Why don't all heavy elements decay to Fe56

hello today we're continuing in our series on nuclear physics and we're asking this very specific question why don't all elements which are above iron-56 spontaneously decay into iron 56 in other words why doesn't the gold ring or the gold necklace that you might be wearing just suddenly change into iron now you might wonder why it should so let me first put the case as to why you might expect that it should and then we'll establish why it doesn't we've already established in the video concerning the binding energy that nature likes a position where the mass energy to start with can be reduced and the way we showed that is we said that if you took Z protons and n neutrons individual neutrons and individual protons and put them as it were in a set of scales this is a thought experiment and then you take a nucleus which consists of Z protons and n neutrons but now bound in a nucleus what you find is that the mass energy of the mass of Z protons and n neutrons is greater than the mass energy of the nucleus containing those protons and neutrons and the differences in mass is converted into energy and that's called the binding energy it's the energy which is used to hold that nucleus together by which we deduced that nature likes to get to a low mass energy and we've also established that Nature has a way of doing this if we plot the famous NZ diagram which we've done many times now where we plot the Neutron number along this axis the proton number along this axis this line is the N equals Z line which is a purely theoretical line and what we said is that the stable nuclei initially have n equal to Z but then they divert off like this such that when you get heavier nuclei the number of neutrons exceeds the number of protons and we explained why and then we said that if you have a nucleus which is just above the line then in this case it's got an excess of neutrons and it decays by beta minus decay and that means that the number of neutrons decreases by one and the number of protons increases by one so a neutron is converted to a proton on the other hand if the nucleus is below the line of stability then you get beta plus decay where a proton is converted into a neutron so you go down by one proton and up by one Neutron if you're well above the line of stability then you decay initially by Neutron emission you just get rid of all the surplus neutrons and till you're into a position where you can do beta decay to get on to the line of stability and if you're well below the line of stability you initially proceed by way of proton emission get rid of excess protons until you're in the position where you can decay by beta decay to the line of stability and if you're anywhere up here then you can do alpha decay where you get rid of two protons and two neutrons in one fell swoop to try to get onto the line of stability but my question is having gone onto the line of stability somewhere here let's say here is iron-56 when you're on this line of stability why don't you decay to iron-56 and you might ask well why should we and here's the reason this is another graph that we plotted many times this is binding energy per nucleon so it's binding energy over a against the a number and you remember that the shape of this graph goes something like this it hits a peak and then it very slowly tails off and this peak is at iron-56 and this is the binding energy per nucleon so if we look at say an element here a heavy nucleus of large a the binding energy per nucleon is less than the binding energy for iron and let's remember what binding energy is binding energy is the depth of the potential well the extent to which the nucleus is bound so iron-56 is the lowest value binding energy in the sense is a negative value it get increases downwards so this is iron 56 and this that's that one and this whatever that happens to be would be a lower level but in other sense because it's a negative level it's it's further up so if you are here you can increase your binding energy by falling to iron 56 which means you go down in terms of binding energy to a deeper level and you give off energy Q so since that is exactly what nature wants to do it wants to go to a higher binding energy which means it goes deeper into the potential well the question is why doesn't this element which might be gold why does that not spontaneously decay to iron where it's mass energy will decrease and it's binding energy will increase in order to get to the lowest level which is essentially the highest value of binding energy in order that you get stability why do not all elements because all elements this side of the iron 56 have got a lower binding energy per nucleon and therefore if nature wants to get to a higher level of binding energy in other words deeper into the binding energy potential well it ought to get decay the same of course applies on this side of the line but we'll deal with that later this time we're simply looking at why these nuclei don't decart decay to iron 56 that's the case suggesting that they should but they clearly don't well we'll start off by asking the and making the case as simple as possible here is the and binding energy per nucleon diagram again and here is the shape dropping off an iron 56 once again is up here we'll start with the nucleus which has a nucleons and we'll ask why it doesn't drop to two nuclei each having a over two okay so our scenario is that we start with a big nucleus which has got a nucleons and let's say n protons in a neutrons and Z protons and we want to know simply why it doesn't just decay into two nuclei each having a over two nucleons and let's say that will be n over two neutrons and Z over two protons we know that it ought to want to do this because if it decays to a over two then there will be a higher binding energy and that means that this amount of energy per nucleon because this is binding energy per nucleon will be released so if you think about this this is typically going to be something of the order of seven MeV per um per nucleon this possibly will be 7 point 5 MeV per nucleon so if this is a is 200 and we're wondering why it doesn't split into two nuclei each of 100 where you would get 0.5 MeV emitted per nucleon and if you've got 200 of them that would be a hundred MeV that you could get out nature surely wants to do this why doesn't it incidentally I should make the point that if you split the nuclei into a half two halves then that those nuclei will be Neutron rich and they will have to get rid of neutrons but that's the second order process just to explain why they need to get rid or neutrons here is the line of stability here is the element with a very high a number it has a certain n it has a certain Z if you split those into exactly half so you now got n over 2 and that over 2 that will be here that's above the line of stability so the two nuclei that you create will have too many neutrons to be stable so they'll have to get rid of excess neutrons in order to get on to the line of stability but that will happen after you've split it initially we're just going to think about why it doesn't split in the first place and then if it wants to subsequently decay it can do so so we're going to think about the liquid drop model and we're going to ask ourselves the question why doesn't this large nucleus which is largely spherical form into two smaller nuclei now one of the things we do need to remember is that nuclear material is incompressible it is as dense as it's going to get so the volume of the large sphere will equal the volume of the two small via Spears okay you can't compress it so whatever that volume is this is the volume it's bit like a liquid drop if you take a liquid drop of water you split it into two drops the volume of the two drops is equal to the volume of the large drop okay and we know that the mass energy of this this one this here these two the mass energy of this is less than the mass energy of this consequently nature wants to get to this position the net then the mass energy is less because the binding energy is higher remember the binding energy represents the difference between the mass energy of the individual constituents of protons and neutrons and the mass energy of the bound nucleus so if the binding energy increases the mass energy decreases so this mass energy is lower than this and nature we think or to get there but we need to think in liquid drop terms how do you get there where you get there by initially deforming your liquid drop to this kind of shape this is moments before it breaks up and gets to a completely separate to separate drops and what I want to ask myself is what is the position when we are almost at the stage of having deformed this nucleus into this kind of shape just before it breaks up to become two separate nuclei and we go back to the video that I did on binding energy and if you remember we said that binding energy is made up of five terms there's a volume term this is all described I'm not going to go over it again but this is all described in the video I did on the binding energy of a nucleus the semi empirical mass formula we said that the five terms were a constant the volume term times a and then minus a surface term recognizing that nuclei near the surface are not so tightly bound and then there's a Coulomb term recognizing that the protons in the nuclei are repelling one another they are effectively acting against the binding energy so we have to subtract as a Coulomb term from the binding energy term and then there was an asymmetric term so we called that a sim n minus Z squared over a and that reflects the fact I hope you see that in - ed can I tip that so you can see oh gosh I hope you can a my Z squared and then there is a pairing term which is the fifth term which I'm going to ignore let's just assume that these nuclei are all even even so the pairing term just cancels out the question therefore is what happens to this binding energy as the nucleus deforms to get here we know that once the nuclei are separate the binding energy is Reiter binding energy per nucleon is greater in these two than it is here that was the chart that we were where if you show binding energy per nucleon against the a number you know that at iron-56 you get the maximum and binding energy so what we're essentially doing just to remind you is we're taking this chart the binding energy per nucleon against nucleon number binding energy per nucleon against a number and we're taking a nuclear nuclei with an a number a and we're asking why it doesn't split into two having a over two and this has a higher binding energy per nucleon than this there are the same number of nucleons of course because all the nucleons in here go into these two so the binding energy will increase the mass energy will decrease so we're asking the question why doesn't it do it and we are looking at this binding AG formula yeah and asking how do these terms change as you deform the nucleus so let's start off with the volume term for this and this well for this the volume term is simply a V times a what will it be for this this is just before the two nuclei break off so that you expect effectively you've got two conjoined spheres of lighter nuclei well the volume term for this is going to be there's going to be two of them so there's going to be two lots of av which is a constant times a over 2 and the twos cancel and so in the left hand corner as it were for the large nucleus you've got AV times a and for the two small nuclei you've got a V times a and that's hardly surprising because the volume doesn't change so the volume term doesn't change as you deform the nucleus no change to the volume term let's get this a symmetric term out of the way so in the left hand corner for the large nuclei you're going to have a sim into n minus Z all squared divided by a and in the right-hand corner the two smaller nuclei conjoined you're going to have to lots of a sim into n over 2 minus Z over 2 all squared divided by a over - well let's enclose the bracket so this 2 can come up here and multiply by 2 to give you 4 these twos can come outside but they're squared so 2 squared is 4 so you get 4 and the denominator takes that out 4 divided by 4 is 1 and what am I left with a sim times into n minus Z all squared over a which is the same as this so the symmetric term doesn't change the symmetric term remember reflects the fact that there are more neutrons than protons in the nucleus that doesn't change as you deform the nucleus now let's have a look at the surface what happens to the surface as you deform the nucleus well let's get these radii the radius of this is capital R the radius of this is little R that's the radius of one of the emerging nuclei so what are the volumes of those two those two elements of those two volumes the large nucleus will have a volume of 4/3 PI R cubed the small nuclei there are two of them so it's going to be two times the volume of each one which is 4/3 PI little R cubed but I said that nuclear volumes are incompressible so those have to be equal in other words the volume of these two must equal the volume of this okay so if they're equal I can cancel out the 4/3 pi on either side and I'm left with that R cubed is equal to 2 little R cubed if I take the cube root of either side I get that capital R is equal to the cube root of two which is one point two six times the cube root of R cubed which is R so the large radius is one point two six times the small radius the large radius one point two six times the small radius so what does that mean for surface area because that's what we're interested in for the surface term well the surface area of the large nuclei or nucleus is going to be 4 PI capital R squared the surface area of the two smaller nuclei are going to be well there's two of them so it's two times the surface area of one of them which is four pi little R squared but R is one point two six little 1/2 so we've now got that four pi times one point two six squared times R squared which is this here and compares with two times 4 which is 8 PI R squared what is one point two six squared call it one point five for cash so one point five times four is six so on this side we've got six PI R squared on this side we've got eight PI R squared this is the surface area of the large sphere this is the surface area of the two small spheres right serves Arab large sphere serves area two small Fears spheres the surface area as you deform the nucleus increases like this this surface area increases which means that this surface term increases and if this term increases because it's a minus the binding energy decreases so as the surface area increases the binding energy decreases and finally we need to think about the Coulomb term which is a C Z into Z minus one minus one over a to the one-third that is the Coulomb term for the large sphere what is it for the smaller spheres well there's going to be two of them so it's two into AC which is constant so that doesn't change into Z over to 0 over 2 minus 1 divided by a over 2 to the 1/3 so those are the two we're comparing this we can rewrite as 2 into AC Z over 2 and then I'm going to rewrite this a little bit here as Z minus 2 all over 2 divided by a to the 1/3 and that 2 to the 1/3 I'm going to bring up here well 2 to the 1/3 is the cube root of 2 we already know what that is is 1.2 6 times 2 now I can bring this 2 and this 2 here 2 times 2 is 4 I can bring that outside so that's now divided by 4 and what I've got is AC Z into Z minus 1 over a to the 3rd and in this square brackets I've got AC times Z into Z minus 2 over a to the 1/3 almost the same almost the same it's just as n minus 1 here and as n minus 2 here and when you think about it said is going to be of the order of 50 so you've got 50 times 49 plays 50 times 48 not a great deal in it what's going to matter is this fraction here this fraction is 1.6 times 2 over 4 which is nor point 6 3 so this is about 2/3 of this consequently the coulomb effect reduces and if I show you as just bring back the formula there is the Coulomb term that is going to decrease and that's the reason for that is pretty obvious as you deform the nucleus the protons within that nucleus are now being pulled further apart so the Coulomb term which is a 1 over R term is going to reduce so consequently the Coulomb effect comes down so what we've shown of the four factors in the binding energy form two of them don't change the surface term increases meaning the binding energy decreases the Coulomb term decreases meaning the binding energy increases so the question is which one of these two wins which is dominant now although it doesn't prove anything it's worth just noticing that AC so is the Coulomb value depending on how you measure it is approximately nor point-7 MPV whereas the surface term is approximately 18 MeV and although that doesn't prove anything in fact and it gives you an indication of what is true and that is that the surface term is dominant when it comes to the Coulomb term so although one is going up on one is coming down it's the surface term that wins so the surface term increase is greater than the Coulomb term decrease and that means that the net binding energy decreases because the surface term is increased and there's a minus sign here so let me just remind you again what the implication of a binding energy increase at binding energy decreases here's our scales is our nucleus and here are our Z protons and our n neutrons individual neutrons and these are the Z protons and the Z neutrons in a nucleus or if you like in two nuclei if you split them in half what this is telling you is that the difference is the binding energy so if the binding energy reduces that means that the mass energy must increase because mass energy of this is this minus the binding energy so if the binding energy decreases the mass energy of this nucleus increases so what we've just shown is that the mass energy of this is certainly greater than the mass energy of this that's why we think nature will want to decay but in order to do so it has to go through a deformed state in the liquid drop model and the mass energy of this is greater than the mass energy of this so if we draw that as a diagram here's the mass energy and this is in a sense a measure of deformation so we've got large nucleus two small nuclei mass energy of the larger nucleus is much larger than the mass energy of the two smaller nuclei we think that nature is going to take this downward course because that's what nature likes to do but it has to go via this deformed nucleus and that deformed nucleus has a higher mass energy so actually what the mass energy curve looks like is this it initially has a barrier and the height of that barrier is called the activation energy and this is what nature doesn't like nature wants to go to a lower mass energy and give off energy as it does so so it's got a deeper and higher binding energy but in order to do so it has to go over the barrier and there's no energy available this could be typically of the order of six MeV that's a large barrier so Nature doesn't have six MeV ordinarily and so it can't get over the barrier and that's why the large nuclei do not spontaneously fall into smaller nuclei and ultimately to iron-56 because they can't get over the barrier and that's why the gold ring on your finger will the sing silver ring of a silver piece of jewelry that you're wearing gold piece of jewelry that's why it doesn't before your very eyes break up into iron nuclei just as well really isn't it but of course some nuclei do break up and we'll find out why that is in the next video

Channel: DrPhysicsA

Views: 44,720

Rating: 4.8954606 out of 5

Keywords: Physics, Binding energy, nuclear physics, Liquid drop model, deformation, semi-empirical mass formula

Id: cMYJKjBHIiA

Length: 26min 28sec (1588 seconds)

Published: Sun Sep 28 2014