WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome to www.educator.com, welcome back to Physical Chemistry.
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So today, 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 one says, plot the probability density vs. the radius for the 2S radial function as seen below.
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Find the values of the radius such that the probability density achieves a maximum.
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We have seen this plot before in a previous lesson.
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This is a plot of the probability density for the radial function.
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That is this thing right here.
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We basically take the radial function, the radial portion of the wave function.
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We square it and of course we have that R² factor because we are dealing in spherical coordinate, that is the Y axis.
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This of course is the radius itself, that is the X axis and it is in increments of the bohr radius A₀.
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For example, A₀, 2A₀, 3A₀, and so on.
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We want to find the values of the radius such that the probability density achieves a maximum.
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In other words, where the most likely places to actually find the electron is far as distance from the nucleus.
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We would be looking for the X value that gives this point and the X value that gives this point.
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We want to find this X value and this X value analytically.
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Let us go ahead and get started.
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Let us go ahead and start over here.
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I will go ahead and stick with black.
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2S means that the N value is equal to 2 and L = 0.
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The radial function is going to be R20, that is the function that we are going to use.
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The integral is the integral of R20 * R 20 R² DR.
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This is the probability integral.
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The normalization integral, if you will.
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It is just a function itself, the conjugate multiplied by the function R² DR.
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This is what gives us the probability.
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This R 20 * × R20 × R², in other words the integrand, it is the probability density.
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We remember this from very early on.
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The function, the conjugate of the function × the function,
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that actually is in this particular case because a spherical coordinates, we have this extra term right here.
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The integrand without the DR that gives us the probability density.
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When we multiply by the DR, that gives us the probability when we integrate it.
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It is the probability density of the radial function.
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This is the one that we need to maximize.
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We need to find this, take the derivative, set the derivative equal to 0 and solve the equation for R.
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The probability density of the radial function.
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Let us go ahead and see.
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R20 is equal to, based on the fact that N = 2 and L = 0.
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Again, we have a formula for the radial function.
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I’m going to do it explicitly here.
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It is going to be - 2 -0 -1!/ 2 × 2 + 0!³ ^½ × 2/ 2 ×
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A₀⁰ + 3/2 R⁰ E ⁻R/ 2 A₀.
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I’m just plugging in all the values for N and L into that equation that we have for the radial function,
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the general equation, you can find it in the previous lesson or you will find it in your book.
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Of course, we have the L, this is going to be 2 + 0 2 × 0 + 1.
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The argument of this particular polynomial is going to be 2R/ 2 A₀.
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When we go ahead and work all that out, in this particular case the L 21, R, A₀,
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is going to equal -2! × 2 - R/ α₀.
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When we actually work all of this out, we get R of 20 is equal to 1/32¹/2 × 1/ A₀³/2
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E ⁻R/ 2 A₀ × 2 - R/ A₀ × 2.
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When we take the conjugate of this, multiply it by this, and then multiply by R²,
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what we are looking for is this, * × R 2₀ R².
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We are going to end up with 1/ 8 A₀³ E ⁻R/ A₀ × 2 - R/ A₀² × R².
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This is the function that we wanted.
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This is the function that we are going to take the derivative of and set equal to 0.
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This is the probability density.
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This is what was graphed in the previous page.
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Maximize means take the derivative with respect to R of this thing, this R20 * R20 R²,
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take the derivative and set it equal to 0.
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We are going to take the derivative of this function.
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Let me see, should I go ahead and do it on this page or should I do it on the next page?
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I can go ahead and just start on the next page here.
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This one I will do in red here.
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The derivative is going to equal 1/8 A₀³.
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It is going be a little long, of course, it is quantum mechanics so it is long.
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/ A₀ × 2 - R/ A₀² R² + E ⁻R/ A₀ × - 2/ A₀ × 2 - R/ A₀ ×
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R² + E ⁻R/ A₀ × 2 - R/ A₀² × 2 R.
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This is just the standard derivative just tends to be a little bit longer, that is all.
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What are we going to do?
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Hopefully, I have not forgotten a + or – somewhere, that is always the case.
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8/ A₀³ E ⁻R/ A₀ × R × 2 - R/ A₀ × - R/ A₀ × 2 - R/ A₀ –
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2 R/ A₀ + 2 × 2 - R/ A₀.
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All of that is the derivative and all of that is going to be equal to 0.
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We have something × something × something × something = 0.
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We have to set each of these factors equal to 0.
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Over here, we have the R equal to 0, that is one possibility.
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We have 2 - R/ A₀ equal to 0 which means that R = 2 A₀, that is another possibility.
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Of course, we have this last factor which we said equal to 0.
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I’m going to go ahead and multiply this out.
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It is going to be - 2 R/ A₀ + R² / A₀² -2 R/ A₀ + 4 - 2 R/ A₀ and that is going to equal 0.
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When I put this together, I'm going to get -2 R A₀ + R².
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I will multiply all of these by A₀² -2 R A₀ + 4 A₀² -2 R A₀ = 0.
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I'm going to end up with R² -6 R A₀ + 4 A₀.
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That is my quadratic equation.
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This is AR² + BR + C =0, basic quadratic equation.
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Let me go ahead and use the quadratic equation here.
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Here, the coefficients are, you have 1, 6 A₀ is the B, and of course you have this.
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That is fine, I will go ahead and actually work this out.
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R is equal to - B so we have 6 A0 + or - √ B² -4 is C so we have 36.
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A₀² -4 × A × C so it is going to be -16.
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A₀²/ 2 A which is going to be 2.
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We are going to end up with 6 A₀ + or – 8₀ × 2 √ 5/ 2.
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This of course, this is going to give us 3 A₀ + or – A₀ × √ 5.
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Our final answer is going to be 3 + or - √ 5 × A₀.
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That is actually going to equal, one of the answers I will do over here.
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One of the answers is going to be 0.76 A₀, that is the 3 – √ 5.
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The other answer is going to be 5.24 A₀.
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These are our two answers and this is the exact answer, if you do not want to do it in decimal form.
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Notice, one of them is R = 0.
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That is just the radius equal 0, that is just use to graph it.
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The graph would like this and like that.
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That is this point.
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R = 2 A₀ is going to end up being the minimum.
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That is where it is going to the probability density actually goes to 0.
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The two maxima, here and here, that is what those are.
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That takes care of that problem.
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Let us see what the next problem is.
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Our model for the hydrogen atom has the electron interacting with a proton via the Coulombic potential.
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The Coulombic potential is basically just the one that you remember from your study of electricity and magnetism.
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Just charges attracting each other, the potential exists between them.
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The Coulombic potential, the potential energy is -E² / 4 π ε₀ R.
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Ε₀ is the permittivity of free space and E is the charge in Coulombs.
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The operator V it just means multiplied by V sub R.
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For the 2P Z electron, we want you to show that the average potential energy = twice
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the over all energy and that the kinetic energy = – the overall energy.
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Let us see what we have.
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We are dealing with the 2P Z electron.
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I think I will go to black.
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For the 2 PZ electron, we have N is equal to 2, we have L is equal to 1, that is the P,
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and we have M is equal to 0.
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The Z subscript always means that M is equal to 0.
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What we are looking at here is the wave function for ψ 210.
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Ψ 210 that is the 2 PZ electron.
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Let us see what that is.
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When we look that up, 210, we get the following.
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We get 1/ √ 32 π × 1/ A₀³ × σ E ^-Σ/ 3 - σ/ 2.
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I think I have written this incorrectly but that is okay.
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Cos θ, and here our σ is equal to R/ α₀.
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Let us see here.
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If I’m not mistaken, I think this is going to be 3/2.
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We are looking for the average kinetic energy, average potential energy, average kinetic energy,
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and we are trying to establish this relationship.
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We know what this is so we need to find the average value, the expectation value.
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The average value expectation value depends on what your teacher calls it, I call it both.
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We know already that the expectation value or the average value,
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the integral is given by the integral of the wave function itself conjugate × the operator operated on the function itself.
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This is the integral that we need to form ψ * V ψ.
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This is a real function so ψ * is just ψ itself.
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All we have to do is multiply this function by itself and multiply by the Coulombic potential.
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When we do that, we get the following.
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We get that the average value of the potential energy is going to equal -E² / 128 π².
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I think it should be to be the 4th π² A₀ A₀³ × the integral.
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The integral was actually a triple integral because we have 3 variables.
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It is going to be the integral from 0 to 2 π of D φ, the integral from 0 to π of cos² θ sin θ D θ.
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It is going to be the integral from 0 to infinity of 1/ R Σ E ^-σ R² DR.
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We are dealing with a triple integral.
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We are dealing with something in spherical coordinates.
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And it is so this is what the total integral looks like.
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The integral from the previous page where we just used ψ, VC.
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When I multiply everything out, this is what I get.
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Again, we have we have seen this several times already.
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This is not a big deal.
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This integral right here is equal to 2 π.
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This integral right here is equal to 2/3.
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I’m not going to go ahead and go through each individual integral.
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When we put that together, we end up with this being equal to, when we take this 2/3 and 2 π and put it all here,
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we end up with -E²/ 96 π E₀ A₀³ × the integral from 0 to infinity of R σ E ^-Σ DR.
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Now, we need to solve this integral.
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We know that σ is equal to R/ α₀ which means that R is equal to σ α₀ D Σ = 1/ α₀ DR,
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which means that DR is equal to D σ α₀.
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When I substitute all of these back into this integral, I end up getting -E² / 96 π E₀
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A₀³ × the integral of σ ^-σ D σ.
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This integral, I have seen it several × before.
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It is just equal to 3!, it is just equal to 6.
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I'm certain I forgot, there is actually A₀² here.
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When I substitute all of these back in, some of these A₀ actually show up here.
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They are pulled out as constants.
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Whet I end up getting is the following.
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I end up getting -6 E² A₀²/ 96 π E₀ A₀³.
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This goes with that, leaving 16.
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This knocks that out and I’m left with -E²/ 16 π E0 A₀.
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The average potential energy is this thing right here.
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Let us go ahead and jump to the next page here.
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Let me write that.
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I have my average, it is -E² / 16 π E₀ A₀.
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Let me go to blue.
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The energy sub N is equal to –E sub²/ 8t π E₀ A₀ N².
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N is equal to 2 so this is going to end up being 32.
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2 × 2 is 4, 4 × 8 is 32, so the energy of level 2 is going to end up being -E²/ 32 π E₀ A₀.
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The average potential energy is this value and the total energy is that value.
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Everything is the same, the only difference is the 16 and the 32.
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What we have is, we have ½ × the average potential energy ½ × this is equal to the energy itself.
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Or V is equal to 2 E, which is one of the things that we wanted to prove.
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Just like actually finding the expectation value for the potential energy.
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The average value of the potential energy using the integral definition of it.
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We just worked out on integral which is really what we do most of time.
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Now, the total energy is equal to the kinetic energy + the potential energy.
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It is equal to the kinetic energy + the potential.
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We just solved the potential, it is twice the energy.
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Now, we have this equation, I’m just going to move this over the other side.
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I end up with is - the energy is equal to average of the kinetic energy.
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This was the other thing that we wanted to prove.
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The relationship ends up being the average potential energy is equal to twice the average energy.
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It is equal to - × the kinetic energy.
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That is the fundamental relationship that exists for this particular orbital.
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Let us see what the next problem says.
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We are taking from example 2, except for the wave function ψ sub 311, we did ψ 210.
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We want to do it for 311, that is find the relation among the average kinetic,
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the average potential, and the average total energy.
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Let us do the same thing, let us solve the integral.
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Let us start off with what ψ 311 is, that is equal to 1/ 81 √ π 1/ A₀³/2.
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Just a more complicated function, that is all, not a big deal.
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6 σ - σ² E ^-σ / 3 × sin of θ × E ⁺I φ.
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This was complex, not a problem though.
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Its conjugate is going to be E ⁻I φ.
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When we multiply ψ conjugate × ψ, the complex parts can actually go away.
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Again, it is always a good idea to write down what it is that you want.
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We want to form the following integral.
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The integral of ψ 311 conjugate × the operator V operating on ψ 311.
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That is the integral that we want to form.
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Therefore, this is the average potential energy.
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Therefore, the average potential energy is going to equal -E² / 4 π ε 0 × 1/ 6561 π × 1/ A₀³ ×
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the integral from 0 to 2 π of the D φ part, the integral from 0 to π of the sin² θ sin θ D θ part.
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We will bring it here because the integral is long.
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× the integral from 0 to infinity of 1/ R × 6 σ.
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This 1/ R came from the Coulombic potential.
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Let us put R into under the integral for the R variable.
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6 σ - σ²² E⁻² σ/ 3 R² DR.
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Let us go to red here.
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This integral right here was equal to 2 π.
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This integral right here, when I solve that one, that integral is equal to 4/3.
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When I put everything together, the 4/3, the 2 π, multiply all this out, I end up getting the following.
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Let me now go to red.
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It is going to equal -4 E² × 2 π/ 3 × 4 π E0 × 6561 π A₀³.
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I left everything there, I had not canceled anything yet.
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0 to infinity of² R, this is going to be R ×, when I square this 36 σ² - 12 σ³ + σ⁴ E ^- 2/3 σ DR.
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Once again, for this integral I have σ is equal to R/ A 0, which means that R is equal to σ A 0.
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D Σ = 1/ A 0 DR which means that DR is equal to A 0 D σ.
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When I substitute all of these into this integral, what I end up getting is the A 0 come out.
00:30:51.500 --> 00:31:06.600
What I end up getting is the following.
00:31:06.600 --> 00:31:13.100
I should go ahead and write it all out this one or do the cancellation.
00:31:13.100 --> 00:31:16.400
I think I will go ahead and write it all to actually see the cancellation.
00:31:16.400 --> 00:31:21.200
When I plug all of these in to this integral, I’m going to leave this alone.
00:31:21.200 --> 00:31:23.100
What I would end up with is the following.
00:31:23.100 --> 00:31:38.000
I end up with -4 E² 2 π A₀ A₀², that came from all the work that I just did with the σ.
00:31:38.000 --> 00:32:10.000
3 × 4 × π × ε₀ × 6561× π A₀³, the integral from 0 to infinity of σ × 36 σ² -12 σ³ + σ⁴ E⁻²/3 σ D σ.
00:32:10.000 --> 00:32:25.400
4 and 4, π and π, A₀ and A₀, that cancels that.
00:32:25.400 --> 00:32:33.500
What I end up with is the following.
00:32:33.500 --> 00:32:49.200
My average potential energy is going to equal - 2 E²/ 3 × 6561 × π ε₀ A₀ ×
00:32:49.200 --> 00:33:06.200
the integral of 36 σ³ -12 σ⁴ + σ⁵ E⁻²/3 σ D σ.
00:33:06.200 --> 00:33:11.000
When I'd taken this integral, can I break it up into 1 integral, 2 integral, 3 integral.
00:33:11.000 --> 00:33:13.200
I have seen this integral several times.
00:33:13.200 --> 00:33:15.400
I end up the following.
00:33:15.400 --> 00:33:50.500
I end up with, 36 × 3!/ 2/3⁴ - 12 × 4!/ 2/3⁵ + 5!/ 2/3⁶.
00:33:50.500 --> 00:34:04.100
All of this ends up equaling to 16 × 81/ 16 -288.
00:34:04.100 --> 00:34:07.600
I hope you do not mind that I’m actually going through all of the arithmetic here.
00:34:07.600 --> 00:34:11.000
I figured that at least do it for this one integral.
00:34:11.000 --> 00:34:14.800
+ 120 × 729.
00:34:14.800 --> 00:34:17.700
I know that you guys are more than capable of doing the arithmetic.
00:34:17.700 --> 00:34:22.600
I can just go ahead and give the answer but figured out that just go through with it here.
00:34:22.600 --> 00:34:41.300
= 17,496/ 16 - 34,992/ 16 + 21,870/ 16.
00:34:41.300 --> 00:34:49.300
That integral = 4374/ 16.
00:34:49.300 --> 00:35:00.700
K so we have 4374/ 16 × -2 E².
00:35:00.700 --> 00:35:07.500
Remember, the constant 3 × 6561 π A₀.
00:35:07.500 --> 00:35:19.600
A₀ actually is going to equal - E²/ 36 π E₀ A₀.
00:35:19.600 --> 00:35:32.100
Our average potential energy is equal to -E² / 36 π E₀ A₀.
00:35:32.100 --> 00:35:42.300
A3, let me do it in blue.
00:35:42.300 --> 00:36:03.400
Energy level 3 is equal to -E² / 8 π E₀ A₀ 3² is equal to - E² / 72 π E₀ A₀.
00:36:03.400 --> 00:36:14.500
Once again, we have ½ × this is equal to this.
00:36:14.500 --> 00:36:23.700
Therefore, the average value of V is equal to twice the value of the energy.
00:36:23.700 --> 00:36:27.300
We ended up with the same result.
00:36:27.300 --> 00:36:38.600
Once again, the energy is equal to the kinetic energy + the potential energy is equal to the kinetic energy.
00:36:38.600 --> 00:36:43.800
The potential is plus twice the full energy.
00:36:43.800 --> 00:36:47.300
Once again, we have this equation that we solve.
00:36:47.300 --> 00:37:01.100
We bring that over there, we end up with - the energy = the average kinetic energy.
00:37:01.100 --> 00:37:20.800
Again, we have that the average potential energy is equal to twice the total energy is equal to - twice the kinetic energy.
00:37:20.800 --> 00:37:28.900
It is the same relation that we had for the 210 orbital, 311 orbital.
00:37:28.900 --> 00:37:33.000
This result is true in general.
00:37:33.000 --> 00:37:36.800
This is where we are going to begin a little bit of general discussion.
00:37:36.800 --> 00:38:12.400
This result is true in general.
00:38:12.400 --> 00:38:25.900
This result is true in general, when the potential energy is the Coulombic potential.
00:38:25.900 --> 00:38:33.200
When V of R is the Coulombic potential.
00:38:33.200 --> 00:38:40.800
If you happen to be dealing with the Coulombic potential, the -E²/ 4 π E₀ R,
00:38:40.800 --> 00:38:48.400
you are always going to get the relationship that the average potential energy = twice the total energy
00:38:48.400 --> 00:38:55.200
= –twice the kinetic energy.
00:38:55.200 --> 00:39:07.300
When V is any potential whatsoever, ½ KX² 3 R³ cos, whatever potential,
00:39:07.300 --> 00:39:12.500
Let me go back to black here.
00:39:12.500 --> 00:39:32.700
When V is any potential whatsoever, then the general expression for the relationship
00:39:32.700 --> 00:40:16.000
between the kinetic energy and the potential energy, is as follows.
00:40:16.000 --> 00:40:26.200
Where V is equal to the function of X, Y, and Z, now we are dealing with all three space.
00:40:26.200 --> 00:40:31.100
This is something called the Virial theorem, the quantum mechanical Virial theorem.
00:40:31.100 --> 00:40:36.500
The earlier result is also called of Virial theorem for the Coulombic potential.
00:40:36.500 --> 00:40:39.500
This is the general expression for the Virial theorem.
00:40:39.500 --> 00:40:55.400
This is called very important, this is called the Virial theorem.
00:40:55.400 --> 00:40:57.500
Let us take a look at what will we get.
00:40:57.500 --> 00:40:59.700
We are dealing with Coulombic potential.
00:40:59.700 --> 00:41:07.500
Potential energy was this -E²/ 4 π E₀ A₀ R.
00:41:07.500 --> 00:41:15.400
When that is the case, the relationship among the average potential energy, the energy, and the average kinetic energy is this.
00:41:15.400 --> 00:41:20.800
In general, for any kind of potential whatsoever, not the Coulombic potential but any potential,
00:41:20.800 --> 00:41:33.500
the relationship is the average value of given some potential energy as a function of X, Y, Z, or R θ φ, whatever.
00:41:33.500 --> 00:41:41.600
If I take X × the partial with respect to X, Y × the partial with respect to Y, Z × the partial with respect to Z.
00:41:41.600 --> 00:41:46.200
If I take the average value of that, in other word solve the integral.
00:41:46.200 --> 00:41:51.100
I'm going to end up getting is going to equal twice the average of the kinetic energy.
00:41:51.100 --> 00:41:54.300
This is the quantum mechanical Virial theorem.
00:41:54.300 --> 00:42:00.800
Another version of the Virial theorem is this one.
00:42:00.800 --> 00:42:05.400
Another version, which might be the version that you actually see in your book.
00:42:05.400 --> 00:42:25.400
Another version of the Virial theorem reads as follows.
00:42:25.400 --> 00:43:00.600
When the potential energy of the particle has the form V is equal to some constant, I’m going to use K.
00:43:00.600 --> 00:43:36.500
Some constant KX⁸, then the average values of kinetic and potential energies are related
00:43:36.500 --> 00:43:46.800
by twice the average kinetic is equal to this constant A × the potential.
00:43:46.800 --> 00:43:51.700
This is that, so this is another version of the Virial theorem.
00:43:51.700 --> 00:43:53.900
This might be the one that you actually see in your book.
00:43:53.900 --> 00:44:03.100
This is the general expression of the Virial theorem or the quantum mechanical Virial theorem for any potential, whatsoever.
00:44:03.100 --> 00:44:06.800
This is specifically if the potential is in this form.
00:44:06.800 --> 00:44:13.600
For the problems that we did for the 210 or the 311 orbital, these are the Coulombic potential.
00:44:13.600 --> 00:44:18.200
In that particular case, this was the relationship that we got.
00:44:18.200 --> 00:44:24.800
This is the general expression right here.
00:44:24.800 --> 00:44:31.200
Whatever the form of the potential is, if you do this and then take the average value of that,
00:44:31.200 --> 00:44:35.000
the integration, you end up twice the average value of the kinetic energy.
00:44:35.000 --> 00:44:38.700
This is a profoundly deep and important theorem.
00:44:38.700 --> 00:44:45.000
Not just in quantum mechanics, in classical mechanics as well.
00:44:45.000 --> 00:44:46.400
Let us a little bit more about this.
00:44:46.400 --> 00:44:57.700
Let me actually go to black here.
00:44:57.700 --> 00:45:22.800
The Coulombic potential is -E² / 4 π ε₀ A₀ × R, which is the same as
00:45:22.800 --> 00:45:33.700
–E₀²/ 4 π ε₀ A₀⁻¹.
00:45:33.700 --> 00:45:35.700
This thing is just a constant.
00:45:35.700 --> 00:45:44.500
Basically, what you end up with is a constant × R⁻¹.
00:45:44.500 --> 00:45:49.100
Based on what we just wrote, that all conversion of the Virial theorem,
00:45:49.100 --> 00:46:07.300
twice the kinetic energy is equal to - 1 × the potential energy, which is exactly what we got before except we want a – here.
00:46:07.300 --> 00:46:16.800
We went ahead and to confirm what it is that we already did.
00:46:16.800 --> 00:46:41.300
If we use the larger expression, the one with the partial derivatives, the Coulombic potential is written as this.
00:46:41.300 --> 00:46:52.100
The Coulombic potential can be written as, here we have R.
00:46:52.100 --> 00:46:59.600
When we are given 3 space, R is just equal to vx² + Y² + Z².
00:46:59.600 --> 00:47:07.000
This is the spherical coordinate designation for something ends in Cartesian coordinates.
00:47:07.000 --> 00:47:12.400
If you have a point in 3 space, there are some vector that goes from the origin to that point.
00:47:12.400 --> 00:47:15.800
That distance is X² + Y² + Z².
00:47:15.800 --> 00:47:38.400
If I put this into here, my potential expressed as X, Y, Z is going to be -E²/ 4 π E₀ A₀ X² + Y² + Z².
00:47:38.400 --> 00:47:54.600
When I form X DB DX + Y, the derivative of this with respect to Y,
00:47:54.600 --> 00:48:06.800
and Z the derivative of this with respect to Z, when I do that and I take the average value of it,
00:48:06.800 --> 00:48:13.300
I end up actually getting that -V = 2K.
00:48:13.300 --> 00:48:24.700
All of these are the same thing, the quantum mechanical Virial theorem.
00:48:24.700 --> 00:48:34.200
Let us go ahead and see what our next example brings.
00:48:34.200 --> 00:48:38.400
Use the quantum mechanical Virial theorem in its full partial derivative form to show
00:48:38.400 --> 00:48:47.300
that the average potential = the average kinetic = 1/2 of the energy for the harmonic oscillator.
00:48:47.300 --> 00:48:55.800
Interesting, once again, the full partial derivative form is this X DDX, Y DDY, Z DDZ = 2K.
00:48:55.800 --> 00:49:01.300
We need to show that for the harmonic oscillator, this is the relationship that exists
00:49:01.300 --> 00:49:07.300
between average potential and the average kinetic and the energy.
00:49:07.300 --> 00:49:26.500
For the harmonic oscillator in 3 dimensions, the potential,
00:49:26.500 --> 00:49:44.500
is equal to the potential X, Y, Z is equal to ½ K1 X² + ½ K2 Y².
00:49:44.500 --> 00:49:49.100
Earlier, we just do one harmonic oscillator, something that is sliding back and forth like this.
00:49:49.100 --> 00:49:52.200
But now in 3 dimensions, it can be oscillating this way or this way.
00:49:52.200 --> 00:49:55.400
We have to include X, Y, Z.
00:49:55.400 --> 00:50:02.800
It is just the potential, we know the potential of the harmonic oscillator is ½ KX².
00:50:02.800 --> 00:50:07.300
We just add the Y potential and the Z potential which happens to be the same.
00:50:07.300 --> 00:50:14.500
½ K₃ Z².
00:50:14.500 --> 00:50:20.800
Let us just go ahead and find DV DX.
00:50:20.800 --> 00:50:25.500
DV DX, that is nothing, that is nothing because we are holding these, we are doing partial derivatives.
00:50:25.500 --> 00:50:33.000
We are holding these constant, we end up with K 1X.
00:50:33.000 --> 00:50:38.400
The partial with respect to Y is going to equal to K 2Y.
00:50:38.400 --> 00:50:45.200
The partial with respect to Z is going to equal K 3Z.
00:50:45.200 --> 00:50:52.400
I’m going to multiply this X DV DX, I’m just forming this thing.
00:50:52.400 --> 00:50:54.900
That is all I’m doing, I’m forming that thing.
00:50:54.900 --> 00:51:07.200
X DVDX + Y DVD Y + Z DVDZ.
00:51:07.200 --> 00:51:18.300
That is going to equal K1 X² + K2 Y², X × K1 X is K1 X².
00:51:18.300 --> 00:51:21.800
Y × K2 Y is K2 Y².
00:51:21.800 --> 00:51:31.800
Z × DV DZ, Z × K3 Z is K3 Z², + K3 Z².
00:51:31.800 --> 00:51:41.200
This is equal to twice the ½ K1 X².
00:51:41.200 --> 00:51:46.300
Twice of this is equal to this, I’m just rewriting it.
00:51:46.300 --> 00:51:54.600
+ 1/2 K2 Y² + ½ K3 Z².
00:51:54.600 --> 00:52:02.300
Well ½ K1 X Squared + 1/2 K2 Y² + 1/2 K3 Z², that is this thing.
00:52:02.300 --> 00:52:04.900
That is V already.
00:52:04.900 --> 00:52:32.200
What we have is this thing that we just did, this X DVDX + Y DVDY + Z DVDZ is equal to twice the potential energy.
00:52:32.200 --> 00:52:37.100
That means this is equal to this.
00:52:37.100 --> 00:52:55.400
We also know this from what we just saw that the Virial theorem, = this 2V = 2K.
00:52:55.400 --> 00:53:05.900
When you to remove the 2, you get V = K, which is the first part of what it is that we have to prove.
00:53:05.900 --> 00:53:16.500
The total energy is equal to the kinetic energy + the potential energy.
00:53:16.500 --> 00:53:26.500
The total energy =, the kinetic energy = the potential energy so this is just twice that.
00:53:26.500 --> 00:53:28.100
Of course, we have the final result.
00:53:28.100 --> 00:53:30.600
We just divide by 2, it does not really matter.
00:53:30.600 --> 00:53:36.000
½ of the energy, this was the other thing that we want to prove.
00:53:36.000 --> 00:53:41.900
That is it, all based on the quantum mechanical Virial theorem.
00:53:41.900 --> 00:53:56.200
Whenever you are given the potential V, if you form X DVDX + Y DVDY + Z DVDZ,
00:53:56.200 --> 00:54:01.300
the average value of that is going to always equal twice the kinetic energy.
00:54:01.300 --> 00:54:05.900
That is the quantum mechanical Virial theorem.
00:54:05.900 --> 00:54:12.500
Let us see what we have got.
00:54:12.500 --> 00:54:15.600
Let us see what is next.
00:54:15.600 --> 00:54:24.200
In problem 5 of the previous lesson, we found that the average value of R is equal to 6 A₀/ Z
00:54:24.200 --> 00:54:27.600
for the 2S orbital, for the hydrogen atom.
00:54:27.600 --> 00:54:32.200
If it is just hydrogen, the Z = 1 that is just 6 A0.
00:54:32.200 --> 00:54:38.700
Find that σ sub R², find the variance for the 2S orbital.
00:54:38.700 --> 00:54:49.600
Recall that σ sub R² is equal to the average value of R² - the average value of R quantity².
00:54:49.600 --> 00:54:55.600
The average value of R 6A/ Z, we already have this one.
00:54:55.600 --> 00:54:56.200
Let me go to red.
00:54:56.200 --> 00:55:08.200
What we need to find is the average value of R².
00:55:08.200 --> 00:55:12.800
We must find the average value of R².
00:55:12.800 --> 00:55:23.000
The 2S orbital means that N is equal to 2, L is equal to 0, and M is equal to 0.
00:55:23.000 --> 00:55:28.200
We are looking for the wave function 200.
00:55:28.200 --> 00:55:44.700
Well , the average value of R² is going to equal the integral of ψ 200 conjugate × R² ψ 200.
00:55:44.700 --> 00:55:51.900
The operator R² just means multiply by R², multiply by R and multiply by R again.
00:55:51.900 --> 00:55:56.400
What we end up getting, I’m not going to go ahead and write the wave function ψ 200,
00:55:56.400 --> 00:55:58.700
you can go ahead and look that up.
00:55:58.700 --> 00:56:01.300
What we end up actually getting is the following.
00:56:01.300 --> 00:56:06.400
We end up with the 1/ 32 π so it is just going to be ψ is real.
00:56:06.400 --> 00:56:19.100
It is just going to ψ 200² × R² 1/ A₀³ × the integral from 0 to 2 π D φ.
00:56:19.100 --> 00:56:25.100
Again we are always working in spherical coordinates here, D θ.
00:56:25.100 --> 00:56:48.200
0 to π of sin θ D θ 0 to infinity of R² × 2 - σ² E ^-Σ R² DR.
00:56:48.200 --> 00:56:51.300
This integral is equal to 2 π.
00:56:51.300 --> 00:56:54.500
This integral is also going to end up equaling 2 π.
00:56:54.500 --> 00:57:12.400
We end up with 1/ 8 A₀³ × the integral from 0 to infinity of R⁴ × 2 -σ² E ^-Σ DR.
00:57:12.400 --> 00:57:23.300
We have the same way, σ = R/ α₀, which means that R = σ × α₀.
00:57:23.300 --> 00:57:33.800
D σ = 1/ α₀ R, that is the same α.
00:57:33.800 --> 00:57:39.800
DR, which means that DR is equal to A₀ D σ.
00:57:39.800 --> 00:57:45.300
When we put all of these in here, under the integrand, we end up with the following.
00:57:45.300 --> 00:58:15.200
We end up with E₀⁴ × A₀/ 8 A₀³ the integral from 0 to infinity of Σ⁴ × 4 - 4 Σ + σ² × E ^-σ D σ.
00:58:15.200 --> 00:58:29.200
All of that is going to equal, when I cancel and work some things out, I'm going to end up with A₀²/ 8 ×,
00:58:29.200 --> 00:58:33.200
I’m going to separate the integrals out.
00:58:33.200 --> 00:58:36.300
That is fine, I will just go ahead and write it here.
00:58:36.300 --> 00:58:55.500
From 0 to infinity of 4 σ⁴ E ^-σ D σ - the integral from 0 to infinity of 4 σ⁵ E ^-σ D σ +
00:58:55.500 --> 00:58:58.900
the integral from 0 to infinity.
00:58:58.900 --> 00:59:02.000
I want to work this entire one myself.
00:59:02.000 --> 00:59:07.700
-σ D σ.
00:59:07.700 --> 00:59:22.000
It is going to end up equaling A₀²/ 8 × 4 × 4! -4 × 5! + 6!.
00:59:22.000 --> 00:59:31.100
When I worked all this out, I'm going to end up with 42 E₀².
00:59:31.100 --> 00:59:41.100
Our σ R² is equal to the average value R² - the average value of R².
00:59:41.100 --> 00:59:43.400
That is going to equal what we just got.
00:59:43.400 --> 00:59:54.500
This is this one, so it is going to be 42 A₀² -6 A₀².
00:59:54.500 --> 00:59:58.800
That is going to equal 6 A₀².
00:59:58.800 --> 01:00:01.100
That is our variance.
01:00:01.100 --> 01:00:05.400
Do I have another page here?
01:00:05.400 --> 01:00:12.000
I do, let me go ahead and go to the next page.
01:00:12.000 --> 01:00:19.700
The general formula, we just found for one particular orbital.
01:00:19.700 --> 01:00:31.400
The general formula for the average value R² is as follows.
01:00:31.400 --> 01:01:02.800
The average value of R² is equal to N⁴ A₀²/ Z × 1 + 3/2 × 1 – L × L + 1 -1/3 / N².
01:01:02.800 --> 01:01:09.300
There you go, that is the general expression for R².
01:01:09.300 --> 01:01:15.300
In the previous lesson, we also have a general expression for the average value of R.
01:01:15.300 --> 01:01:23.800
We have those based on just the quantum numbers N and L.
01:01:23.800 --> 01:01:27.300
Everything else is very easily taken care of.
01:01:27.300 --> 01:01:29.300
Thank you so much for joining us here at www.educator.com.
01:01:29.300 --> 00:48:33.000
We will see you next time, bye.