 Hello everyone. In the last lecture, we discussed the alpha decay and we observed that alpha decay is seen in the heavy nuclei, particularly the masses more than 200, but of course up to 150 to 200 also alpha decay has been seen with very very long half-lives. And then we also discussed that the half-lives for alpha decay would be explained by considering the barrier penetration formula. And also we discussed the systematics of alpha decay among the isotopes and the isobars of a particular element of isobaric chain. Today, we will discuss the beta decay. As all of you know that beta decay happens along an isobaric chain, because the mass number does not change in the beta decay. Beta decay could be beta minus decay, wherein the atomic number increases by 1 or it could be beta plus decay, wherein the atomic number decreases by 1. So, I had just given the schematic for beta decay. In beta minus decay, the atomic number increases by 1 and along with that a beta minus which is an electron and an antinutron is emitted. For example, sodium 24 emits a beta minus to give you sodium magnesium 24. Beta plus decay atomic number decreases by 1 emitting a positron and a neutrino. And this reaction has got a threshold of 1.02 MeV. So, why this condition has come? Why is there a threshold for beta plus decay? And why there is no threshold for beta minus decay? So, as you can see here that when the atomic number increases from z to z plus 1 by beta minus decay, there is one extra electron. In the example, you take sodium going to magnesium. Now, this sodium, so when it comes to magnesium, it is actually one electron more, but there is no electron gain. So, it will pick up electron. It should be, it will become actually positively charged. When sodium converts to magnesium, it will be positively charged. It can easily pick up an electron from the surrounding. So, one electron was emitted by sodium 24 and magnesium picks up an electron. So, in terms of electron, there is no loss or gain. But in case of beta plus decay, when the atomic number decreases by 1, for example, phosphorus goes to silicon, phosphorus 30 goes to silicon 30. So, from 15, it becomes 14. Then when a electron, see here is a, it is what is happening, a positon is coming out and there is a, so this atomic number has decreased. So, there is one extra electron, there is one electron is actually gone. So, it is one electron less than the parent because of that the atom loses an electron from the atomic cells. So, instead of, so because of that, there is a pair, a positon is going out and electron from the atomic cells is going out. So, because of this condition, the mass difference between the parent and daughter has to be more than the rest mass of a pair of electron positon that is 1.02 MeV. I hope it is clear. In beta plus decay, since the atomic number has decreased, the electron, the positon is anyway imitated for the electron which is accessed now in silicon 30, that electron also will be imitated. And so, the mass difference between the parent and daughter has to be more than or equal to the rest mass of a pair of electron positon that is 1.02 MeV. So, whenever the q beta value is more than 1.02, then only the beta plus decay is possible otherwise beta plus decay not possible and in such situation when q beta is less than 1.02 MeV, electron capture is the mode of decay. So, electron capture competes with beta plus decay when q beta is less than 1.02 MeV. Just to give an example, beryllium 7. So, it is electron capture. What does it mean? The nucleus captures the electron on the atomic orbitals. So, it could be k electron or l electron depending upon the q value for the reaction. So, electron is captured by the nucleus and you have here lithium 7 plus nutrient. So, this is in this process there are no particles but when the electron is captured by the beryllium, lithium 7 will have a hole in the k shell and there will be emission of gamma x rays from the atoms. So, electron capture is always accompanied by emission of x rays and also the OZ electrons. So, these particles that are limited in the beta decay, it could be electron or positron and also the butyno and antinutrients. Let us first discuss the energetics of the beta decay. As we have seen previously, the beta decay occurs along the isobaric chain say particular A. For example, it is odd A, then the lower z like this and tin antimony tellurium they are undergoing beta minus decay and the higher z's are undergoing beta plus decay to stabilize at the most stable isobar. And as you can see here the q beta values are decreasing from both sides. So, as the whatever nucleus is away from the stability has got a higher q value. One of the important observations of beta decay was that beta decay is beta spectrum is continuous. Just I have given an example here, the phosphorous 32 has a spin of 1 plus undergoing beta minus emission to sulfur 32 which has a spin of 0 plus. So, this gamma decay the beta decay taking place between two discrete states ground state of phosphorous 32 that of the sulfur 32. But then if you see the beta spectrum, the beta spectrum what is this energy of electron versus the number of electron or dN by dE. The beta spectrum is continuous, the blue one is beta minus and the red one is the beta plus. So, beta minus beta plus have a continuous spectrum. So, this cannot be explained simply if you consider the decay of a state of 1 plus 2 another state both are they are discrete. So, apparently there looks to be a violation of convention of mass energy. How to explain this mass the continuous spectrum beta plus beta spectrum we will see shortly. Another violation that appears to be here is the angular momentum convention. You see here phosphorous 32 is 1 plus sulfur 32 is 0. If only electron was emitted or then electron spin is half. So, 1 plus minus half can be half or 3 by 2, but the doctor product spin is 0. So, again the initial spin and final spin are not matching and apparently there is a violation of the conservation of angular momentum. So, this puzzle was actually solved by Wolfgang Pauli in 1930 when he proposed the neutrino hypothesis. You see it was a hypothesis. Wolfgang Pauli proposed that there has to be one more particle emitted along with the beta particle in beta. And that particle he named as neutrino. So, this is that another particle is accompanied with a beta particle. So, two particles are emitted along with the residue nucleus that is formed and this the properties that Pauli predicted it has to have a spin of half no charge and no mass. So, once you include one more particle neutrino having spin half no charge and new mass then you can explain the conservation of mass and energy as well as that of the angular momentum. So, you can see here when you have the another particle. So, there is no there are three particles now. So, a neutron is getting converted into positron plus electron for proton electron mind plus a neutrino. So, the q beta of this reaction is shared among the three particles and when the q is shared among three particles there can be infinite solutions. Therefore, see that the number that the energy of electron and neutrino can be very interesting. So, in this particular one when the energy of electron is very high then energy of neutrino will be low when energy of electron is very low energy of neutrino will be high. So, that is how you can explain the continuous spectrum of beta. Similarly, the angular momentum also can be explained now. So, you have the plus plus 30 to 1 plus charge state and electron goes out and a neutrino also is emitted both having spin half spin half. So, the angular momentum can couple in a way plus minus half plus minus half and so, it can be a combination of L and S1 S2 here S1 S2 electron and neutrino can give you 0 spin. For example, if you have 1 minus half minus 1 half can be 0. So, it is now possible to explain the conservation of angular momentum by spin half neutrino and it also possible to explain the continuous nature of the beta spectrum. So, this in fact, this neutrino was discovered in 1956 you can see here and the reaction was that in the reactors there are a lot of neutrino and there is a fission there is a neutrino in the beta decay there is a hydrogen. So, the neutrinos interact with the hydrogen to give you a neutron and a positron and a triple coincidence. So, you detect the neutrino in a detector neutron in a neutron detector and the positron will annihilate give you 2 511 kV gamma ray. So, the triple coincident between 1 neutron and 2 511 kV gamma ray is a unambiguous proof of the existence of neutrino. So, that is how the neutrino was discovered after almost 25 years of its prediction by Pauli. So, we can explain the energetics. Now, let us see how to explain the decay constant. So, that the theory of the beta decay was in fact, given by Enrico Fermi in 1934 and the first question it comes to mind is whether this electron that is emitted it comes from nucleus and how it comes. So, that is the question and we will see later on that actually it is not present in the nucleus, electron is not present in the nucleus, but it is generated whenever there is a weak interaction we will see that. So, let us see how it is not present in the nucleus because wavelength of the electron, if you have 1 mv electron coming out in beta decay and what is the wavelength? Lambda is given by H upon P momentum the de Broglie wavelength or P can be written as root 2 mE by mass and energy and if you substitute the value of the Planck's constant in Jolt second and then the mass of the electron and then the conversion factor for Jolt mev it comes to 10 to the power minus 13 meters. So, the wavelength of the electron having energy of about beta particles is much larger than the dimensions of the nucleus. Therefore, it is that the hypothesis is that if electron is present in nucleus is not positive. So, essentially the beta particle is created during the conversion of a neutron into a proton or a beta plus is emitted is created when a bound proton because the proton mass is less than that of neutron. So, proton cannot decay into neutron free proton, but a bound proton in the nucleus can be converted into a neutron by emission of a beta plus. So, these are the expressions give you how the conversion between neutron proton happening in the beta. So, these particles electron and positron are created when the one nucleon is converted into another in this process. Now, to calculate the decay constant we will go a little bit into the more details though I will not able to go to the complete theory of complete derivations of the expression for the decay constant, but I will give you a flavor by considering what are the major factors that Hermit took in the determining the lambda for the decay the beta decay. So, Hermit golden rule for any transition given by this kind of an expression here is only for is for the beta decay that is the probability that an electron is emitted with momentum between P e and P e plus D P e. That means P e to a small increment in a P e plus delta e delta P e what is the probability that electron is emitted and corresponding to that probability there is a electron in neutrino also emitted. So, when there is a D P e there will be D P nu also. So, N P D P e N P D P e is given by Fermi golden rule 4 pi square by h. The probability of finding electron at the nucleus psi e psi nu probability of finding the neutrino at the nucleus this is a matrix element between initial and final state for beta decay g is the g factor which Fermi introduced to characterize the weak interaction between the three particles nucleus electron and the neutrino. And in fact, this g is analogous to electronic charge in electromagnetic interaction. So, it is essentially a particle of weak interaction and then this is the density statistical factor which is density of states. So, when a say for example, a particular isotope is decaying by beta. So, it will be populating some states. So, what are the density of states in that nucleus suppose it emits gamma. So, there is the density of states D n by D e. So, when the electron is emitted and along with that neutrino is also emitted then the density of states in the residual nucleus will be given by D n by D e 0. And e 0 the total energy of kinetic energy of total kinetic energy equal to energy of electron plus energy of the. So, e 0 is shared in a infinite number of ways between electron and neutrino. So, let us see how we can derive the final expression for the decay constant. So, a little bit more details, the D n the density factor is nothing but the D n e into D n nu density for the electron and neutrino and it is you can see the phase space in momentum space for the electron and neutrino or pi p square dp by hq. These are the momentum space for electron and neutrino. So, it can be it is calculated to be 16 pi square by hq into e nu square by c square. So, you can say p equal to e by c or neutrino and D e 0 by c and so D e 0 because e nu is e 0 minus e e. So, since e e is constant you can say D e nu is D e 0. So, you actually try to eliminate the term corresponding to the momentum and energy of neutrino because what we are observing is the electron. So, p nu can be eliminated using p nu equal to e nu by c and so that is equal to e 0 minus e e by c and accordingly then D p nu will be D e 0 by c. From here D p nu will be D e 0 by c. So, you so final expression will be D n by D 0 will be the term momentum space term then p e square e 0 minus e e square into D p. So, we have now the terms in energy and momentum of electron and now we can make a small substitution that is called the in the relativistic domain you can write the relative momentum in terms of the momentum of electron and n 0 c relativistic momentum of electron. So, p e upon n 0 c similarly the energy can be written as energy kinetic energy of electron upon the S mass energy. So, it is the relativistic energy W. So, these two terms eta and W we introduce in to get this simplified version D n by D e 0 which is the the term corresponding to this 16 pi square m 0 5 c 4 upon h 6 into W square minus 1 W 0 minus W square into W d W where W is nothing but the relative energy and eta is the relative momentum and you can also find out that W square is nothing but 1 plus eta square or eta square is W square minus 1. So, finally you can see this expression actually the density term which does not explains the shape of the beta spectrum and W becomes 1 that means eta equal to 0. So, that corresponds to the 0 momentum of electron and W becomes W 0 then this term become 0. So, at both ends first end 0 energy term the end W equal to 1 and the highest energy term W 0 equal to W. So, the the bell shape spectrum of beta can be explained by this fermi d t theory. But the observation is that you see the bell shape is observed for beta plus but for beta minus the there is a at low energy there is a rise so more electrons of high low energy are observed than beta plus. So, this difference is there. So, this is explained by a Coulomb correction factor that means when the electron or positon is coming out of the nucleus electron is attracted by the nucleus where a positon is repelled by the nucleus in simple terms to explain. So, there are more high energy positrons than electrons there are less there are more low energy electrons than positrons. That is why the electron spectrum has a lower energy component more than the positron. So, this Coulomb correction factor was given by fermi it depends upon the atomic number of the nucleus and the energy and the energy W is the relative energy and it is given by 2 pi y upon 1 minus here is 2 minus 2 pi y where y is z e square by h v where z is the atomic number of the residual nucleus and v is the velocity of beta particle at infinity. And this term y is actually positive for electrons and negative for positrons. And if you substitute this value this plus minus y plus minus z e square by h v in this the factor here then we can explain the shape of the beta ionized spectrum where the minus spectrum is more having high low energy component beta plus spectrum has got high high energy. So, that is how the Coulomb correction term can explain the shape of the beta minus n beta plus. The fermi theory in fact got a boost when it was successfully validated by Curie plot. So, Curie in fact came out with an excellent idea of plotting what is called as the this factor let us see what is this factor. So, the in the again when I am saying momenta it is the momentum relative momentum eta. So, n eta d eta proportional to f z Coulomb correction factor eta square w 0 minus w we can write it from the previous expressions. And then you can rearrange this equation that n eta upon the f factor to the power half equal to a constant into w 0 minus w. So, actually it has to be v square term here. So, this term now if you plot this left hand side against w then you get this. So, i w nothing but n eta we can write in terms of w or eta w is related w square equal to 1 plus eta square upon the Coulomb factor eta square to the power half is was found to follow a straight line with the negative slope because of this negative term and very nicely these plots Curie plots in fact were used to identify if there are if there is a single beta or there are multiple beta particles. So, like if there are more than one beta group then you have this two linear paths there are two more than two you have three linear paths and so on. So, the Curie plot was very handy when the mass spectrometers were utilized to determine the electron spectrum in momentum space mass spectra will resolve them as per their momentum and when you plot the data this particular expression i w upon f z w then you find that the each beta group followed a straight line with the slope that you can find from this equation. So, this was the success of Curie plot that validated the Fermi theory of beta decay. Now, let us see how to compare the half lives for beta decay. So, again the same expression I am writing again in more details n w d w is the term which is came from the momentum space g factor the the transition matrix is square column correction factor and the w energy factors. Now, the decay constant essentially is the we discussed previously the decay constant lambda nothing but the probability of decay per unit time probability of decay of an atom per unit time. So, if you integrate this equation over from w equal to 1 that means energy from 0 to i s energy then you integrate this n w d w this expression. So, you get you can be final expression I am giving k a constant term m f i x square the transition matrix square f z comma w 0 and this you can lambda you can mention it 0.693 or l n 2 upon t half. So, t half into f is called f t in fact it is very popularly called as f t then that is f t is proportional to constant upon m f square where n is the matrix element for transition. So, you can see here f t is inversely proportional to m i x square smaller the value of log f t greater the value of f square essentially it means that if a transition is allowed it will have higher value of m i f. So, lower value of log f t is an indication of a allowedness of this. So, for allowed transitions log f t is varying from 3 and it goes on to higher values we will see it very soon. So, the selection rules for beta decay are in fact there were two selection rules one by Fermi because depending upon the spins of electron and neutrino whether they are perpendicular to each other or parallel to each other whether they are parallel or anti-parallel. So, if electron and in the Fermi selection rule electron and neutrino are imitated anti-parallel to each other. So, the spin component is 0 whereas in the Gamoteller selection rule electron and neutrino are imitated with parallel spin. So, the spin factor is 1. So, Fermi selection rule says i i equal to i f plus l, l is the angular momentum carried away by the electron and so s because s is 0, but Gamoteller selection rule say that i i equal to i f plus l plus 1, 1 is due to the spins of electron and neutrino which add up to 1. For allowed transitions we say l equal to 0 and there is no change in the parity because this is the s-way pattern particles. So, we can say l plus s equal to l i i minus i f to i i plus i f where i i and i f are the spins of the parent and daughter at set groups. Delta pi the change in parity is given by minus 1 to the power n. These are the selection rules. The super allowable transitions occur between the pairs of mirror nuclei. You can see from neutron to proton or trichium to glian3 or fluorine 17 to fluorine 17 to oxygen 17. You can see here that the orbitals of proton and neutron in the two nuclei are exactly same and the result of that delta i is 0, delta l is 0 and delta pi is 0 that means there is no change in parity and the log of t values are of the order of 3. Whereas in the case of allowed transitions the delta i delta i is 0, but the the proton and neutron need not necessarily occupy the same order. Therefore, delta i is 0, l equal to 0 and no change in parity, but the log of t values are higher than 3 in the range of 4 to 7. First forbidden now you can see delta i 0, 1, 2 and value 1, 1, 1 and there is a change in parity. So, log of t values are higher and second forbidden again delta i is 2 and these two no change in parity 10 to 13 log of t, third forbidden and so on. So, as you go to higher and higher forbiddenness there is a rise in the value. Just to give an example of this decay sodium 24 having spin state 4 plus decays to magnesium 24 and you see the ground to ground transition is forbidden because it is delta i equal to 4. Whereas the 100 percent transition goes to 4 plus state. So, 4 plus to 4 plus is allowed transition delta i is 0, delta pi is no change in parity. 4 to 2 see this second forbidden transition delta i is 2 and delta pi is no change in parity whereas there is no ground to ground transition. So, that is how we can explain that to transition and lastly just to before I conclude that I was telling about the how electron and neutrino are coming out. So, from the quark structure of nucleons a quark a neutron is UDD 1 up quark and 2 d quark a proton is UUD 2 up quark and 1 d quark and so conversion of a d quark into u quark happens during beta minus t. So, this is actually nicely explained by this Feynman diagram a neutron is converted into a proton in the process a W boson is emitted and then this W boson decays to a electron and a antinutrient. So, this is how in fact the advanced theory of beta decay in terms of the weak interaction whereby the neutron when the neutron is converted into proton then a W boson is emitted and in the process W boson breaks into electron and a and the forward backward acidity of beta particles in fact will observe that explain that in the beta decay or in the weak interactions parity is not concerned. So, this bosons W bosons and Z bosons in fact they are the actually the particles that are mediating the weak interactions in beta decay and you can see the masses of these particles they are very high of the order of G E V. So, this is in fact the ultimately all these interactions beta decay weak interaction explained in terms of the weak interactions wherein the W bosons or the Z bosons are involved in the process. So, I will stop here and next I will take the gamma t. Thank you very much.