1 00:00:17,880 --> 00:00:23,900 welcome back to the video for solving problems in elementary spectroscopy this continues 2 00:00:23,900 --> 00:00:31,519 from what i just did for four problems in the video tutorial two and the we shall start 3 00:00:31,519 --> 00:00:38,040 with next problem namely problem five n which we are asked to do the following determine 4 00:00:38,040 --> 00:00:45,830 the fraction of molecules ah in the v is equal to one two and three state relative to the 5 00:00:45,830 --> 00:00:53,500 ground state v is equal to zero for the molecule bromine monoflouride the i think the last 6 00:00:53,500 --> 00:01:03,030 some type of graphical things here we frequency of the ah transition is six six nine point 7 00:01:03,030 --> 00:01:15,530 seven centimeter inverse ok so what the problem say is at t is equal to thousand kelvin at 8 00:01:15,530 --> 00:01:23,600 t is equal to thousand kelvin and when nu is given by six six nine point seven centimeter 9 00:01:23,600 --> 00:01:32,479 inverse ah for the molecule bromine monoflouride ok you are asked to calculate the number of 10 00:01:32,479 --> 00:01:37,560 molecules in the v is equal to one relative to the number of molecules in the v is equal 11 00:01:37,560 --> 00:01:43,249 to zero state ok this is very elementary problem in the sense it looks at the boltzmann population 12 00:01:43,249 --> 00:01:49,520 at equilibrium and since vibrational states for this single degree of freedom or non degenerate 13 00:01:49,520 --> 00:02:06,600 vibrational states are non degenerate the ratio is given by the simple exponential 14 00:02:06,600 --> 00:02:16,030 factor namely exponential minus e v is equal to one by kb t divided by exponential minus 15 00:02:16,030 --> 00:02:24,130 e v is equal to zero by kb t and you know the difference between v is equal to one and 16 00:02:24,130 --> 00:02:36,730 v is equal to zero the energy the answer is h nu ok and nu is given as sixty nine therefore 17 00:02:36,730 --> 00:02:50,000 hc nu bar therefore the delta e e to the minus delta e by kb t for this ration n is equal 18 00:02:50,000 --> 00:02:57,720 to v is equal to one to n at v is equal to zero this delta e is given by hc nu bar ok 19 00:02:57,720 --> 00:03:02,980 and the nu bar is given as six sixty nine point ah seven centimeter inverse therefore 20 00:03:02,980 --> 00:03:09,820 what you have to do is to ensure that we use the right units namely its given as six sixty 21 00:03:09,820 --> 00:03:17,700 nine point seven into ten to the two meter inverse times six point six two six into ten 22 00:03:17,700 --> 00:03:25,340 raise to minus thirty four this is h this is nu bar times three into ten to the eight 23 00:03:25,340 --> 00:03:35,380 meters per second and this is joules and the meter inverse and so what you see is the delta 24 00:03:35,380 --> 00:03:53,100 e is given as ah ah joule second ok joule second times that therefore the answer is 25 00:03:53,100 --> 00:04:01,730 given in terms of joules so this can be calculated and kb t is ah one point three eight times 26 00:04:01,730 --> 00:04:10,550 ten to the minus twenty three joules per kelvin and its calculated at thousand kelvin so that 27 00:04:10,550 --> 00:04:17,310 is given as joules ok therefore delta e by kb t is a numerical exercise 28 00:04:17,310 --> 00:04:23,780 ok for you have to do that and take the exponential of minus that in order to calculate the ratios 29 00:04:23,780 --> 00:04:29,569 of n between the two different levels and the only thing that you have to do for two 30 00:04:29,569 --> 00:04:37,560 and three is that for two the energy difference between e zero and e two is two h nu and for 31 00:04:37,560 --> 00:04:42,750 ah three it is three h nu therefore you will see that there are relative populations in 32 00:04:42,750 --> 00:04:50,530 the first second and third excited state drastically come down to very very small values thats 33 00:04:50,530 --> 00:04:56,210 the purpose of this exercise to show that in vibrational ah spectroscopy the number 34 00:04:56,210 --> 00:05:01,870 of molecules in the higher energy levels is much less than the number of molecules in 35 00:05:01,870 --> 00:05:09,220 the ground state and this number s less if the frequency is even more ok that was the 36 00:05:09,220 --> 00:05:16,290 idea for this problem ok therefore you can calculate that ok then next problem is formal 37 00:05:16,290 --> 00:05:21,840 ah process is basically asking you to write all the simple harmonic oscillator wave functions 38 00:05:21,840 --> 00:05:28,500 for n equal to zero to n equal to three is a text book exercise and most important for 39 00:05:28,500 --> 00:05:34,720 this is determine why the selection rule for vibrational spectrum of a harmonic oscillator 40 00:05:34,720 --> 00:05:40,690 contains only one line with the selection rule delta v is equal to plus minus one thats 41 00:05:40,690 --> 00:05:59,669 important ok problem seven and you are asked to write the harmonic oscillator functions 42 00:05:59,669 --> 00:06:07,680 so text book exercise please remember that the psi n is a normalization constant which 43 00:06:07,680 --> 00:06:16,990 is a function of n and then it is e to the minus alpha x square by two times hn of root 44 00:06:16,990 --> 00:06:21,900 alpha x ok therefore what is alpha alpha is equal to 45 00:06:21,900 --> 00:06:30,240 square root of k mu by h bar square so k is the force constant mu is the reduced mass 46 00:06:30,240 --> 00:06:34,771 and those are the only two parameters of this problem therefore this is the vibrational 47 00:06:34,771 --> 00:06:49,510 wave function and h zero is one h one of root alpha x is two root alpha x h two of root 48 00:06:49,510 --> 00:07:00,510 alpha x is four alpha x square minus two and the third h three root alpha x is eight alpha 49 00:07:00,510 --> 00:07:09,800 root alpha x cube minus twelve root alpha x please remember this makes the whole expression 50 00:07:09,800 --> 00:07:15,940 dimensionless because the alpha has a dimension of one by length and therefore you see that 51 00:07:15,940 --> 00:07:22,389 the harmonic oscillator x is the length parameter its distance from the equilibrium ah for vibration 52 00:07:22,389 --> 00:07:28,040 and therefore this all quantity polynomials are dimensionless so these are the corresponding 53 00:07:28,040 --> 00:07:33,310 forms and each one of them you have to multiply by to the minus alpha x square by two and 54 00:07:33,310 --> 00:07:37,810 also put down the general formula for n n these are available in text books 55 00:07:37,810 --> 00:07:48,850 but the second part of the problem is important namely why the selection rule delta nu is 56 00:07:48,850 --> 00:08:00,229 equal to plus minus one ok ok this is a mathematical statement that the selection rule for transition 57 00:08:00,229 --> 00:08:14,580 between e n with the wave functions psi n and e lets write n double prime n prime psi 58 00:08:14,580 --> 00:08:22,199 n prime n double prime the selection rule for vibrational transition is due to the dipole 59 00:08:22,199 --> 00:08:35,940 moment matrix element which connects these two states in fact the transition probability 60 00:08:35,940 --> 00:08:41,490 this is the statement you have to take it from me transition probability for ah transition 61 00:08:41,490 --> 00:08:52,960 between n prime and n double prime n prime and n double prime is given by the absolute 62 00:08:52,960 --> 00:08:59,980 square of the integral from minus infinity to plus infinity psi n double prime psi n 63 00:08:59,980 --> 00:09:09,820 prime star mu which let me not write ah yeah mu is he dipole moment here dipole moment 64 00:09:09,820 --> 00:09:21,260 ok not the reduced mass mu dipole moment psi n double prime dx 65 00:09:21,260 --> 00:09:36,580 therefore mu is essentially mu naught times ah x and this x keeps changing therefore during 66 00:09:36,580 --> 00:09:42,450 vibration the dipole is created and changed destroyed and so are increased and decreased 67 00:09:42,450 --> 00:09:49,010 therefore this x is the operator the position operator and the mu naught is a ah is the 68 00:09:49,010 --> 00:09:57,800 constant per for the permanent dipole moment position operator ok therefore the dipole 69 00:09:57,800 --> 00:10:04,690 keeps changing from mu naught to mu naught times x during the process of vibration and 70 00:10:04,690 --> 00:10:12,910 the integral that you needed to calculate is psi n prime of x from minus infinity to 71 00:10:12,910 --> 00:10:27,620 plus infinity star mu naught times x and then you have psi n double prime x dx ok this is 72 00:10:27,620 --> 00:10:33,360 the this is what is call the transition probability amplitude and the absolute square of a of 73 00:10:33,360 --> 00:10:40,830 this is the transition probability ah per unit time whatever units that you wanted to 74 00:10:40,830 --> 00:10:46,450 use but the absolute square is the one which is connected with the actual process of transition 75 00:10:46,450 --> 00:10:51,990 between vibrational levels and therefore it is important for this bit what is called the 76 00:10:51,990 --> 00:11:07,870 matrix elements of the dipole moment operator 77 00:11:07,870 --> 00:11:20,920 between the two states x n prime and n prime ok n double prime and n prime so the calculation 78 00:11:20,920 --> 00:11:26,460 essentially boils down therefore to writing down in the case of vibrational spectroscopy 79 00:11:26,460 --> 00:11:32,420 the psi n are given by the exponential minus alpha x square 80 00:11:32,420 --> 00:11:39,470 so let us write that minus infinity to plus infinity the normalization constant associated 81 00:11:39,470 --> 00:11:48,630 with psi n prime which is en n prime and to the function is e to the minus alpha x square 82 00:11:48,630 --> 00:11:59,170 by two times h n prime root alpha x this is psi including this his is psi n prime ok and 83 00:11:59,170 --> 00:12:07,610 then you have x times the mu not and then you have e to the minus alpha x square by 84 00:12:07,610 --> 00:12:17,950 two h n double prime root alpha x dx so it turns out to be from minus infinity to plus 85 00:12:17,950 --> 00:12:23,560 infinity proportional to let me drop the constants out there is also n double prime for the wave 86 00:12:23,560 --> 00:12:29,910 function associated with this state ok thats the normalization concept for that state so 87 00:12:29,910 --> 00:12:38,490 this is the integral that needs to be calculated its called the transition moment integral 88 00:12:38,490 --> 00:12:47,390 transition moment integral now without any further explanation i will 89 00:12:47,390 --> 00:12:56,440 just make the statement that this integral will be non zero only if when you write that 90 00:12:56,440 --> 00:13:02,300 like that the alpha x square by two the alpha x square by two add up and then this integral 91 00:13:02,300 --> 00:13:19,830 will be non zero only if n prime and n double prime differ by plus or minus one because 92 00:13:19,830 --> 00:13:30,339 it involves x if it involves x square then the dipole moment is now being considered 93 00:13:30,339 --> 00:13:36,190 as the square of the amplitude for vibration that i mean those kind of things do happen 94 00:13:36,190 --> 00:13:41,320 when you talk about the unharmonic states of the molecule and so on the dipole moment 95 00:13:41,320 --> 00:13:46,851 in the lowest level of approximation is simply proportional to the length and therefore what 96 00:13:46,851 --> 00:13:51,180 you see is the length x itself and this is harmonic oscillator model and therefore the 97 00:13:51,180 --> 00:13:56,920 harmonic oscillator model this integral is non zero only if n double prime is n prime 98 00:13:56,920 --> 00:14:01,490 plus or minus one because of the nature of the hermite polynomial please understand the 99 00:14:01,490 --> 00:14:09,579 hermite polynomials are odd and even and so what you see here as that the integral becomes 100 00:14:09,579 --> 00:14:15,980 even only when n prime and n double prime and x these three products give you an even 101 00:14:15,980 --> 00:14:20,870 function but that argument should be very carefully stated because this does not mean 102 00:14:20,870 --> 00:14:27,490 n and n prime can differ by three now doesnt work this integral is still zero it cannot 103 00:14:27,490 --> 00:14:37,149 differ by other than one and therefore its a property of the wave functions which results 104 00:14:37,149 --> 00:14:46,300 with the transition moments being connected only in the vibrational energy levels if i 105 00:14:46,300 --> 00:14:52,740 have to write the potential energy and then simply write the vibrational energies as something 106 00:14:52,740 --> 00:15:02,480 like that n equal to zero n equal to one n equal to two and so on n equal to three the 107 00:15:02,480 --> 00:15:09,040 only transition that possible that can be measured experimentally are this because of 108 00:15:09,040 --> 00:15:12,980 the nature of the wave functions therefore the transition moment integral needed to be 109 00:15:12,980 --> 00:15:21,829 introduced in order for you to understand this problem ok this will not happen k will 110 00:15:21,829 --> 00:15:28,140 not happen so vibrationally there is only one line for a harmonic oscillator corresponding 111 00:15:28,140 --> 00:15:37,810 to this frequency ok they are all the same therefore there is only one line nu bar ok 112 00:15:37,810 --> 00:15:42,579 thats the important point next problem ok i miss problem six 113 00:15:42,579 --> 00:16:00,579 so let us go back to problem six ok problem six is on a morse oscillator model ok please 114 00:16:00,579 --> 00:16:19,700 remember the morse oscillator model the energies are given as h omega e h bar omega e v plus 115 00:16:19,700 --> 00:16:29,660 half minus ok here i am using omega e in terms of wave numbers therefore let me write to 116 00:16:29,660 --> 00:16:44,930 this as h omega e and then minus h omega e xe v plus half whole square this is the morse 117 00:16:44,930 --> 00:16:57,860 this is the expression for the morse oscillator energies ok ah if you are writing this in 118 00:16:57,860 --> 00:17:04,459 terms of centimeter inverse lets do one more thing lets get rid of this and lets write 119 00:17:04,459 --> 00:17:22,390 this as sum nu e as omega e xe minus omega e times e plus half minus omega e xe times 120 00:17:22,390 --> 00:17:33,090 v plus half whole square therefore all are in wave number units ok 121 00:17:33,090 --> 00:17:43,240 now the problem that is given to you is br f let me now read the problem the problem 122 00:17:43,240 --> 00:17:50,110 given to you is that the vibrational frequency omega e and unharmonocity constant omega ex 123 00:17:50,110 --> 00:17:59,059 e have been given from have informed from experiments to be about centimeter inverse 124 00:17:59,059 --> 00:18:04,860 to be six six nine point seven and three point eight six nine centimeter inverse and you 125 00:18:04,860 --> 00:18:10,360 are asked to calculate the first four morse oscillator energy levels for those model molecule 126 00:18:10,360 --> 00:18:16,560 and also the transition frequencies for the transitions v is equal to zero to two and 127 00:18:16,560 --> 00:18:24,640 v is equal to one to three ok so lets look at that so the nu corresponding 128 00:18:24,640 --> 00:18:35,330 to the vibrational quantum number v is omega e times v plus half minus omega e xe this 129 00:18:35,330 --> 00:18:46,270 is the unharmonocity v plus half whole square so when v is equal to zero nu zero is half 130 00:18:46,270 --> 00:18:59,540 omega e minus one by four omega e xe nu one is three by two omega e minus this is three 131 00:18:59,540 --> 00:19:07,170 by two square therefore nine by four omega e xe and like wise you are asked to do that 132 00:19:07,170 --> 00:19:19,299 for four levels isnt it so nu two is five by two omega e minus twenty five by four omega 133 00:19:19,299 --> 00:19:29,299 e xe and the last one omega three is seven by two omega e minus forty nine by four omega 134 00:19:29,299 --> 00:19:34,910 e xe because the second term is the square v plus half square so everything is a square 135 00:19:34,910 --> 00:19:46,170 here so this is these are the expressions for the four energy levels of the morse oscillator 136 00:19:46,170 --> 00:19:59,340 ok now you are given omega e and omega e xe so this is given as six six nine so if you 137 00:19:59,340 --> 00:20:08,210 write nu zero it is six six nine point seven by two centimeter inverse and the other is 138 00:20:08,210 --> 00:20:17,190 one by four minus one by four times three point eight six nine centimeter inverse so 139 00:20:17,190 --> 00:20:22,490 you know that therefore in a similar way you can calculate nu one nu two nu three 140 00:20:22,490 --> 00:20:32,070 now what is important is the next step which is v is equal to zero to v is equal to two 141 00:20:32,070 --> 00:20:42,179 so what is delta nu e zero to two if you look at that its five by two omega e minus one 142 00:20:42,179 --> 00:20:55,410 by two omega e thats the first term five by two omega first term minus one by two this 143 00:20:55,410 --> 00:21:03,090 is between two and zero and the next one is minus twenty five by four omega e xe minus 144 00:21:03,090 --> 00:21:12,799 twenty five by four omega e xe minus one by four omega e xe and so what you have is two 145 00:21:12,799 --> 00:21:25,799 omega e minus six omega e xe so this is the second frequency and likewise delta nu e from 146 00:21:25,799 --> 00:21:35,860 one to three is seven by two omega e minus one to three is one is three by two omega 147 00:21:35,860 --> 00:21:48,220 e minus seven by two corresponds to forty nine by four omega e xe minus omega one corresponds 148 00:21:48,220 --> 00:21:55,830 to one by no ah nine by fourth three by two whole square therefore nine by four omega 149 00:21:55,830 --> 00:22:07,970 e xe so what you have is two this uis also two omega e xe minus forty by four so it is 150 00:22:07,970 --> 00:22:18,730 ten ten omega e xe what do we gain from this process of calculating thats important is 151 00:22:18,730 --> 00:22:23,929 not just the problem what is important from this problem is that if you look at it he 152 00:22:23,929 --> 00:22:31,970 morse oscillator model which has a potential energy r v of r if you remember has something 153 00:22:31,970 --> 00:22:40,950 of this kind we saw in another problem if you look at it the lowest energy level is 154 00:22:40,950 --> 00:22:51,850 half omega e minus one by four omega e xe the next energy level if you look at it s 155 00:22:51,850 --> 00:23:03,790 three by two omega e minus nine by four omega e xe the difference between the two is two 156 00:23:03,790 --> 00:23:07,830 omega e xe the unharmonocity term is two omega e xe 157 00:23:07,830 --> 00:23:18,470 but if you go to the next level the levels are closing in thats a purpose of doing this 158 00:23:18,470 --> 00:23:29,960 exercise the levels are closing in that is successive energy differences are smaller 159 00:23:29,960 --> 00:23:38,340 because you remember this omega e xe term is negative to any omega e term because you 160 00:23:38,340 --> 00:23:44,930 see any near by energy level say v is equal to zero v is equal to one if you look at the 161 00:23:44,930 --> 00:23:52,320 transition between two nearby levels you see this is zero to one is omega e minus two omega 162 00:23:52,320 --> 00:24:00,940 e xe but from one to two if you look at that v is equal to two it is again omega e but 163 00:24:00,940 --> 00:24:09,169 one corresponds to nine by four and two corresponds to ah twenty five by four thats is five by 164 00:24:09,169 --> 00:24:17,340 two whole square therefore the difference between the two is four omega e xe so the 165 00:24:17,340 --> 00:24:26,860 second energy level difference v is equal to one to two is smaller than the first energy 166 00:24:26,860 --> 00:24:33,039 level difference therefore the morse oscillator energy levels actually they converge and so 167 00:24:33,039 --> 00:24:37,810 what you see is that there are many lines that you can see in the vibration spectrum 168 00:24:37,810 --> 00:24:42,750 because not every one of them is not equal to any every other thing they are different 169 00:24:42,750 --> 00:24:47,100 and also in morse oscillator this possible for you to have multiple transitions v is 170 00:24:47,100 --> 00:24:51,720 equal to zero to two v is equal to one to three and so on therefore you see morse oscillator 171 00:24:51,720 --> 00:24:58,850 gives you a real model of what is called the vibrational spectrum a spectrum not one line 172 00:24:58,850 --> 00:25:07,940 thats the purpose for doing this exercise lets go to problem eight now problem eight 173 00:25:07,940 --> 00:25:14,500 talks about molecule boron trifluoride and it asks you to calculate the three principle 174 00:25:14,500 --> 00:25:20,940 moments of inertia and identify the time and it also calculates ask us to calculate all 175 00:25:20,940 --> 00:25:26,990 the energy levels of this molecule for the rotational quantum number j up to and including 176 00:25:26,990 --> 00:25:42,000 three ok first of all this is a symmetric top bf three so is actually a an equilateral 177 00:25:42,000 --> 00:25:57,360 triangle that you can draw b the three florents and these are the bond distances between the 178 00:25:57,360 --> 00:26:04,820 three fluorines they are identical ok this is what you have 179 00:26:04,820 --> 00:26:16,409 now you are asked to calculate all the three moments of inertia is a very simple exercise 180 00:26:16,409 --> 00:26:24,690 so what are these axes so we can choose one axis let me use a different color let me choose 181 00:26:24,690 --> 00:26:31,230 one axis as the axis that passes through one of the bonds ok the center of mass is right 182 00:26:31,230 --> 00:26:36,860 way the boron is because of the symmetry of the system and therefore the second axis which 183 00:26:36,860 --> 00:26:42,789 is perpendicular to that is that axis again passing through the boron and the third axis 184 00:26:42,789 --> 00:26:50,690 is perpendicular to the plane of this ah screen ok thats perfect thats a third axis so you 185 00:26:50,690 --> 00:27:03,000 have what is call the ix iy and iz axis ok now is easy to see that with respect to these 186 00:27:03,000 --> 00:27:11,740 two axis this axis as well as this axis the moment of inertia that you calculate are identical 187 00:27:11,740 --> 00:27:18,790 lets do that ok the bf bond distance is given please remember these angles are one twenty 188 00:27:18,790 --> 00:27:26,340 degrees therefore you can see that for if we calculated for this axis we call it as 189 00:27:26,340 --> 00:27:33,570 the yeah we will call it as the x y and z will be up the other so lets call this as 190 00:27:33,570 --> 00:27:41,419 a x axis one fluorine is on the axis and therefore it doesnt contribute to the moment of inertia 191 00:27:41,419 --> 00:27:51,669 mu i mu one r one square call it as one thats zero r is zero the other two fluorine atoms 192 00:27:51,669 --> 00:28:03,330 if you think about that this is sixty degrees this is sixty degrees and this is the fluorine 193 00:28:03,330 --> 00:28:14,230 boron bond distance so if you write this as r b f this distance as r b f then this perpendicular 194 00:28:14,230 --> 00:28:27,789 distance of the fluorine atom from the bond axis is r b f times sine sixty ok so you get 195 00:28:27,789 --> 00:28:41,490 the ri the r one r two this is two this is three so m two r two square plus m three r 196 00:28:41,490 --> 00:28:48,140 three square we want to calculate this for the moment of inertia i along the x axis because 197 00:28:48,140 --> 00:28:57,820 m one r one square is zero the mass of the fluorine atom being m of f we have r two which 198 00:28:57,820 --> 00:29:11,049 is the perpendicular distances given as the ah r b f sin sixty degree whole square and 199 00:29:11,049 --> 00:29:19,390 into two because this atom is also exactly at the same distance from the axis therefore 200 00:29:19,390 --> 00:29:27,720 this is what you have sin sixty is root three by two therefore you have mf three by fourth 201 00:29:27,720 --> 00:29:37,340 three by two so it is three by two mf r bf square ok thats the first one 202 00:29:37,340 --> 00:29:46,919 now what about this axis lets look at this axis if you look at this axis all the three 203 00:29:46,919 --> 00:29:54,289 atoms are away from this axis the distance of let me now remove some of these things 204 00:29:54,289 --> 00:30:04,970 make it easier the distance is this the distance of this 205 00:30:04,970 --> 00:30:11,220 atom is this they are both identical and the its easy to see that this distance is and 206 00:30:11,220 --> 00:30:27,929 also this one ok so let me redraw the boron trifluoride this is ah the bond this is the 207 00:30:27,929 --> 00:30:33,220 bond so the axis that we have is this so we wanted to calculate this distance and we wanted 208 00:30:33,220 --> 00:30:44,779 to calculate this distance they are both the same ok and then you have this is simply rbf 209 00:30:44,779 --> 00:30:54,390 whole square it contributes m f one this is f one this is f two this is f three this is 210 00:30:54,390 --> 00:31:01,690 boron ok therefore this you can see immediately as this is again ah thirty degrees so this 211 00:31:01,690 --> 00:31:24,340 distance will turn out to be r ah bf two times sin thirty whole square times two of them 212 00:31:24,340 --> 00:31:33,340 because f two and f three so that would be the answer so this gives us as rbf square 213 00:31:33,340 --> 00:31:40,520 m f one plus all the rbf are the same therefore this is sin thirty is one by two so its one 214 00:31:40,520 --> 00:31:55,210 by four times two it gives you rbf square times one by two mf so this is the total moment 215 00:31:55,210 --> 00:32:07,549 of inertia i and this is also three by two rbf square mf so this is identical to the 216 00:32:07,549 --> 00:32:11,399 moment of inertia about this axis thats also three by two 217 00:32:11,399 --> 00:32:22,340 now what about the perpendicular axis you talk about the perpendicular axis it is 218 00:32:22,340 --> 00:32:29,340 something which is like that therefore you see all the three atoms are away from this 219 00:32:29,340 --> 00:32:37,010 axis bu exactly their bond distances therefore the ah perpendicular distance from this axis 220 00:32:37,010 --> 00:32:47,710 is actually the bf bond distance so you have three times mf times the rbf square so this 221 00:32:47,710 --> 00:33:00,029 is what you get this is for iz now is interesting that ix plus iy is equal to iz ok because 222 00:33:00,029 --> 00:33:04,940 you have three by two here you have three by two for the previous case and this is three 223 00:33:04,940 --> 00:33:10,091 this is a planarity in fact this is a planarity condition for the moment of inertia the personal 224 00:33:10,091 --> 00:33:15,520 reason for including this as an example therefore you see you have to sit down and try to calculate 225 00:33:15,520 --> 00:33:20,790 the moments of inertia from simple geometries and you must know he actual structure the 226 00:33:20,790 --> 00:33:25,210 equilibrium structure of the molecule the bond is the bond angles and then go back and 227 00:33:25,210 --> 00:33:30,520 calculate the rotational constants using that and see in experiments whether thats the rotational 228 00:33:30,520 --> 00:33:35,240 constants you get and so on that is a reason for this particular problem 229 00:33:35,240 --> 00:33:42,130 next the next problem talks about rotational spacing between successive lines of hc l as 230 00:33:42,130 --> 00:33:46,550 twenty point eight centimeter inverse and its asking you to calculate the bond distance 231 00:33:46,550 --> 00:33:53,960 ah its also asking you to determine the number of molecules in the level j is equal to four 232 00:33:53,960 --> 00:34:00,140 relative to j is equal to three and then one more thing namely determine the level in which 233 00:34:00,140 --> 00:34:06,240 there is maximum population that is t is equal to three hundred k may be assumed ok its important 234 00:34:06,240 --> 00:34:17,710 to remember that in the case of rotations the line intensities actually increase as 235 00:34:17,710 --> 00:34:26,229 you go from j is equal to zero one two three etcetera and if you talk about the intensity 236 00:34:26,229 --> 00:34:33,429 here the rotational line intensity is increases something like that and there is a maximum 237 00:34:33,429 --> 00:34:42,950 and this we remember is due to the fact that nj divided by nj lets write it as prime for 238 00:34:42,950 --> 00:34:48,490 the upper level double prime for the lower level this is given as the nu j prime the 239 00:34:48,490 --> 00:34:55,039 degeneracy of j prime and the degeneracy of nu j double prime therefore the ratio times 240 00:34:55,039 --> 00:35:08,849 exponential minus delta e from j prime to J double prime divided by kb t so the difference 241 00:35:08,849 --> 00:35:15,140 between vibrational population and the rotational population in this case is that here is a 242 00:35:15,140 --> 00:35:21,869 degeneracy factor associated with nu ah nu j prime and j double prime and in the absence 243 00:35:21,869 --> 00:35:28,059 of any external field the j prime level has a degeneracy j prime level has a degeneracy 244 00:35:28,059 --> 00:35:33,559 of two j prime plus one and like wise the j double prime level has a degeneracy of two 245 00:35:33,559 --> 00:35:48,099 j double prime plus one therefore this problem ask you to calculate the energy level n j 246 00:35:48,099 --> 00:35:55,779 is equal to four divided by n j is equal to three this is a second subdivision the first 247 00:35:55,779 --> 00:36:04,430 division is the bond distance itself for r h cl this is easy please remember the successive 248 00:36:04,430 --> 00:36:14,059 rotational lines when you write the intensities the successive rotational lines are all differ 249 00:36:14,059 --> 00:36:24,130 by two b ok this is intensity this is the wave number they all differ by two b and therefore 250 00:36:24,130 --> 00:36:31,069 two b is equal to twenty point twenty point eight centimeter inverse and you know b is 251 00:36:31,069 --> 00:36:39,210 given by the formula h by eight pi square i c and therefore h by eight pi square ic 252 00:36:39,210 --> 00:36:47,380 is ten point four centimeter inverse and since i is equal to mu r square and in the case 253 00:36:47,380 --> 00:36:56,849 of hcl mu of hcl is one point zero zero eight into thirty five point four five six divided 254 00:36:56,849 --> 00:37:06,670 by thirty six point four six four times one point six six six one into ten raise to minus 255 00:37:06,670 --> 00:37:14,160 twenty seven kilograms ok therefore mu is known r square is what you 256 00:37:14,160 --> 00:37:25,279 are asked to calculate so you write h by eight pi square mu r square c is equal to ten point 257 00:37:25,279 --> 00:37:32,769 four into ten raise to two meter inverse make sure that you use the same si units because 258 00:37:32,769 --> 00:37:41,089 we are using kilogram and r is in meters therefore you see immediately r is given as square root 259 00:37:41,089 --> 00:37:54,670 of h by eight pi square mu c times ten point four into ten raise to two ok this is and 260 00:37:54,670 --> 00:38:00,039 h of course is known mu is known and c is known therefore you can calculate r therefore 261 00:38:00,039 --> 00:38:03,519 this is a very straight forward substitution formula ok 262 00:38:03,519 --> 00:38:11,369 now the second is of course n j is equal to four divided by nj is equal to three please 263 00:38:11,369 --> 00:38:19,809 remember e j is equal to four if you recall for the ah microwave spectrum of rigid diatomic 264 00:38:19,809 --> 00:38:28,920 molecules it is hc b j into j plus one ok therefore for j is equal to four it is twenty 265 00:38:28,920 --> 00:38:39,010 hc b in terms of joules ok thats why the h and c are here for e j is equal to three it 266 00:38:39,010 --> 00:38:48,309 will be four three into four so it will be twelve hcb therefore the delta e between j 267 00:38:48,309 --> 00:39:01,119 is equal to four to j is equal to three is eight hc b and you are given two b as twenty 268 00:39:01,119 --> 00:39:09,339 point eight centimeter inverse therefore eight the delta e is given as eight into hc into 269 00:39:09,339 --> 00:39:18,789 this is two b therefore i would say four times four hc times twenty point eight into ten 270 00:39:18,789 --> 00:39:28,959 raise to two meter inverse so delta e is known and nj is equal to three to four to three 271 00:39:28,959 --> 00:39:37,119 yes nj is equal to four divided by n j is equal to three is nine to j plus one j is 272 00:39:37,119 --> 00:39:44,609 equal to four two j plus one seven and then you have e to the minus delta ej is equal 273 00:39:44,609 --> 00:39:54,329 to four to j is equal to three divided by kb t where kb is one point three eight into 274 00:39:54,329 --> 00:40:02,979 ten raise to minus twenty three minus twenty three joule per kelvin and t is three hundred 275 00:40:02,979 --> 00:40:08,749 kelvin therefore you can calculate this number by doing the numerical work ok 276 00:40:08,749 --> 00:40:18,440 now the last section of this problem is to ask you what is the maximum what is the value 277 00:40:18,440 --> 00:40:33,199 of j for which the intensity is maximum the intensity is maximum thats what is you are 278 00:40:33,199 --> 00:40:41,709 asked to do and 9i think you must remember this formula that the nu ah the j max this 279 00:40:41,709 --> 00:40:50,359 was discussed in the in the lectures j max is approximately the nearest integer to kb 280 00:40:50,359 --> 00:41:03,900 t by two hc b minus one half you have already been given b you know t you know hc therefore 281 00:41:03,900 --> 00:41:08,180 whatever is the integer nearest to this number this will not be an integer because the square 282 00:41:08,180 --> 00:41:14,640 root factor will not give you ah round integer value or half integer value for you to substitute 283 00:41:14,640 --> 00:41:30,390 therefore take the nearest integer to this number as the j for which the population is 284 00:41:30,390 --> 00:41:47,029 for which the intensity is maximum ok thats that so we will come to the last problem this 285 00:41:47,029 --> 00:41:54,289 is purely a classification of the molecules as spherical symmetric or asymmetric tops 286 00:41:54,289 --> 00:42:02,059 which is important for ah understanding the molecular spectroscopy and therefore ah we 287 00:42:02,059 --> 00:42:08,400 need to make sure that the geometry of the molecules are visualized and that you see 288 00:42:08,400 --> 00:42:14,249 that the moments of inertia about the three different axis whether they are the same or 289 00:42:14,249 --> 00:42:17,940 whether two of them are the same and different from the third or if all the three of them 290 00:42:17,940 --> 00:42:22,699 are different you have to have a mental picture and its very clear from some of these 291 00:42:22,699 --> 00:42:31,710 for example the tetrahedron now cbr four you should be immediately obvious that this is 292 00:42:31,710 --> 00:42:37,660 the spherical top the reason is tetrahedron is the half the symmetry of the cube so you 293 00:42:37,660 --> 00:42:45,089 know hat if you place this molecule in the cube ok and then you start worrying about 294 00:42:45,089 --> 00:42:49,400 what those moments of inertia are you will immediately see that the moments of inertia 295 00:42:49,400 --> 00:42:54,859 are identical about any three axis because this is one br this is another br and the 296 00:42:54,859 --> 00:42:59,339 third br is going to be here and the fourth br is going to be here and the carbon is right 297 00:42:59,339 --> 00:43:06,160 on the middle c therefore you see that this is a tetrahedral molecule inside the cube 298 00:43:06,160 --> 00:43:11,059 if you want to do that it has half the symmetry of the cubes therefore you see that with respect 299 00:43:11,059 --> 00:43:16,430 to the structure whether the axis as this or whether the axis is this or whether the 300 00:43:16,430 --> 00:43:21,749 axis is perpendicular to this phase the moments of inertia is same because the molecules are 301 00:43:21,749 --> 00:43:28,390 ideally displaced from these axis by the same distance therefore something should be immediately 302 00:43:28,390 --> 00:43:34,950 seen as based on the geometrical conservations that it is a spherical top in this case now 303 00:43:34,950 --> 00:43:43,059 ch cl three again you should see that this is actually a symmetric top the reason is 304 00:43:43,059 --> 00:43:52,509 the three chlorines actually form the base of an equilateral triangle and then the carbon 305 00:43:52,509 --> 00:43:57,920 is the hydrogen bond is above tetrahedral molecule if you put the three molecules as 306 00:43:57,920 --> 00:44:03,869 the base of a ah of an equilateral triangle then it does not matter whether the axis is 307 00:44:03,869 --> 00:44:09,749 this whether the axis is that all the three atoms are away from hose two axis by he same 308 00:44:09,749 --> 00:44:14,200 distance and so is the hydrogen because its in the plane therefore two of the moment of 309 00:44:14,200 --> 00:44:19,190 inertia right away equal and the third moment of inertia which is the moment of inertia 310 00:44:19,190 --> 00:44:26,579 along the ch bond the the symmetry axis of the molecule is different ok so this is the 311 00:44:26,579 --> 00:44:33,729 symmetric top and s of six octahedron when i talked about 312 00:44:33,729 --> 00:44:39,250 cube and symmetry of the tetrahedron is half of this symmetric of the cube octahedron has 313 00:44:39,250 --> 00:44:43,799 the same geometry of as that of the cube because those are the six mid points of the faces 314 00:44:43,799 --> 00:44:49,119 therefore it does not matter whether you draw the axis about any two of these atoms the 315 00:44:49,119 --> 00:44:55,329 moments of inertia for all these three axis will be identical so s of six will be a spherical 316 00:44:55,329 --> 00:45:04,650 top ok ch two cl two we notice that it is an asymmetric top because it has only the 317 00:45:04,650 --> 00:45:10,160 symmetry of a two fold axis molecules which have only two fold symmetry as a rule you 318 00:45:10,160 --> 00:45:16,880 can also try and remember will have to be asymmetric top molecules need a minimum of 319 00:45:16,880 --> 00:45:22,440 a three fold axis in order for them to be asymmetric top o zone has only a two fold 320 00:45:22,440 --> 00:45:28,309 axis symmetry the bond ax the bisecting the bond of the oxygen oxygen oxygen and therefore 321 00:45:28,309 --> 00:45:39,910 that you can see right away that o zone is these two are asymmetric tops hcn no problem 322 00:45:39,910 --> 00:45:45,729 its a linear molecule therefore two of its moment of inertia are the same about the axis 323 00:45:45,729 --> 00:45:54,079 both axis are perpendicular to the hcn bond its a linear molecule therefore there is he 324 00:45:54,079 --> 00:46:00,249 carbon hydrogen so i would say that the ah center mass is here and the other axis is 325 00:46:00,249 --> 00:46:07,670 that but about this axis the moment of inertia is zero therefore its a linear molecule and 326 00:46:07,670 --> 00:46:15,789 last c two h four also has only two fold axis symmetry so despite the fact that it looks 327 00:46:15,789 --> 00:46:29,769 to be very symmetric this is an asymmetric top because the axis here and the axis here 328 00:46:29,769 --> 00:46:35,390 the atoms are at different distances compared to those and the third axis is of course perpendicular 329 00:46:35,390 --> 00:46:38,799 to the plane of this molecule therefore its an asymmetric top 330 00:46:38,799 --> 00:46:44,729 so the purpose of this video tutorial is to tell you a little bit about how to look at 331 00:46:44,729 --> 00:46:49,930 problems from the point of view of not just solving them but in terms of relevance of 332 00:46:49,930 --> 00:46:56,180 these problems in understanding the concepts and so on and therefore i suggest that you 333 00:46:56,180 --> 00:47:01,309 solve many such problems by yourself and if you have questions please feel free to write 334 00:47:01,309 --> 00:47:04,770 that in the forum i wish you all the best thank you