WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to continue our discussion of the quantum mechanical harmonic oscillator.
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Let us dive right on in.
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A vibrating molecule can be modeled by the harmonic oscillator.
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This is all that is happening, this is going back and forth like this.
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A vibrating molecule can be modeled by the harmonic oscillator.
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A diatomic molecule has two masses that are moving, not one fix end and the other one moving back and forth like that.
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A diatomic molecule has two masses that oscillate and no fixed end.
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In general, if you have like a really huge atom and a tiny atom, for all practical purposes, the huge atom is not going to move very much.
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You can consider that fixed end and the tiny atoms itself would vibrate back and forth like that.
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We cannot always guarantee you that so we need something a little bit more sophisticated than just a fixed wall and one mass moving.
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The situation we have is this.
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Something like that where you have mass one and mass two.
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There are some modifications that we make actually to the equations of motion.
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Instead of one equation of motion, we have two equations of motion, one for each mass.
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We combine them and we come up with a new differential equation.
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When I can to go through that process but will say this much.
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We will not discuss the modifications due to the equations of motion but we end up with this.
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But we end up with a differential equation that we end up with is μ D² X DT² + KX equal 0.
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This is exactly the same equation that we came up with for the classical harmonic oscillator
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with one fixed end of the wall and one body, one mass moving back and fourth.
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The only difference is instead of the mass of the single ball moving back and forth, this is μ something called the reduced mass.
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I will tell you what it is.
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Where μ is called the reduced mass and is given by the μ is equal to 1/ M1 + 1/ M2.
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It is also equal to M1 +,
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I'm sorry to not μ, this is 1/ μ.
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M1 + M2/ M1 M2.
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If you like μ directly is M1 M2/ M1 + M2.
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Basically, what we have done when we modify the equations of motion,
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we have come up with this way of combining the masses into something called reduce mass.
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By combining I do not just mean adding, it is this, this is the relationship.
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The reduced mass is 1/ M = 1/ M1 + 1/ M2.
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We have combined them and we have basically taken this two body problem and we have turned it into a one body problem.
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This is a single equation, this is a single number.
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This happens a lot in classical mechanics and classical physics.
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In fact, we can do this, we can take a two body problem or three body problem and turn it into a one body problem.
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It is actually quite extraordinary that we can do this.
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It is all we have done.
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Everything is the same, the only difference is this thing called the reduced mass which is a combination of the masses of the two objects.
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In this case, the two atoms or maybe the two blocks of cement, whatever it is you have to be dealing with.
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This is possible to do but you will end up with the same equation.
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You end up with the same general solution, that is what makes this beautiful.
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Let me go to blue actually.
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This notion of reduced mass has allowed us to treat two body problem as a one body problem.
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The same as before.
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The equation, μ D² X DT² + KX is equal to 0, is exactly the same as before.
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The general solution is the same, everything is the same.
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Everything that we did in the previous lesson applies here.
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The general solution is the same.
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We have X of T equals C1 × the cos of ω T + C2 × the sin of ω T.
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Where ω is the angular velocity.
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Ω is K/ M, the only difference is ω is K.
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This time instead of /M it is /μ.
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Everything else is the same, nothing is changed.
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Wherever we see mass before, we are replacing it with a reduced mass.
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Let us take a look, we are talking about molecules.
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We are talking about atoms and atoms that we come together back and fourth, they oscillate like this.
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Let us look at how good an approximation of the harmonic oscillator potential is to the actual intra molecular potential.
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This intermolecular should be inter atomic potential but that is fine.
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I will just leave this as intermolecular.
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They are potential of the vibrating molecule.
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These are one of those terms that are stuck.
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We are talking about potential that exist between the two atoms of the molecule.
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Intermolecular means between two molecules.
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Basically, you have something that can look like this.
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I will go ahead and draw something like that.
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In this axis, we are going to have the energy.
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On this axis, we are going to have the distance that one atom is to the other atom.
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I will go ahead and mark S of 0 here.
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And this is just S, this is the distance that the two atoms are away from each other.
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We are end up getting something like this.
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Let me go ahead and put a minimum mark here.
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I will go ahead in red.
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I will go ahead and draw something like that, sort of magnified.
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This right here, we will go back to blue.
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This curve right here is the potential energy, that is the ½ KX².
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That is the parabola, that is the potential energy of our harmonic oscillator.
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As we pull something further apart the potential energy increases.
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They want to be pulled back together.
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As we squeeze them together, the potential energy increases.
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They want to push themselves apart.
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There are someplace where the equilibrium position is actually perfect where the energy is minimized.
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That is this point right here.
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This is the potential energy, this is the actual potential energy curve of
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what happens when you take two atoms and you squish from together.
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Over here, there are an infinite distance apart from each other.
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As you start bring them together, bring them together,
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as we are bringing the atoms from an infinite distance apart from each other, the closer and closer and closer.
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The attraction that they feel toward each other is actually going to drop the potential energy.
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The potential energy is going to drop.
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It is going to hit a point where they are as close as they are going to be to each other where energy is minimized.
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If you push them any closer together, the energy rises again.
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This point right here, this little minimum that is the actual bond length.
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That is the length at which the atoms are comfortable.
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They are as close as they are going to be to each other.
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It pull them apart, the potential energy gets bigger.
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You push them together, the potential energy gets bigger.
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This is an approximation based on the harmonic oscillator.
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Again, harmonic oscillator you pull them apart, the potential energy increases down to 0.
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It push them away from equilibrium position goes to 0.
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For very small displacements which is pretty much what happens when a molecule vibrates and a molecule is not going like this.
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There are not huge displacements away from the equilibrium position.
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They are very small and for very small displacements this parabolic approximation to the actual potential energy curve is very good.
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In fact, you can see them overlap beautifully and that is what we are doing.
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We are a approximating this inter nuclear potential.
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It is not intermolecular, it is inter nuclear potential.
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Sorry, earlier in the day I was thinking about intermolecular forces and I think I’m getting them mixed up.
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The inter nuclear potential, this is what actually happens when you take atoms and bring them closer together.
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This pair of parabola is an approximation, this is the harmonic oscillator.
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For very small displacements, they are right on top of each other.
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It was a perfect match.
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It is only when we get it to huge oscillations that they start to deviate from each other
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or our parabolic model and the potential energy does not fit with the data anymore.
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But it is never going to be that way.
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For normal vibration of frequencies the displacement is very small.
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We are good, let us go ahead and write this down.
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For small vibrations about the equilibrium position it is an excellent approximation.
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Of course that place which is the minimum, that distance, that distance they are from each other, that is the bond length.
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It makes sense.
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Now the Schroeder equation and we want talk about a quantum mechanical harmonic oscillator.
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The Schrӧdinger equation for the 1 dimensional quantum mechanical harmonic oscillator looks like this.
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It is going to be –H ̅²/ 2 and again we use the reduced mass.
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D² ψ/ DX² + the potential energy × ψ equals the energy × ψ.
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We know what the potential energy is, it is ½ KX².
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When we put B equals ½ KX².
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When we put that into here, rearrange this equation, we get D² ψ DX² + 2 μ/ H ̅² × energy - ½ KX² × ψ equals 0.
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This is the equation that we end up solving.
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This is the Schrӧdinger equation for the harmonic oscillator.
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When we solve this, when we solve this equation which is actually not an easy thing to do
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because the coefficients are no longer constant.
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We will start with the energies first.
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We get the following quantized values for the energy, that right there.
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For the energy E of the system, we get that E is equal.
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Let me do this in blue.
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The energy of the system was equal to H ̅ × K/ μ ^½ × R + ½,
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Where R is equal to 0, 1, 2, and so on.
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R is another quantum number.
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It is another quantum number so are 0, 1, 2, it can only take on those values.
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The energy of the system is quantized, it cannot be any energy.
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It is going to be one energy and then it is going to be another energy.
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It does not make a nice smooth transition jumps.
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It is quantized.
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K/ M ^½ is just ω.
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H ω R + ½ those are the energies for the quantum mechanical harmonic oscillator.
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Since frequency equals ω/ 2 π and E is also equal to planks constant × frequency × R + ½.
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Again, ω and K/ μ ^½ power and frequency equals ω/ 2 π,
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Which is going to equal 1/ 2 π K μ ^½.
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In most books you will see that the energy is equal to H ̅ ω × something that looks like a μ + ½.
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Μ= 0, 1, 2, and so on.
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These V are modified μ, what ever it is that they call this thing in the books.
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Let me go ahead and write it out.
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This and this, look too much alike in a book.
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I reserve that symbol for frequency and R for the quantum number.
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When we plot the energies, we end up doing something like this.
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We have our parabolic.
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This is R = 0, R =1, R=2, R=3.
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Again, as we are getting further and further away, our potential is increasing.
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This one, our energy₀, our 0 state energy is equal to ½ H ̅ ω.
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Our energy for level 1 is equal to 3/2 H ̅ ω.
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Our energy 2, is equal to 5/2 H ̅ ω.
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Our energy 3, and so on, equals 7/2 H ̅ ω.
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Again, we are just plugging them into these values.
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Notice that the successive energy levels ½ H ̅ W, 3/2 H ̅ W, 5/2 H ̅ W, 7/2 H ̅ W.
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They are the same, the jump is the same.
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It is H ̅ ω.
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Notice that the successive energy levels have equal spacing.
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That equal spacing is H ̅ ω.
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Notice, specially that the ground state energy is E0 ½ H ̅ ω.
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It is not equal to 0.
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This is very different from the classical harmonic oscillator.
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The classical harmonic oscillator, when the mass is sitting at its equilibrium position, it is not moving, it is not stretched or compressed.
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There is no potential energy, there is no kinetic energy, it is 0.
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The ground state of the classical harmonic oscillator is 0.
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The ground state of the quantum mechanical harmonic oscillator is not 0.
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There is always some vibration going on, that is what is happening here.
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The ground state, the one with a quantum number R=0 is called the 0 point energy.
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The 0 point energy does not mean it is 0.
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Sometimes it will be, sometimes it will not be.
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In the case of a quantum mechanical harmonic oscillator, it is not.
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The fact that it is not 0 it actually comes from the uncertainty principle.
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It does not come from as a result of the uncertainty principle.
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It is a result of the uncertainty principle and we will show you how.
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The energy of the system is equal to its kinetic energy + its potential energy.
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The kinetic energy can be written as P²/ 2 × the mass.
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½ mass × velocity² is the same as the momentum² / twice the mass + ½ KX².
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Let me go to red here.
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We have momentum and we have position.
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In order for the energy to be 0, I have to be able to make that position arbitrarily 0 and the momentum arbitrarily 0.
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And we know from previous work that we cannot do that.
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These two operators do not commute.
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You cannot specify to an arbitrary degree of precision or accuracy both the position and the momentum simultaneously.
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As you make one better, the other one gets worse.
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But you cannot arbitrarily make them both.
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You can bring the error of both to 0, which would make the 0.
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As a result of that, because the energy is kinetic + potential, you have the momentum and
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the position showing up in the same expression, you are never going to get something which is 0.
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That is where it comes from.
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Let us go back to black here.
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If we take the quantum mechanical harmonic oscillator as the model for an oscillating diatomic molecule or a vibrating diatomic molecule,
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then the vibration energy levels of the molecule are given by what we said before E sub R = H ̅ × ω × R + ½.
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It is giving me the different energy values of the different vibrational states of this vibrating molecule.
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R goes from 0, 1, 2, and so on.
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Molecule can transition from one energy level to the next if it absorbs or emits radiation energy of frequency μ.
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It is vibrating, its molecule is vibrating.
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If I hit it with some radiation, the frequency of the radiation matches the energy change, the H ̅ ω,
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the system is going to start vibrating at the next level up.
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Or if it is vibrating to the next level up, it releases enough energy that happens to match certain frequency which is given by,
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if the energy of that particular frequency happens to be H ̅ ω, then it will go from a higher energy state to the next lower energy state down.
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It is just making transitions between energy states.
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A molecule can transition from one energy level to the next, if it absorbs or emits radiation of frequency μ.
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The change in energy is equal to planks constant × μ.
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The radiation of a given frequency has energy equal to planks constant × that frequency.
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Later in the course, we will demonstrate that the quantum mechanical harmonic oscillator
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only allows transitions between successive energy levels, that is the δ R =+ or -1.
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In other words, if I'm at the energy level 1 and if I want to get to the energy 5, I cannot just go directly from 1 to 5.
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I have to go to 2, to 3, to 4, to 5.
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In can only jump go up or down in individual stages.
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Successive energy levels, I cannot make a huge leap like an electronic transition or something.
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In this particular case, I have to pass through the successive stages.
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This right here, this DR is equal to + or -1.
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In other words, R 12345 54321 is called the selection rule.
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Let us write here it is called the selection rule.
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+ or - 1 is called the selection rule and you are going to see quite of them.
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The change in energy from one state to another is the energy of R + 1 - the energy of R, the one stage above,
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- the one stage that you are coming from.
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Let us do it this way, H ̅ × K of μ ^½ × R + 1 + ½ - R × K/ μ ^½ × R + ½.
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This is going to equal H ̅ × K/ μ ^½ × R + 3/2 - R - ½.
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R and R, you end up with H ̅ K/ μ ^½ 3/2 - 1/2 is 2/2 = 1.
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It equals that, and that equals H ̅ ω.
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The change in energy, the transition from one energy level to the next one, either up or down is going to be H ̅ ω.
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The change in energy is equal to planks constant × the frequency.
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You see that the change in energy between successive energy levels of the quantum mechanical harmonic oscillator is H ̅ ω or H ̅ × K/ μ ^½.
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Let us set them equal to each other.
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We have H μ is equal to, this H ̅ is equal to H/ 2 π.
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I’m just going to write it that way.
00:32:05.100 --> 00:32:15.600
It is equal to H/ 2 π × K/ μ ^½.
00:32:15.600 --> 00:32:29.600
The H's cancel and you are left with μ is equal to 1/ 2 π × K/ μ ^½.
00:32:29.600 --> 00:32:33.400
In order for it to actually make the transition from one energy level to the other,
00:32:33.400 --> 00:32:40.000
the quantum mechanical harmonic oscillator it has to absorb or emit energy of frequency that is given by this.
00:32:40.000 --> 00:32:44.700
Nothing more than the spring constant and the actual reduced mass.
00:32:44.700 --> 00:32:48.000
We are going to express this in terms of something called a wave number,
00:32:48.000 --> 00:32:55.700
which is something that is what you actually the scales that you see in spectroscopic data are actually expressed in wave numbers,
00:32:55.700 --> 00:33:01.300
not necessarily in frequencies, more often than not.
00:33:01.300 --> 00:33:06.500
Anything with a little tilde sign over it is a wave number.
00:33:06.500 --> 00:33:10.600
It is nothing more than the actual thing divided by the speed of light.
00:33:10.600 --> 00:33:20.300
In this particular case, it is going to be 1/ 2 π × the speed of light K / μ ^½.
00:33:20.300 --> 00:33:22.500
That is it, just take the frequency and divide by ψ.
00:33:22.500 --> 00:33:26.900
That is all you are doing.
00:33:26.900 --> 00:33:36.500
Where this μ tilde is the wave number.
00:33:36.500 --> 00:33:47.500
It is called the wave number in units of inverse centimeter.
00:33:47.500 --> 00:34:28.800
Since successive energy states are separated by the same energy which is H ̅ ω δ E is the same for every transition.
00:34:28.800 --> 00:34:34.300
Δ E is the same for every transition so when we actually irradiate something or when we measure the radiation that actually being radiated.
00:34:34.300 --> 00:34:40.600
We measure the radiation that actually gives off when it is dropping back down to is ground state.
00:34:40.600 --> 00:34:47.800
It is all the same energy level from 5 to 4, 4 to 3, 3 to 2, 2 to 1.
00:34:47.800 --> 00:34:50.500
All we measure is one value.
00:34:50.500 --> 00:34:58.500
The spectrum for this only gives us one line at one frequency or one particular wave number.
00:34:58.500 --> 00:35:22.400
Spectrum predicted by μ or μ tilde consists of a single line.
00:35:22.400 --> 00:35:34.500
We see the spectrum predicted by that, predicts that it should be represented by a single line.
00:35:34.500 --> 00:35:46.100
The prediction is good and fits well with the actual data that we correct.
00:35:46.100 --> 00:36:18.300
The prediction is good and this single line is called the fundamental vibration of frequency.
00:36:18.300 --> 00:36:49.100
For diatomic molecules, this line appears at around 10⁻³ inverse cm.
00:36:49.100 --> 00:37:05.300
Or somewhere in the range of 10⁻⁴ 10⁻⁵ Hz, if you want to express it in terms of frequency.
00:37:05.300 --> 00:37:10.800
This falls into the infrared range.
00:37:10.800 --> 00:37:31.600
Now we can use IR spectra data and this thing.
00:37:31.600 --> 00:37:45.400
We can use that to actually find force constant defined values of K for individual diatomic molecules.
00:37:45.400 --> 00:37:48.500
How stiff is the bond.
00:37:48.500 --> 00:37:49.900
That is what we are doing, we are finding K.
00:37:49.900 --> 00:37:59.200
How stiff is the bond when it vibrates, is the atom going a lot or it is going a little really fast, what is it?
00:37:59.200 --> 00:38:02.800
Let us go ahead and do an example.
00:38:02.800 --> 00:38:08.300
Let me go ahead and do this in blue.
00:38:08.300 --> 00:38:47.100
Our example is going to be, the IR spectrum of H 35 CO has a single line at 2886 inverse cm.
00:38:47.100 --> 00:38:55.300
Calculate K, the force constant, for this molecule.
00:38:55.300 --> 00:38:57.800
Let us go ahead and do it.
00:38:57.800 --> 00:39:23.700
Let us see, we start with our basic equation which is μ tilde = 1/ 2 π C × K / μ ½.
00:39:23.700 --> 00:39:38.300
We are going to get, 2 π C that² is equal to K/ μ.
00:39:38.300 --> 00:39:48.700
Therefore, our force constant is going to be μ × 2 π C that².
00:39:48.700 --> 00:39:54.800
We just have to work out all the values.
00:39:54.800 --> 00:40:03.000
Let me write it again, K = μ × 2 π C μ tilde².
00:40:03.000 --> 00:40:07.700
Let us go ahead and calculate μ.
00:40:07.700 --> 00:40:20.100
Μ we said it is going to be M1 M2/ M1 + M2.
00:40:20.100 --> 00:40:31.600
Hydrogen is 1 atomic mass unit or dealing with HCL 35, the chlorine is 35.
00:40:31.600 --> 00:40:53.100
The atomic mass units/ 1 + 35 atomic mass units × 1.661 × 10⁻²⁷ kg/ atomic mass unit.
00:40:53.100 --> 00:41:04.600
And we end up with μ = 1.615 × 10⁻²⁷ kg.
00:41:04.600 --> 00:41:35.200
Therefore, K is equal to 1.615 × 10⁻²⁷ kg × 2 π × 3.0 C, which is 3.0 × 10⁸.
00:41:35.200 --> 00:41:48.100
This is going to be m/ S × 2886 inverse cm.
00:41:48.100 --> 00:41:50.900
It is going to be ×, I’m dealing with meters and cm.
00:41:50.900 --> 00:42:03.800
I have to go 100 cm/ 1m and all of that is going to be².
00:42:03.800 --> 00:42:09.000
This whole thing, that is going to be².
00:42:09.000 --> 00:42:21.400
When I actually do this, I get K is equal to 478 kg/ s².
00:42:21.400 --> 00:42:28.800
A Newton is a kg m/ s².
00:42:28.800 --> 00:42:37.600
Therefore, a N/ m is equal to kg m/s² m.
00:42:37.600 --> 00:42:43.600
The m cancels leaving me kg/ s² which is exactly the unit that we got.
00:42:43.600 --> 00:42:46.100
We got N/m.
00:42:46.100 --> 00:43:01.700
K =478 N/ m that is the force constant of the bond between Hydrogen and Chlorine 35.
00:43:01.700 --> 00:43:03.600
Thank you so much for joining us here at www.educator.com.
00:43:03.600 --> 00:43:04.000
We will see you next time, bye.