The Semi Empirical Mass Formula

hello today we're going to look at the semi empirical mass formula or I could have called this the binding energy of the nucleus or I could have called it what holds the nucleus together or I could have called it why can't you have any number of protons and neutrons in a nucleus so this is a subject based on nuclear physics and if you haven't already done so I advise you to watch my videos in the a level playlist on the nuclear radius nuclear physics and nuclear fission and fusion because that will give a background to what we're now going to cover this material is more than a level this is essentially university level physics we know that an atom consists of a central nucleus which is positively charged with and I'm talking classically here orbiting electrons we know that the nucleus itself consists of protons and neutrons protons are positively charged so they contribute to the positive charge of the nucleus neutrons have no charge at all and don't contribute to the charge but the net charge on the nucleus is positive and there are an equal number of positive charges that is protons in the nucleus as there are negative charges electrons orbiting the nucleus now as we saw a level if you make a plot of the number of neutrons versus the number of protons the number of neutrons we put as n the number of protons we use Z if you make a plot this firstly is just going to draw a line that line is n equal Z that would be the number of neutrons equals the number of protons but if you look at all the stable nuclei that there are what you find is that at the low end n does indeed equal Z but then the line die Earth's off toward the Neutron end and you find that stable nuclei with high values of Z have more neutrons than they have protons so there is an excess of neutrons in stable nuclei at the high end in other words when Z gets larger the number of neutrons exceeds the number of protons you also find that there are unstable nuclei surrounding this line and I'll talk about what I mean by unstable in just a moment so there are unstable nuclei both above and below this line so the stable nuclei represented by the crosses unstable nuclei are represented by the circles and if you like we could draw a dotted line around all of this this is nothing more than a container ah and that dotted line shows you all the nuclei that you can find naturally-occurring you don't find any here you don't find any here and you don't find any above this point here and what we're going to ask in this video is why well let's first consider what happens with these unstable nuclei now as I've said in the series of circles which are above the stable line basically you've got excess neutrons and below these stable line you've essentially got excess protons and what happens with those unstable nuclei is that they will decay they will change in order to get onto the stable line so if you've got an excess of neutrons what that means is that a neutron is going to decay into a proton so you get a neutron becomes a proton plus an electron plus an anti-electron neutrino if on the other hand you've got an excess of protons then a proton will convert into a neutron so proton converts into a neutron plus a positron plus an electron neutrino we've seen those formula before at a level and what is actually happening is that if you're here at this unstable nucleus here what you're going to do is you've got too many neutrons so if you convert a neutron to a proton the neutron number will fall by one but the proton number will increase by one so what you'll essentially do let me just draw it here if that's your unstable position that's this one here you will fall down one on the neutron scale but you'll increase one on the proton scale so effectively you'll fall down in that direction onto a stable nucleus so you'll become stable by decreasing the number of nuclei that number of neutrons by one but increasing the number of protons by one similarly if you're on this position here where your proton rich and you're going to change protons to neutrons what you're going to do then if you are at this position here let's say is you're going to decrease the number of protons by one and increase the number of neutrons by one so essentially your pattern let's move this up a little bit like this your pattern will be to decrease the number of protons by one increase the number of neutrons by one so you're moving diagonally in this direction onto the line of stability so in both cases you are changing either neutron to proton a proton to a neutron and what you do is you'll notice you emit either electrons or positrons now before anybody knew what electrons were this emission was given a name it was called beta in fact people who were studying X studying nuclei and their decay found that there were three types of emissions alpha beta and gamma alpha was positively charged beta was negatively charged and gamma wasn't charged at all that's how they knew that there was a difference and these emissions are called beta decays a beta particle is simply an electron a beta plus particle is simply a positron so this decay happens by what's called beta decay which is simply if we were changing a neutron to a proton you emit an electron if you're changing a proton to a neutron you emit a positron and this is the way that unstable nuclei get back on to the stable line if you get to the very high end you don't get electron emission anymore you get alpha particle emission suddenly you've got to start reducing by significant numbers of neutrons and protons and alpha particle is two protons and two neutrons and if you've got an alpha particle decay it means that an unstable nucleus will reduce its noop Neutron number by two and its proton number by two and move on to a stable position in that way now the question I'm essentially asking is why is it that some nuclei are stable some nuclei are unstable and some nuclei don't seem to feature at all what is causing this and if we are speculating that there is a nuclear force and we are so we'll say here is the nucleus and there's some kind of force that's holding it together there must be because there are protons and there are neutrons in this nucleus and protons are positively charged so there will be a Coulomb force and electrostatic force between them which will be a repulsive force so it will be trying to push the protons apart and make the nucleus unstable if the nucleus is at is stable there must as it were be a stronger force which we call the nuclear force that's holding it together and that force must be stronger than the Coulomb force that is trying to push it apart now if there is this nuclear force which is essentially putting its arms round the nucleus and say I'm going to hold you together I'm going to bind you together then what does it matter how many protons or neutrons you have in the nucleus if there's a nuclear force that's binding them well if there is a nuclear force that's binding them together then there must be some binding energy that is responsible for holding the nucleus together and what is the source of that binding energy well if you were and this is just a kind of thought experiment if you were to take a set of scales and take the total number of protons in the nucleus and the total number of sorry that should be in the total number of neutrons in a nucleus so we've got Z protons and n neutrons but you just take the in the number of protons and the number of neutrons just take each individual proton out of a box and the right number of neutrons are out of a box and put them on these scales and on the other scale you put the actual nucleus itself which contains Z protons and n neutrons what you find is that the mass of the individual protons and neutrons is greater than when those protons and neutrons are bound in a nucleus there is what is called a mass deficit some of this mass has been lost when all these protons and neutrons are bound in the nucleus and how does mass get converted into energy well e equals mc-squared so what we're saying is that this very tiny amount of mass that is the deficit between the individual constituent particles and when they actually get bound together in a nucleus that is responsible for the binding energy in other words of the binding energy is the mass difference or the mass defect multiplied by C squared now the mass of a proton is of the order of 10 to the minus 27 kilograms and this mass difference will be much smaller than that so we are talking tiny tiny amounts of mass but look that tiny amount of mass is multiplied by C squared C squared is 9 times 10 to the 16 so you're already there's almost came to the 17th you're already converting that and that will be in joules and if you want to convert joules to electron volts you might be not quite this but you you multiplied by 10 to the 19 you've actually divided by one point six times ten to the minus 19 but we'll just look at orders of magnitude so you can see that this is essentially 10 to the 17 this is 10 to the 19 all of this combined is a massive 10 to the 36 to get an energy in liquor tender 36 to get an energy in electron volts so even though you've got a very very small mass smaller than the mass of a proton you multiply that by 10 to the 36th to get Evie and what you actually find is the binding energy for a nucleus will not be any electron volts but it will be in mega electron volts MeV as we should see well how are we going to measure this binding energy well on the face of it is pretty straightforward because we know that we've just got to find the mass deficit and multiply it by C squared well define the mass deficit what we need to do is to multiply the mass of a proton by the number of protons plus the mass of a neutron which we know times the number of neutrons and subtract from it the nuclear mass of the atom trouble is measuring nuclear masses is more difficult than you may think whereas measuring atomic masses is very easy John Dalton for example the chemist was measuring atomic masses way back in 1803 long before a hundred years before anybody even knew what a nucleus consisted of John Dalton was measuring atomic masses so the real question is can we find a way of measuring binding energy of a nucleus from atomic masses and the knowledge of the mass of a proton and the mass of a neutron which we can measure so let's consider an atom which I will give the symbol X it doesn't stand for anything it's just any atom which has Z protons and a nucleons nucleons means protons plus neutrons so essentially a is equal to protons plus neutrons so if you want to know how many neutrons are they are obviously a minus Z neutrons so what is going to be the Constituent bits of this atom well there are going to be Z protons so we multiply Z times the mass of a proton gives us the total mass of protons plus n times the mass of a neutron gives us the total mass of neutrons and then of course there will be electrons and there are always the same number of electrons as there are protons so there will be said electrons so we multiply the mass of the electron by Zeta and that gives us the total mass and to convert that to energy we have to multiply by C squared so all of that is a mass multiplied by C squared gives you an energy and that is going to be equal to the atomic mass so I'll put em at an atomic atomic mass of this element here X Z a plus the binding energy of the nucleus plus if I'm getting in the binding energy of the atom because remember not only is the nucleus bound but the atom is bound a hydrogen atom for example the binding energy of a hydrogen atom it's at its lowest level is minus thirteen point six electron volts in other words you have to provide you've got me electric if you've got a hydrogen atom the electron is bound to that atom and if you want to get that atom to escape you have to provide it with more than thirteen point six electron volts in order to give it enough energy to escape this is of course nothing more than the photoelectric effect here is the hydrogen atom it's got one proton it's got one orbiting electron the positive and the minus charges are attracting and that means that the electron is bound to the atom and it is bound by thirteen point six electron volts you have to provide that to get it out the photoelectric effect doesn't use hydrogen it uses a metal and what it says is if you shine light it usually has to be ultraviolet light on the metal you provide energy and that will cause electrons to come off and what is the energy of ultraviolet light its Planck's constant times the frequency of the light so in other words the higher the frequency you need the higher sorry the higher the energy you need the higher the frequency of the light you have to use and for hydrogen for example thirteen point six bv which is what this energy would have to be turns out to be a frequency in the ultraviolet range which is why you'd have to use ultraviolet light and the same applies to metals although the binding energy for metals will be slightly different nonetheless you still need ultraviolet light to give enough energy to kick the electron out because the electron is bound so we've got two terms for binding energy the binding energy of the nucleus which is actually much larger because that will be in MEA and the binding energy of the atom so I'm just going to slightly rewrite this formula here and I'm gonna write that as Z into MP plus MP e because we've got the Z protons and Z electrons times C squared that's an energy plus you'll remember the mass of the neutron multiplied by the number of neutrons C squared is equal to the atomic mass of the atom a said X plus the binding energy nucleus plus the binding energy atomic now the reason I rewrote that let's just show you the old formula all I've done is to take these two terms out and then that term is left so essentially I've just rewritten it like that the reason I've done that is that if you take this term here that is a rewriting MP plus MP C squared is not that the energy of the mass of a hydrogen atom so that is the Constituent bits of a hydrogen so I'm adding the two message to get the multiplied by C square to get an energy and that is going to be the atomic mass of hydrogen multiplied of course by C squared plus the binding energy of the hydrogen atom because the hydrogen atom is bound so in other words we're saying that there's a very tiny mass difference between the individual constituent elements of the atom and when the atom is bound and the difference between the two the mass difference is what provides the the atomic binding energy so if I take this formula here I can say Zed now instead of MP plus M II C squared I can now write this term here so I'm going to put Z times the mass of atomic hydrogen times C squared plus Z times the binding energy that's this term here of a hydrogen atom that's that's that now rewritten in this way then I've got to add my n m n C squared the of neutrons times the mass of a neutron time C squared is equal to what we had before the mass of the atomic a X plus the binding energy of the nucleus plus the binding energy of the atom right now where can we go from there I want to know what the binding energy of the nucleus is that's what we're trying to find here and so if I rearrange this formula I will get that the binding energy of the nucleus is equal to well before we do that let's just think for a moment here we've got Z times the binding energy of the hydrogen atom and here we've got the binding energy of this atom X with Z protons the binding energies of atoms are in E V electron volts whereas the binding energy of the nucleus is millions of times bigger in mega electron volts when you bring this term over to this side because you want binding energy alone on this side you're going to get Z times the binding energy of a hydrogen atom minus the binding energy of this atomic atom X the difference of two very small binding energies is going to be very very small consequently those two terms really are irrelevant they are so close that when it comes to a binding energy in the of the order of MeV they're really not going to matter so now we can rewrite our formula with binding energy is equal to Z times the atomic mass of hydrogen times Z squared plus n MNC squared so the number of neutrons times the mass of neutron C's times C squared minus this term here the atomic mass of the element X with Z protons and a nucleons and that's it now even John Dalton could measure the atomic mass of X we know and indeed he measured the atomic mass of hydrogen and we know the mass of a neutron so we know these atomic masses we know the mass of a neutron we know the number of protons we know the number of neutrons so we can calculate the binding energy of the nucleus and all of this comes from experimental data so for example we showed you that the binding energy of a hydrogen atom was thirteen point six electron volts the binding energy of a helium nucleus for example is twenty eight point three MeV it just shows hugely greater than the binding energy of an atom what this means is that by measuring the atomic movie you only need to measure the atomic mass of hydrogen and the atomic how or the mass of a neutron once you've got those two terms forever you've just got to measure the atomic mass of all the elements and for each element using this formula you can calculate the Associated binding energy of the nucleus so the nuclear binding energy can be calculated from experimental results for all nuclei so the binding energy is as it were the energy that is holding the nucleus together and the nucleus consists of protons and neutrons and we know that were there to be no binding energy the protons would force the nucleus to disintegrate because they would repel and push the nucleons apart and the binding energy overcomes this and holds them together but now let's ask what is the actual source of the nuclear force that is providing this clinging this as it were cementing of the nucleons together well let's postulate for a moment that the force is actually simply a force that acts between any two nucleon so whether it be a proton and another proton proton and a neutron or neutron and a neutron we're going to speculate that there's a force that acts between them that holds them together so in other words the more pairs of nucleons you have the stronger will be the force because each pair is contributing one bit of the force well how many pairs do you get well let's think about the number of nucleons if you've got say three one two three how many pairs are there there are 1 2 3 pairs if you've got 4 nucleons 1 2 3 4 then you've got 1 2 3 4 5 6 pairs and for a nucleons and this is just purely mathematical you can show that that is equal to a into a minus 1/2 pairs which is of course equal to a squared minus a divided by 2 pairs I mean work it out when a is 3 that would be 3 times 2 is 6 divided by 2 is 3 pairs when a is 4 that's 4 times 3 is 12 divided by 2 is 6 pairs so it seems to work so the total number of pairs when you've got a nucleons which is total number of protons and neutrons is a squared minus a over 2 so this means we're saying that since each pair is contributing towards the nuclear force the total force or if you like total binding energy is going to be proportional to the number of pairs which is a squared minus a over 2 so very roughly you can say that the binding energy the total binding energy of the nucleus is proportional to a squared particularly as a gets larger this a term is of significance very roughly the binding energy of the nucleus is proportional to a squared make sense as a increases you'd expect the binding energy to increase so if I divide both both sides by a I get that the binding energy of the nucleus divided by a is proportional to a a itself which would mean if I drew a graph of binding energy over a the total number of nucleons against the number of nucleons I would expect that to be a straight line then I what the gradient would be but it would be a straight line now in practice we know how to measure the binding energy of just shown you you do it from atomic masses and you can you can deduce the binding energy of the nuclear binding energy for all elements and what you find is if you actually plot the experimentally determined binding energies per nucleon against hey that what you actually get is something that looks like this is a very steep rise initially but then it Peaks and there's a very slow fall off and it peaks at a equals iron-56 that is where you get the maximum binding energy per nucleon and that is roughly 8 MeV so 8 MeV per nucleon there are 56 nucleons so there'll be 56 times 8 MeV total nuclear binding energy and it's it there's a very slight fall off but it's almost a constant and right at the beginning of the periodic table the hydrogen's and the helium's they are much less bound the binding energy per nucleon is much smaller as we'll see in a future video so very roughly you could say that the binding energy per nucleon is equal to 8 MeV because that's roughly what the position is at this stage of the business that means that B E is proportional to a and not a squared as we speculated up here so we've obviously made a mistake if the binding energy is only proportional to a rather than a square so where have we gone wrong well our problem probably is in the assumption of that the binding energy is caused by all nucleon pairs here's a nucleus here's a nucleon and here's another nucleon and we were speculating that no matter where the nucleons are they all exert a nuclear force between them and the total number of nuclear forces is the total amount of pairs multiplied by the nucleon force for each pair of nucleons but this in a large nucleus this might be five Fermi's across suppose the nuclear force doesn't have that range in fact we do know that nuclear force is very short-range suppose two nucleons on either side of a nucleus actually don't have any interaction because in a sense they're too far apart let's suppose that actually this is a very short range force so that actually the nucleons can only interact with the nucleons immediately around them so they each have an interaction panel but this one doesn't interact with this one so we're no longer saying that the binding energy is proportional to the total number of pairs we're essentially saying that the binding energy per nucleon is a constant because for each nucleon is only the immediate nucleons around it that are contributing to the binding energy for that nucleon now the total number the total binding energy will obviously therefore be proportional to the atomic number a but not a sweat because as the atomic number grows so you're going to have more nucleons and each nucleon will have it little cluster and that little cluster is providing The Binding energy so now you can say that the binding energy is proportional to a the binding energy per nucleon also if you say B divided by a is proportional to a constant so it is actually a straight line and that is because doesn't matter how many nucleons there are each nucleon only sees as it were or only interacts with the ones in its immediate vicinity so we can now say that the binding energy the total binding energy of the nucleus is proportional to a and that means let's call it is equal to a constant which I'm going to call a V standing for volume times a in other words the binding energy of the nucleus will increase as the number of nucleons increases but not as the square but I'm afraid we're going to have to make some Corrections to this you see the first thing we've got is we've forgotten that there is a service to the nucleus again I'm talking in classical terms in quantum mechanical terms that surface would be very fuzzy but that nucleon there is different from this nucleon here this nucleon has got nucleons all around it in its interaction mode but this one has got these but the one that would be there isn't there because it's at the surface so the total binding energy for that nucleon is slightly less than the nucleons that are in the middle because there's not so many local pairs so the binding energy is reduced because of this surface effect so I've got to put a minus here because I'm reducing the binding energy as a consequence of the surface what is the surface well it's 4 PI R squared where R is the radius and we learnt in our a level physics that the radius is equal to some constant which we call R naught times a to the third where and again is the nucleon number so obviously r-squared is going to be a proportional to a to the two-thirds you just square both sides consequently we can say that the correction term that we need to put into this is going to be proportional to a to the two-thirds and what we actually say is there's a constant of proportionality which we'll call the surface term that's a constant times a to the 2/3 and that term is reducing the binding energy because the nucleons on the surface aren't experiencing quite as much binding energy as those in the middle so that's why it's a minus term it reduces the amount of binding energy but there's another correction term that we need here's the nucleus it's made up of protons and neutrons the neutrons don't have any Coulomb force but the protons certainly do they are trying to push one another apart and break up the nucleus so they are essentially acting in the opposite direction of the binding energy the binding energies trying to keep them in the Coulomb force is trying to push them out so we need another minus term here because we've got a Coulomb force that is reducing the binding energy what is the nature of that force well we know that the force is equal to q1 q2 the charges on the two protons divided by 4 PI epsilon nought R squared that's the force we're talking about energy the potential energy is given as q1 q2 over 4 PI epsilon naught R so it's just a 1 over R dependence which of course is just a constant because these are going to be the charges on the protons which is constant 4 PI epsilon naught is a constant so the only variable here is R and as we've seen R is proportional to a to the 1/3 so we just need to put a to the 1/3 here that is going to be the potential energy between any two protons how many pairs of protons are they going to be well just as before the total number of pairs if there are zeg protons the total number of pairs is Z into Z minus 1 over 2 so V so as it were the potential energy term that we're going to need to put in will have a Zed into Z minus 1 divided by a to the third the 2 we can wrap up in this constant so we now need to have a because this is the total number of proton pairs this is the contribution to reducing the binding energy of each pair so you multiply this by this and you and you take the 2 and you wrap it up in the constant so now we need this minus term here which is Z into Z minus 1 times a constant which we'll call AC that's the Coulomb constant divided by a to the 1/3 and that is the classical that is the classical bit done but there are two more as it were quantum mechanical corrections to be made the first recognized is the Pauli exclusion principle that you cannot have all these nucleons at the same energy level they are going to occupy energy levels the way it works is that one proton spin up and one proton spin down and one Neutron spin up and one Neutron spin down can occupy the lowest level but then that level is full so further protons have to occupy spin up spin down and neutrons spin up spin down the next energy level and so on until you've used up all your protons and let's say that we use up all our protons when we get to this level here but for as we showed there is an excess of neutrons over protons when you get to higher values of a so now you've got to accommodate the excess neutrons and they are going to build up a spin up spin down spin up spin down and so on until all the neutrons have been accommodated in the energy levels but that you will see produces an asymmetry you know there's a low level energy here there's a higher level energy here these neutrons up here are at a as it were lesser binding energy it's easier to kick one of those neutrons out than it is one of these protons because this is binding energy goes down in this way energy was zero is here consequently this is nearer to zero than this is and you need less energy to kick this out than you do to kick this one out it's rather similar to you know the electron in the atom this is minus thirteen point six Evy for a hydrogen atom but if that electron is at a higher energy level where this of course is P equals zero then you don't need thirteen point six Evie to kick that electron out you mean you need much less well what we want to know is what is the extra energy associated with these kind of excess neutrons let's assume that the energy difference is Delta for each energy level then what you're going to have is an energy excess of caused by these neutrons which is going to be two neutrons which are Delta higher so that's two times the Delta plus another two neutrons which are now two Delta higher because there are two levels above so that's two into two Delta and then if there were more it would be 2 into 3 Delta because you'd have two more neutrons that were three Delta higher than the proton level all the way up to 2 into x over 2 Delta where X is the number of new trongs in excess of the number of protons so X is the number of excess as it were neutrons over protons and it's x over 2 because you will only have half the number of energy levels because each energy level can accommodate two neutrons now this is a geometric series and you can simply reduce it to e that's the excess energy of the neutrons because they're in this asymmetric arrangement is equal to 2 Delta into x over 2 into x over 2 plus 1 divided by 2 that's the geometric series those two twos cancel out so you've now got Delta into x squared over 4 which is those two terms plus X over 2 so I've rewritten that formula that the as it were the energy excess consequence of having these asymmetric neutrons is Delta into x squared over 4 plus X over 2 now what is the size of Delta well what we're arguing is that Delta is the separation between the energy levels and generally speaking the greater a is the less Delta will be so a as a increases the energy levels are squeezed together that means that Delta is proportional to 1 over a so I've got for my Delta term I can replace that by a 1 over a term what is X well X remember was just the surplus of neutrons over protons is n minus Z X will be small compared to x squared when we've got larger numbers so again we can ignore the X term and essentially we can wrap up all the constants into a single constant the only things we've got our a delta which is a 1 over 8 and an x squared which is an N minus Z squared term so essentially the correction term we're going to need here is a constant which we'll call the symmetric constant times n minus Z all squared which is essentially this x squared term divided by a which is essentially the Delta term and now I'll rewrite our binding energy formula starting over here so the binding energy of the nucleus is equal to this volume constant times the total number of nucleons and then we have to subtract from that the surface term then we have to subtract from that the Coulomb term said into Z minus one over a to the one third and now we have to minus this symmetric term a symmetry n minus Z all squared yes I've just got the in brilliantly over a now there is one final term that we need to consider and that is what's called the pairing term because it turns out that if you've got two protons they form a pair and clear binding energy is much greater than if you've got an isolated proton similarly for neutrons so if you've got an even number of protons and an even number of neutrons they can all form pairs and you get a high binding energy if you've got an odd number of protons and an odd number of neutrons then when they form pairs there's always going to be one poor old proton and one poor ol Neutron left over without a partner that actually reduces the binding energy because those isolated proton and neutron are easier to get out of the nucleus they're not bound by this pair so they're binding energy goes down and basically if you've got an even and an odd so an even number proton odd number of neutrons or other way round kind of evens out so what we've got to do with this binding energy term here is to add just one more term to it which I'm going to put down here I'm just going to call it Delta it wasn't the Delta we used before this is now the the representing the proton pair issue and that Delta is going to be since it contributes a plus term so if it's positive it will increase the binding energy so even even will be positive because that increases the binding energy odd-odd will be negative because that decreases the binding energy and odd even or even odd will be zero okay so when we come to Delta will happen up either a plus term for even even a minus term for odd odd or zero if it's odd even or even odd so now we can go back to the basic equation we came up with right at the beginning where we said that the nuclear mass of an atom which is Z a so the nuclear mass of this particular atom x times C squared because that converts it into energy is equal to the mass of the individual components so it's Z times the mass of a proton C squared plus n times the mass of a neutron C squared that's the individual masses of all the protons plus the individual masses of all the neutrons minus the binding energy because when you form a nucleus you lose some of this mess some of this mass is taken away and is converted into energy called the binding energy but we've just calculated that's the nuclear binding energy course we just calculated that nuclear binding energy is this great long formula here so if we were to put that term into here we end up the semi empirical mass formula and there it is let's go through it again the nuclear mass of any atom but we're just looking at the nuclear mass not the electrons times C squared convert it to energy is equal to the total number of protons times the mass of each proton times C squared so that gives you the proton energy the total number of neutrons multiplied by the mass of each Neutron gives you the total Neutron energy minus the binding energy which is the volume term the surface term the Coulomb term the symmetric term of the pairing term and that is known as the semi empirical mass formula empirical means that you fit things with experimental results but you don't have very much by the way of theory behind it semi empirical means you fit with experimental results but you've got a loosely based Theory the you know this was all fairly hand-waving but it's at least based on some kind of theory well what that means is that we ought now to be able to calculate the mass energy of the nucleus of any atom but before we do that let's just have a look and see how that binding energy formula effects the graph binding energy per nucleon divided by sorry plotted against the nucleon number if you take just the first term binding energy equals a constant times a that means that the binding energy per nucleon is a constant you would just get a straight line if you take the first two terms binding energy is av times ad minus the surface term but you divide per nucleon what you start to get is something that looks like that so the binding energy is still increasing with with nucleon number but it's now slightly different from just being a straight constant if you now add this Coulomb term and what you get is something that looks a bit like this and if you add the symmetric term you get a slightly declining line and that is not dissimilar from the experimental results with iron-56 at the maximum and a very slightly tailing away line here and this is all at about 8 MeV which is precisely what experimentally you observe so by putting in those first four terms I haven't bothered with the pairing term but certainly by putting in those four terms you get a binding energy per nucleon which accords pretty much with the experimental results that I showed you earlier well this is all very well but we haven't found the values of the constants IVAs ACA sim and Delta well I told you earlier that we can now measure the binding energy of the nucleus by knowing the atomic mass of hydrogen the mass of a neutron and the atomic mass of each element so as long as you measure those things experimentally you can calculate the binding energy of all the nuclei so for all nuclei there are awful they're 92 atoms but you can have more than one nucleus you can have a stable you can have unstable nuclei for all of those you can calculate the binding energy you know for each of those atoms how what the a number is you know what the Z number is you know what the N number is so essentially you can create 92 equations if you want to one for each element with each in each case a Z and n would be known and you can find the nuclear energy and so you get 92 of these equations of which everything is known apart from these five constants well 92 equations and 5 constants 5 unknowns pretty easy you can just get the constants falling out problem is that those constants vary depending on precisely how you do this so I'm just going to quote you one set of constants if you are being taught a different set that simply means a slightly different experimental method or a slightly different actual calculation method but the the values are not quite so important as knowing roughly what they look like and what you find is they all of course measures in energy a levy the volume term for my point of view is 15 point 8 MeV a s the surface term I should put equals there is 18.3 MeV the Coulomb term very much smaller nor point 7 1 4 MeV the symmetric term is 23.2 MeV and then Delta which you'll remember was the pairing term well that turns out to be 33 point 5 MeV if you're talking even even it's minus 33 point 5 MeV as we said odd odd because that is actually acting against the binding energy and it's zero if it's odd even or even odd so now we know it all we've got a formula for the binding energy we know the values of the constants so for you give me any atomic nucleus you like with values of a Z and N and I can calculate from that the binding energy of the nucleus so the question remains why then can you not have a nucleus with any old number of protons and nutrient and nucleons and have stability after all the nuclear forces are holding all these things together you should be able to have any number of nucleons and any number of protons you like put those in that formula and derive a binding energy what's the problem why can't we do that why are some nuclei unstable and why don't some nuclei exist at all and the answer is very simple it is nature's insistence that it always wants to occupy the lowest energy that's essentially all there is to it so here is our semi empirical mass formula which we derived earlier and yes in theory you could put any number of Zed's and any number of ends and don't forget a is constrained a is a plus n but any old eggs any old games into this formula and you would generate the nuclear mass of The Associated atom but nature says are when you give me a certain number of nucleons I want to occupy the lowest possible mass energy that's the one that will be stable in other words for any given value of a and I'll just remove the semi empirical mass formula for a moment we'll put it back but let's say for example that we fix a equals 100 how many combinations of Z and n can we have well Z plus n must equal a hundred but we could have everything from ze was a hundred and N equals naught to ze was a naught and N equals 100 that's a hundred and one combinations of N and Zed or which add up to a hundred because I have fixed a at 100 so let's bring back the semi empirical empirical mass formula so I could fit all the different values a would be a hundred in each case but I could put Z and with all the values such that Z and n always equal 200 but you start with teddys 100 and it was naught then you lose n is 99 n equals 1 and you calculate the mass energy for all those atomic nuclei and what nature says is I want to occupy the lowest energy that you calculate so for all the possible values of Z what is the value of Z that will give us the minimum value of mass energy to find a minimum you differentiate and set equal to zero so we differentiate the mass energy by Z which is going to be our variable and set that equal to zero so take a deep breath and let's do it follow closely so hang on to your hat here we go we're going to take the differential of the mass with respect to Z and then we're going to set that equal to zero so let's do it our first term is Z MPC squared differentiate that with respect to said you just get MP C squared here we've got any men that C squared well n is their dependent because n is simply a minus Z so if you replace n by a minus Z you get a MN C squared which of course is not so dependent so that will go but then you'll get minus z MN C squared and when you differentiate that with respect as edge you're going to get minus MN C squared and what we've got here before we move on is the difference between the proton energy proton mass energy and the neutron mass energy that is less than one MeV for these purposes we can ignore it the difference between the two is fairly trivial in these terms so we'll ignore that that took care of that particular term now we come to the binding energy we start with minus AV times a that is not Z dependent because a is fixed remember so that gives us nothing the second term is minus time for minors which is plus as8 of the two-thirds again not Z dependent so that gives us nothing now we've got a Z dependency here we'll start with - well it's minus times a minor so it's plus AC over a to the 1/3 AC over 1 to the a to the 1/3 and then we've got a Z squared minus a Zed Zed squared minus Z differentiated with respect to Zed gives you 2 Z 2 Z minus 1 right that's differentiating Z squared minus ed with respect to Z and then we're going to get a minus times a minus which is a plus this asymmetric term now n is a minus Z and therefore n minus ed is a minus 2z I'll just write that down here a minus 2z that is n minus n because n is a minus dead this is squared and that's going to give us a squared minus 4a Z plus 4z squared that is the square of this and now we need to differentiate that term with respect to Z the a squared term will go and that's going to give us minus 4a plus differentiate 4z squared you get 8z so that was all by way of calculation so I'll just put a ring around that to show that that was just a kind of a a calculation so we now know that the symmetric term is going to be and I'll put this up here so this comes down here to become the symmetric term which is plus a sim over a which is this term here multiplied by the differential here which we just calculated as minus 4a plus 8 Z and then finally we've got Delta Delta is not Z dependent so that's going to give us another 0 when we differentiate so when you look we've now got differential of mass with respect to Z is equal to nothing nothing we've got a Coulomb term and we've got a symmetric term and that's it so now what we do is to set that equal to 0 a 1 what I'm effectively going to do is to jump well here let's set that equal to 0 there you are set to equal to 0 now I'm going to rewrite that equation bringing the zegs on one side and the just pure nan Zi constants on the other side so from this term here I've got an a Z into 2 AC over 1 to the third so I've got Z into 2 AC divided by a to the 1/3 that's that Z term there and the Z term here is a Taysom over a so 8a sim over a that's + +8 a sim over a those are the only two zip terms that is a term there and this said term here on the other side I'm going to have well - AC over a to 1/3 but put it over the other side becomes + AC over a to the one-third and then down here I've got on - for a a CM over a so the A's are going to cancel I've got a minus for a sim take it over the other side it becomes a plus for a sim and that means that I can write that Z is equal to this term here AC over a to the 1/3 plus for a sim / this term here - a/c over a to the 1/3 plus 8 a sim over a now if we observe this formula what we say is we know all of these terms a was fixed you can choose any value of a you like but having done so it's fixed in my example I chose a hundred we've already ever established what the constants are so if you plug this in you will get a value of z which will be the preferred value of Z that will be the value of Z that gives you the stable nucleus if you did simplify this further you would find that this reduces to Z equals a over 2 as I say there's no need to do it because you can just fit in you just fill in the necessary values here in a spreadsheet and it calculates it for you but if you insist on going further you will get very roughly that Z is equal to a over 2 into 1 over 1 plus naught point naught naught 8 a to the 2/3 and if we examine that what you'll find is that when a is very small this term is very small and therefore this whole term is 1 so Z is a over 2 in other words if you've got a small number of nucleons a half of them will be protons and of course the other half will be neutrons and that's exactly what you find if you look at carbon-12 that has 6 protons nitrogen 14 has 7 protons oxygens 16 as 8 protons in each case it has half the number of protons of the overall nucleons but as a yet so larger this term will increase and consequently this fraction will be less than 1 if it becomes less than 1 that means that Z will be less than a over 2 less than half of the nucleons in a stable atom where a is larger we we protons and the remainder will be neutrons so that explains why there are more neutrons than protons when you get to the higher values of a now if we look at my now corrected semi empirical mass formula you will observe that the mass of the nucleus is a function of well what is the actual terms in some cases you've got constants where there is no Z dependence at all in some cases you've got single values of Z and in some cases you're going to get Z squares you're going to get these n squares from here you're going to get Z squares from here in other words you can say that the mass the nuclear mass is going to be equal to a formula which looks something like a plus B Z plus C Z squared right you could rearrange this formula so all the constants were brought together as a all the values of Z on its own were brought together with a constant B in front and all the values of said squared were brought together with a con with a value C in front that is a quadratic equation in said and it is of course a parabola what that means is that if you plot the parabola here I'm going to plot the mass-energy against different values of Z and what am I going to find I'm gonna find a parabola that looks like this now if this is an odd a that is to say the nucleon number is odd 117 79 no not even what you're going to find is that for all the values of Zed you're going to get different energies and those energies will fall on a parabola and what nature says is I want to occupy the lowest level so for any give a which in this case is odd this is the one this value of Z is the one that's going to give you that stable nucleus what happens if you create this one with a lower value of Z well then that will decay by dropping in energy terms to the lower energy one how does it do that it has to increase the proton number by one so it has to convert a neutron into a proton and that means you get electron or beta minus emission if you create this nucleus nature says aha I can go down a bit to a lower level by reducing the number of protons that means a proton gets converted to a neutron that means you get positron emission or beta plus emission so this way down the scale you're getting beta minus emission this way down you're getting b2 plus emission and nature insists that she will only be satisfied when she's at the stable level which is when Z is equal to this value for any given value of a and that of course means that although you can have any theoretical value for Z only one of them will be stable and only one or two on either side will be unstable what you will find is once you get too far away from this equilibrium position some of these nuclei won't even exist at all and we'll find out why in a moment now I said that this was for an odd number of nuclei which means it's an odd even or an even odd arrangement and in that circumstance you remember the pairing constant equals zero what happens if we've got an even number of nucleons so we're now going to do an even value of a if we look at this quadratic term up here the Delta term that features in the semi empirical mass formula is not as a dependent so that sits here so for an even even arrangement a will be higher than for an odd odd arrangement and what that does to the parabola is it simply says there will be two parabola depending on whether for this even number of of nucleons you can have an even number if you have even plus even or odd plus odd both will give you an overall number but what you find is that the even even parabola is lower than the odd odd parabola because it's the value of a in the quadratic equation which changes and all that does is to push the parabola up or down it doesn't change its shape so if I should have drawn to equal shapes and what you will find is that this is a value of Z on the odd odd parabola but you may get this and this on the even even parabola and then you'll get another value here and another value here on the odd odd right what happens when the nucleus gets to this value of Z is that the end of the story not a bit of it because nature says I can either go here or I can go here either by beta minus or beta plus emission in order to get to lower energy levels and consequently these two nuclei will be stable now you might say oh but what happens if one of them is a slightly greater energy level than this one it doesn't look it on my chart but it could be does that mean that this would decay to this well the answer is there is a theoretical possibility that it could but look in order to do that it has to jump not one proton but two protons because these are one two protons apart and that would require a double beta decay and once that's not impossible it's significantly unlikely so what nature does is it settles and it says okay both of those can be stable and that's how you can find two nuclei with the same even a number that have stability now you can also have almost the same arrangement this incident you just remind you is the mass energy versus said and once again we're having an even a so we can have even plus even neutrons plus protons or odd plus odd neutrons plus protons and once again we get the same parabola this is the even even parabola this is the odd odd parabola they're both exactly the same they're just displaced because of that Delta function which was the pair pairing function in these semi empirical mass formula and now you might have a Z value here on the even even parabola now that clearly is going to be the very lowest energy you can possibly get so that's definitely stable then you might get your next values of Z would be here so you get two elements there and then you might get other elements here now this one is definitely stable because this is certainly at the lowest energy level but what happens if you pop this one here this can get to decay in two ways it can either decay down to that one which is the most stable of all but it can decay to that one and this here can decay to the most stable of all or it can be k2 that one in both cases you're reducing the energy which nature likes but suppose you do Decatur this one will nature then say this one must decay to this one well no because we are now two protons apart again and again you'd need double beta emissions and what's not impossible is unlikely so now nature says on this occasion there can be three stable nuclei because in order to get to single one I'd have to have a lot of bb10 emissions so you can see that when you have an odd number of nucleons there's usually only one possible stable nucleus when you have an even number of nucleons you might have two stable nuclei or in some cases you might have three stable nuclei but all the rests will be unstable because by decaying they can get to lower mass energy levels so let's go back to where we started which was the neutron proton chart where we said this was essentially the neutronic was proton line but now we know why for stable nuclei neutrons equals protons at the early stage but we now know why there are more neutrons than protons as the nucleon number increases we also know now why there are unstable nuclei on either side because they are the ones that are further up the parabola in terms of energy and therefore they are going to decay until they get to the lowest mass energy so we know that but the question we want to ask ourselves is why is it all constrained in that single element in that single box if you like why is there nothing here why is there nothing here and why is there nothing up well let's take this region here first so let's fix a z-value okay and let's find these stable nucleus for that value of x sorry that value of Z if you put more neutrons in you get to a neutron rich environment in which the neutron will decay to a proton in order to get back on to the line of stability that we know but suppose you try and put so many neutrons in that you end up here then a new form of decayed becomes favorable energetically right nature says I want to operate at the lowest mass level and what happens is you don't just get beta decay you get Neutron emission you've got too many neutrons it becomes energetically favorable for the neutron just to come out and so what happens is you've got this element that you think you've created with XZ a you push that up XZ a but that is sitting up here so it's a very high number of neutrons and what happens is that one of those neutrons will be kicked out the protons aren't changed so it's the same element it still has the same number of protons but one of the neutrons goes so the a number drops by one and you get a whole Neutron emitted and why will that happen very simple the mass energy of this must be greater than the mass energy of this so the mass energy on this side is m1 the total mass energy on that sided m2 if n1 is greater than M 2 then Neutron emission will occur in this case and what you find is that the line that separates the position where you can have unstable nuclei but at least they exist and can be K into stupid into stable nuclei and a position where the nuclei can't exist because they will immediately emit a neutron and fall backwards you know if you emit neutrons you fall back down this line into the safety zone as it were anything in this region you're going to get neutron emission and the line that separates them is called the neutron drip line what about the other way why can't we have here a putner nucleus here well once again if we take a fitzy there's the stable value for the number of neutrons or if you like we can go that way that would be stable we've got far too many protons so now we can get what's called proton emission so we start with our element I'll take the same element and that will decay to and now we're going to decay by a proton so if you lose a proton you're going to change the element X will now change to Y because zyg will be Zed minus 1 and a will be a minus 1 because a proton has gone so that affects both numbers plus a proton and why will that happen because the mass energy on that side is greater than the mass energy on that side and if that's the case the proton is spontaneously emitted and so you won't ever get anything here because the proton just is chucked out of the nucleus because it's energetically favorable to do so and the line that separates the position where you have unstable nuclei too many protons but nonetheless they will decay to the position where the nucleus just can't keep hold of the proton that is called the proton drip line and finally as you might say why then is there nothing up here why can't the staple nuclei continue to glow with increasing values N and Z what you would tend to find is that when Z is about 110 and when n is about 160 roughly you won't be able to go beyond that and the reason for that is that now alpha decay is happening and alpha decay is the spontaneous emission of an alpha particle which is two protons and two neutrons so now you're gonna have the element which would be up here with Z protons and n neutrons but what it will do is quickly decay into a new element because the number of protons goes down by 2 the number of nucleons goes down by 4 because you lose two protons and 2 neutrons and you get a alpha particle which of course is just a helium nucleus being emitted or energetically you may even find that the on our destroy it the heavy nucleus because this is a nucleus with over 100 protons over hundred and 50 neutrons this heavy nucleus instead of just emitting an alpha particle of becoming a slightly smaller nucleus could well fission which means it becomes too much smaller nuclei and then very often Plus an alpha particle as well and why will either this or this happen it is because the mass energy of the left-hand side is greater than the mass energy of the right-hand side nature's desire to get to the lowest energy level means that if you're starting mass energy is greater than your finishing mass energy nature will contrive away either by beta decay or by Neutron or proton emission or by alpha emission to get to that lowest mass energy level and all of that came out of the now corrected semi empirical mass formula which whilst it certainly isn't perfect by any means but at this level it gives a very good indication of why you have only some stable nuclei why you have some unstable nuclei and why some nuclei just don't seem to exist or final point to make is that in general the greater the mass difference between the start point and the end point the faster the decay will happen so if you have a large mass difference the half-life for the that is the time for half of it to change will be very fast indeed

Channel: DrPhysicsA

Views: 63,748

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Keywords: Physics, Semi Empirical Mass Formula, Nuclear Binding Energy, stable, unstable, beta decay, alpha decay, proton emission, neutron emission, fission, nuclear physics, nuclei, nuclear force, nuclear potential, nucleon nucleon interaction, proton, neutron, nucleon

Id: oJaq8g8nFZU

Length: 82min 23sec (4943 seconds)

Published: Fri Jul 04 2014