 So, we discussed alpha decay, the key points I can summarize once again all nuclei with mass number capital A greater than certain value somewhere around say 150 or so. They are energetically they can decay through this alpha decay thing because that product nuclei daughter nucleus and alpha particle rest mass energy of that is lower than the rest mass energy of the parent once this capital A is more than 150 or so. So, energetically it is favorable, but then there are many stable nuclei or very nearly stable nuclei beyond 150 and we saw that the key feature is that alpha particle forms inside this bigger nucleus of course, that pre formation probability is also there not that all heavy nuclei all already have alpha particle all the theory developed by gamma assumed something like that the pre formation is also something which will depend on the nucleus, but it is an overall lowering of this total probability of decay as far as the variations are concerned it does not have much effect and this alpha particle when it tries to come out of the nucleus it encounters the Coulomb potential barrier. So, that is the key feature that is the central feature this alpha particle when it tries to come out it encounters this Coulomb barrier the barrier height is around say 25 m e b or so and the energy of the alpha particle which one can obtain experimentally once it comes out that energy is somewhere around say 4 to 9 m e b that is the q value and this is that barrier and this barrier tunneling probability which is a purely quantum mechanic effect is so almost solely responsible for this alpha decay and this probability is very very sensitive with this q value if q value is raised slightly the width of the barrier to be crossed reduces considerably and the probability enhances. So, about 1 mega electron volt raise in this q value will lead to 100,000 times greater probability of this tunneling. So, that explains how the variation of q from say 4 m e b to 9 m e b leads to some 25 orders of magnitude variation in the life time from say giga years to micro seconds or so. We also saw that for even even nuclei the ground state of the parent is 0 plus and the ground state of the daughter is also 0 plus and so the alpha decay when it takes place from ground state to ground state at alpha is l equal to 0 angular momentum of that alpha particle is l equal to 0, but if it decays to higher states then you have l which is more than this will correspond to l equal to 0 this will correspond to l equal to 2 and so on. So, if alpha particle comes out with a non-zero angular momentum then this potential barriers is raised because of that centrifugal term l l plus 1 h cross square by 2 m r square. So, that will reduce the probability of tunneling and that will increase the life time. So, the branching what is the probability of going to ground state and what is the probability going to the excited states that probability will be guided by this barrier paneling tunneling with raised potential because of this angular momentum. This alpha particle when it forms normally these paired neutron and paired proton take part information of this alpha particle and therefore, for odd a nuclei it will be slightly energy less energy efficient because when pairs are combined to form alpha particle that last unpaired nucleon that is left in higher energy state. So, total q value available decreases for this daughter nucleus will have that raised energy because of that unpaired nucleon. So, that probability is less. So, all these things we discussed if you the parity should also be conserved. So, 0 plus 2 2 plus is all right because plus 2 plus parity change and l equal to 2 is positive parity. So, that is perfectly all right, but suppose you have some state where this is J i and this is J f daughter is J f daughter state where it is decaying is J f and the parent state from where it is coming is J i then the l which has to come that must be between this J i minus J f to J i plus J f and it has to also conserved the parity in strong interactions the parity is conserved. So, this has to be satisfied as well as parity has to be satisfied. So, for example, if you have say from 0 plus you are looking at transition to 2 minus suppose this daughter has a state 2 minus and the parent is in 0 plus and you are looking for this transition of alpha particle that will be forbidden because 0 to 2 if you use this inequality this l must be 2 and if l is equal to 2 and this parity is plus here and parity is minus here. So, that is not possible 1 minus 1 to the power 2 is plus. So, this is this transition is forbidden absolutely forbidden. So, these are some of the things that are interesting in alpha decay. So, now I move on to the next type of decay radioactive decay that is beta decay. So, the basics of this beta decay you know from a school days and also we have discussed a lot in the beginning about the basic processes which leads to this beta decay. So, once again that stability Z n curve you can bring in your mind and there are some specific combinations of a neutron number n and proton number z for which you have those stable dots black dots in that Z n diagram and if you create a nucleus or a nucleus is created somehow with a neutron number slightly more than that then it is possible that 1 neutron of that nucleus converts itself into a proton giving an electron. So, in that case this x a z if a neutron is converted into proton the proton number is increased by 1 and if neutron converts into proton it also creates an electron and a particle called antinutrino plus energy. So, that is one type of decay which we call beta minus decay another possibility is if proton number is more than what is required for the stability then one proton of that nucleus can decay can convert itself into a neutron in that case a x z will become y z minus 1 a and if that happens if a proton converts into neutron then you have E plus positron and this neutrino this is known as beta plus decay and then the nucleus in any real experiment or phenomena nucleus is part of the atom and you have atomic electrons. So, neutron this nucleus can capture one of these electron electrons from the orbits in that case this a z x will take up an electron and that will make z minus 1 a and plus neutrino this is known as electron capture both of these will occur if you have more protons then what is required for stability. So, one proton is trying to converts itself into neutron. So, here proton is converting into neutron emitting or creating a positron here proton is converting into neutron by combining with this already existing electrons. So, these are the processes. So, most of the time when proton number is more than what is required both these channels are open and depending on the situation certain probability for this type of decay and certain probability for this type of decay. Now, let us take the Q values. So, for this equation beta minus decay take this beta minus decay first beta minus decay for this how do I get the Q values from the known masses the Q value will be mass of the initial parent nucleus here that is mass of a x z. Let me write it n here this n is for nuclear mass we are talking of not atomic mass nuclear mass. So, Q value of this reaction is mass of this parent and minus mass of the product. So, minus mass of nuclear mass of this a y z plus 1 and then minus mass of this electron z plus 1 and then minus mass of this electron and minus mass of neutrino times c square. A neutrino mass is something very very interesting and lots of recent experiments have enhanced our knowledge about neutrino mass. We will talk about that later in somewhat more detail, but for the time being this neutrino mass is going to be extremely small this m nu c square will be say hardly few tens of electron volts. This is rest mass energy of electron 511 kilo electron volts which is 0.511 mega electron volts and this is only some 10 electron volt or 20 electron volts or 7 electron volts like that. So, we will just neglect this part. Now, if I write in terms of the atomic masses because atomic masses are which are readily available which are measured which are tabulated we can get the values of those atomic masses very from those tables. So, convert it into atomic masses this will be once again forgetting that atomic binding energy this will be atomic mass if nothing is written it is atomic. So, A x z and minus z times mass of electron this is nucleus and this is atom. So, atom is nucleus plus electrons there are z electrons. So, z times mass of electrons that if I subtract from the atomic mass I get this nuclear mass neglecting the atomic binding energies which are again few electron volts. Then here it is minus this is one part and then minus here it is other part mass atomic mass of A y z plus 1 and minus z plus 1 times mass of electron and then this is mass of electron and I neglect this mass of nucleus c square. So, now we can look at this electron masses you have minus z m e here and minus m e here. So, that makes it minus z plus 1 m e and here you have plus z plus 1 m e. So, all these electron masses will cancel out and you have only the difference between this parent nucleus mass atomic mass and then minus m A y z plus 1. Now, it is atomic masses that we are talking of time c square. So, to calculate that q value in a beta minus d k scheme you only have to look at the atomic mass of the parent and atomic mass of the daughter and subtract to get this q value. Now, the second equation beta plus d k we can do a similar analysis q value of so equation 2 which is beta plus d k. So, q value will be similarly nuclear mass of this parent A x z minus nuclear mass of the product. So, mass of the rest mass of the product m of this A y z minus 1. So, that is it minus mass of positron which is same as mass of electron and neutrino mass we neglect. So, this into c square and if I convert it into atomic masses this will be atomic mass of this A x z and minus z times m E. So, this is this one then minus this one will be atomic mass A y z minus 1 and minus z minus 1 times m E. So, it is this one and then minus m E. Now, look at the electron masses your minus z coming from here and plus z minus 1 coming from here. So, this plus z minus 1 and minus z will be minus m E this term and this term is minus m E and here also you have another minus m E. So, it is twice electron mass. So, that electron mass has to be carefully seen and the rest is difference between the parent and daughter atomic masses. So, it is m times A x z and minus m times A x z and minus A y z minus 1. So, this is parent atomic mass minus daughter atomic mass in case of beta minus d k we just had to stop here and into c square, but for beta plus you have 2 times m E c square. So, be careful if you have to calculate the q value of this beta d k the expressions are different for beta minus d k and beta plus d k. In beta minus d k you only look at the atomic masses of the parent and daughter and subtract, but for beta plus d k you also have to subtract this double of electron mass and it is quite significant because the beta d k q values are again typically in mega electron fuel mega electron volt range 1 m 1 m E v 2 m E v 3 m E v type and twice of m E c square will be more than 1 mega electron volt. So, it is quite sizable this q value and for electron capture if you do similar thing third equation electron capture. So, here the q value will be mass of the initial constituents. So, it is nuclear mass of A x z and plus m E that is the mass of this part left hand side and minus the rest mass of the product side. So, it is minus nuclear mass of A y z minus 1 neglect neutrino mass this into c square convert to atomic masses it will be atomic mass A x z and minus z times m E that is this part then plus m E and then minus atomic mass A y z minus 1 and minus z minus 1 m E times c square one more bracket somewhere. So, now if I look at the electron masses I have minus z and plus 1 from here and here this minus and this minus product is 1. So, z minus 1 m E. So, z minus 1 m E from here and negative of that minus z plus 1 m E here. So, they will all cancel out and you will only have difference between the atomic masses of parent and dot parent. So, if we compare these two same nucleus can decay through beta plus and electron capture because in both of these cases proton is converting into neutron, but then the q value of electron capture will be larger by this two times m E c square then the q value of this beta plus. So, you can find in many cases that this electron capture is energetically possible, but beta plus is not because q becomes negative there in electron capture it is q is positive. So, those nuclei will only go through electron capture and not through beta plus. So, these are the energy considerations. So, the most interesting observation about these beta decay process which puzzled physicists for two decades or so is the energy distribution of these beta particles. So, if I consider some beta decay then q values we have to consider have calculated this q is the available energy and this available energy this extra energy reduction in the rest mass energy this appears as the kinetic energy of the products as usual and here the products are the daughter nucleus the beta particle itself and the neutrino. Now, this neutrino and anti neutrino when I say neutrino here it could be anti neutrino it could be neutrino we will talk more about neutrinos separately as I said. So, consider for example, beta minus decay. So, you have this daughter nucleus and electron and anti neutrino these three particles share this energy. Now, this daughter nucleus require that energy take up due to this is very small because daughter nucleus is much heavier mass is very high as compared to the mass of electron or mass of neutrino. So, that with that part you can safely approximate to 0 and neglect that. So, this is now shared by electron and this anti neutrino. Now, how in what proportion it should be shared for that you need to go into quantum mechanics and we will be doing that right in this lecture itself. But during 1911 when this beta spectrum was measured by some physicist and in 20s or so that time this neutrino thing was absolutely not known. So, experimentally what people observed was only these electrons not these neutrinos or anti neutrinos and these electron kinetic energies were measured and it was found that this kinetic energy has a distribution. So, if I do not know about neutrino anti neutrino what I will expect is that q value is almost entirely taken up by the electron and therefore, electron kinetic energy should be very definite should have very definite value depending on the decay scheme the rest mass of parent rest mass of daughter and so on. And if the daughter nucleus is not in ground state it is in excited the state then the q values will be different the kinetic energies will be different, but then they will take some discrete values as we saw in alpha particle alpha decay. But the experimental result which I am trying to show you qualitatively this is from 64 copper. Now, 64 copper is a very interesting nucleus. So, it can go through positrons emission and it can also go through electron emission. So, here is a energy spectrum what do I mean this is the kinetic measured kinetic energy of electron. So, and this is the number of electrons. So, it is decaying to 64 zinc and plus beta minus and plus of course, in neutrino. So, this these electron these beta particles which are measured in nuclear detectors energies are measured. So, one can count that at this energy take a very small interval whatever is allowed by that detection system in that channel in that energy range d e at this e how many beta particles are detected in a given time. And the same time how many these beta particles are detected in some other d e centered at some other e. So, that is how this distribution is plotted number of electrons as a function of energy and that turns out to be of this type. I am just trying to reproduce this as nicely as possible. So, this is the scale in M E V's and this side the points are going like this somewhere here here here here like this and then decaying the last value is somewhere here. So, these are the kinds of points obtained in the experiment. The experiment is in physical review 76, 1949, 1, 7, 2, 5 and for this this beta plus decay when where it goes to 64 nickel and neutrino. Here also one can measure the energy in M E V and this side is number of these positrons. And here once again the you have that scale. So, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6 scale energy is the kinetic energy of the this beta particles in mega electron volts. And this side you have these numbers and here it goes like. So, it is somewhere maximizes here and then on this side it falls and it goes up to here. So, something of this sort. So, if you do not know about neutrino and try to understand what should be the kinetic energy of this beta plus or beta minus from these decays. You only have to calculate the rest mass energy of this parent and daughter and perhaps if it is positron, if it is beta plus decay minus twice of the electron energy and you can calculate this should be the Q value. And neglecting the kinetic energy taken by the recoil of the daughter nucleus this whole thing should come with this beta particle and you should have almost mono energetic type of spectrum or if there are excited states may be a discrete spectrum. So, this kind of continuous distribution in energy or in momentum was a great puzzle that is time and where is the missing energy. If electron is coming with energy much is less than capital Q many of the electrons are doing that or many of the positrons are doing that where is that missing energy and similarly momentum and similarly angular momentum and so on. So, this was a big puzzle that time when finally, Pauli suggested that perhaps there is a third particle a neutral very light small particle which is coming out together with this beta which is sharing the energy and later on that particle was named by for me as neutrino and experimentalists finally, searched that neutrino. So, that is fine the neutrino is taking the remaining energy if the Q value is this final point here is related to the Q value that we calculate. So, if the electron is coming out with smaller energy less energy than the Q value then rest of it is being taken by the neutrino. So, qualitatively the existence of this third particle can tell how this distribution is coming, but then the shape the functional form how much energy will neutrino take and how much energy will this beta particle take what are different probabilities for different compositions different proportions in which this energy is divided between neutrino and beta particle. So, for that one needs to develop a theory and once again in the early days of quantum mechanics this was a wonderful success very successful theory could be developed based on quantum mechanics which could almost reproduce this type of shape. So, that theory known as Fermi theory derived sometimes 1934 or so that I will try to describe some steps some features of that, but that also needs some quantum mechanical equations quantum mechanical formula and here the mechanism is very different from what it was in alpha d k in alpha d k the central point was that barrier coulomb barrier and tunneling through that barrier that was deciding the main features. Here for electron beta minus d k there is no coulomb barrier because beta minus is a negatively charged particle and the nucleus will anyway attracted for beta plus d k the it will it will repel the nucleus will repel and you will have some kind of coulomb repulsion, but there also the barrier is not the main feature the barrier crossing is very easy that probability is almost one. So, here the features are very different and there is no nuclear interactions electron or positron these are so called leptons and do not take part in nuclear interactions nuclear forces strong forces. So, here the situation is very different and the theory is also very different and that is in terms of what we call time dependent perturbation theory in quantum mechanics. What is that time dependent perturbation theory in quantum mechanics the main result is that if you have a system and the initial wave function is psi i and it makes a transition to this state where the wave function is psi f then the probability of this transition is given by this transition rate equation lambda 2 pi over h cross and then this is h i f square and d n by d e f. This is known as this is as for me golden rule this is the main equation of this time dependent perturbation theory. Now what is this h h is that interaction Hamiltonian this transition is taking place because of certain Hamiltonian because of certain interaction. So, that is here and h i f means psi f h psi i and this these brackets for space dependent parts it is just an integration of this psi f complex conjugate and then this h and this psi i over the volume of interest d tau. So, that is this h i f d n d f is something which I will talk more and this decides in fact the shape of this energy distribution known as density of states at final energy e f. So, first let me settle this part what should be this Hamiltonian and this is somewhat involved because you know this beta decay processes are because of the weak interaction and at a fundamental level if you think it is a conversion of one quark into another quark neutron you know it is d d u and proton you know it is u u d up quark and down quark down quark has a charge of minus 1 by 3 up quark has a charge of plus 2 by 3. So, when the conversion takes place at that level if you talk to a neutron to proton for example, then this d is to be converted into u and plus what you call exchange particle w minus and then this w minus decays into this electron and neutrinos. So, all these kinds of processes go on. So, writing Hamiltonian for that is an involved thing, but then what happens this interaction is indeed a very very short range this w boson which mediates this weak interaction that is some 80 giga electron volt rest mass energy or so interaction range is in few tens of picometers. So, one can take that as a point interaction in that case one can write we will just keep it here and we will just keep it here. Take this integration and assuming that is a kind of point interaction this Hamilton can be written there. This psi f and psi i look at this final state final wave function and initial wave function. So, here you have this parent nucleus. So, this psi i is the wave function of parent nucleus whereas, psi f you have three particles one is daughter nucleus one is this beta particle and one is this neutrino. So, if I write the this matrix element it is integration of psi f star psi f star will be psi d star and psi e star and psi nu star and then that h and this side it is only parent d tau. So, these are nuclear wave functions psi p and psi d are nuclear wave function and psi e and psi nu are electron and neutrino wave functions which you can treat to some extent as free particles they are not exactly free particles as neutrino is more closely free particle, but for positron or electron it interacts with this daughter nucleus through Coulomb interaction that is that will have its own correction term, but no nuclear interaction or that. So, first approximation at least treat that as a free particle. So, this psi e written as some volume nuclear volume you can take e power i p e momentum dot r over h cross free particle wave function. Similarly, nu 1 over square root of that volume e power i p nu mu mu mu mu mu mu mu mu mu mu mu dot r over h cross. Now, this quantity p e dot r over h cross or p nu dot r over h cross turns out to be small we can make an estimate of it. For example, here p e linear momentum of electron this which comes out of this decay if I take that kinetic energy of the beta particles say 1 m e v what should be its p cannot use half m v square is equal to k because the rest mass of energy of electron is 511 kilo electron volt something like 0.5 m e v and kinetic energy 1 m e v means double of the rest mass energy. So, one has to use relativistic expressions and for this the relativistic expression is kinetic energy. If you want to write kinetic energy will be total energy which is p square c square plus m naught square c 4 this is the square root of this. This is the total energy e square is p square c square plus m naught square c 4 this is the total energy and minus the rest mass energy this is kinetic energy. This is a relativistic expression of kinetic energy simple if you remember that basic equation e square is equal to p square c square plus m naught square c 4. If you remember this basic expression then this is the total energy and from the total energy we subtract the rest mass energy and the extra is kinetic energy. So, if I use numbers kinetic energy is 1 m e v here. So, this is 1 m e v and then this is take it this side plus m naught c square is 0.5 m e v. So, it is this 1.5 m e v n square this whole thing is equal to p e square c square plus m naught square c 4 is 0.5 square m naught square c 4 is 0.5 square that is 0.25 m e v square. So, this is 1.5 whole square is 2.25. So, where do I do the calculations? Let me do it here. So, 1.5 square is 2.25 m e v square keep track of the units also is equal to and what is this 0.5 m e v square right 0.5 m e v m naught c square whole square is 0.5 m e. So, that is 5 this is equal to p square c square and plus 0.25 m e v square. So, p e square c square is like 2 m e v square p e c is like square root of 2 m e v. So, p is say 1.4 m e v by c. So, if I look at this term p e dot r by h cross p e dot r by h cross and everything is happening inside the nuclear volume. So, r you can take something like nuclear radius which will be few femtometers. So, p e is 1.4 m e v by c and r is 1.4 m e v by c. So, it is a femtometer and h cross and h cross c this is about 200 m e v f m. So, you can see it is a very small number 1.4 divided by 200. So, 7 into 10 power minus 3 dimensionless of course, unitless. So, this is a small number and therefore, this exponential term can be expanded and you expand that you can write this as 1 over root v and then 1 plus p dot r over h cross and so on. Similarly, you can expand this psi nu and psi nu will be similarly, you can write this as 1 over square root of v and then 1 plus p e dot r over h cross and higher terms. It turns out that it is not a bad approximation if you just keep this 1 and neglect everything else here as well as here. So, it will be just reduced to 1 by square root of this volume and 1 by square root of this volume. So, if we do this what happens in this expression which this is the expression for this h i f this expression. So, this is daughter nucleus. So, this is nuclear wave function this is parent nuclear wave function these 2 it is just 1 by root v here and 1 by root v here. So, this will be 1 by volume coming out. So, now I can erase this and it is 1 by volume coming out this v is for volume and it is psi d star h psi p d tau. This is now only nuclear wave functions and this interaction which is a very short range interaction this you can write as m i f square m i f. And this is it is known as nuclear matrix element in make the assumption that this momentum dependence of lambda or energy dependence of lambda will not come from here. This is all in all that nuclear wave functions not electronic wave functions are not coming here. So, this electron momentum or electron kinetic energy distribution will not get affected by this. So, it is just a nuclear matrix element which is here 1 by v coming from here. So, this is only giving me 1 by v square this is just giving me 1 by v square and this is a constant factor. So, ultimately it comes out to this d n d f. So, let me talk on this d n d f what is d n d f? It is density of states at final energy E f. So, what does that mean? So, yes. So, this final state which is here to which you are looking for that transition from initial to state to final state. So, there is some energy E f for this final state, but then the available energies around that this is a system. In this system it will occupy some energy level, but then there are other energy levels which are available and it is occupying this one. So, how much is the density? That means, in an interval of say d e around that E f how many quantum states are there? So, that is d n. So, d n by d f. So, if I am treating this electron and neutrino as a free particle in that volume nuclear volume. So, it is confined in that nuclear volume and for a particle confined in a volume v, one can write the density of states easily. If you remember your quantum mechanics relation in one dimensional deep square well potential, the energies are given as n square h cross square y square 2 m l square n equal to 1 2 and so on. In terms of p, you can write this each quantum state is characterized by this value n and this is n h cross n then how much p h cross k in fact pi over l. So, this is p n and energy is in terms of that each n corresponds to one quantum state. So, pi by you can write this as n times h cross by l by pi. So, equal intervals of this quantity n going 1 2 3 4 you are getting this quantum states. If you go in three dimension a similar thing will appear. So, in three dimensional box also if you have a three dimensional box with length l there also the quantum states will be given by E equal to n 1 square plus n 2 square plus n 3 square each one for each dimensions h cross square pi square 2 m l square similar and in terms of p it will be characterized by these three numbers and here once again in three numbers you will have similar expressions and each quantum state will correspond to one set of these integers n 1 n 2 n 3. So, anyone of these going by one unit will give you one more quantum state. So, how to count first you have some energy E and corresponding to that energy you have some p and then you have to count that around that E if you construct E to E plus d E if this is the range E to E plus d E in that range number of quantum states that you have to count number of quantum states. So, we will take from here and we will tell how to make this calculation this these counting and once I get this d n number of energy states then we will be able to calculate this d n by d f and from there we will be able to find the distribution as predicted by this theory the energy distribution of beta particles stop here.