 We are discussing hydrogen atom. We have talked about a couple of quantum numbers. And now we are going to discuss the quantum number that we are perhaps most familiar with and that is principal quantum number. Remember all the discussion of not just hydrogen atom but also quantum mechanics. One of the beginnings of quantum mechanics is this transitions from one level to the other of hydrogen atom which showed up as line spectra of hydrogen atom and from there different experiments revealed different features of the spectrum and that is what led to the idea of quantum mechanics anyway. So, today we talk about principal quantum number. But before that let us do a quick recap as usual. What we have done so far in hydrogen atom is that we have been able to write Schrodinger equation and separate perform separation of variables and hence generate three different equations. One in R, one in theta, one in phi. And we have solved the phi part and we said that it is exactly the same as our treatment of rigid rotor and from the phi part we got this wave function 1 by root over 2 pi e to the power plus minus i m phi. And we showed that when Z L Z operator operates on phi it gives an eigenvalue of m h cross. So, phi is what phi contains information about the Z component of angular momentum. And from here by using cyclic boundary condition we get the magnetic quantum number. Magnetic quantum number takes values of 0 plus minus 1 plus minus 2 plus minus 3 and so on and so forth. And what we have not solved explicitly but given you the results is the solution of the theta part which yields another quantum number L. The solution of theta part capital theta turns out to be an associated legendary polynomial in cos theta. And what we see is that this beta turns out to be L into L plus 1. And when L square operator operates on the angular part of the wave function we get h cross square multiplied by L multiplied by L plus 1 multiplied by the same wave function. So, it is an eigenvalue equation. This is the value of square of angular momentum. And hence we had worked out that the value of total angular momentum of electron in hydrogen atom is square root of L into L plus 1 multiplied by h cross. Now, let us start talking about the R part. Once again this is a differential equation solution to which was already known by the time people started working on shorting an equation of hydrogen atom. So, there was no need for them to really work it out all of I mean from the beginning. So, I mean in the present day with this COVID-19 outbreak one term that has become very familiar with many of us is repurposing of drugs inventing a new drug is not so easy it takes many many years. So, what people are trying to do is that they are trying to use already existing drugs and explore whether they have any efficacy against this novel coronavirus. So, similarly already existing equations were sort of if I may use the same term repurposed to get the solution. So, with little bit of digression but I mean let us not forget the times we live in live in. So, this equation solution of this was already known and the solution was when put in context of this hydrogen atom problem solution turned out to be R it was a function of two quantum numbers n and L we know about already n is the new quantum number that we are going to talk about that is a constant the constant is factorial n minus L minus 1 divided by 2n into n plus L factorial to the power 3 the whole thing under square root sign that multiplied by 2z divided by na what is a a is 4 pi epsilon 0 h cross square by mu h square if you remember or if you just go back and have a look at what we had said while discussing both theory in very brief this is the expression that we had got for the radius of the atom. So, this is essentially Bohr radius and very soon it will boil down to Bohr radius of hydrogen atom. Now this is a constant we do not really have to remember all this all we need to know is that this is a constant then the next part is important you have R to the power L L. So, R is the radius that will be raised to the power L. So, for if L equal to 0 that will just be 1 L equal to 1 it will be R L equal to 2 it will be R square and so on and so forth. So, something that increases as a function of R that is multiplied by e to the power minus Z R by na 0. So, that is an exponential decay in terms of R. So, you have unless L equal to 0 you have an increasing function of R multiplied by a decreasing function of R but that is not all you also have that multiplied by a once again another kind of polynomial the solutions are all so sort of power series solutions we always end up getting this polynomials and this polynomial that we get here these are called associated Laguier functions. So, associated Laguier functions you do not really need to know their form but just remember that this associated Laguier functions are functions of R multiplied by some constant that constant is 2 Z divided by na 0 but we do not need to remember all that all we need to know is some constant multiplied by R to the power L this I would like you to remember constant multiplied by R to the power L e to the power minus constant into R multiplied by Laguier function in constant into R that is what that is the general form of the R part of the wave function. Now let us have a look at this function by function but before that let us also state something that we have said already that n takes up values of 1, 2, 3, 4 so on and so forth it can go up to infinity it also puts a restriction on the value of L remember beta is the bridge between the R dependent part and the theta dependent part and theta dependent part gives rise to the values of L anyway. So restriction on L and the restriction essentially is L has to be less than n that comes from applying the boundary conditions here one more thing energy when you do the full quantum mechanical treatment turns out to be exactly the same as what was predicted by Bohr theory does that mean that Bohr theory was correct and all this discussion that we are doing is rubbish not really Bohr theory is not correct it is just that it gives the right answer it gives the right answer because the physical assumptions and considerations in many cases are actually valid but we still cannot really use it because it is a 100% classical theory it violates uncertainty principle and also it uses classical and quantum maybe it is not right to say it is 100% classical theory it uses classical and quantum theories reversibly whenever whichever is more convenient that is not such a good thing to do most importantly it violates uncertainty principles we cannot use it this quantum mechanical treatment is the way to go but the beauty is even though Bohr theory is not really correct it gives the correct values of many physical quantities energy being the most prominent of them you might remember that one in one of our earlier modules we had said that the value of Rydberg constant that we can predict from Bohr theory turns out to be fairly accurate when compared with the Rydberg constant determined experimentally why is that so because Rydberg constant essentially comes from difference in energies of energy levels so that 1 by n square is there everywhere so interestingly we get the same value of energy as that of Bohr theory but maybe it is not such a big surprise because you knew that Bohr theory does give the right value of energy anyway but let us not put too much of weightage to that let us move ahead one more thing that we should say is that interestingly in this expression for E the only quantum number that makes a contribution is n and not L remember what we are doing here we are working out Schrodinger equation for a hydrogenic atom hydrogenic atom means a one electron atom so for a one electron atom what we are saying is that L does not contribute to energy however when we have more than one electrons in an atom L is going to have a rule to play so we will talk about it when the time comes for now let us have no confusion about this that different L values for the same n value are associated with exactly the same energy minus 13.6 by n square electron volt for hydrogen atom and hydrogenic atoms L comes in only when you have more than one electron and your phenomena like screening shielding so on and so forth so the only thing that is there is the radial the radial part is the only part that makes contribution to total energy. So now we have reached a stage where we have talked about four of the three quantum numbers that we know n L and m and we have come to a reconciliation with what we had said using old quantum theory that n specifies the total energy of the electron in the atom L talks about orbital angular momentum m talks about z component of the orbital angular momentum what we have not obtained and what we will not obtain at all from the Schrodinger treatment is the spin angular momentum quantum number spin is something that arises out of relativistic quantum mechanics direct treatment. So that is something that has to be brought in externally if you remember and we are going to say this again later on if you have a free electron then it is not associated with any n or L or m n L m these quantum numbers arise due because the electron is in the field of a nucleus however even for a free electron you do have s you do have what is called m s m s has values of plus half and minus half once again they denote the contribution of the spin angular momentum towards the z axis. So z component of spin angular momentum we will talk about that in more detail when we discuss multi electron atoms for now let us move on to the next section with the understanding that we develop a concepts of n L and m interestingly in the opposite sequence remember in both theory n came first then some of it modification brought in L and Zeeman effect necessitated bringing in m in Schrodinger treatment however m comes first followed by L followed by n because that is a sequence of a solution of Schrodinger equation s comes from some place else right but with that understanding let us now continue with our discussion of the R dependent part capital R. So let us have a look at the functions as we had said few minutes ago this here is a general form what we will do is we will write rho for 2 z r by n a because that is something that keeps arising everywhere we will write a for this 4 pi epsilon 0 h cross square by mu square and since it is hydrogen atom when mu equal to m e we are going to write a equal to a 0. Now this is the wave function radial part that we get for n equal to 1 l equal to 0 and just for the record it is the only part remember when l equal to 0 then this l part and phi part are just constant. So that gets a subsumed in the normalization constant and this is what we get. So it is a simple exponential decay in R remember l equal to 0 so r to the power l equal to 0 and this Laguerre function that also is a constant. So it is for n equal to 1 l equal to 0 that is a simple exponential decay in R. Now even though we have not written it in the slides we are going to I think talk about it in the next module let me just say something here n equal to 1 l equal to 0 denotes an orbital I think we know that it denotes a 1 s orbital. So even though I might be jumping the gun a little bit let me pop this question right now what is an orbital and let me give you the answer. An orbital is an acceptable solution for Schrodinger equation for a 1 electron atom some of us might be taken by surprise by this definition because in many textbooks especially in 11-12 level it is written in bold letters that an orbital is a region of space in which the probability of finding the electron is maximum well that is a very popular long definition that region of space where probability of finding the electron is maximum can be worked out using the orbitals but they are not orbitals. An orbital by definition I will repeat is an acceptable solution of Schrodinger equation for a hydrogenic atom which is a 1 electron atom and these are solutions we can get directly like what we are getting right now we are talking about only the R dependent part we will come back to this orbital business in the next module. Let us go ahead n equal to 2 l equal to 0 we get once again a constant multiplied by e to the power minus r by 2 0 no harm done multiplied by now you see 2 minus r by a 0. So this R to the power l what would R to the power l be l is still equal to 0. So R to the power l is 1 so what is this 2 minus r by a 0 where does it come from it comes from the Laguerre function 2 minus r by a 0 into e to the power minus r by 2 a 0. Now that brings us to a an interesting something maybe I just draw it even though we have talked about it a little later. Let us say I want to plot this function what will it look like I can call it I will call it capital R because it is only function of R capital R for 1 0 if I plot it as a function of R it is just some constant multiplied by e to the power minus r by a 0. So this is what it will look like and we have a better looking diagram in the next slide what will this look like here you have the same e to the power minus r this time by 2 a 0 so it will be a faster decay but multiplied by 2 minus r by a 0. So this 2 minus r by a 0 can become 0 somewhere this whole thing where does it become equal to 0 where R is equal to 2 a 0. So this factor becomes 0 at R equal to 2 a 0 and then it does not matter what the value of e to the power minus r by 2 a 0 is the product will be a 0. So if I try to plot R 2 0 against okay I forgot to write the x axis here this is of course R please believe me that I have written R what will happen at R equal to 2 a 0 this will become 0 so it will go down become 0 and we are going to show this many many times so please forgive my poor artistic skills for now. So the point is there is a node in the function remember nodes what is a node a node is where a wave function changes sign through a value of 0. So here the wave function changes sign so this is plus this is minus okay of course plus and minus are related. So that is where a node is and what is the locus of that node R equal to 2 a 0 R equal to 2 a 0 is the equation of a sphere is not it. So in 3 dimensions this sphere is the node for the 2 s orbital okay. So along this spherical surface the 2 s orbital changes sign goes from plus to minus okay. So remember 1 s orbital has no node 2 s orbital does have a node and since the node arises from equating the radial part to 0 this is called a radial node okay more about that in a while for now let me show you n equal to 2 l equal to 1 here there is a theta phi part okay in fact phi parts will be for m equal to 1 0 minus 1 okay but if you look at only the radial part then what we have is once again some constant multiplied by R by 2 a 0 multiplied by e to the power minus R by 2 a 0. What will this look like R by 2 a 0 that is increasing in R obviously it is a straight line multiplied by e to the power minus R by 2 a 0 that is a decreasing function. So what you get here is essentially a function that goes through a maximum. So if you plot capital R for 2 1 n equal to 2 and l equal to 1 so that is a 2 p orbital remember against R what do you get it never at R equal to 0 what do you get at R equal to 0 obviously it is 0 e to the power minus R by 2 a 0 is 1 but R equal to 0 so at R it is 0 at sorry at R equal to 0 the function is 0 at R equal to infinity also it is 0 so it goes through a maximum. What is the value of R where the maximum arises well you just differentiate this and equate to 0 you will find what is the value of R where this I think sorry where this function goes through a maximum it is very simple differentiation. Let us move on n equal to 3 l equal to 0 now see we have here things are getting little more interesting again R to the power l right R to the power l l equal to 0 so R to the power 0 is equal to 1 here R to the power 1. So here we have 1 minus 2 by 3 R by a 0 minus 2 by 27 R by a 0 to the power 2 so a second order polynomial multiplied by e to the power minus R by 3 a 0 what will this function look like if I plot against R what is this function going to look like I am plotting R 3 0 against small R now see this is a second order polynomial if you equate that to 0 you will get a quadratic equation and that quadratic equation is going to have 2 roots and just believe me when I say the roots are going to be real. So in 2 places for 2 values of R the wave function will become 0 so to start with if it is plus then it should come down become 0 here become negative but again it has to become 0 here so it will increase like this becomes 0 here becomes positive then of course it has to go down somewhere and become equal to 0 asymptotically what is this point what is this point what is this point these 2 can be obtained but from the quadratic equation and this point and this point you can find out by differentiating the function the whole function and equating to 0 not very difficult thing to do hence we can go ahead and we can write down different functions so by looking at it and I leave you to figure this out we get something well this is absolutely empirical just by looking at the functions we see that the number of radial nodes turns out to be n minus l minus 1 so when n goes up number of radial nodes actually goes up but then when l goes up it goes down because it is minus l that is why you have this alternate not alternate you have this functions going through a node and then not going through a node also as you increase n so this is the number of radial nodes and this expression is again something that we know from our high school days right so this then is a brief discussion of the radial functions of hydrogen atom here now this slide has become not very important I am going to show some prettier pictures of this when we talk about contour representations of orbitals just remember 1s and 2s orbitals are functions of r only and now we come to this interesting question that take the s orbitals look at say 1s orbital or 2s orbital 2pz has a value of well psi of 2pz has a value of 0 when r equal to 0 but what about 1s what about 2s what about 3s for all of them the maximum value of psi is at r equal to 0 so the maximum value of psi square these are real functions of course maximum value of psi square is also equal to r equal to 0 what is the meaning of r equal to 0 r equal to 0 is the position of the nucleus so are we saying that the maximum probability for finding an s electron is at the nucleus because if we do then we are back to square 1 square 0 because once again if the maximum probability of finding the s electron is at the nucleus then it is very similar to rather fort situation where the electron and nucleus would be at the same position so plus and minus charges would annihilate right the atom sort of gets short circuited that cannot be the case so the fallacy here actually we do not it is very important to remember that psi psi star is not probability psi psi star gives us probability density so it is true that s orbitals have maximum probability density in the nucleus at the nucleus that does not mean that probability is also maximum because you might remember well you do remember that see if I ask you a question what is heavier iron or cotton wool what would your answer be the correct answer would be that the question is incomplete makes no sense well iron is more dense than cotton wool that is true but suppose I drop an iron pin from a fifth floor and then I drop 5 kg of cotton which one will have more impact on anybody or anything that it falls upon definitely cotton so density of cotton is much less but then we are using a higher volume that is why the mass is more density of iron is more but we are using a smaller volume so mass is less so let us not confuse the extensive quantity extrinsic quantity probability with the intrinsic quantity probability density for s orbitals it is true that maximum probability density of finding the electrons is at the nucleus but volume is 0 as we are going to see in the shortish next module that we are going to have