0:14

We talk about infrared spectroscopy we're gonna talk about

Â the frequency, and we're also going to talk about the wavenumber.

Â And again, I remind you of these important equations,

Â E = h nu, that's the famous Planck Equation.

Â Or E = hc over lambda.

Â C is the speed of light, and lambda, if I write it properly,

Â is this Greek letter, lambda, which we use for wavelength.

Â Or E = Page C, nu bar 1 over lambda

Â lastly is also called new bar.

Â Nu bar is called wavenumber.

Â So the vibrations frequency for

Â that diatomic oscillator had these

Â = half k and r minus re squared.

Â Again using mathematics that we're not gonna go into,

Â you can work out [COUGH] what is the frequency for that vibration.

Â And the frequency is given by this equation here,

Â which you'll need to remember.

Â It's 1 over 2 pi, [COUGH] the square root of k, which we've met.

Â K is the force constant we talked about.

Â 1:40

Divided by mu, and mu is what's know as the reduced mass.

Â And the reduced mass, worth keeping in SI units, and

Â we want the reduced mass In kilogram.

Â 1:57

Now, as we see here, m1, if we go back to our instant, m1 is just a mass of atom 1.

Â We're talking about a diatomic here at the beginning, and m2 is the mass of atom 2.

Â So to reduce mass it's the product of the masses divided by the sum of the masses.

Â And for quantitive purposes, you need to quote that in kilograms.

Â And we'll be doing an exercise on that in a few minutes,

Â which will hopefully make it clear for you.

Â So the vibrational frequency for the diatomic oscillator is given by this.

Â And you know all the terms here.

Â K's the force constant, and mu is the reduced mass, and that's given in hertz.

Â And again hertz, is the same as seconds to -1.

Â 2:48

Right, so hertz is seconds -1 for nu.

Â The force constant and the SI units we want to keep

Â it in is newtons per meter and then we want to quote

Â the reduced mass in kilograms per molecule for mu.

Â 3:14

So we'll again do a little exercise on this in a minute,

Â how do we work out the reduced mass.

Â The reduced mass you see the periodic table,

Â it's usually quoted in grams per mole.

Â 3:30

So you have say hydrogen is 1, carbon is 12,

Â oxygen is 16, the atomic masses.

Â And to work out the reduced mass you simply

Â put the atomic masses in gram for mol, like this here.

Â And then you multiplied them by a conversion factor,

Â which is called a unified mass constant, u.

Â And this is always given usually in the back of text books or

Â you guys remember it.

Â Or if you're asked in the exam,

Â you've been given this in the data sheet that would accompany the exam.

Â 4:11

But it's not that, sometimes even if you forget it, you're quoting here,

Â you want to convert.

Â So you had these units here.

Â And you wanted to quote them in kilograms as I say here they're

Â expressed in grams per mole.

Â 4:41

Avogadro's constant, so you divide by that.

Â So that would give you grams, wouldn't it?

Â Grams per molecule.

Â You divided these atomic mass weights here by,

Â that will give you grams per atom or grams per molecule.

Â And then you want to convert to kilograms, so

Â you'd just divide by 1,000 wouldn't you.

Â 5:13

Right, okay.

Â So another thing, what you we go on to practical applications of spectroscopy.

Â The frequency or the energy is almost always given in wavenumber,

Â which is one over the wavelength but

Â it's almost always quoted in si units it should be meters -1.

Â But traditionally it is given to centimeters to the -1.

Â Because somebody asked me that in the workshop and

Â it's generally, usually if the number is easy to deal with.

Â Sometimes it gives nice whole numbers in this region and

Â therefore people like to quote it in that centimeters to -1.

Â So that's how it has stayed, if you like.

Â But you then need to be able to convert, you see, from that.

Â So again, reminding you here, that nu bar, the wavenumber,

Â is 1 over the wave length C = lambda times nu.

Â So, you do the manipulation you can also get nu bar = nu over c.

Â So, as I said the SI unit should be meters -1, but centimeters -1 is more common.

Â So, you need to remember this.

Â That nu bar, given in meters -1 = the nu bar and

Â centimeters -1 by a 100.

Â So people will always tend to get that wrong.

Â And one way I think of it is nu bar is wavenumber,

Â so it's number of waves in a meter.

Â So say you had a number of waves as we have here, say in a centimeter.

Â 6:56

Then the number of waves in a meter is gonna be greater.

Â Usually, people divide here.

Â If you think about converting to meters,

Â the number of waves in a meter is gonna be 100 times that.

Â Sorry about that.

Â So that's going to be, so nu bar in meters -1 = nu bar in centimeters -1 by 100.

Â And then we have given the.

Â 7:24

We get the vibration frequency before, and frequency,

Â and I think it was nu = 1 over 2 pi the square root of k over mu.

Â That's in, if you remember, that was in seconds -1.

Â 7:43

So if we quoted nu bar, nu bar, which is the wavenumber,

Â you have to divide by c here, to convert from nu bar to nu.

Â So nu to nu bar.

Â So you had 1 over 2 pi c, the square root of k over nu.

Â And that's in the meters -1.

Â You divide it by the speed of light c.

Â And now we want to go to centimeters -1.

Â So we need to divide by 100 to convert it to centimeters -1.

Â So these are all kinds of manipulations that you should be able to do.

Â 8:29

All right.

Â Sorry, I've written over here a bit.

Â So the frequency, After

Â vibration is directly proportional to the force constant.

Â We can see that from this equation here.

Â Or the wavenumber, it's on the top line so it's going to be as the force constant

Â increases the frequency of the vibration increases and so also does the wavenumber.

Â 8:59

And what you also see is that it's inversely proportional to reduce mass.

Â So the reduced mass is below the line.

Â So as you increase that you will reduce the frequency,

Â or the wavenumber of that vibration.

Â So I think what we have here is, Some examples of that.

Â So here you have hydrogen and here you have the force constant.

Â So this is 500, get rid of that.

Â Okay.

Â 9:47

So H2, the force constant in Newtons meters to the -1 is 509.

Â And then you have the reduced mass here is this 8.31 times 10 to the -28.

Â So it's got a wavenumber given in centimeters -1 of 4159.

Â And then you go to HCl, it's got the slightly lower force constant but

Â you see that the reduced mass is quite high so

Â it's higher -27 as opposed to 28.

Â So both of these factors the lower force constants will reduce the wavenumber.

Â And also the higher reduced mass will reduce the wavenumber as well.

Â So you can see the value of the wavenumber of that bond is 2890.

Â And then you go on to Cl2, okay the force constants again,

Â down a bit so you'd expect a lower value for the wavenumber.

Â And the reduced mass is quite a bit bigger, it's 10 to the -26.

Â So now the wavenumber has gone down as well.

Â And then lastly, you have nitrogen, we talked about nitrogen already.

Â Nitrogen of course has a triple, has a triple bond, so

Â the force constant is quite high.

Â And therefore that will govern the frequency, at least compared with Cl2.

Â Cl2 and nitrogen have fairly similar reduced masses but

Â you can see the force constant of the nitrogen is quite high.

Â So the frequency of the vibration you observe depends on the force constant and

Â inversely proportional to the reduced mass and

Â directly proportional to the force constant.

Â