 Alright, so we've seen that we can solve the Schrodinger equation and we can write down the wave functions for the hydrogen atom. So there's one thing left that we haven't done for the Schrodinger equation for the hydrogen atom and that's to talk about the energies. So Schrodinger equation and its full form is written out here for the hydrogen atom. We've got the kinetic energy term in spherical coordinates and we've got the potential energy term which is just the Coulomb interaction, the 1 over r Coulomb interaction for the electron interacting with maybe one proton or maybe a proton with a nucleus with more than one proton in it. That kinetic plus potential energy terms must sum to give the energy times the wave functions so that's Schrodinger's equation and we can write down wave functions, any of an infinite number of wave functions, they might be simple or they might be complex but we haven't yet talked about what the energies of those wave functions are and of course those are the important things as we go on and begin to do statistical mechanics. So as an example let's just take the simplest wave function and we'll confirm again that it does in fact obey the Schrodinger equation and in the process we'll learn what the energy of that 1, 0, 0 wave function is. So the n equals 1, l equals 0, n equals 0 wave function and I've reminded us here that wave function which just looks like e to the minus atomic number times r divided by the Bohr radius. I've reminded us here that that Bohr radius is just this particular collection of constants. So to find the energy we'll plug the wave function into this expression so we'll plug in psi 1, 0, 0 here and here and here and confirm that this does in fact obey the Schrodinger equation. So the fact that we're using an l equals 0, n equals 0 wave function and there's no angular dependence in the wave function at all, there's r's in the wave function but no theta's or phi's in the wave function. That means that this term and this term which involve theta and phi derivatives, those are going to completely disappear once I take theta and phi derivatives of the wave function so that makes it a little bit easier. The r derivative of the wave function and e to the minus zr over a naught just pulls down the exponent so that's the r derivative of the wave function. If I multiply by r squared I'll just go ahead and throw that in there so that's the r derivative multiplied by r squared. Next thing we need to do is take the r derivative again of that so the r derivative of r squared is 2r so the 2 I'll put out front and the r squared becomes an r, an r not an r squared and if I focus on the exponential term I've got z over a naught with the negative sign and r squared, derivative of the exponential pulls down another negative z over a naught so negative becomes positive and those coefficients get squared. So that's what I have for the r derivative of r squared, r derivative of the wave function. If I then take 1 over r squared including this term so I'll just extend this out to include the 1 over r squared then the r becomes a 1 over r and the r squared just disappears. So that's what I have for this first term inside the brackets. Expanding out to the full Schrodinger equation now, h psi is going to equal this collection of constants h squared over 8 pi squared mass times, let's see we've got 2z over a naught 1 over r and an exponential that's these terms times this term the negative signs have cancelled one another. I've also got same collection of constants h squared over 8 pi squared mass times z over a naught squared times and the exponential that's the full kinetic energy term. Lastly I can include the potential energy term z e squared over 4 pi epsilon not 1 over r times the wave function that should equal energy of the 1 o o wave function multiplied by that wave function. Notice also that everywhere I have n e to the minus z r that's equal to the wave function so that we're looking pretty good so far the things on the left are all things multiplied by a wave function things are on the right are all things multiplied by a wave function the derivatives have all been taken. I do have some 1 over r times wave function in the first and the third terms and I have just some constants times the wave function in the second term and on the right hand side so we better make sure that the first and the third term cancel each other which they should so if that's true then the first and the third terms when I combine them will look like h squared over 8 pi squared mass 2z I've got a 1 over a naught and this is why I reminded us what the value of a naught is if I take 1 over a naught I'll get epsilon not h squared over or under pi mass e squared so that's this first term where I've replaced the Bohr radius with the constants that it equals and then for the third term so this is all one large fraction for the third term I've got minus z e squared over 4 pi epsilon not again each of those terms multiplies a 1 over r and a psi so if we double check what cancels h squared top and bottom cancels the mass of the electron top and bottom cancels we've got a 2 over an 8 which is the same as a 4 we've got a z on top in both terms we've got an e squared in both terms we've got a pi squared on the bottom which cancels the pi on top so the pi on the bottom and the epsilon not so each of these terms is exactly the same one of them minus the other does in fact equal zero as we expected so what that means is the entire first term cancels the entire third term all we have left now is that the second term this collection of constants times the wave function so minus h squared over 8 pi squared mass z over a naught squared z over a naught quantity squared is equal to energy so those constants times the wave function are equal to energy times the wave function so the value of the energy is in fact this collection of constants which I suppose I can rewrite as minus z squared h squared over 8 pi squared mass of an electron a naught squared so that is in fact the energy of this one zero zero wave function and again if we're interested in the numerical value we can plug in constants the only things we need to know are Planck's constant pi the mass of an electron and the Bohr radius so all those are constants that we can calculate we need to know whether we're talking about a hydrogen atom or a helium plus nucleus or a lithium 2 plus what the atomic number is of this hydrogen like ion and we can calculate the energy of the wave function in fact this collection of constants is common enough that we can define a constant called e sub h or the Hartree so this value one Hartree is equal to the collection of constants that is h squared over not an 8 but a 4 pi squared mass of an electron Bohr radius squared so when I combine the constants in that particular combination that happens to be a value of about 27.2 in units of electron volts it's a considerably larger number in other units like Kcals per mole but 27.2 electron volts is defined to be a Hartree this collection of constants so when I do that then this e100 because there's an 8 here instead of a 4 works out to be minus one half times z squared times the value of a Hartree and in fact we haven't proven it the energy for anything but this 100 wave function but in a more general sense the energy of any nlm wave function if we were to plug in the wave function that we know how to write down now plug that into Schrodinger's equation what happens to be true is the energy is still proportional to a Hartree and a half and z squared but when I do n equals one wave function I get exactly this result when I do an n equals two wave function I divide it by 4 n squared any n wave function I divide that energy by n squared so the energy ends up proportional to 1 over n squared and notice here that the energy doesn't depend on l and m there even though I can calculate the energy of a 2 1 0 or a 2 or 3 2 1 any any collection of l's and m's that I want the value of l and m don't affect the way the energy of the wave function at all all I need to know to calculate the energy is the value of n