Right, so go ahead and just do a, because we're talking about a lot of new things perhaps to you, and for some reduced mass, I am going to do a calculation and it usually helps to make it a bit clearer. So let me, Move on to a, right, so we have this example here. So you have the CO group, as you all know that's part of a peptide bond. So what you can regard that as isolated from the rest of the molecule. And we move into this a bit later. But some CO stretching, vibration, and most molecules, including peptides, is quite separate. There's no other motion going on. So it's not a bad approximation. And here you've been given that the force constant in a carbonated group is 908 Newtons per meters, and what you've been asked then in the question, you want to calculate vibration frequency of a, Carbon-12, Oxygen-16. So that's the natural isotopes, Carbon-12 is 99% of the carbon. You also then want to compare that with the frequency you calculate for carbon-13. And this is a very simple example this is a common technique in spectroscopy, you can change the isotopes of certain groups and sometimes what you're doing in practical in spectroscopy is you're trying to assign a different band. Sometimes it can be difficult to assign as we'll see later on. But if you can make the isotopic substitution like this, what's going to happen is you 're changing the reduced mass for that vibrator and that will change your frequency. So it's an often technique used In biological systems to assign infrared bands to particular groups within a complex module. So let's just go through this. So you have to calculate the vibration frequency. We're going to calculate it in hertz. I should have said hertz there, we're going to calculate it in hertz. So we go from the fundamental equation, which you need to remember, that mu in hertz is equal to one over two pi, square root of K over mu. And you know now that K is the force constant and Mu is the reduced mass. Right so let's do the first one first. A, you have carbon 12, and it's 16 O. So we know pretty much everything here. We know two pi, and then we're given the force constant minus one. So this is just an exercise really in calculating the reduced mass which we talked about. So the reduce mass is m one. It's the product of the masses divided by the sum. That will give you grams per mole and then you need to multiply it by this u value, which is, that's ten minus three times, Avogadro, one over Avogadro's number. So you have the masses here, you just take the most approximate masses, so it's 12.01 multiplied by 16.00, and then you divide it by the sum of these. So it's 12.01 plus 16.00, and that'll give it to you in grams per mol, but we were asked to do this is Hertz. So you need to then multiply that by this u value, this unified atomic mass unit again, which you will get in tables. And that is 1.661 x 10 to the -27. And that could give it to you in kilograms, which is the SI unit of weight. So you put that into your calculator, and at least I come out with 1.14 by 10th to the minus 26 kilograms. Okay? So, I've calculated my mu, so I've got everything I want from the above equation into that. So you just plug it in, so now you say mu is equal to one over two pi square root of first constant, which is given in the question as 908 Mutants. These are -1. And it's divided by this reduced mass mu, which is 1.14 by 10 to the -26 kilograms. Right, so it we want the answer, the answer should be in seconds minus one. So you can actually break this down. You should be able to do this. So Newton is force. And if you break it down into fundamental constituents, that's going to be kilograms, meter, seconds to the minus two. Force is equal to mass by acceleration. Mass is in kilograms, acceleration is meters per second squared. So, that's kilograms, meter, seconds to minus two, meters to minus one. So, what you'll find is you'll have, let's just write this little change to kilograms. Seconds to minus two. All right? So we'll have kilograms to minus two above the line. We'll have kilograms below the line. The kilograms will cancel. We're left with seconds minus two. We get the square root of seconds minus two is seconds to minus one, so that's where you get your second minus one. So as I say when you're doing calculations you always need to be able to quote the units. A calculation without units Is meaningless. And you should really be able to do this type of manipulation with the units. Okay. So we've worked out, we put that in. And again, as you plug that into your calculator, what I get, I get 4.49 by 10 to the 13 seconds minus 1. So that's your bond that peptide is vibrating. What that means it's vibrating 4.49 by 10 to so 13 times per second. So it's pretty quick. Let's quickly go through the second bet, the second bet is just pretty much the same. The second bet was just where you replaced carbon 12 with carbon 13. So again, you had the oxygen 16 isotope. So, again, all you had to do was work out the mu for that. And the mu, of course, is going to be different, because now you've got the carbon 13. So, it will be, we'll just go through it quickly, it's 30.0 by 16.0 mass of the oxygen atom, and then it's the sum of these at the bottom, and then you multiply by it by this atomic mass unit which is 1.661 by 10 to the -27. And that'll give it to you in kilograms. So work that out, plug it in. You should get 1.19 x 10 to the -26 kilograms. So that gives you reduced mass. You plug that then into your equation, so you get Nu = 1/2 pi, square root of 908 Newtons meters minus one, divided by 1.19 by 10 to the minus 26. Kilograms this occasion, and if you work that out, you'll get 4.39 x 10 to the 13 per second mass. Okay, so what you can see is, it's smaller isn't it, than the frequency for that is smaller than it is for the 12. Because you've increased, that's due to the fact that you've increased the reducement.
