 All right, so one of the other things we can spend some time understanding about hydrogen atom wave functions is what the energies of the electrons are when they're in those particular wave functions. So we've seen previously that the energy of an electron in a wave function of the hydrogen atom or a hydrogen-like ion is this relatively simple expression, minus a half atomic number squared divided by the n, that's the quantum number of the wave function, in fact I could write the energy of an nlm state is the same value, so the value of l, the value of m, don't matter at all. We just divide by the value of n squared and then multiply that by some energy, some collection of constants that have the units of energy that we call a Hartree. That unit of energy is 27.2 in units of electron volts which are pretty common for working with atomic energies or if you prefer it in SI units it's this very small number of joules, 4.36 times 10 to the minus 18th joules. So for one thing we can draw an energy ladder and ask ourselves what it looks like that the atoms have this energy of minus some constants times n squared, so I guess let me actually put zero since the energies are negative, let me put a marker for zero up here, energies are rising in this direction. The lowest energy an electron and a hydrogen atom can have, if it's a hydrogen atom in particular with z equals one, the lowest energy can have is when n equals one, so the energy is going to be minus half of a Hartree. So negative 13.6 ev, so that would be for the n equals one state for the electron. The next state up the ladder is going to be the n equals two state. Now the energy is going to be negative a half times one over two squared for a hydrogen atom, so half times a quarter is an eighth, so 75% of the way up towards zero we find the n equals two state. So I'll stop labeling these with specific values but that's minus an eighth of a Hartree. So if this was the n equals one state, n equals two is a quarter as large, is a quarter as negative, the n equals three state is going to be three squared or one ninth as negative, so that's going to be somewhere around here. We're also going to have n equals four state, n equals five state, n equals six state, so those states get packed tighter and tighter together as we go to higher and higher n values and then in the limit where n becomes very large, the energy is becoming zero. We're dividing this number by a very large number, so it's a very small negative quantity. So our energy ladder looks like this, there's a big gap from the first state to the second and then a smaller gap and a smaller gap still and those gaps get smaller and smaller as we asymptotically approach an energy of zero. The other thing we can do with those energies is do some numerical calculations and one of the things we know the energies tell us is how likely those states are to be occupied. So let's consider this problem. We know now, let's say, let's calculate the probability of occupying a 2s orbital, so remember the 2s orbital is the same as the n equals two spherical orbital, so the two zero zero orbital and to make things relatively simple we can say what's the probability of occupying the 2s wave function relative to the probability of occupying the 1s wave function? So what's the probability of being in this excited state relative to the probability of being in this ground state? And we know that Boltzmann tells us that that's just going to be energy of the 2s state or the two zero zero state over kt with a negative sign in the exponent. We divide that by e to the minus energy of the 1s state over kt and then because I'm taking this ratio I don't have to worry about partition functions. The actual probability of the 2s state is e to the minus energy over kt divided by a partition function but if I divide by a partition function on the top and on the bottom then those two will cancel. So all together this is e to the minus difference in energy between those two where the difference in energy is the energy of the numerator minus the energy of the state referred to in the denominator. So let's think about that difference in energy first so that e of the two zero zero state minus energy of the one zero zero state we know how to calculate those energies. Energy of the two zero zero state we've just thought about that a minute ago is minus a half times one over four so minus an eighth of a hard tree and I guess I should specify since I'm doing this with z equals one since I've plugged z equals one into this equation I'm doing this calculation for a hydrogen atom if I were doing it for a helium plus ion it would be different numbers because I would use a different atomic mass. So energy of the two zero zero state is minus an eighth of a hard tree energy of the one zero zero state as we've said is minus a half of a hard tree so all together that adds up to positive three eighths of a hard tree negative an eighth plus a half and if we multiply let's do the calculation in SI units if we multiply this number by three eighths what we find is the difference in energy between these two states this is the delta e that we're talking about between the n equals two and n equals one state that particular difference in energy is one point six times ten to the minus eighteenth joules. That number by itself is not terribly meaningful what's meaningful is the ratio of that energy to kt so we'll need to know what temperature we're interested in so let's say we've got some hydrogen atoms at room temperature 298 Kelvin and we'd like to know how likely it is that they're in the 2s orbital relative to the probability of being in the 1s orbital so the value of kt as our benchmark for energies is Boltzmann's constant multiplied by the temperature we're interested in and that number works out to be again something small four times ten to the minus twenty one joules so those are both small numbers but notice that the small number that's kt is quite a bit smaller ten to the minus twenty first quite a bit smaller compared to the small number that's the delta e that's still small but it's ten to the minus eighteenth joules so this number is several hundred times larger than kt so that ratio when we calculate the ratio of delta e over kt that ratio is 397 and if we convert that to a power of ten rather than a power of e we find that that ratio is ten to the minus 173 so that's an almost unimaginably small number the probability of finding an electron in the 2s state is so small that if we have a universal of hydrogen atoms lying around at 298 Kelvin chances are we would not see a single one of them in the 2s state they're all gonna prefer greatly prefer to be in the in the 1s state have a higher probability of being in the 1s state so what that means is this energy difference at least at room temperature is so much larger than kt that all of essentially all of the electrons occupy the n equals one state they never spontaneously find themselves in the 2s state just due to thermal excitation what that means is that's the reason why for example in general chemistry if I ask you tell me the electron configuration for a hydrogen atom you would say hydrogen is a 1s one electron configuration the single electron in a hydrogen atom is found in the 1s orbital you didn't need to consider the possibility that well sometimes it might be in the 2s orbital or the 3s or the 5d orbital or something else the probability of ever finding at room temperature a hydrogen in an orbital other than the 1s the electron in the orbital other than the 1s orbital is vanishingly small so we can with very much confidence say the electron configuration is just 1s one and leave all the other possibilities alone