WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com, welcome back to Physical Chemistry.
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Today, we are going to continue with our example problems for molecular spectroscopy.
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Let us jump right on in.
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Under a correction for anharmonicity, the vibrational term is this thing right here.
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G of R equals ν sub E × R + ½ - X sub E ν sub E × R + ½².
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This is the correction for anharmonicity for just the vibrational term.
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Deriving expression for δ G of R and recall from the image of part B,
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recall from the image of the true potential energy curve, this thing down here, that δ G goes to 0 as R increases.
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In other words, δ G is the difference between the energy levels.
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As you go up, the energy levels, the difference between them decreases until you are at 0.
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Show that the maximum vibrational quantum number is given by this expression.
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In other words, our max is the number of vibration states before the molecule actually flies apart.
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Let us get started.
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Part A, this is going to be just an algebra problem.
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G of R, we have G of R is equal to ν sub E × R + ½ - X sub E ν sub E × R + ½².
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Δ G of R is equal to G of R + 1 – G of R, the upper state - the lower state.
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That is going to equal ν sub E × R + 1 + ½.
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There is a reason I cannot do a ½ symbol, that saved my life.
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- X sub E ν sub E × R + 1 - ½ - ν sub E, + ½ not – ½.
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Crazy, + and -.
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This one is – X sub E ν sub E × R + ½².
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This is going to equal ν sub E.
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Actually you know what, I wonder if I should do it that way?
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Ν sub E × R + 3/2.
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This is just a big long algebra problem.
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We are looking for δ G, we are looking for the frequency of absorption or emission
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when you go from 1 vibrational level to another vibration level.
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- X sub E ν sub E R + 3/2² – ν sub E × R + ½.
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- and -, this is going to be + X sub E ν sub E × R + ½².
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Let me go ahead and expand that.
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We get ν sub E × R + 3/2 ν sub E – X sub E ν sub ER² - 3 X sub E ν sub E.
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I hope you got my algebra correctly, please confirm it for me.
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-3X sub E ν sub E × R -9/4 X sub E ν sub E - ν sub E × R - ½ ν sub E + X sub E ν sub ER²
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+ X sub E ν sub ER + ¼ X sub E ν sub E.
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This goes with that, this goes with that.
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When I put all the terms together, we are going to end up with ν sub E - 2 X sub E ν sub E × R -2 X sub E ν sub E.
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Δ G is equal to ν sub E - 2 X sub E ν sub E × R + 1.
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Or I could take out a ν sub E and I can write it is 1 – 2X × R + 1.
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Either one of those is just fine.
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That is the expression that we are looking for.
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This gives me the frequency of absorption or emission for that transition.
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Part B, as R increases δ G goes to 0.
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Let us just set δ G in this equation equal to 0 and solve the equation.
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Let us set δ G equals 0 and solve for R max.
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We have ν sub E - 2 × X sub E ν sub E × R max + 1.
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Use this equation first one, this version of the equation, either one is fine, equal to δ G which is 0.
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We get 2X sub E ν sub E × R max + 1 is equal to ν sub E, what we solve the ν sub E cancels.
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We are going to end up with R sub max equals ν sub E / 2X sub E ν sub E – 1.
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Or if you want to cancel the ν sub E, you may or may not actually prefer not to cancel because ν sub E
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and X ν sub E are actually individual spectroscopic parameters and I like to see them.
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Or you can write it as R sub max equals 1/2 X sub E – 1.
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Again, either one is fine.
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This one happens to be my preference, not a big deal.
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Example 2, using the equations we derived in example 1, derive an expression for D sub E,
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the dissociation energy of a molecule.
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Recall that the dissociation energy of the molecules taken from a minimum
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of the potential energy curve not a 0 point energy of the oscillator.
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D sub E is from the minimum to the energy of dissociation.
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The D₀ that is from the 0 point to the dissociation.
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It is that little difference.
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The difference is the 0 point energy of the harmonica oscillator.
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Part B, looking at the equation for δ G very carefully, is there a way to find this ν sub E and X sub E ν sub E from a graph of some sort.
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Let us go ahead and do part A.
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Let u see, the dissociation energy is equal to the energy of R max.
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It is just equal to ν sub E × 1 / 2X sub E - 1 + ½ - X sub E ν sub E × 1/ 2 X sub E - 1 + ½².
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This is equal to ν sub E × 1/ 2 X sub E – ½ - X sub E ν sub E × 1/ 2 X sub E – ½².
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When I expand this I get, and multiply it out, I get ν sub E / 2X sub E - ν sub E / 2
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– X sub E ν sub E / 4X sub E + X sub E ν sub E or 2 X sub E² 2X sub E – X sub E ν sub E all / 4.
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That gives me ν sub E / 2X sub E - ν sub E / 2 - ν sub E / 4 X sub E + ν sub E / 2 - X sub E ν sub E/ 4.
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These go away and I'm left with, this is dissociation energy, my final answer,
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my expression for dissociation energy equals ν sub E / 4 X sub E – X sub E ν sub E/ 4 or ν sub E / 4 × X sub E × 1 - X sub E².
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This is the expression that I was looking for.
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Notice if you make dissociation energy in terms of ν sub E and X sub E, this is extraordinary.
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Now, this X sub E spectroscopic parameter.
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This X sub E is a lot less than 1, it is very small.
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If you look and if you remember some of the previous example problems that we did, it is very tiny.
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It is much less than 1.
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The second term here or this one right here is 0.
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The second term is close to 0.
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We usually just take the first term and we rewrite the dissociation energy as approximately equal to ν sub E / 4 X sub E.
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Or if you want because X sub E ν sub E tends to be a single spectroscopic parameter.
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At least that has listed a lot of data, you can just do ν sub E² / 4X ν sub E X sub E, that is fine too.
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There we go, that takes care of that.
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Part B, we had δ G is equal to ν sub E - 2X sub E ν sub E × R + 1.
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Let us rearrange this.
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Let us write δ G as a function of R is equal to - 2X sub E ν sub E × R + 1 + ν sub E.
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Y equals MX + B, a linear equation.
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We can plot δ G.
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And again, I have values for δ G.
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I can pick out a δ G values.
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Δ G vs. R + 1.
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0 + 1, 1 + 1, 2 + 1, 3 + 1, I have all these values, I can tabulate them.
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In this particular case, the ν sub E is the one you intercept when I get it, when I plot it.
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And -2 X sub E ν sub E, left of the slope.
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I can go ahead and divide by -2 to make it the X sub E ν sub E.
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This type of graph is called a rich mono plot, you should look it up on www.google.com.
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Now as R increases the plot starts to deviate from linear behavior, deviate from linearity.
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Interesting is when you look at a plot, you will see the plot.
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From your perspective looking at a plot, it is going to be like this and it is going to start curving down.
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The plot deviates from linearity as R increases.
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If you get values of R, I will start to get 8, 9, 10, 11, and so on.
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It is going to start to deviate from linearity because this energy for the vibrational term,
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the vibrational energy which we have which is ν × R + ½ - X sub E ν sub E × R + ½².
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It is actually part of an infinite series.
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When we solve the Schrödinger equation for the anharmonic oscillator,
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what we get is we get an infinite series for our energy.
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We only took the first two terms because really all that we need.
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It is really all that we need because we are not often going to find ourselves in the 15 or 20 vibrational state.
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It is extremely high temperatures for something like that.
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We are just going to take that.
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As R increases, these other terms of the series start to become important.
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It starts to deviate from linearity.
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What you end up actually getting is because R increases the plot deviates from linearity because of this.
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Yes, we only take the first two terms of the series solution to the energy of the anharmonic oscillator.
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The graph itself will look something like this.
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It is going to start pretty linear.
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It looks nice but then you know it starts to deviate.
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This is δ G, this is the R + 1, this axis.
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How many vibrational states are there for BR2 before the molecule dissociates.
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In other words, it breaks apart.
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For BR2, the ν sub E is that and X sub E ν sub E equals that.
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We have an expression.
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We have that R max is equal to ν sub E / 2X sub E ν sub E – 1.
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We are just going to put it in and find R max.
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The number of vibrational states before the molecule falls apart.
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R max, I’m just going to put the values in.
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The ν sub E 325.321 divided by 2 × X sub E ν sub E is 1.0774 inverse cm -1.
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When I do that, I get R sub max is equal to 149 vibrational states.
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In other words, I can excite this molecule BR2 through the 149 vibrational states.
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It is not going to hit the 150th.
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At that point, it is just going to fly apart.
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Let us calculate the dissociation energy for this too, we have an expression for that.
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Approximately equal to ν sub E / 4 X sub E ν sub E.
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We have ν sub E / 4X sub E which we said was equal to ν sub E²/ 4 X sub E ν sub E.
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We get 325.321 inverse cm² divided by 4 × 1.0774 inverse cm.
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We are left with the dissociation energy equal to 24,557.067 inverse cm.
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Of course, if you want to convert that to Joules, you can go ahead and do that, not a problem.
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If the R equals 0 to R equals 0 vibronic transition.
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This is not vibrational, this is vibronic.
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This is the transition that takes place between vibrational states from 1 electronic state to another electronic state.
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If the R = 0 to R equals 0 vibronic transition for carbon monoxide equals 64748.5 inverse cm, find the value of T sub E.
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This is the difference between the two minima of the potential energy curves below.
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T sub E is the difference between this level and this level.
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That is P sub E, the difference between the potential energy minima.
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The data below gives us, for the ground state our ν sub E is this and our X sub E ν sub E is this.
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For excited electronic state, the excited state ground state 1518.2, 19.4.
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Let us see what we have got here.
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Let us do this one very carefully.
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Let me go ahead and go back to black here.
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For the 0 to the 0 vibronic transition, we have the electronic energy, the rotational energy,
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and the electronic vibration rotational energy.
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For the 0 to 0 vibronic transition that means R equal 0 in both the upper and lower states, lower electronic states.
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The equation for the total energy of a molecule, we have the total energy
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is equal to the electronic energy + the vibrational energy + the rotational energy which is F of J.
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For electronic transitions, the rotational energies are small compared to the vibration and electronic but we are going to ignore it.
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We are going to drop that.
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Our expression for total energy is just going to be our electronic energy of a molecule in a given state + the vibrational energy.
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We have an equation for the observed, the calculated frequency of absorption,
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that is going to equal the energy of the total energy of the upper state – the,
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we will use a single prime for the total energy of the lower state I will use a double prime for that.
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And we have expression for that, it is going to be the electronic energy prime + ν sub E prime × R prime + ½ - X sub E prime ν sub E prime × R + ½² - the total energy for the lower which consists of the electronic energy of lower state + ν sub E double prime × R double prime + ½
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- X sub E double prime ν sub E double prime R double prime + ½².
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When we expanded and solve this, we get it in a previous lesson we got ν equal to the upper electronic energy - the lower electronic energy + ½ ν sub E upper -1/4 X sub E ν sub E upper - ½ ν sub E lower -1/4 X sub E ν sub E lower
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+ ν sub E upper R upper – X sub E upper ν sub E upper R upper × R upper + 1.
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We end up with this equation right here.
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This gives us the frequency of the observed spectral line for the transition between vibrational states when lower to upper state.
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Now, because R single prime equals 0, in other words the lower and the upper are both 0,
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these terms right here, let me go to blue now.
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These terms, they dropped out.
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We get just this term, this term, this term, and this term.
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I’m going to call this term 1, I’m going to call this term 2.
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I’m going to call this term 3, I’m going to call this term 4.
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We get that the ν observed is equal to 1 - 2 + 3 – 4.
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The difference in the electronic energies, the electronic energy of the upper –
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the electronic energy of the lower that is our TE.
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TE equals the upper electronic - the lower electronic.
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It equals that.
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What we end up with is and because it is a 0 to 0 transition vibronic, it is actually the ν 00 transition frequency
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that equals to T sub E + ½ ν sub E upper - 1/4 X sub E upper ν sub E upper - ½ ν sub E lower -1/4 X sub E lower ν sub E lower.
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We have all of these numbers already.
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We know what ν 00 is.
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It actually gave it to us, the problem gave it to us.
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64,748.5 is equal to T sub E + 1/2 1518.2.
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1518.2 – ¼ × 19.4, this is the data that was in there.
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- ½ 2169.81 – ¼ × 13.29, I get 64748.5 is equal to T sub E + 754.25 - 1081.58.
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When I do the math, I get my T is equal to 65,075.33 inverse cm.
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This is the difference in energy between the two electronic states.
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Example problem 5, the following data were obtained from an analysis of the rotational spectrum of CLO.
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Use this data to find B sub E and Α sub E.
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We have R values 0, 1, 2, 3, and we have B sub R values.
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You do not need all the data, we are just going to use R equals 0 and R equals 1.
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I can choose any that I want.
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I can do 0 2, 0 3, 1 2, 1 3, or 2 3, it does not really matter.
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Using the data for R equals 0 and R equals 1, we will take those two.
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Our equation is B sub R, this is the vibration rotation interaction correction.
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It is equal to B sub E - Α sub E × R + ½.
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We are looking for α sub E and B sub E.
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B₀ that is 0.62054.
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It is equal to B sub E - R is 0.
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It is - ½ α sub E.
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For B₁, I have 0.61474 that is equal to B sub E - 3/2 Α sub E 1.
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When I put 1 into here, 1 + 1/2 is 3/2.
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When I subtract one from the other, I end up with Α sub E is equal to 0.0058 inverse cm
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and I go ahead and put this into equation 1.
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And I end up with B sub E is equal to 0.62344 inverse cm.
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There you go.
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Now, if I want to I can take 0 1, 1 2, 2 3, 0 2, I can take any combination.
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I can take 0 1, 1 2, 2 3, I can get three different values and I can take the average.
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It is not really necessary, you can just take two data points and go ahead and find that value.
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Thank you so much for joining us at www.educator.com.
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We will see you next time, take care, bye.