WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com, welcome back to Physical Chemistry.
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Today, we are going to continue our example problems in molecular spectroscopy.
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Let us jump right on in.
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Our first example is the force constant for carbon monoxide is 1857 N/m.
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The equilibrium bond length = 11.83 pm.
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Using the rigid rotator harmonic oscillator approximation,
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construct a table of energies for the first 5 rotational states for the vibrational levels R = 0 and R = 1.
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The 5 rotational levels for R = 0, 5 rotational levels for R = 1.
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Specify which transitions are allowed and calculate the frequencies of each transition,
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associating each with its appropriate branch R or P.
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Let us see what we can do.
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Let me go ahead and work in blue today.
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Under the rigid rotator harmonic oscillator approximation,
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our energy equation is going to be energy RJ is equal to ν₀ R + ½ + this rotational constant × J × J + 1.
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That is it, it is the vibrational energy + the rotational energy.
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We have to find ν₀ that, and this, before we can actually start running J from 0 to 5.
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Let us see what we can do.
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Ν₀, the definition is 1/2 π C × the force constant divided by the reduced mass ^½.
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We have to find the reduced mass.
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The reduced mass is equal to the product of the weights divided by the sum of the weights.
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12 × 16, 12 for carbon 16 for oxygen, divided by 12 + 16.
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This is in atomic mass units, I’m going to multiply that by 1.661 × 10⁻²⁷ kg per atomic mass unit.
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My atomic mass units cancel and I'm left with a reduced mass in kg.
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When I do this calculation, I should get,
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I hope that you are confirming my arithmetic, I am notorious for arithmetic mistakes.
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The arithmetic is less important, the process is important.
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The equations that you choose, that is what is important.
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I may clear my throat a few more times than usual during these few lessons today.
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× 10⁻²⁶ kg.
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I want to keep writing these numbers over and over again.
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When I go ahead and put this value of K into here, when I put this value ν into here and run this calculation,
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I'm going to get a ν₀ equal to 214354.49.
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The answer is actually going to be in inverse meters because this is N/ m.
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When I convert that, I will just move that decimal over a couple of times,
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I'm going to get 2143.54 inverse cm.
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The problem is, the issues with spectroscopy are like many issues throughout quantum mechanics and thermodynamics.
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They are just for conversion issues, just watch your units.
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That is really all you have to watch out for.
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We have that number.
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Let us go ahead and find the rotational constant.
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Let me go ahead and do that on the next page here.
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Our rotational constant is equal to, the definition H/ 8 π² C × the rotational inertia.
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We need I to put it into here.
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I = that, that².
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Therefore, I = R 1.139 × 10⁻²⁶ kg × RE is a 112.83 × 10⁻¹² m.
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What we get is I = 1.45 × 10⁻⁴⁶ kg/ m², the unit of rotational inertia.
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We put this value in here with all of the other values.
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I will go ahead and write this one out, it is not a problem.
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E = 6.626 × 10⁻³⁴ J-s divided by 8 π² × 2.998
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× 10⁸ m/ s × 1.45 × 10⁻⁴⁶, that is the I.
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When I do that, I will get 193.04 but this is going to be in inverse meters because a joule has meters.
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That is going to give us, when I convert this, move the decimal over twice 1.9304 inverse cm.
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Remember, inverse cm and inverse meters, it is the other way around.
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It is going to be 100 inverse meters in 1 inverse cm, which is why this one goes to the left.
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We have that and we have that.
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We said that the energy was, let me go ahead and rewrite the equation.
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The energy was E= sub RJ equal to ν₀ × R + ½.
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This is the rigid rotator harmonic oscillator approximation + B ̃ J × J + 1.
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Let me make my J a little more clear here.
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For the R = 0 vibrational state, that is going to be E⁰ and J is going to run through these quantum numbers.
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That is going to equal, when I put R and 0 into this equation, it becomes ½ ν₀ + B ̃ × J × J + 1.
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And for the R = 1 vibrational state, I just put 1 into this equation and I get the E₁ J is equal to 3/2 ν₀ + B ̃ J × J + 1.
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Let me double check that.
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Therefore, our E0 J, when I actually put the ν 0 that I got and the B that I got in here,
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I'm going to get equation that says 1071.77 + 1.9304, which is the rotational constant, × J × J + 1.
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Now, I just set up a table of values.
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I have J over here and I have my energy 0 J over here.
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I’m going to take 0, 1, 2, 3, and 4.
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I get 1071.8, I get 1075.6, I get 1083.4, I get 1094.9, I get 1110.4.
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These are energies not frequencies of absorption or emission.
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These are not spectral lines, these are not frequencies.
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These are the energies of the actual rotational states in the R = 0 vibrational state.
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Let us go ahead and do the energies for 1 vibrational state.
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When I put the values in, I get 3, 2, 1, 5.
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In other words 3/ 2, the ν₀.
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3215.31 + 1.9304 which is the rotational constant, × J × J + 1.
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Let us go ahead and set up my table of values again.
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This is going to be E1 J.
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I have 0, 1, 2, 3, and 4.
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I have 3215.3, I have 3219.2, 3226.9, 3238.5, and 3253.9.
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Let us go ahead and let me do it on one page here.
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Let me see if I want to just go ahead and do this one in red.
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I got J and I got E0 J.
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Over here, I have J and I have E1 J.
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I have 01234, 01234.
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These are the first to 5 table of values.
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I had 1071.8, 1075.6, 1083.4, 1094.9, and 1110.4.
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Over here, I have 3215.3, 3219.2, 3226.9, 3238.5.
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Again, I want them on the same page so I can see them together.
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3253.9, the energies of the 0 vibrational state, the first 5 rotational for the 1 vibrational state.
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The allowed transitions, they also asked about that.
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The allowed transitions, δ J is + or -1.
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The allowed transitions, the selection rule is δ J = + or -1.
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Therefore, for the 0 state to the 1 state, the 0 vibrational state to the 1 vibarational state, I can go 0, 1, 2, 3, 4.
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I have 0, 1, 2, 3, 4.
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I can go from 0 to 1 that is + 1.
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1 to 2, 2 to 3, and 3 to 4.
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Let me do this one in black.
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I can go from the lower 0, 1 to 0, to the 1, 3 to 2, 4 to 3.
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Those are my allowed transitions.
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The 0 state to 1 state, I can go 0 to 1, 1 to 2, 2 to 3, 3 to 4.
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Or the 0 to 1 vibrational state, I can go rotational 1 to 0, 2 to 1, 3 to 2, and 4 to 3.
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Those are my allowed transitions, the frequencies of those particular transitions.
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The frequencies of these transitions, in other words the frequencies of the spectral lines of the R and P branches,
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we have ν sub R branch is equal to ν₀ + 2B × J + 1, where J takes on the values 0, 1, 2, and so forth.
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The P branch is going to be ν₀ -2B J.
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Here, J takes on the values 1, 2, 3, and so on.
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My ν sub R is going to be 2143.54 + 2 × 1.9304 × J + 1.
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My ν sub P, those branches are going to be 2143.54 -2 × 1.9304 × J.
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The R is 0 branch that represents the 0 to 1 transition, that was going to take place at 2147.4.
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I will just put the J values in here.
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J = 0, J = 1, J = 2, J = 3, J = 4.
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That is the R0, let me be label it with the particular J value.
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We put the R branch for J value 0 and it is going from 0 to 1.
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When we do the P branch, it is actually going to be a P1 because it is going to be going from 1 to 0.
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The R1, that represents the 1 to 2 transition.
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The R2 represents the 2 to 3 transition.
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The R3 represents the 3 to 4 transition.
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These are the frequencies that we have.
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We have 2151.3, we have 2155.1, and we have 2159.
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We have the P1 transition, the P1 line, the P2 line, the P3 line, and the P4 line.
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This one represents the 1 to 0 transition.
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This one represents the 2 to 1 transition.
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These are subscripts or the beginning energy level.
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The energy level of the lower vibrational state.
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For the lower rotational state, on the lower vibrational state.
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In other words, the departure state not the arrival state.
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I like that better, departure and arrival.
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I think the mathematical terms are actually a lot better.
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This one represents the 3 to 2 transition and this one represents the 4 to 3 transition.
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These are 21392139.7, 2135.8.
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These are going down, these are going up.
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R branch increases, P branch decreases.
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2132 and 2128.
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There you have it, we found the energies, we found the allowed transitions.
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These are the frequencies of the actual transitions that are allowed.
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These are the labels for them representing this particular transition.
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Let us go back to red here.
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Be very careful to distinguish, especially on test.
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On a test, you are going to be stressed out and your mind is going to be racing.
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You need to really slow down when dealing with spectroscopy because they can be asking about energy,
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they could be asking about absorption, emission, frequency.
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Those are not the same.
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Be careful to distinguish between the energy of a given level and the energy of transition,
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the energy of transition between two energy levels.
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The transition between two energy levels that is what we see in spectra are the transition energies from one level to another.
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One final thing to notice, we found the energy levels for the E₀ J and E₁ J.
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We used equations, the ν sub R, ν sub P to actually find those frequencies.
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What you could have just done is take the difference between energies in the table.
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Notice that the frequencies of transition,
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instead of using equations for ν R branch and ν of the P branch,
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we could have just taken δ E between the two vibrational states
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for the appropriate J values.
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For example, if I wanted the R1 line,
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Let us go ahead and start with R0 line.
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The R0 line that represents the transition from 0 to 1.
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We have the energies already in the table.
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We could have just taken the energy of 1 1 - the energy of 0 0.
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This is the vibrational number, that is the vibration number.
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Upper lower, the transition going from the 0 to 1 transition.
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Upper lower, we could have just done that, just subtract the two values.
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I will do that to the problem a little bit later on in this lesson.
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Let us see what is next.
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Using the table of parameters below and equations which correct for anharmonicity and vibration rotation interaction,
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construct a table of energies for the first 5 rotational states of hydrogen iodide, for vibrational levels R = 0 and R = 1.
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Specify which transitions are allowed and calculate the frequencies of these transitions,
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associating each to its appropriate peak in the R and P branches.
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The same sort of thing except now we need to make adjustment to the rigid rotator harmonic oscillator
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and we are correcting for anharmonicity and vibration rotation interaction.
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Notice, we are not correcting for centrifugal distortion.
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We can always correct for all three, for two of those things, or for one of those things.
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It just depends on what the problem is asking for.
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We have our table of parameters, our rotational constant, our Α sub E,
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our fundamental frequency and our correction for anharmonicity.
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We said that the energy for the rigid rotator harmonic oscillator approximation
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that was E sub RJ = ν₀ × R + ½ + B × J × J + 1.
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This is the rigid rotator harmonic oscillator approximation.
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It does not really matter.
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The dependence of B on R, B sub R is equal to B sub E - Α sub E × R + ½.
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This thing goes in where B is, that is the adjustment for vibration rotation interaction.
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The adjustment for harmonicity, the energy is ν sub E × R + ½ - this X sub E ν sub E × R + ½².
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This is the adjustment for anharmonicity.
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Now when we put this together, when we put both of these into this,
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we get a new equation for the energy which is RJ is equal to,
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It gets a little complicated, a little long, no doubt about that.
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R + ½ - X sub E ν sub E × R + ½² and the rotational term which is going to be B sub E - Α sub E × R + ½ × J × J + 1.
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This is the ν for the two corrections that we have to make.
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Notice again, this problem does not ask us to account for centrifugal distortion.
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If it did, we have one more – that D thing, D × J² J + 1², but that does not matter here.
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For R = 0, we get E of 0 J is equal to ½ ν sub E -1/4 X sub E ν sub E + this B sub E - Α sub E.
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Again, these are just a bunch of parameters that we just have to account for.
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It is not terribly a big deal.
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R is 0 here so we can just go ahead and put the -1/2 Α sub E × J × J + 1.
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This ends up being E₀ J is equal to 1154.
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I will just put the numbers in.
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In other words, I will put the ν sub E in, we put the X sub ν E in, the BE, the Α E, we just put it in.
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That is it, nothing strange.
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9.911 + 6.4275 × J × J + 1.
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For the R = 1 vibrational state ,we get E of 1 J that is going to equal 3/2 ν sub E - 9/4
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X sub E ν sub E + B sub E – 3/2 Α sub E × J × J + 1.
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And we end up as far as numerically is concerned, we end up with 3463.521 - 8189.199.
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This is nothing more than just a bunch of tedious arithmetic, that really is L comes down to.
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+ 6.2585, which is why I personally love theory as opposed to the tedium.
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That is why we work with symbols, you do not want to do those arithmetic.
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× J × J + 1.
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This equation right here, let me go to blue, we use this equation and we use this equation.
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We create a table of values.
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We have J, J is going to take on the values 0, 1, 2, 3, 4, that is the first 5 states.
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E₀ J, all the values are in inverse cm.
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E1 J we have 1144.6, 1157.5, we have 1183.2, we have 1221.7,
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we have 1273.1, we have 3374.3, we have 3386.8, we have 3411.9, 3449.
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My head is already spinning, believe me.
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I completely understand, this is not something that you actually get used to.
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After a long time doing this, you lose your mind from what are all these numbers.
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Your head starts to spin.
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3449.4 and this is going to be 3499.5.
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The allowed transitions are δ J = + or -1.
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We are going to have the 0 to 1, 1 to 2, 2 to 3, 3 to 4.
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Or we are going to have the 1 to 0, 2 to 1.
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Basically, all you have to do when you need to find the actual frequencies of the absorption,
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the frequencies of the spectral lines, you are just going to take for example the 0 to 1 transition,
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you are just going to take this number - this number.
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The 1 to 2 transition, this number - this number.
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You do not actually need the equations, which is why the energy equation is really the most important.
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If you have that, you can get everything else.
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The δ J = + or -1, when we do that, let us go ahead and create a new table of values here.
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Once again, let us say δ J = + or -1.
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The R0, R1, R2, R3, that represents the 0 to 1 transition, this is the 1 to 2 transition,
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this is the 2 to 3 transition, this is the 3 to 4 transition.
00:30:37.700 --> 00:30:42.600
This is going to be at 2242.2.
00:30:42.600 --> 00:30:48.500
It is just the second number the second column - the first number the first column.
00:30:48.500 --> 00:30:52.600
The energy of the 1 - the energy of the 0.
00:30:52.600 --> 00:30:54.500
The energy of the 2 - the energy of the 1.
00:30:54.500 --> 00:30:58.000
For the two vibrational states.
00:30:58.000 --> 00:31:02.900
We do not necessarily need equations for Ν sub ρ and ν sub P.
00:31:02.900 --> 00:31:15.300
2254.4, 2266.2, 2277.8, and we have the P branch.
00:31:15.300 --> 00:31:20.500
This is going to be P1, P2, P3, P4.
00:31:20.500 --> 00:31:23.200
This represents the 1 to 0 transition.
00:31:23.200 --> 00:31:26.400
This represents the 2 to 1 transition.
00:31:26.400 --> 00:31:28.300
This represents the 3 to 2 transition.
00:31:28.300 --> 00:31:32.700
This represents the 4 to 3 transition.
00:31:32.700 --> 00:31:37.300
This is going to be 2216.8.
00:31:37.300 --> 00:31:41.300
This is going to be 2203.6.
00:31:41.300 --> 00:31:45.400
This is going to be 2198.2.
00:31:45.400 --> 00:31:58.600
And this is going to be 2176.3.
00:31:58.600 --> 00:32:00.000
I really like the black.
00:32:00.000 --> 00:32:14.100
If we wanted specific equations for ν sub ρ and ν sub P, basically the observed frequencies.
00:32:14.100 --> 00:32:24.900
These numbers, we have to do some algebra, that is the problem.
00:32:24.900 --> 00:32:32.200
We would have to do a lot of algebra actually.
00:32:32.200 --> 00:32:34.600
We have to do a lot of algebra.
00:32:34.600 --> 00:32:38.700
Essentially, what you have been doing is the following.
00:32:38.700 --> 00:32:43.100
For ν sub R, you take the energy of the upper states and subtract the energy of the lower state.
00:32:43.100 --> 00:32:44.200
That is what you are doing.
00:32:44.200 --> 00:32:54.800
You are going to end up with, it is going to be the energy E R + 1 J + 1 - ERJ.
00:32:54.800 --> 00:32:58.000
You have to do all that and come up with some final equation.
00:32:58.000 --> 00:33:09.700
For the P branch, that is going to end up being E R + 1 J - 1 ERJ.
00:33:09.700 --> 00:33:13.700
Based on the equation above, that long equation that we had for the energies,
00:33:13.700 --> 00:33:18.900
you have to take one for the upper state then subtract that long equation, one for the lower state.
00:33:18.900 --> 00:33:24.000
Go through all the algebra and come up with some equations, which is essentially what you see in your books.
00:33:24.000 --> 00:33:30.300
Most of what you see in your books, the equations, the derivations of the spectroscopy section are just that.
00:33:30.300 --> 00:33:33.000
They are just taking one energy and subtracting other energy.
00:33:33.000 --> 00:33:36.200
It tends to look complicated but it is just a lot of algebra.
00:33:36.200 --> 00:33:40.800
In order to come up with equations for the spectroscopic lines that we see,
00:33:40.800 --> 00:33:43.200
based on whatever correction we are making it.
00:33:43.200 --> 00:33:48.700
It is either going to be the harmonic oscillator rigid rotator approximation for the energy.
00:33:48.700 --> 00:33:52.700
You can either make a correction for vibration rotation interaction.
00:33:52.700 --> 00:33:55.400
You can make a correction for anharmonicity.
00:33:55.400 --> 00:33:58.100
You are going to make a correction for centrifugal distortion.
00:33:58.100 --> 00:34:01.100
Either 1, 2, or all 3 of those.
00:34:01.100 --> 00:34:05.700
The equation is longer for every correction that you make.
00:34:05.700 --> 00:34:10.100
You remember all this from the first lesson, the initial lesson that we did in spectroscopy
00:34:10.100 --> 00:34:13.300
where all these things are coming from.
00:34:13.300 --> 00:34:16.800
That it is not this ocean of equations that you are dealing with.
00:34:16.800 --> 00:34:28.000
It is just the energies of the upper state - the energies of lower state that give you the frequency of the spectral line.
00:34:28.000 --> 00:34:30.100
Let us do some more examples here.
00:34:30.100 --> 00:34:37.700
I think we are on example number 3 now.
00:34:37.700 --> 00:34:44.000
Wait, I think we are missing some data here.
00:34:44.000 --> 00:34:45.700
We do not have this data but that is not a problem.
00:34:45.700 --> 00:34:48.400
I actually have it right here with me, I will go ahead and write it.
00:34:48.400 --> 00:34:51.100
For some odd reason, it did not end up on this page, my apologies.
00:34:51.100 --> 00:35:00.600
Calculate B₀, B₁ of the equilibrium bond length for the R = 0 vibrational state and
00:35:00.600 --> 00:35:18.600
the equilibrium bond length for R = 1 vibrational state, the B sub E and Α sub E for carbon monoxide.
00:35:18.600 --> 00:35:22.600
I apologize, it did not show up on this slide.
00:35:22.600 --> 00:35:32.600
Ν observed, we observed an R 0 line of 2173.81.
00:35:32.600 --> 00:35:39.900
We observed R1 line at 2177.58.
00:35:39.900 --> 00:35:46.400
We observed a P1 line at 2166.15.
00:35:46.400 --> 00:35:52.900
And we observed a P2 line at 2162.27.
00:35:52.900 --> 00:36:02.100
Given this data for two R lines and the first two lines of the P branch, how can I calculate all of these?
00:36:02.100 --> 00:36:24.000
We see B0 and B1, that equations I want to deal with are going to only account for the vibration rotation interaction.
00:36:24.000 --> 00:36:25.100
I do not see anything else here.
00:36:25.100 --> 00:36:27.200
I do not see any X sub E, ν sub E.
00:36:27.200 --> 00:36:29.900
I do not have to worry about the anharmonicity.
00:36:29.900 --> 00:36:33.400
I do not see a D here so I do not have to worry about centrifugal distortion.
00:36:33.400 --> 00:36:36.900
I see B₀ B₁, the equations that I’m going to be looking at,
00:36:36.900 --> 00:36:43.100
I'm going to be looking at the rigid rotator harmonic oscillator corrected for vibration rotation interaction.
00:36:43.100 --> 00:36:45.000
That is how you choose your equation.
00:36:45.000 --> 00:36:50.700
You take a look at what constants are at your disposal, what they want, and I will tell you what to choose.
00:36:50.700 --> 00:37:14.500
We see B0 and B1, we consider only the vibration rotation correction.
00:37:14.500 --> 00:37:18.500
I always like to begin with the basic equation and then add to that.
00:37:18.500 --> 00:37:25.600
It is just a nice way to keep sort of refreshing yourself and see that the equation over and over and over again.
00:37:25.600 --> 00:37:42.900
Our harmonic oscillator rigid rotator equation E sub RJ, the energy is ν R + ½ + B × J × J + 1.
00:37:42.900 --> 00:37:47.500
I do not need that parenthesis at J × J + 1.
00:37:47.500 --> 00:37:59.400
The vibration rotation correction, that one says that this B, the rotational constant depends on R.
00:37:59.400 --> 00:38:09.100
B sub R is equal to B sub E - Α sub E × R + ½.
00:38:09.100 --> 00:38:17.200
This we put into here and get a new equation.
00:38:17.200 --> 00:38:22.400
The corrected equation for the vibration rotation interaction is equal to,
00:38:22.400 --> 00:38:25.100
Sometimes, I will put a tilde over the E, sometimes not.
00:38:25.100 --> 00:38:28.300
It is in wave numbers, do not worry about it.
00:38:28.300 --> 00:38:43.600
That is equal to, it is going to be ν R + ½
00:38:43.600 --> 00:38:58.200
+ this thing B sub E - Α sub E R + ½ × J × J + 1.
00:38:58.200 --> 00:39:17.300
Unless specifically stated otherwise,
00:39:17.300 --> 00:39:25.200
always take the R = 0 to the R = 1 vibration transition.
00:39:25.200 --> 00:39:27.700
That is the one that you want to take.
00:39:27.700 --> 00:39:34.400
At normal temperatures, room temperatures, most molecules are actually going to be in their ground vibrational state.
00:39:34.400 --> 00:39:37.700
They are also going to be in their ground electronic state.
00:39:37.700 --> 00:39:44.500
However, at normal temperatures, it is the rotational states that molecules tend to be the higher states,
00:39:44.500 --> 00:39:46.200
they cannot be the ground state.
00:39:46.200 --> 00:39:52.900
J = 0, they can be anywhere from like J = 2 or 3, above that.
00:39:52.900 --> 00:40:01.300
At normal temperatures, most molecules are in the vibrational ground state and electronic ground state.
00:40:01.300 --> 00:40:05.400
It is only the rotational state that there are in excited states at normal temperatures.
00:40:05.400 --> 00:40:19.200
We will talk more about that when we do statistical thermodynamics next.
00:40:19.200 --> 00:40:23.000
When we look at this, our new observe, what we are looking for is this.
00:40:23.000 --> 00:40:33.100
We need an equation that is going to be the energy of the upper state - the energy of lower state.
00:40:33.100 --> 00:40:35.500
The energy being the equation that we just had.
00:40:35.500 --> 00:40:47.700
Our R branch is going to be E of 1 J + 1 - E of 0 J.
00:40:47.700 --> 00:40:59.600
Our ν sub P is going to be E of 1 J-1 E0 J.
00:40:59.600 --> 00:41:04.300
When I do the algebra for these and I put the energy equation that I just had in the previous page,
00:41:04.300 --> 00:41:06.500
one for the upper state and one for the lower state.
00:41:06.500 --> 00:41:14.900
When I work out that algebra, here is what I get.
00:41:14.900 --> 00:41:31.100
When we work out these differences, we get a rather daunting looking equation actually.
00:41:31.100 --> 00:41:52.500
Ν₀ + B1 - B0 J² + 3 B1 - B0 J + 2 B1.
00:41:52.500 --> 00:41:57.700
The J takes on the value starting from 0, 1, 2, and so on.
00:41:57.700 --> 00:42:09.000
For the P branch, I get ν₀ + this same term B1 - B0 J².
00:42:09.000 --> 00:42:20.100
The third term is different, this is going to be B1 - B0 × J.
00:42:20.100 --> 00:42:22.800
Here, the J takes on the values of 1, 2, 3.
00:42:22.800 --> 00:42:25.300
And again, for the R branch 0, so on.
00:42:25.300 --> 00:42:41.600
The P branch is 1 and so on.
00:42:41.600 --> 00:42:44.800
Let me go to blue here.
00:42:44.800 --> 00:42:57.500
The R is 0 line that represents the J = 0, that is going to be ν₀ + 2 B1.
00:42:57.500 --> 00:43:03.700
In other words, I put J = 0 into this equation and I see what I get.
00:43:03.700 --> 00:43:08.000
I get ν₀ + 2 B1.
00:43:08.000 --> 00:43:22.100
The data gave us the frequency of absorption, the ν₀, the R₀ line is at 2173.81.
00:43:22.100 --> 00:43:42.400
The R1 line that is where J is equal to 1, that is ν₀ + 6 B1 – 2 B0 is equal to 2177.58.
00:43:42.400 --> 00:43:47.000
The P₁ line that represents that J = 1.
00:43:47.000 --> 00:43:50.800
I’m going to use this equation, the one for the ν sub P.
00:43:50.800 --> 00:44:02.100
I get ν₀ -2 B0 = 2166.15 inverse cm.
00:44:02.100 --> 00:44:07.600
For the P2 line that is the J = 2, I get ν₀.
00:44:07.600 --> 00:44:14.000
I get + 2 B1 -6 B0.
00:44:14.000 --> 00:44:20.500
And that one was observed at 2162.27.
00:44:20.500 --> 00:44:27.800
These are the equations that I’m going to work with now.
00:44:27.800 --> 00:44:34.300
I’m going to take the R1 equation - the P1 equation.
00:44:34.300 --> 00:44:44.300
When I do that, I get 6 B1 = 11.43.
00:44:44.300 --> 00:44:54.400
Therefore, B1 is equal to 1.905 inverse cm.
00:44:54.400 --> 00:44:56.300
I found B1.
00:44:56.300 --> 00:45:00.900
Now, I take the R0 line and I subtract from it the P2 line.
00:45:00.900 --> 00:45:10.900
When I do that, I get 6 B0 = 11.54.
00:45:10.900 --> 00:45:21.700
Therefore, B0 is equal to 1.923 inverse cm.
00:45:21.700 --> 00:45:26.000
We found B1, we found B0.
00:45:26.000 --> 00:45:39.000
BR, this B sub R that we just found is equal to H/ 8 π² C × I.
00:45:39.000 --> 00:45:44.400
I is the reduced mass × this, whatever this is².
00:45:44.400 --> 00:45:51.200
Now, we need to find the equilibrium bond length for each of these.
00:45:51.200 --> 00:45:53.900
When I put the values in, I will get the following.
00:45:53.900 --> 00:46:11.400
When I rearranged this for RE², R sub E vibrational state 1² is equal to 6.626 × 10⁻³⁴.
00:46:11.400 --> 00:46:13.900
I’m not going to put the units in and I’m just going to put the numbers in.
00:46:13.900 --> 00:46:22.300
8 π² 2.998 × 10¹⁰.
00:46:22.300 --> 00:46:26.300
I went ahead and use cm directly.
00:46:26.300 --> 00:46:37.100
1.139 × 10⁻²⁶ that was the reduced mass and then the 1.905 that is this number right here.
00:46:37.100 --> 00:46:39.900
This goes down here, this comes up here.
00:46:39.900 --> 00:46:52.000
When I solve for this, I get that it is equal to 113.6 pm.
00:46:52.000 --> 00:47:12.000
The same for R0, when I do the same for R0, R sub E for the 0 state, I end up with a value of 113 pm.
00:47:12.000 --> 00:47:16.600
We found RE 1, RE, we found B0, we found B1.
00:47:16.600 --> 00:47:30.200
Let us go ahead and find for B sub E and Α sub E.
00:47:30.200 --> 00:47:42.600
We said that b sub R is equal to B sub E – α sub E × R + ½.
00:47:42.600 --> 00:47:46.100
We know what B0 is.
00:47:46.100 --> 00:47:54.200
B0 is equal to B sub E – ½ Α sub E.
00:47:54.200 --> 00:48:00.700
We know what B1 is, let us put that 0 into here, 1 into here for R.
00:48:00.700 --> 00:48:07.700
We get B sub E - 3/2 Α sub E.
00:48:07.700 --> 00:48:11.800
We know what B0 is, we already found it 1.923.
00:48:11.800 --> 00:48:23.400
1.923 = B sub E – ½ Α sub E.
00:48:23.400 --> 00:48:33.700
And we know that 1.905 which is B1, that = B sub E - 3/2 Α sub E.
00:48:33.700 --> 00:48:36.400
Two equations and 2 unknowns.
00:48:36.400 --> 00:48:41.600
Let us go ahead and take, this is equation 1 and this equation 2.
00:48:41.600 --> 00:48:47.800
Let us go ahead and take equation 1 - equation 2.
00:48:47.800 --> 00:49:04.000
When you do that, you end up with Α sub E is equal to 0.018 inverse cm.
00:49:04.000 --> 00:49:17.500
Put this into one of these other equations and you end up with B sub E is equal to 1.932 inverse cm.
00:49:17.500 --> 00:49:22.200
There you go, I hope that make sense.
00:49:22.200 --> 00:49:29.700
Let us see what is next.
00:49:29.700 --> 00:49:42.700
The IR spectrum of ML shows a fundamental line at 1877.62 inverse cm and the first overtone at 3728.66 inverse cm.
00:49:42.700 --> 00:49:53.300
Find the values of ν sub E and X sub E ν sub E for ML.
00:49:53.300 --> 00:49:58.400
Let us go ahead and work in black here.
00:49:58.400 --> 00:50:27.100
The frequencies of the overtones for the anharmonic oscillator
00:50:27.100 --> 00:50:36.900
are given by G of R – G of 0, just looking at vibrational transitions right now.
00:50:36.900 --> 00:50:43.400
G of R to G of 0, the anharmonic oscillator.
00:50:43.400 --> 00:50:46.400
Under normal temperatures, most molecules are in the ground vibrational states.
00:50:46.400 --> 00:50:50.400
They are all going to start at the 0 vibrational state.
00:50:50.400 --> 00:50:53.400
But δ R is no longer + or -1.
00:50:53.400 --> 00:50:56.400
It can be + or -1, + or -2, + or -3.
00:50:56.400 --> 00:50:59.300
The + or -1 is the fundamental.
00:50:59.300 --> 00:51:02.000
The + or -2, those are the first overtone.
00:51:02.000 --> 00:51:05.800
The first overtone, second overtone, third overtone.
00:51:05.800 --> 00:51:15.800
The G of R equation that is equal to ν sub E × R + ½.
00:51:15.800 --> 00:51:17.700
The anharmonic oscillator – X sub E ν sub E R + ½².
00:51:17.700 --> 00:51:32.900
When we take the G sub R – G of 0, when we do the algebra,
00:51:32.900 --> 00:51:49.400
what we get is the observe line equal to ν sub E × R – X sub E ν sub E × R × R + 1.
00:51:49.400 --> 00:51:56.700
R1 runs from 1, 2, 3, and so on.
00:51:56.700 --> 00:52:11.000
The fundamental transition is the transition from 0 to 1, that is the one that we see.
00:52:11.000 --> 00:52:15.300
That is the brightest one that we see.
00:52:15.300 --> 00:52:21.800
That is equal to, I put into this equation, fundamental 0 to 1.
00:52:21.800 --> 00:52:37.200
I put R = 1, it is going to be ν sub E × 1 - X sub E Ν sub E × 1 × 1 + 1.
00:52:37.200 --> 00:52:58.300
We end up with ν sub E - 2 X sub E ν sub E.
00:52:58.300 --> 00:52:59.600
It tells us what that is already.
00:52:59.600 --> 00:53:04.000
The fundamental is 1877.622.
00:53:04.000 --> 00:53:11.500
1877.622 that is one of the equations that we are going to use.
00:53:11.500 --> 00:53:18.500
The first overtone represents the transition from 0 to 2.
00:53:18.500 --> 00:53:21.800
It is a weaker line.
00:53:21.800 --> 00:53:28.100
That is going to be ν sub E × 2.
00:53:28.100 --> 00:53:34.600
Just using this equation right here, taking care of the R = 1.
00:53:34.600 --> 00:53:47.300
Now the first overtone R = 2 - X sub E ν sub E × 2 × 2 + 1.
00:53:47.300 --> 00:54:00.000
What we end up with is 2 ν sub E – 6 X sub E ν sub E, it tells us what the first overtone is 3728.
00:54:00.000 --> 00:54:06.700
3728.66 that is the second equation that we are going to use.
00:54:06.700 --> 00:54:08.900
When you solve simultaneously, I will not go through the process.
00:54:08.900 --> 00:54:10.300
I will let you go ahead and do that.
00:54:10.300 --> 00:54:13.200
This equation and this equation, when you solve simultaneously,
00:54:13.200 --> 00:54:34.300
you are going to get a value of ν sub E is equal to 1904.20 inverse cm and X sub E ν sub E is equal to 13.29 inverse cm.
00:54:34.300 --> 00:54:38.900
That is it, nice and easy.
00:54:38.900 --> 00:54:41.900
I know it is not nice and easy, believe me I do.
00:54:41.900 --> 00:54:48.700
I have been there, the stuff still gives me grief too.
00:54:48.700 --> 00:54:51.400
Let us take a look at our final example.
00:54:51.400 --> 00:54:57.100
I’m actually wondering whether I should leave this example off because the first two lessons took care of it.
00:54:57.100 --> 00:55:03.900
Of course, we had an example from the previous lesson that did this but we will see.
00:55:03.900 --> 00:55:12.000
In examples 6 of the previous lesson, we asked you to calculate the frequencies of the first two lines of the R and P branches
00:55:12.000 --> 00:55:19.800
for the vibration rotation spectrum of HBR under the harmonic oscillator rigid rotator approximation.
00:55:19.800 --> 00:55:22.800
Here we ask you to repeat the problem.
00:55:22.800 --> 00:55:26.600
But instead of the harmonic oscillator rigid rotator approximation,
00:55:26.600 --> 00:55:35.200
we asked you to do so by accounting for all three corrections we made to the HIR equations.
00:55:35.200 --> 00:55:40.400
We accounted for vibration rotation interaction for centrifugal distortion and for harmonicity.
00:55:40.400 --> 00:55:43.600
Use the following spectroscopic parameters as necessary.
00:55:43.600 --> 00:55:49.900
This is a great example because we get a chance to put everything together,
00:55:49.900 --> 00:56:06.900
the HRR approximation and all of the corrections that we made.
00:56:06.900 --> 00:56:08.500
I’m not going to write all the equations here.
00:56:08.500 --> 00:56:12.600
I can do so in the last couple of lessons, basically what we are going to be doing is,
00:56:12.600 --> 00:56:18.300
in this one you are going to take the harmonic oscillator approximation.
00:56:18.300 --> 00:56:20.700
You are going to make the corrections for all three of these things.
00:56:20.700 --> 00:56:25.600
The vibration rotation interaction, the centrifugal distortion, and the anharmonicity.
00:56:25.600 --> 00:56:28.300
You got to find an equation for the total energy.
00:56:28.300 --> 00:56:31.600
You are going to take the upper level - the lower level.
00:56:31.600 --> 00:56:34.800
In other words, you are going to find δ E.
00:56:34.800 --> 00:56:37.800
That is going to give you your equation.
00:56:37.800 --> 00:56:41.800
You do not necessary have to go through the algebra, the equations are actually have been done for you.
00:56:41.800 --> 00:56:44.500
They are in your books, I think we actually did in our lessons too.
00:56:44.500 --> 00:56:51.600
When you do it, the equations that you come up with are the following.
00:56:51.600 --> 00:57:13.900
Ν sub R, that is going to be ν₀ × 1 – 1.
00:57:13.900 --> 00:57:44.900
2648, we are going to have ν sub E × 1 - 2 XE + 2 × BE × J + 1 - Α sub E × J + 1 × J + 3 - 4 D × J + 1³.
00:57:44.900 --> 00:57:51.200
The J values are going to run from 0, 1, 2, and so on.
00:57:51.200 --> 00:57:55.200
This is the equation that you get when you take the total energy,
00:57:55.200 --> 00:58:02.300
the correction for all three things and the energy of the upper state – the energy of the lower state.
00:58:02.300 --> 00:58:04.400
It is a lot of algebra but this is the equation you come up with.
00:58:04.400 --> 00:58:12.000
This is the frequency of the spectral line that we should see.
00:58:12.000 --> 00:58:45.000
The ν sub P branch that is going to be ν sub E × 1 - 2X sub E -2 B sub E × J - Α sub E × J × J - 2 + 4D J³.
00:58:45.000 --> 00:58:49.600
Here J = 1, 2, and so on.
00:58:49.600 --> 00:58:51.800
These are the two equations that we actually get.
00:58:51.800 --> 00:58:57.200
Notice the equation for the spectral line changes depending on what corrections we are making.
00:58:57.200 --> 00:59:01.800
In this case, we corrected for all three.
00:59:01.800 --> 00:59:03.700
They give us ν sub E, we have that one.
00:59:03.700 --> 00:59:06.600
We are going to need that parameter.
00:59:06.600 --> 00:59:08.800
We are going to need this parameter.
00:59:08.800 --> 00:59:13.700
We are going to need this parameter.
00:59:13.700 --> 00:59:15.100
We do not need those others.
00:59:15.100 --> 00:59:18.900
We also need the X sub E but they gave us N sub E ν sub E.
00:59:18.900 --> 00:59:20.800
Actually, you have to take this.
00:59:20.800 --> 00:59:21.600
We would need that.
00:59:21.600 --> 00:59:26.800
We would have to take this parameter divided by that, in order to get just the X sub E.
00:59:26.800 --> 00:59:30.600
Let us go ahead and do that first.
00:59:30.600 --> 00:59:46.000
The data gives ν sub E and X sub E ν sub E.
00:59:46.000 --> 00:59:51.100
Let us find X sub E alone.
00:59:51.100 --> 00:59:58.700
The X sub E is equal to X sub E ν sub E divided by ν sub E.
00:59:58.700 --> 01:00:17.400
We end up with 45.217 divided by 2648.975 and we end up with X sub E = 0.01707 inverse cm.
01:00:17.400 --> 01:00:21.400
We put all these parameters into the equations.
01:00:21.400 --> 01:00:35.700
What we end up with, ν of the R₀ line.
01:00:35.700 --> 01:00:53.100
That is going to equal 2648.975 × 1 - 2 × 0.01707 + 2
01:00:53.100 --> 01:01:23.600
× 8. 465 × 0 + 1 - 0.2333 × 0 + 1 × 0 + 3 - 4 × 3.457 × 10⁻⁴ × 0 + 1³.
01:01:23.600 --> 01:01:32.800
When I calculate that, I end up with 2574.768 inverse cm.
01:01:32.800 --> 01:01:36.100
This is my R0 line.
01:01:36.100 --> 01:01:38.800
I should see it there.
01:01:38.800 --> 01:01:46.400
Notice, we expected something lower than what we get with the rigid rotator harmonic oscillator approximation.
01:01:46.400 --> 01:01:48.100
Exactly what we got.
01:01:48.100 --> 01:01:54.300
Now, when I do the same for the R1 line, I will write all this out.
01:01:54.300 --> 01:02:03.800
What I end up with is 2590.522 inverse cm.
01:02:03.800 --> 01:02:07.700
Notice the spacing between the lines.
01:02:07.700 --> 01:02:09.300
The spacing between the spectral lines is this line - this line.
01:02:09.300 --> 01:02:15.500
This line - this line, the absolute value there.
01:02:15.500 --> 01:02:30.100
The spacing is ν of R1 - ν of R0, that is equal to 15.755 inverse cm.
01:02:30.100 --> 01:02:42.800
2B E = 2 × 8.465 that is equal to 16.93 inverse cm.
01:02:42.800 --> 01:02:46.400
You see that the spacing, the corrections that we have made.
01:02:46.400 --> 01:02:49.600
On the actual spectrum, the spacing is less than 2B.
01:02:49.600 --> 01:02:51.500
2B is 16.93.
01:02:51.500 --> 01:03:00.100
The corrections that we made give us the spacing which is about less than 2B, which is exactly what we expect.
01:03:00.100 --> 01:03:03.300
For the R branch, the spacing is smaller.
01:03:03.300 --> 01:03:06.900
For the P branch, the spacing will actually be bigger than 2B.
01:03:06.900 --> 01:03:12.600
The 2B is the rigid rotator harmonic oscillator approximation.
01:03:12.600 --> 01:03:14.500
Let us go ahead and do the P branch.
01:03:14.500 --> 01:03:21.200
I think I have one more page, If I’m not mistaken, yes I do.
01:03:21.200 --> 01:03:41.800
I got ν of P1 and when I put into the equation for the P1, I end up with 2541.8.
01:03:41.800 --> 01:03:43.700
Let me write this one, at least.
01:03:43.700 --> 01:04:03.700
It is going to be 2648.975 × 1 - 2 × 0.01707 - 2 × 8.465
01:04:03.700 --> 01:04:22.700
× 1 - 0.2333 × 1 × 1 - 2 + 4 × 3.457 × 10⁻⁴ × 1³.
01:04:22.700 --> 01:04:30.500
This will give me 2541.843 inverse cm.
01:04:30.500 --> 01:04:44.800
And when I do the same for the P2 line, I end up with 2524.690.
01:04:44.800 --> 01:04:51.000
Now, the difference here, the δ ν.
01:04:51.000 --> 01:05:02.900
In other words, one of them - the other, that is going to equal 17.153, which is greater than 2B,
01:05:02.900 --> 01:05:09.700
which is exactly what we expect for the P branch.
01:05:09.700 --> 01:05:13.000
That is it, thank you so much for joining us here at www.educator.com.
01:05:13.000 --> 01:01:05.000
We will see you next time.