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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In the last lesson, we started our discussion of molecular spectroscopy.
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In this lesson, we are just going to go ahead and continue our discussion.
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Let us get started.
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This is the vibration rotation spectrum for HCL.
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Here, we have the R branch, sometimes you going to see it like this
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and sometimes you are going to see it the other way when the peaks are actually pointing up.
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It does not really matter, the thing you will remember those, the R branches with the frequencies are increasing,
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the P branch is the one with the frequencies are decreasing.
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R is left and right, in this case the frequencies increasing this way, the R branch is over here.
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This is a decreasing frequencies of P branch over here.
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This is not unnecessary bit of information.
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Some of the things that we want to notice here.
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I will stick with black.
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Some things to notice, the first thing you want to notice is that the distance between the lines in the 2 branches are not equal.
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In other words, the distance from here to here is not the same as the distance from here to here.
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We would expect 2B, 2B, 2B, 2B, 2B in the P branch, in the R branch but that is not the case.
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These are narrower and as we actually increase, go further along the R branch, the gap tends to get smaller.
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And as we go further along the P branch, the gap actually gets bigger.
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We will say 1 the distance between lines in the 2 branches are not equal.
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The distances are not equal.
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For the R branch, the gaps get smaller as J increases this way.
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For the P branch, the gap get larger, the gaps get wider as J increases.
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We have the following for the harmonic oscillator rigid rotator approximation.
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We had that the energy RJ, R is the vibrational quantum number.
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J is the rotational quantum number.
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The total energy is equal to the vibration term + the rotational term and we had prime R + ½ + this constant × J × J + 1.
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R goes from 0, 1, 2.
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J = 0, 1, 2, and so on.
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In this particular case, B the rotational constant is equal to planks constant over 8 π² ψ × the rotational inertia.
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The rotational inertia happens to be the reduced mass × the equilibrium bond length².
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This is the expression for the energy under the harmonic oscillator rigid rotator approximation.
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Let us take a look at this next image here.
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This green line represents the potential curve for the harmonic oscillator, that is what we based it on.
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You remember the potential energy was equal to ½ KX², a parabola.
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The actual potential curve is not parabolic, that is the blue line.
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This is the actual potential curve, that is 2 molecules come together.
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From an infinite distance, the energy decreases and may reach a certain length between the two.
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If you push them any closer together, the potential energy rises.
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If you pull them apart, the potential energy rises.
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This down here, that point of lowest energy is what we call the bond length.
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R sub E equilibrium bond length far as the distance between them.
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You will notice that the harmonic oscillator approximation is good for the lower vibration states.
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By the way, these ν right here, these are the quantum numbers or what I call R.
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For our purposes, the reason I use R for the vibrational quantum number instead of the ν,
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you have probably seen in your books is I want to confuse it with the actual frequency absorption
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or transmission or other frequencies that we happen to talk about.
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These are the vibrational states R = 0, R = 1, R =2, R =3, 4 and 5.
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For the lower vibrational states, the parabolic, the harmonic oscillator
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actually is a pretty good approximation for the non harmonic oscillator.
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The actual, the real potential energy curve.
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Very very good approximation.
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Under normal temperatures, most of the molecules are going to be in the R = 0 vibrational state
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and we will show you why that is later when we discuss statistical thermodynamics.
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You know that that is the case, for that vibrational state R = 0, the harmonic oscillator is a great approximation.
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As the vibrational state increases, in other words, as R increases, the value of R sub E, the equilibrium bond length,
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I use a capital here, I just happen to use a small letter.
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The value of RE, the bond length also increases.
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You can see this on the potential energy curve.
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Notice, if I stay just for the harmonic oscillator, if I just go straight up,
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I'm right down the middle of in between the potential energy curve along the energy level.
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But as R starts to increase, the vibrational states go up.
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You go higher and higher levels.
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The halfway mark on the blue line tends to start to shift further and further to the right.
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What happens is that the equilibrium bond length tends to go further and further to the right.
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In other words, the equilibrium bond length have to get wider and wider and wider.
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The bond length increases as we move up the vibrational states.
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As the bond length increases, the rotational constant B~ decreases precisely because the definition of that is equal to,
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Remember in the previous page, 8 π² ψ, that RE², this is the rotational inertia.
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As something in the denominator increases, the value itself is defined by this ratio decreases.
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We express the independence of the rotational constant on R, the vibration quantum number as B sub R.
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We put that little subscript there to let us know that this B actually depends on which vibration we happen to be in.
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Is it 0, 1, 2, 3, 4, things like that.
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We can rewrite the energy equation E sub RJ equal to + ½ + B sub R × J × J + 1.
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Nothing is different, all that we have done is actually put this little BR here.
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There is an equation that actually relates to the BR to something.
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We will get to that.
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We call this dependence of the rotational constant B on R, the vibration rotation interaction.
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The vibration rotation interaction is going to be one of the corrections that
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we are going to have to make to our harmonic oscillator rigid rotator approximation equation.
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There would be 3 corrections that we are going to make.
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The first one is this, the vibration rotation interaction.
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Later, we are going to correct for anharmonicity of the anharmonic oscillator and we are going to correct for the non rigid rotator.
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This is the first of the corrections, the vibration rotation interaction.
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What we are going to do is we are going to examine the transition from R0 to R1,
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that is the equation that we are going to look at.
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Let us examine the R = 0 to the R = 1 transition.
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Therefore, the δ R is equal to + 1, 0 to 1.
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The first case we have, we said that the δ J can equal + or -1.
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Δ R = that and δ J is equal to + or -1.
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The first case that we are going to look at, case 1, we are going to take a look at the δ J = + 1.
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We would be making the transition from the 0 vibrational state to the 1 vibrational state.
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Within those vibrational states, you have a bunch of rotational levels.
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We are going to see the transitions that take you from 0 to 1, 1 to 2, 2 to 3, things like that,
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between the rotational states.
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E sub R, the frequency that we see on the spectrum or that we expect to see on the spectrum,
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is going to be for the R branch because the δ J + 1 represents the R branch.
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It is equal to upper energy - the lower energy.
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It is going to be E1 J + 1 - E0 J, this is the upper energy level, this is the lower energy level.
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I will go through the algebra, at least for the first one here.
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This is going to equal, R is 1, 1 + ½ + E₁.
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I’m specifying my R, my quantum number, × J + 1 × J + 2 - the lower energy.
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This is going to be 0 + ½ + B0 × J × J + 1.
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All I'm doing is I'm actually putting these quantum numbers 1 and J + 1, 0 and J into our equation for the energy.
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The upper energy level, the lower energy level, and subtracting the lower from the upper.
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That gives me the frequency of the absorption or the transmission that I see.
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This is going to equal, after a little bit of algebra which we will actually do right here.
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+ B1 × J² + 3J + 2.
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And I hope that you are confirming my algebra because I am notorious for algebraic errors.
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- the - distributes over everything.
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-B0 J² – B₀ J = ν + B1 J² + 3 B1 J + 2 B1 – B₀ J² - B₀ × J.
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Our final equation that we have is this going, the frequency that we observe for the R branch
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is going to be the fundamental frequency + B1 - B0 J² + 3 B1 - B0 × J + 2 B0.
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This is one very important equation.
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Under this dependence that the rotational constant has on the quantum number or the vibration quantum number R,
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we have come up with an equation from the 0 to 1 vibration level for the R branch.
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This actually gives me the frequencies of absorption that I should see.
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Something very very important to remember.
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In this case, J is the initial rotational state for lower.
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It is the lower the initial, we are talking about absorption here.
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We went from a lower to a higher, J is the lower rotational state.
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It is a quantum number of the lower rotational state.
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When we do one for P branch, that is going to equal, the upper energy level
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which is going to be vibrational level 1 and this is δ J = -1.
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We are still going from vibrational level 0 to 1.
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The rotational levels are going to go from 1 to 0, 2 to 1.
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The J values are going to go down by 1.
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What we have is the following.
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We have the P branch is going to equal the upper energy level which is 1 J-1 - the lower energy level which is 0 J.
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We can go ahead and go through the algebra.
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I’m not going to go ahead and go through it.
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I will just go ahead and write out the equation that you get.
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P is equal to ν + B1 - B0.
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I should we stop using these tilde because it is really a lot of excess of symbolism that makes you crazy.
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But I guess that is the nature of quantum mechanics, symbols.
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-B1 + B0 × J, this equation.
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J is the lower rotational state, the quantum number of the lower rotational state, those two equations.
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Now, as R increases like we said, B sub R decreases.
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This implies that B1 is less than B0.
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B1 being less than B0, if we look at these equations that we got, these two equations.
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The one for the P branch, and 1 for the R branch.
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As J increases in the two equations above, the R branch gaps decrease and the P branch gaps increase.
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In other words, once I put J = 0, 1, 2, 3, 4, into the R branch J = 1, 2, 3, 4, 5 in the P branch,
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when I take the differences between 1 rotational state and another, the R branches, the difference is going to get smaller.
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When we started this lesson, we notice that the R branch actually tend to decrease.
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The P branch, the lines of the spectrum are going to widen as J increases.
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These two equations explain that behavior.
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Vibration rotation interaction explains the appearance of a lines or spectra.
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The equation, the dependence of B on R is given by an explicit relation B sub R
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is equal to something called B sub E - Α sub E × R + ½.
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B sub E and α sub E are 2 part spectroscopic parameters that are tabulated for different diatomic molecules.
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The values for B sub E and Α sub E are available in tables of spectroscopic data.
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And we will see some those tables in the example problems later on.
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Let us go ahead and do a quick example here.
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Basically, in the problems you are going to get two source.
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You are going to be given some spectroscopic data and you are going to be asked to actually find the spectroscopic parameters.
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Or you are going to be given the spectroscopic parameters and you are going to be as to find the lines spectroscopic data.
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It goes both ways.
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We will see examples of that when we did example problems.
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Let us go ahead now and take a look at an example.
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Example 1, this is going to be only example for this particular lesson.
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I'm collecting all of the example problems for afterward when we do them all at once after the other.
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I want you to see at least one here. The numerical values spectral lines are put into tables.
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The lines of the P and R branches are labeled with a branch they belong to,
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+ the initial rotational state of the transition line it represents.
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For example, for the R branch, the first line is labeled R₀.
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The reason is because it belongs to the R branch, it represents the transition from the 0 to 1 rotational state.
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The second line is R1.
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It is in the R branch and repressed transition from 1 to 2.
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These subscript are the lower rotational states, the initial rotational state.
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The third line R2, be very careful to keep the indices very clear and distinct.
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For the P branch, the first line is P1.
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The P branch because the transition is going from the initial state rotational state of 1
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to the upper rotational state which is actually the 0.
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This represents the 1 to 0 transition, δ J = -1.
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The second line is P2, it goes from 2 to 1.
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I hope that makes sense.
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Using the equations above, like what we just did for ν sub R and ν sub P,
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and the data table of frequencies for the HCL vibration rotation spectrum below,
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or the numerical values from the spectrum directly, your choice what you want to use the spectrum,
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with numbers or from the table that we provide.
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Find the rotational constant for the 0 vibrational state, the rotational constant for the 1 vibrational state,
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the equilibrium bond length for the 0 vibrational state, and equilibrium bond length for the 1 vibrational state.
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Use 1.63 × 10⁻²⁷ kg for the reduced mass of HCL.
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Here, in this particular case, we have given you both.
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We have given you the spectrum and here is the table.
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The table looks like this.
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Basically, just read and write off the spectrum.
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The R is 0 branch that is this one right here, it is 2906.
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The R1 branch was 2926.
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When I go to the P branch over here, this is P1, this is R0.
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This is R1 P1, 2865.
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I’m sorry, it is actually not that clear here.
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The P2 branch 2844.
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Again, these are in inverse cm.
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We would be working almost exclusively in inverse cm.
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Some of the problems might give you data in Hertz, megahertz, gigahertz, things like that.
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We will always be converting to inverse cm because for spectroscopy, inverse cm is the standard unit.
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Sorry about that, this is what I just have at the table.
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It was later that I decided to put the spectra in.
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In general, sometimes you need a table, sometimes you are going to be given a spectrum.
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You have to be prepared to work from either one.
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Let us go ahead and start this problem.
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Let us go ahead and rewrite our equations.
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The R branch spectral line that we see is going to be the ν,
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the fundamental vibration frequency + B1 - B0 J² + 3 B1 – B 0 J + 2 B1.
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That is the equation for the frequency for the R branch.
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For the P branch, what we are going to see is ν + B1 - B0 × J² - B1 + B0 × J.
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These equations and we are going to be working with.
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Let us go ahead and do the ν for R 0 and R₀.
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R branch, the first line is 0 to 1 transition.
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Our R value = 0, this small R value = 0.
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That is going to be, I will just put them in.
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I'm sorry, let me start again here.
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And so we have ν, this is going to be R₀.
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The 0 is the J value.
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We are going from R branch, we are going from the 0.
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Let us do it this way.
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R0 represents the 0 to 1 transition for the J.
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That is a little bit better.
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This is going to be ν + this thing initial J0, that is going to be 0.
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J is 0 that is going to be 0 + 2 B1, that one is equal to 2906.
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We have the R1 transition, that represents the transition from J = 1 to J = 2.
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I just put 1 in for J in here, I get this + B1 - B0 × 1² which is 1 + 3 B1 - B0 + 2 B1.
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I combine all like terms and I get ν + 6 B1 – 2 B0 and that is equal to 2926.
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The two equations we have are this one and we have this one.
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Let me go ahead and work in red.
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I got B0 + 2 B1 = 2906 and I have ν + 6 B1 -2 B0 = 2926.
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This represents the R branch.
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Let us go back to black and go ahead and do the P branch.
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Our P1 that represents the transition from 1 to 0.
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In this particular case, the initial J value is 1.
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I’m going to put 1 into this equation, the 1 for ν sub P.
00:31:12.900 --> 00:31:32.100
What we end up with is ν + B1 - B0 × 1 - B1 - B0.
00:31:32.100 --> 00:31:41.800
And that we say is equal to 2865.
00:31:41.800 --> 00:31:56.700
The P2 line that represents the transition from J = 2 to J = 1, that is going to be that.
00:31:56.700 --> 00:32:04.600
It is going to be + 2 into this equation, 2² is 4 so it is going to be +4.
00:32:04.600 --> 00:32:24.900
B1 -4 B0 -2 B1 -2 B0 and that is going to equal 2844.
00:32:24.900 --> 00:32:36.800
What I have here, the other equations that I have are ν B1, that goes away.
00:32:36.800 --> 00:33:00.300
- 2 B0 = 2865, and the second equation here I have that, + 2 B1 – 6 B0 = 2844.
00:33:00.300 --> 00:33:13.000
This represents the P branch.
00:33:13.000 --> 00:33:30.100
This is the R branch, let me go ahead and rewrite.
00:33:30.100 --> 00:33:38.500
I will just make sure my numbers are correct here.
00:33:38.500 --> 00:33:46.300
What I'm going to do, I’m going to rewrite these equations on the next page so that I actually have them.
00:33:46.300 --> 00:34:01.800
My R branch, my R is 0 is ν + 2 B1 = 2906.
00:34:01.800 --> 00:34:13.700
My R1 = ν + 6 B1 – 2 B0 = 2926.
00:34:13.700 --> 00:34:26.100
My P1 =, I have ν – 2 B0 = 2865.
00:34:26.100 --> 00:34:41.200
And I have my P2 branch is ν + 2 B1 -6 B0 = 2844.
00:34:41.200 --> 00:34:56.900
We want B0 and B1, the rotational constant for the 0 vibration level and for the 1 vibration level.
00:34:56.900 --> 00:35:02.300
Let us go ahead and take, I'm going to take these 2 equations.
00:35:02.300 --> 00:35:05.900
Let me actually do this one in blue.
00:35:05.900 --> 00:35:15.300
I'm going to take these two and I’m going to form R0 - P2.
00:35:15.300 --> 00:35:17.800
I’m going to take the first equation.
00:35:17.800 --> 00:35:21.300
I’m going to subtract this second equation.
00:35:21.300 --> 00:35:34.800
What you end up with is when you take this equation - this equation, this and this go away.
00:35:34.800 --> 00:35:38.600
2 B1 – 2 B1 go away.
00:35:38.600 --> 00:35:50.500
This – this, what you end up with is 6 B1 = 2906 – 2844.
00:35:50.500 --> 00:35:53.700
You should get 62.
00:35:53.700 --> 00:36:04.600
Therefore, the value of the one that we end up getting is 10.33 inverse cm.
00:36:04.600 --> 00:36:11.600
I'm going to go ahead and take R1 - P1.
00:36:11.600 --> 00:36:24.600
I’m going to take R1 - P1, this - that cancels.
00:36:24.600 --> 00:36:34.100
Did I make a mistake here?
00:36:34.100 --> 00:36:41.400
Let me go back, these indices floating around is enough to make you crazy.
00:36:41.400 --> 00:36:48.200
I have something written on my piece of paper but I just want to make sure that my indices are actually correct here.
00:36:48.200 --> 00:36:50.600
Ν + 2B1 = 2906.
00:36:50.600 --> 00:37:02.500
Ν + 6 B1 – 2 B0 = 2924.
00:37:02.500 --> 00:37:14.800
Again, this is 2926.
00:37:14.800 --> 00:37:24.000
I have my P1 which is ν 0 -2 B0, that is correct, 2065.
00:37:24.000 --> 00:37:29.700
I have ν + 2 B1 -6 B0 = that.
00:37:29.700 --> 00:37:31.300
I think everything should be okay.
00:37:31.300 --> 00:37:32.500
Let us try this again.
00:37:32.500 --> 00:37:43.300
Let us take R0 - P2.
00:37:43.300 --> 00:37:46.900
That was my mistake, sorry, a little bit of a notational error.
00:37:46.900 --> 00:37:47.600
We just have to be very careful.
00:37:47.600 --> 00:37:52.200
We are taking R0 – P2.
00:37:52.200 --> 00:38:01.700
The ν cancel, the 2 B1 – 2 B1 that cancels 6 B0, my apologies for that.
00:38:01.700 --> 00:38:13.200
This is 6 B0 is equal to 62, 2906 -2044.
00:38:13.200 --> 00:38:15.900
Again, I hope that I have done my arithmetic correctly.
00:38:15.900 --> 00:38:22.400
B0 not B1, is what should be 10.33 inverse cm.
00:38:22.400 --> 00:38:27.000
There you go, sorry about that.
00:38:27.000 --> 00:38:30.200
We will go ahead and take the R1 - P1.
00:38:30.200 --> 00:38:39.400
R1 - P1, the ν they cancel.
00:38:39.400 --> 00:38:54.300
The B0 cancel and we are left with 6 B1 = 2926 - 2865 =59.
00:38:54.300 --> 00:38:57.400
I think again, please check my arithmetic.
00:38:57.400 --> 00:38:59.100
Again, the arithmetic is unimportant.
00:38:59.100 --> 00:39:01.300
It is the process that is important.
00:39:01.300 --> 00:39:09.400
B1, other value of 9.83 inverse cm.
00:39:09.400 --> 00:39:22.400
Note that B1 is less than B0, 9.83 is less than 10.33.
00:39:22.400 --> 00:39:24.000
We need to find the equilibrium bond lengths.
00:39:24.000 --> 00:39:39.700
We need to find the R, E, for the R = 0 and find the RE for the R = 1.
00:39:39.700 --> 00:40:06.700
B, we know that B is equal to planks constant over 8 π² × ψ × the reduced mass RE².
00:40:06.700 --> 00:40:11.300
I wonder if I should go ahead and do this in terms of,
00:40:11.300 --> 00:40:24.300
B1 is equal to 6.626 × 10⁻³⁴ J/ s.
00:40:24.300 --> 00:40:27.000
Here is what we have to watch our units.
00:40:27.000 --> 00:40:32.800
We are working in inverse cm but joules, kilograms, and meters, and stuff like that,
00:40:32.800 --> 00:40:36.600
you have to make sure the your units match.
00:40:36.600 --> 00:40:48.200
We have 8 π², I suppose we can go ahead.
00:40:48.200 --> 00:40:57.700
2.998 × 10¹⁰ centimeters per second.
00:40:57.700 --> 00:41:04.500
The reduced mass is 1.63 × 10⁻²⁷.
00:41:04.500 --> 00:41:12.700
We are looking for R₀², this is for B1 R1.
00:41:12.700 --> 00:41:18.600
We said that one was actually equal to 9.83.
00:41:18.600 --> 00:41:25.400
We found that value when we solve this equation for R sub E.
00:41:25.400 --> 00:41:38.900
We get R sub E for the R = 1 value = 1.705 × 10⁻²⁰ m.
00:41:38.900 --> 00:41:50.900
I will stick with what it is that I have written here.
00:41:50.900 --> 00:42:11.000
This is meters, therefore equilibrium bond length is going to equal 1.305 × 10⁻¹⁰ m which is 138.5 picometers.
00:42:11.000 --> 00:42:13.500
It tends to be expressed in terms of picometers.
00:42:13.500 --> 00:42:44.300
B0, that is equal to 6.626 × 10⁻³⁴ divided by, we have 8 π² × 2.998 × 10¹⁰ cm/ s.
00:42:44.300 --> 00:42:57.000
This is joule seconds × 1.63 × 10⁻²⁷ kg × R sub E for 0².
00:42:57.000 --> 00:43:01.600
And we said that one was equal to 10.33.
00:43:01.600 --> 00:43:18.600
We will get this equal to 1.662 × 10⁻²⁰ m which means the equilibrium bond length is equal to 1.289.
00:43:18.600 --> 00:43:32.700
1.289 × 10⁻¹⁰ meters which means you are looking at 128.9 picometers.
00:43:32.700 --> 00:43:40.000
Notice that the one for the vibration state R = 0 is 128.
00:43:40.000 --> 00:43:57.300
When we moved up a vibrational state to R = 1, the equilibrium bond length increased.
00:43:57.300 --> 00:43:59.500
Let us do one last thing here.
00:43:59.500 --> 00:44:17.300
We said that the relationship B sub R is equal to its equilibrium bond length - this Α sub E × R + ½.
00:44:17.300 --> 00:44:42.200
From the example, we got a B1 value = 9.83 and we got a B₀ value = 10.33.
00:44:42.200 --> 00:44:49.200
I think we can use these values in this equation for that different R.
00:44:49.200 --> 00:44:56.500
We can find the actual parameters that are tabulated in the data.
00:44:56.500 --> 00:45:16.800
We can now use B0 and B1 to find B sub E and Α sub E.
00:45:16.800 --> 00:45:29.500
B₁ = B sub E - Α sub E × 1 + ½.
00:45:29.500 --> 00:45:43.300
We said that that one is equal to 9.83 and the P₀ that is equal to be B sub E - Α sub E × 0 + ½.
00:45:43.300 --> 00:45:46.700
We said that that = 10.33.
00:45:46.700 --> 00:45:52.400
There you go, you have two equations and two unknowns.
00:45:52.400 --> 00:45:55.400
The two unknowns are B sub E and Α sub E.
00:45:55.400 --> 00:46:01.600
This is just 3/2, this is ½, you have these.
00:46:01.600 --> 00:46:05.400
I will leave that to you.
00:46:05.400 --> 00:46:08.900
Two equations and two unknowns.
00:46:08.900 --> 00:46:19.200
The two unknowns are B sub E and Α sub E, the spectroscopic parameters.
00:46:19.200 --> 00:46:21.500
Thank you so much for joining us here at www.educator.com.
00:46:21.500 --> 00:46:22.000
We will see you next time, bye.