 Welcome back to the sessions on solving numerical problems. In the previous session or previous lecture, we discussed this problem that is for a three level system in which different non degenerate energy states are equally spaced by E which out of the following expressions is correct for fractional population of excited state. We talked in terms of three different states, your ground state, first excited state, second excited state and then based upon our discussion we showed that the logarithm of fractional population is given by this expression. Now based upon the same question that is for the system in the previous question that previous question was these three energy states 0, E, 2, E equally spaced. Calculate the population of each state when t approaches 0 Kelvin and when t approaches infinity. So, that means now we need to write expressions for populations. So, populations of ith state is equal to exponential minus beta E I over partition function here is 1 plus exponential minus beta E plus exponential minus 2 beta E. This is how the fractional population can be calculated. So, let us write down P I is equal to exponential minus E I upon k t divided by 1 plus exponential minus E upon k t plus exponential minus 2 E upon k t. Note that the numerator has E I, the denominator which is partition function is written based upon this set of energy levels energy states expression will be because the ground state energy is 0. So, exponential 0 is 1. So, 1 plus exponential minus beta E plus exponential minus 2 beta E and we have to now use the given conditions. Let us say when t approaches 0, when t approaches 0, your P 0 is going to be for 0th energy is 0. So, exponential 0 right exponential 0 is equal to 1 divided by 1 plus when t approaches 0 it is exponential minus infinity which is 1 by exponential infinity. So, this term is going to be 0. So, 0 plus 0 that means P 0 is equal to 1. So, you see when temperature approaches 0 all the molecules are going to be in the ground state. Now, let us calculate P 1, P 1 is going to be when t approaches 0 exponential minus infinity it is 0. So, P 1 is equal to 0 and similarly P 2 is equal to 0 ok. So, when t is equal to 0, P 0 is 1, P 1 is 0, P 2 is equal to 0. This is the answer to the first part. Now, second is when t approaches infinity. Now, let us say when t approaches a value of infinity. When t approaches infinity then this becomes exponential 0 which is 1. When t approaches infinity that means exponential minus 0 or exponential 0 which is equal to 1. So, what do we have then? We have P 1 is equal to 1 over 1 plus 1 plus 1 because each term is contributing here 1 which is equal to 1 by 3 ok. I wrote for P 1 or you can also say this is P 0 for P 0 this is 1. For P 1 again the term upper 1 and lower 1 are all going to be 1 over 1 plus 1 plus 1 because each term is contributing 1 which is 1 by 3. And similarly when you calculate P 2 you also will get 1 by 3 same arguments. So, what do we conclude from this? When temperature approaches infinity the ground state fractional population is 1 by 3, first excited state fractional population is 1 by 3, second excited state fractional population is 1 by 3, 1 by 3, 1 by 3. There are 3 states ground first excited, second excited and each one is having equal fractional population. So, therefore you know let us revisit one of the conclusions that we have drawn in one of the earlier lectures that when the temperature approach is 0 all the molecules are most likely to be found in ground state. When the temperature approaches infinity all the available energy states are going to be equally populated. It is not that all the molecules from the ground state will be pushed to the upper excited states and the ground state will have 0 occupancy. No, you can see even the calculations here what do they suggest? When the temperature approaches infinity each ground state or first excited state or second excited state has equal fractional population. Let me introduce one more aspect here which is basically similar to this question. For example, now let us take a case not connected with this question. Let us say I have ground state, I have first excited state, I have second excited state which is doubly degenerate and this value is 2 E. In this case you should yourself try to find out when the temperature approaches infinity what is going to be fractional population of the ground state first excited state and second excited state. Remember that in this case your Q is going to be recall your formula g j exponential minus beta E j. So, that means Q is going to be 1 plus exponential minus beta E plus 2 exponential minus 2 beta E. Pay attention to how we have written 1 comes from the ground state exponential 0 degeneracy is 1. First excited state degeneracy is 1 exponential minus beta E. Second excited state degeneracy is 2 2 times exponential minus beta E and how do you get fractional population n i upon n that is fractional population is equal to exponential minus beta E i upon Q. Now you see in this case Q when the temperature approaches infinity then all these terms you know exponential minus E upon k t temperature approaches infinity this is going to contribute 1. This factor is going to contribute 1 that means the denominator is going to be 1 plus 1 plus 2 that is 4. In that case the fractional occupancy here it is going to be 1 by 4 here 1 by 4 here and 2 by 4 here. 1 by 4 1 by 4 and 1 by 4 into 2 because there are 2 states which are corresponding to the same energy. So, therefore, it is it makes sense that we can introduce something over here which represents the degeneracy of a particular energy level. We can introduce degeneracy term even in the fractional population, but that we will discuss in the next numerical problem. Now let us look at a numerical problem which is based upon translational partition function. The question is calculate percent increase or decrease in the value of translational partition function of a molecule of 50 picometer thermal wavelength when it is allowed to move on a surface of a square of 5 centimeter side from allowing movement only in one dimension of 5 centimeter length. That means we are considering here 2 scenarios. One scenario is that the molecule is allowed to move only in one dimension that is a length of 5 centimeter. Second scenario is molecule is allowed to move only on the surface, surface of a square and that side is also 5 centimeter. So, that means we need to now think about the partition function for a particle for a molecule which is free to move in one dimension and partition function for a molecule which is free to move in two dimension and of course, at this time we will be concentrating on translational contribution. Let us write down the expressions q for square let me write square by sq is equal to area upon lambda square and q for 1 d is going to be length upon lambda. Remember that we have derived expression for partition function for a particle for a molecule which is free to move in one dimension and it was x upon lambda. Then we derived expression for three dimension it was v upon lambda q and at that time we said that you can similarly derive expressions for 2 d and that is going to be area a upon lambda square. So, we need to now find out how much is the difference when the molecule is allowed to you know move from 1 d movement to 2 d movement and 2 d here is square. So, what is the change? Let us try to find out I am interested in q square minus q 1 d divided by q 1 d and I am interested in knowing the percentage therefore, I will multiply by 100. What is q square? q square is area divided by lambda square 1 d is l upon lambda and 1 d is l upon lambda into 100 this is equal to what? This is equal to area minus l times lambda divided by lambda square into l into lambda into 100 which is equal to area is nothing but length square minus length times lambda divided by lambda square into length into lambda into 100. So, common factor of l and l cancel this lambda and this lambda cancel. So, therefore, what I have is this expression l minus lambda divided by lambda into 100. Now, let us substitute the numbers. So, this is equal to what now? This is equal to l length is 5 centimeter. So, I will put 5 into 10 raise to the power minus 2 meter minus lambda lambda is 50 picometer 50 into 10 raise to the power minus 12 meter divided by lambda 50 into 10 raise to the power minus 12 meter into 100. I can ignore this 50 into 10 raise to the power minus 12 meter this is much less compared to 5 into 10 raise to the power minus 2 meter. So, if I ignore this and then multiply by 100 my answer is going to be this 100 and 10 raise to the power 2 minus 2 cancel. So, 5 by 50 is 0.1 this is 5 into 10 raise to the power minus 11. So, that means, the answer is going to be 10 raise to the power 11 this much you see the increase in the value of translational partition function when a molecule is just allowed to have a transition from movement in one dimension to movement in two dimension. The points to be noted in this problem is how to write the partition function translational partition function for a particle for a molecule which is free to move on the surface of a square which is a is area over lambda square. And the partition function for a molecule which is free to move in one dimension is simply l upon lambda and then simply the next step is simple mathematics and what we find out that the percentage increase when the molecule is allowed to move from one-dimensional movement to two-dimensional movement that see the number percent increase is used 10 raise to the power 11 alright. Now, let us discuss one more problem. The next question is calculate the translational partition function at 300 Kelvin and 400 Kelvin of a molecule of molar mass 120 gram per mole in a container of volume 2 centimeter cube. In fact, this is a revisit of a similar problem that I have earlier done, but I am deliberately bringing one more problem over here. So, that everything become clear that the various parameters their units are to be taken care of and for example, the mass mass has to be that of one particle. This is the expression q is equal to 2 pi m k t over h square raise to the power 3 by 2 into v which is basically v upon lambda cube where lambda is equal to beta h square of 2 pi m I am just you know reading for the sake of your connection with the previous lectures. The point to be noted over here m we are given 120 gram per mole. So, therefore, to convert 120 gram per mole into the mass of one particle you need to divide by Avogadro constant which is 6.023 into 10 raise to the power 23 and to convert gram into kilogram another factor of 10 raise to the power minus 3 is required. This is Boltzmann constant and I am keeping t as general and Planck's constant is 6.626 into 10 raise to the power minus 34. So, when you calculate this q turns out to be 4.94 into 10 raise to the power 23 multiplied by temperature in Kelvin raise to the power 3 by 2. And now we can look at we were asked to calculate at 300 Kelvin and 400 Kelvin and you can substitute t here as 300 Kelvin and 400 Kelvin and you can get the result. Let us take a look at the result at 300 Kelvin substituting this t equal to 300 over here we see the value is 2.57 into 10 raise to the power 27 when it is 400 Kelvin the value increases to 3.95 into 10 raise to the power 27. So, what kind of general conclusions can be drawn over here one conclusion is that look at the order 10 raise to the power 27 both at 300 Kelvin and 400 Kelvin 10 raise to the power 27 quantum states are very high. So, therefore, first comment large number of quantum states are accessible 10 raise to the power 27 and when you compare at 300 versus 400 there is an increase in the number also that means more quantum states are accessible at higher temperature. Let us now have another problem discussed the question is what are the relative populations of the states of a two level system when the temperature is infinite. In fact, I have already discussed these kind of problems by taking a two level system and by taking a three level system, but why I am deliberately putting this question again on this slide is that remember that in the previous lecture I said that if there is a particular level which has a degeneracy that means two or three states have the same energy then that level is degenerate g fold degenerate. So, therefore, you remember that I said that you can include a degeneracy into n i upon n let us rewrite that n i upon n is equal to exponential minus beta e i upon q and what I said that if there are more than one state corresponding to one particular energy that will form a level and that degeneracy factor can be included over here. So, therefore, if I put g i over here and then I write n i upon n j I end up with this kind of result and obviously when temperature approaches infinity 1 over k t approaches 0 n i upon n j is equal to 1 both states are equally populated. So, this again leads to the same conclusion that when the temperature approaches infinity all the available states are equally populated the only difference that I brought here in discussion is the inclusion of the degeneracy of the two states. A similar question is being discussed in the next problem that is a certain molecule has a non-degenerate excited state lying at 540 centimeter inverse above the non-degenerate grounds at what temperature will 10 percent of the molecules be in the upper excited state if these are the only states we need to consider. You are considering you are comparing here two states n i and n j and the question is at what temperature will 10 percent of the molecules be in the upper excited state that means we need to act upon n i upon n j and we have just discussed that n i upon n j is equal to g i upon g j into exponential and then we need to just consider the differences in the energies of ith state and j state. And if we consider i that is the ground state which will have 90 percent and upper excited state 10 percent then the corresponding energies when they are put in and you solve this for temperature you get a value of 354 Kelvin that means at 354 Kelvin 10 percent of the molecules will be in the upper excited state where the difference between these two states is 540 centimeter inverse. So, in this session of solving the numerical problems we have used the expressions for molecular partition function we have used the expression for translational partition function and we have also used the expressions for the fractional population of ith state. With this knowledge of partition function and translational contribution to partition function now we will discuss about the internal energy in the next lecture. Thank you very much.