> 00:35:45.900
I can use one, in fact I do.
00:35:45.900 --> 00:35:47.900
Let us go back.
00:35:47.900 --> 00:35:53.400
In general, we generalized all of this.
00:35:53.400 --> 00:35:59.700
It is nice to go through the problem but in general you have the following.
00:35:59.700 --> 00:36:30.200
In general, for the harmonic oscillator, for ψ sub R, the average value of X² is going to equal 1/ α × R + ½.
00:36:30.200 --> 00:36:46.200
It is always going to equal that, which is equal to H ̅/ K × μ¹/2 × R + ½.
00:36:46.200 --> 00:37:17.700
The average value of the square of the momentum is going to equal H ̅² Α × R + ½ which is equal to H × K × μ¹/2 × R + ½.
00:37:17.700 --> 00:37:23.200
In this particular case, this particular problem that we just did R = 3. So we just put it in.
00:37:23.200 --> 00:37:26.200
3 + ½ is 7/2, you end up with the answer.
00:37:26.200 --> 00:37:34.700
If you work ψ₆, it would be 13/2, that is it.
00:37:34.700 --> 00:37:39.500
This is the general expressions and if you remember from a previous discussion that
00:37:39.500 --> 00:37:43.700
the average value of X for the harmonic oscillator is actually equal to 0.
00:37:43.700 --> 00:37:48.500
The average value of the momentum is actually equal to 0.
00:37:48.500 --> 00:37:51.500
Now, we have the average value of X and its momentum.
00:37:51.500 --> 00:37:54.700
The average value of X², the average value of the momentum².
00:37:54.700 --> 00:37:56.700
These are all very important values.
00:37:56.700 --> 00:38:08.300
You remember the relationship between this and this, and this and this, gives you the uncertainties.
00:38:08.300 --> 00:38:15.000
Let us go ahead and finish off with a reasonably simple problem.
00:38:15.000 --> 00:38:17.000
It seems to be a lot is actually quite simple.
00:38:17.000 --> 00:38:22.500
For this example, of the root mean square displacement is just a square root of what it is that we just found.
00:38:22.500 --> 00:38:26.500
We found this X² term, the average value.
00:38:26.500 --> 00:38:30.300
The root means square displacement is just the square root of this number.
00:38:30.300 --> 00:38:36.000
It is just this raised to the ½ power.
00:38:36.000 --> 00:38:42.000
Given the following table of fundamental vibration frequencies and equilibrium bond lengths,
00:38:42.000 --> 00:38:49.800
calculate the root mean square displacement of the molecules for the R = 0 state for the harmonic oscillator.
00:38:49.800 --> 00:38:55.500
Then, compare them to their respective equilibrium bond lengths.
00:38:55.500 --> 00:39:05.300
We have a molecule, we have N14 and this 16.
00:39:05.300 --> 00:39:07.800
I can erase it.
00:39:07.800 --> 00:39:13.100
You should actually be superscript it, sorry about that.
00:39:13.100 --> 00:39:20.300
Basically, this is just the inter molecule, these two should be down below.
00:39:20.300 --> 00:39:25.800
This is N2 and this is O2, there is nothing more than N2 and O2 molecules.
00:39:25.800 --> 00:39:34.100
To the inter molecules, the fundamental vibration frequency is 23, 30 and this is in wave numbers.
00:39:34.100 --> 00:39:39.900
It looks like I forgot about the 7 superscript of the numbers.
00:39:39.900 --> 00:39:43.100
The equilibrium bond length of the bond is 109.4.
00:39:43.100 --> 00:39:50.900
You remember the N2 is a triple bond, something like that, remember from general chemistry.
00:39:50.900 --> 00:39:53.200
For the oxygen, it is not 14.
00:39:53.200 --> 00:39:57.400
I’m just having a great time here, this is oxygen 16.
00:39:57.400 --> 00:40:07.200
The oxygen molecule, its fundamental vibration frequency is 1556 inverse cm.
00:40:07.200 --> 00:40:11.900
Its equilibrium bond length is 120.7.
00:40:11.900 --> 00:40:16.700
These are actually in pm.
00:40:16.700 --> 00:40:23.200
And if you remember, a pm is actually a 10⁻¹² m.
00:40:23.200 --> 00:40:27.700
Milli, micro, nano, and pico.
00:40:27.700 --> 00:40:31.000
Let us see what we have got.
00:40:31.000 --> 00:40:34.700
They give us the fundamental frequency.
00:40:34.700 --> 00:40:41.700
They want us to find the root mean square displacement of molecules.
00:40:41.700 --> 00:40:53.500
I guess the first thing we probably need to do is find K.
00:40:53.500 --> 00:40:57.200
Because we want us to find this, they want us to find that.
00:40:57.200 --> 00:41:20.700
We said in the previous example, when we generalize that, that is just equal to H ̅/ K μ¹/2 × R + 1/2.
00:41:20.700 --> 00:41:29.500
We need to go ahead and find what this K is and we are also going to be probably end up finding μ also.
00:41:29.500 --> 00:41:31.500
Let us go ahead and do that.
00:41:31.500 --> 00:41:35.700
We are going to go ahead and do the nitrogen molecule first.
00:41:35.700 --> 00:41:37.500
Let us take care of that.
00:41:37.500 --> 00:41:43.000
We will go ahead and do nitrogen first.
00:41:43.000 --> 00:41:54.700
The fundamental vibration frequency is equal to 1/ 2 π C × K/ μ¹/2.
00:41:54.700 --> 00:42:04.700
When I rearrange this I, end up with K, a force constant is actually going to be equal to
00:42:04.700 --> 00:42:17.200
2 × π × C × this fundamental vibration frequency² × μ.
00:42:17.200 --> 00:42:19.200
Let us see what we have got.
00:42:19.200 --> 00:42:30.800
We said from the previous problem, we said that X² is going to equal 1/ Α R + ½.
00:42:30.800 --> 00:42:41.500
For R = 0, we are just going to get 1/ 2 Α.
00:42:41.500 --> 00:42:54.000
This is going to actually end up equaling H ̅/ 2 × K μ¹/2.
00:42:54.000 --> 00:42:58.300
Let us go ahead and find some things here.
00:42:58.300 --> 00:43:03.000
We just said that K is equal to this.
00:43:03.000 --> 00:43:08.000
We need to find this and I would end up taking the square root of it.
00:43:08.000 --> 00:43:16.600
Μu for this was just, it was mass 1 × mass 2/ mass 1 + mass 2.
00:43:16.600 --> 00:43:24.100
14 × 14/ 14 + 14, you are going to end up with actually 7 atomic mass units.
00:43:24.100 --> 00:43:27.300
Let me go ahead and do explicitly here.
00:43:27.300 --> 00:43:34.100
It is going to be 14 × 14 atomic mass units/ 14 + 14.
00:43:34.100 --> 00:43:43.900
I will multiply that by 1.661 × 10⁻²⁷ kg/ atomic mass unit.
00:43:43.900 --> 00:43:52.600
We end up of μ equaling 1.163 × 10⁻²⁶ kg.
00:43:52.600 --> 00:43:58.400
I hope that the arithmetic is correct, I’m not exactly sure if I did.
00:43:58.400 --> 00:44:31.100
Our K is going to equal 1.163 × 10⁻²⁶ kg × 2 × π × 3 × 10¹⁰ cm/ s × 2330 inverse cm².
00:44:31.100 --> 00:44:39.600
When I do that, I get a K is equal to 2243.
00:44:39.600 --> 00:44:52.400
Our X² is going to equal H ̅/ 2 K μ¹/2.
00:44:52.400 --> 00:45:21.700
That is going to equal, H ̅ is 1.055 × 10⁻³⁴ J-s divided by 2 × 2243 × 1.163 × 10⁻²⁶¹/2.
00:45:21.700 --> 00:45:25.900
I hope that I do my arithmetic correctly.
00:45:25.900 --> 00:45:33.200
It is going to be 1.033 × 10⁻²³.
00:45:33.200 --> 00:45:57.400
When I take the square root of that, I'm going to end up with 3.214 × 10⁻¹² m or 3.214 pm.
00:45:57.400 --> 00:46:02.700
There you go, from the previous problem we had a formula for that.
00:46:02.700 --> 00:46:05.700
Let us say it is R = whatever.
00:46:05.700 --> 00:46:10.700
We found that, and we realized that we have to find K and we have to find μ.
00:46:10.700 --> 00:46:15.000
We used the fundamental frequency that they gave us to actually find a value of K.
00:46:15.000 --> 00:46:26.500
We put it in, we found the average value of X² and we took the square root in order to get the root mean square displacement.
00:46:26.500 --> 00:46:32.500
We found that its 3.214 pm for the nitrogen molecule.
00:46:32.500 --> 00:46:37.500
As far as the oxygen molecule is concern, I'm going to go ahead and let you do that one yourself.
00:46:37.500 --> 00:46:40.200
It is the exact same process.
00:46:40.200 --> 00:46:42.200
Thank you so much for joining us here at www.educator.com.
00:46:42.200 --> 00:46:43.000
Take care, I will see you next time, bye.