0:08

So today we're going to move on to another type of spectroscopy,

Â called infrared spectroscopy.

Â And if you took chemistry in school,

Â I think you have done some applications of infrared spectroscopy.

Â So what we're going to do today,

Â is we're gonna get more involved in the theory behind the spectroscopy.

Â And then we're gonna to move on to look at how it's

Â useful in in analyzing various types of molecules.

Â 0:39

So, as i say here in the first line of the slide, is that the atoms in a molecule,

Â you don't think of them as fixed, that they're constantly,

Â constantly moving.

Â And this, this should be automated to move but, yeah there we go.

Â As we can see here, we have some simple molecules.

Â And molecules, the atoms in them, are constantly vibrating, if you like,

Â moving, back and forth.

Â 1:24

So if you have a simple molecule, a diatomic simple molecule like,

Â say, H-Cl, then the only motions that the atoms can do in there

Â is they can move back and forth.

Â They can have what they call a stretching of the bond, a lengthening and

Â compressing.

Â So when you get more complex molecules, like you get in biological systems.

Â Then you have other types of modes.

Â And likely two of the molecules here on the right,

Â you can see that the stretching there of all the bonds.

Â And also, on the bottom, you can see bending of the atoms.

Â So, they're stretching, and there's bending going on.

Â And when you measure an infrared spectrum,

Â you're measuring the energy of these movements really.

Â So the vibrational motions, like these.

Â The energy needed to excite them, from their ground state or their excited state,

Â which is what all spectroscopy is about.

Â That energy that you need for that,

Â comes within the infrared region of the electromagnetic spectrum.

Â So, if you remember again,

Â support the electromagnetic spectrum from the first day.

Â You have the high energy regions, the gamma rays and the X rays.

Â Then you come along to what we talked about the last day, the UV and

Â then you move into the visible region.

Â And then lower energy from that, you have the infrared region.

Â So this is the region of the electromagnetic spectrum,

Â that you observe vibrations of molecules.

Â And of course in practical applications as we move to later on,

Â what you're trying to do, you're trying to gain from spectroscopy,

Â you're just trying to get some information about the structure, the geometric

Â structure and sometimes maybe the electronic structure of the molecule.

Â So that's the main points on that slide there.

Â So now we're gonna move into the, the theory behind it, because you need to have

Â an appreciation of that as well as just being able to assign assign spectra.

Â 4:10

So what you have is something like that there.

Â So there you have Simple diatomic system.

Â You have one mass m2, m1.

Â And then, it's joined by a spring and from that spring,

Â you're going to have what we call the equilibrium bond length,

Â 4:40

But if we let it go, it will go back to it's equilibrium value.

Â We can push it together and

Â if we let it go again it will come back to it's equilibrium value.

Â So you imagine bond lengths are moving back and

Â forth between the compression and lengthening,

Â but they have an equilibrium bond length which is at lowest energy.

Â 5:07

So there's a very simple law that relates the energy,

Â the force that you exert on such a system.

Â And this is known as Hooke's Law, and

Â some of you have done Physics might have come across this.

Â But the force, the F(r) here as a function of r,

Â which r is the distance, is given by this here.

Â It's minus k times r minus re and r, so

Â you lengthen that spring then r is the r that you get when you lengthen it.

Â And re is just what we call the equilibrium value, the one,

Â the lost energy bond length.

Â So F(r) = minus k.

Â You need to remember the minus,

Â because what you have to remember is if you're forcing it,

Â If you are pulling the atoms apart, increasing the distance.

Â Therefore the force is actually against it, so

Â the force if you compress it the same way the force is actually against you.

Â So that is why you have this minus sign here.

Â So F(r), is equal to minus k or min re.

Â So r is the actual length as I said,

Â re is the equilibrium length and then we have [COUGH] this k here.

Â So what we're saying here is that F(r) is proportional to r- re, and

Â then you have a proportionality constant and that proportionality constant is k.

Â And we call that the force, the force constant, and

Â if you like k is a kind of a measure of how difficult it is to do that,

Â or the strength of the bond between the two atoms.

Â 6:56

So the force constant, the way you measure it in Newtons at N,

Â Newtons per meter because force is Newtons, and

Â then if you divide the cross by R, or minus R E as in distance.

Â SI units we like to keep it in meters, so it's Newtons divided by meters or

Â Newtons to the meters, -1.

Â So here you have again this analogy between the spring.

Â If you have a large force constan,t then you have a strong spring, or strong bond.

Â If you have a weak force constant, small force constant.

Â Then you have a relatively at least, weak spring.

Â 7:47

Okay. Okay, so let's just talk again,

Â it's best just to talk about the simple system of diatomic to get these

Â theoretical ideas In your heads.

Â So we're talking about diatomic molecules.

Â So we've already said that the strong bonds have large force constants.

Â So how do we know that?

Â So here we just show.

Â For some very simple diatomic molecules, hydrogen,

Â HCl, chlorine, and nitrogen.

Â And here on this column here, we have the force constants and

Â then here we have the bond energy that you

Â measure for that bond.

Â So here we have all of these above.

Â The nitrogen are single bonds.

Â 8:39

So what you can see is that

Â the force constant is directly proportional to the bond energy.

Â The bond energy's the amount of energy you need to break that bond.

Â So, as you can see, the force constant varies with that.

Â And then you can see you have nitrogen here, which has a triple bond, and

Â you should know that if a triple bond is going to be a very strong

Â bond system between the two nitrogen atoms.

Â And you can see that's reflected in the bond energy,

Â the amount of energy required to break the bond, and

Â then that's reflected in the large forced constant for that bond.

Â 10:02

can remember that for any of you that have a mathematics knowledge,

Â you can also know that f of r Is related

Â to the energy by (-dv4)/(dr),

Â or the force is equal to the negative of the change

Â of the energy as a function of distance.

Â That's from basic, basic physics.

Â And then from that, if you integrate across each side, what you'll get is that

Â the, I try to write it in here, so we're taking the integral of each side so

Â we get that the integral of F(r) Is

Â equal to -V(r), okay?

Â That's just some simple algebra.

Â Or we could say, let's make it a bit easier,

Â and we can just bring the minus sign The minus sign over there.

Â 11:11

Now we know that F(r), we've already saw it on the last slide.

Â F(r) is equal to -k(r-re).

Â So we've got to integrate that, and then we've got to change the sign.

Â So for those of you who know a bit of calculus.

Â If you integrate r(re),

Â you increase the power by two, and you divide by the new power.

Â So you get minus a 1/2 k, or n or r- re, V squared.

Â And of course we know there's a negative sign outside it, so

Â therefore v of r is equal to plus a half k into r minus r [INAUDIBLE].

Â So that's just, if you have a bit of mathematics it's nice

Â to be able to do that, but that's where this equation comes from.

Â Otherwise you just have to try and remember that equation.

Â So another thing is that this equation then, if you plotted it out and

Â here it's shown here.

Â If you pull out it gives you a parabola.

Â All right, it's like a half KX squared,

Â it gives you a parabola like we've shown here.

Â And so what it means is here you have the equilibrium value

Â