 Hello, welcome to lecture number 37 of the course quantum mechanics and molecular spectroscopy. As I told you in the previous lecture that you know we are going to look at some solved problems. So today I am going to talk about the problems related to rotational spectroscopy okay. So when you record a rotational spectrum, when you record a rotational spectrum you get you know numbers, B you know things like that. How do we interpret the spectra? And what you have to do to get the spectroscopy correct? One of the things that you do in the rotational spectroscopy, record the rotational spectra of isotope embers, what do I mean okay. Now if I have molecule HCl okay, now there are various isotopes that I can think of. So if I have HCl molecule then there are I can have one hydrogen 35 HCl or I could have one hydrogen 37 Cl. So 37 and 32 are isotopes of the chloramide or one could have its hydrogen and 35 Cl. So by the way this is also called D okay, deuterium is nothing but the isotope of hydrogen with two molecular atomic mass or you could have two hydrogen 37 Cl okay. One could look at various isotopes and you see when you look at the rotational constant B or let us call it as B0 because at the equilibrium distance this is given by H by 8 pi square where I is the rotational constant, I is the moment of inertia sorry moment of and B0 is the rotational constants. Now how do you get P0 can be measured by experiment. So when you record the rotational spectrum we know that the separation between the lines so this is 2B, 4B, 6B, this is 0 to 1, 1 to 2, J values, J is equal to 1, J is equal to 1, J is equal to 2, 2, 3, 3, 2, 4, etc. And the separation between them is 2B okay this is pure rotational spectrum. Now when you have rotational constant B0 so that you can 2B or 2B0 can be directly measured from the experiment okay. So this is experiment but generally so this is B0 if you look at this thing it will come out to be in Hertz or second inverse or Hertz the in this case B0 is given the unit of B0 is second inverse or Hertz okay. But generally B0 is not measured in also measured in also measured in centimeter inverse so B0 is also measured in centimeter inverse when I do so you can do a quick conversion. So if I measure B0 in centimeter inverse then my B0 will now be equal to h pi 8 pi square c i where c is speed of light and what is i moment of inertia and i is given by mu r naught square. So where mu is the reduced mass r naught is the equilibrium distance okay so this is the necessary. So in experiment it turns out after doing the experiment in one experiment what they have found that 1 h 35 cl the value of B0 was equal to 10.44 centimeter inverse and similarly value of 1 sorry 2 h cl 35 the value of B0 is equal to 5.39 centimeter inverse. So you see when I substitute the hydrogen vitrioturium there is a drastic change in the value of the B0 okay this is what I call as isotope effect. Now let us use this and try to evaluate what is the length or the equilibrium length of the h cl molecule or d cl molecule. Now we know that B0 in centimeter inverse is given by h by 8 pi square c i which is also given by h by 8 pi square c mu r naught square. So I can slightly rewrite this equation in r naught square is equal to h by 8 pi square c mu B naught. So I am going to use this formula and plug it in and see what I will get. Now h is nothing but 6.626 into 10 power minus 34 joule second you can see this you can verify if the formula is dimensionally correct okay. Now 6.626 into 10 power minus 34 is the value of h that will be in joule second is not it? So it is an action constant joule second 8 pi square c is 2.997 since I am looking at this centimeter inverse I am looking at speed of line in centimeters per second. So that will be 2.10 power 10 centimeters per second into mu okay mu I need the reduced mass. So if I look at 1 h 35 cl then my reduced mass will be nothing but 35 by 36 into 1.66 into 10 power minus 27 kilograms. So that will be nothing but 35 divided by 36 into 1.66 into 10 power minus 27 into value of B naught which is 10.44. So I can do little bit of equations and change I mean do algebra. So this 34 and this 27 will cancel and then you will get 10 power minus 7 and this 10 power 10 and that will this I can cancel and then write 10 power 17 okay. So what I will get is that 6.626 divided by 3991.031 into 10 power minus 17. So that is going to be the value just you can do the math I have already done it. So this will be now be equal to 1.660 into 10 power minus 20 so r naught square. So I take a square root r naught will be now equal to 1.88 into 10 power minus 10 meters. You know 1.10 to 10 power minus 10 meters is nothing but your Armstrong. So this is r naught will be nothing but 1.288. Similarly for if I take the same value r naught square is equal to 6.626 into 10 power minus 34 divided by 8 pi square into 2.997 into 10 power 10. Now instead of I take 2h 35cl so dcl so that will be nothing but into 70 divided by 37 because 2 into 35 is 70 2 plus 35 is 37 into 1.66 into 10 power minus 27 into now the value of the rotational constant is 5.39 centimeter inverse. If I do the math what I will get is r naught will be equal to 1.286 Armstrong. So what I get is the following for 1hcl 35 that is hcl with 35 set of r naught will be equal to 1.288 Armstrong and 2hcl 35 that is dcl r naught is equal to 1.286 ok. So minute changes in the isotopes do affect the lens ok here the changes in third decimal ok. So you should have that kind of calibration that will allow you to differentiate the bond lens at third decimal ok and mind you that the rotational spectroscopy is the only spectroscopic technique by which one can measure bond lens ok. Now this is just an example so let me just tell you quickly ok how we can use this problem to look at something else ok. Now the other problem that one can look at in the rotational spectroscopy is let us suppose a carbon monoxide molecule CO. Now CO can have various isotopes one of the isotopes the standard one is 12 CO 16 ok. The other one will be 13 CO 16 the third one will be 12 CO 18 of course the fourth one will be 13 CO 18 there is also oxygen 17 but will not get into that but these are the four possible isotopes that one can isotopomers. So these are isotopomers of CO. Now I do not have data for this so I will go to ignore ok. So for 12 CO 16 and 13 CO 16 and for 12 CO 18 the B naught in hertz is given by 57898.4 mega hertz and this is 55346.3 mega hertz and this is 55135.3 mega hertz. Now I can always converts mega hertz or hertz into centimeter inverse so if I convert this will cannot to be 1.9318 centimeter inverse and this will be nothing but 1.8467 centimeter minus 1 and this will be nothing but 1.8396 centimeter inverse. Now if you look at if you look at the difference this is less than 0.09 centimeter inverse and this is less than 0.01 centimeter. Now if I have to distinguish between this and this then my resolution of my spectrometer should be less than 0.01 centimeter inverse or at least half of them or this should be 0.09 centimeter inverse. So if I have to distinguish the isotopomers of my carbon monoxide the microwave spectrometer the spectrometer that records this rotational spectra ok should have very high resolution so that I can separate out this isotopomers. By the way oxygen it is in not available but carbon 13 is 1% natural abundance ok. In fact there are spectrometers in which you can record the spectrum of this using natural abundance ok. Now when I use this data and convert ok so I can use this data to convert I know the values of B naught so these are B naught in centimeter so by the way when you got in centimeter it is also called B naught bar ok. So I have this in megahertz and also in centimeter inverse I can use the value B naught is equal to H by 8 pi square I this is nothing but H by 8 pi square mu R square or naught square or B naught bar that is in centimeter inverse is H by 8 pi square C I R H 8 pi 8 pi square C mu R naught square I can use one of them and plug in the values of mu of course when you have these the values of mu will keep changing then it turns out when you calculate all this will get R naught for 12 C 16 O is equal to 1.1285 angstroms and you will get R naught for 13 C O 16 is equal to 1.2735 angstroms and R naught for 12 C 18 will be 1.2 sorry 127, 1.127, 1.1285 I might have done some mistake here you can recheck. So this is how one can use the rotation spectroscopy data to calculate the bond lengths ok and one also has to calculate bond lengths of several isotope amounts to get the statistically correct answer ok we will stop it here thank you very much.