 coming back to most of spectroscopy. Now we said that over here that this source and sample, although they have the similar transition happening 3 by 2 to half or vice versa, that energy is not actually matching. And the loose explanation I was giving that because of the electronic environment that is different. Now the question is why the electronic environment is affecting the nuclear scale. So that is we are going to cover in the final section of today's class. So electron nucleus interaction. So generally when you look into a nuclei, say this is the nucleus, it is actually covered by an electronic cloud. And this electronic cloud it is nothing but we can say it is going to create an electrical field. So you can imagine there is a positively charged system that is nucleus is sitting inside an electric field. So that is why there is going to have some electron nucleus interaction which we can define as a monopole. What do we mean monopole? With respect to the nucleus, if I see in the perspective of the nucleus, the nucleus is sitting as a positively charged and we put it in an electrical field. So there is no other pole over there. So it is actually taken as a monopole. And this monopole is actually situated inside an electrical field and there I can measure the columbic force, the columbic force that monopole is facing. Now before I go there, there are certain things I need to understand. First thing we need to understand that this interaction will happen once the electron and the nucleus share a common place. That means they have to situate it at the same place. Now what do I know? That electron cloud is surrounding the nucleus but the electron cannot generally go inside the nucleus. But unless the electron goes inside the nucleus, there shouldn't be any common space but they are actually sharing the space. Otherwise there will be like a totally different field. There will be an interface between them. But what I want to understand is if they can share a common space. So for that I have to understand is there even a finite possibility that electron can penetrate the nucleus? Is it possible for the electron to go inside the nucleus? So any suggestion on that? Is it even possible or what do you think? Is it possible the electron can penetrate the nucleus? Yes sir, for the radioactive nuclei it happens sometimes because electron capture is not taken place from the K cell. So I think that is possible. So yes, so that is true. Yes, anyone else want to add anything? So what you are talking about is a little bit different. When you talk about the electron capture, the electron at the end is actually losing its identity. It is going inside the nucleus and become a part of the nucleus. But what I am trying to understand is the electron can still hold its identity just stay inside the nucleus for a bit and they can come out unarmed. That is possible. And the answer over there is actually yes, it is possible. So now, if I look into the difference kind of electron and if I define it like what is the different orbitals they stay in, these are the four different orbitals we generally actually exposed to, A, B, D, L. And among them, only S orbital can actually go through the nucleus. The rest of them cannot go through it. Now the question is how do I know it? So that you know from a particular parameter which is known as radial distribution function. I am quite sure most of you have gone through that in your quantum mechanical class. So what is a radial distribution function or idea? Now when we have an electron, we define the electron in quantum mechanics as a wave function, as a wave form, psi. And then if I take the square of that psi square, that actually defines the probability of finding that electron in that particular orbital and which is not says over a particular space. Now how if I can draw that in a graph, we can draw the orbitals fine. But if I want to draw that with respect to a graph, over there I am drawing psi square. That means what is the probability of finding the electron? And in this X axis I am drawing R. R is nothing but distance from the nucleus. Now when I look into what we found for one S orbital, I am going to draw that for each and one orbital. For one S orbital it looks like this. What does it mean? That over there I have the maxima, after a while it is almost non-zero existence. So there is everything fine. But one important thing we found over there. Here we found at R equal to 0, psi square value is actually not equal to 0. Why? I am not going into the details. It depends on how I define the psi function. Look into the psi function of one S orbital. And over there if you put R equal to 0 we will find your psi square value is not going to be 0. It is a non-zero value. What that means? That means there is a finite possibility. Even for some time that electron can stay inside the nucleus. R equal to 0 means you are inside the nucleus. That is what is actually happening. Now if I draw the 2S orbital, I am drawing the radial distribution function for the 2S and the 2P set of orbitals. Again I am drawing the psi square and the internal nuclear distance from the electron distance from the nucleus. And if I draw the 2S orbital, the 2S orbital looks like this. So over there there is a particular point where psi square becomes 0 at a finite R value. This is something known as radial node. So in 1S orbital you do not see such kind of R equal to psi square equal to 0 value over here. But over here you see that. That is called known. And how the radial node looks like? See if I want to draw 1S orbital it looks like a sphere. And if you cut down the sphere in half you will find it is still a total sphere. But if you look into the 2S orbital from outside it looks like a sphere. If you cut it half you will find this polynomial. At one particular place there is 0 electron density. And outside there is 1. So there is a hollow region at particular R value. There is no electron density present. So that is called as a node and a radial node. And this is present in the 2S orbital. So that is why when you are drawing the RDF you found this is 0 electron. What about the 2P orbital? 2P orbital looks like this. So this is 2S, this is 2P. And now over here you can see that 2P value is saying that it has a psi square equal to 0 value and R equal to 0. Because look into the wave function of psi equal to 2P. And then if you look into the wave function of 2S you find psi square equal to non-zero equal to R equal to 0. Again look into the wave function of psi equal to 2S and compare that with 2P. You will find that if you put R equal to 0 psi square not equal to be 0 for 2S. Similarly that means 2S has finite probability to be inside the nucleus. But 2P is not. Now very quickly I will go to 3S, 3P and 3D. So 3S looks like this. So now it has two nodes. You probably have learned earlier that how many nodes you expect. It is N minus L minus 1, N equal to primary quantum number, A equal to azimuthal quantum number, all those things. So there will be two nodes. So that means if I take a 3S orbital which looks like a sphere. If I cut it in half you will find two regions where there is no electron density present. That is how it is going to look like. Now if you look into 2P that is how it is going to look like. So 3P is going to have one node. And then you probably can have 3D that is how 3D is going to look like. So that is how it is going to look like. And over here again you are going to see for 3S at R equal to 0. It is non-zero. For 3P and 3D value they are equal to 0. So they cannot be present inside the nuclei but 3S can. And that is why we suggested that yes it is possible to have some electron density present inside the nucleus. If you have S orbital but not P or D or A orbital. Now with respect to that we come to this thing that okay so we are talking about a nucleus. We have some electron density outside and we are trying to find out how E is the effect of this electron density on the nuclei. And over there the electron density we are concerned about is only the S electron density. Now because as we discussed earlier it is a monopole. And we are trying to find what is the effect of this electron field on the nucleus. And if I want to draw the Hamiltonian because that is how we define the surrounding. How this electrical field is affecting the nuclei. We are going to find an equation. And before finding the equation I am taking two assumptions. First assumption is this nucleus is ready. We are talking about is spherical. And the radius of this nucleus is R. These are the two assumptions I am taking. I am taking it as a sphere not any other particular tradition. Which is possible but at this moment we are taking only spherical. If that is the case the Hamiltonian will be given by this equation. 2 by 5 Z is square psi 0 square R square. Now I will go slowly what each of these terms believe. So over there the Z is square term is actually a mixture of Z E. Which actually define what is the nuclear charge. Z is the number of protons present in the system. So it will surely define the monopole energy with respect to what is the nucleus I am talking about. And then the electron you have the electron charge density. That we also have to think about and that is given as minus E. That is the charge of an electron into psi 0 square. What is psi 0 square? Psi 0 square is that electron density inside the nucleus. As I just said so S electron density possible to spend some time inside the nucleus. How much electron density is present inside the nucleus? That is given by psi 0 square. Psi 0 is the function inside the nucleus. Psi 0 square is the probability. And if I multiply that with the minus E sign that will be the electron charge density. In the Hamiltonian previously it should have a negative sign because I am talking about a positive charge nucleus with an electron. So there should be a negative charge in the beginning but that negative sign is cancelled by this electron charge density. So that is why it is not there. And R is the radius of the system. So that is how the equation comes. So now I can have an Hamiltonian how the system is behaving with some electron density present inside the nucleus. Now imagine this is not going to be the same for the sample and the solution. And that is why this Hamiltonian is going to change. That means it is very similar to some perturbation. And that is going to affect my energy states. Now how they are going to be happening? What will be the parameters we have to look into? That we will cover next week on Friday. So we are going to stop it over here. Any questions up to this point please go ahead.