 Hello everyone. So, so far we have discussed the radioactivity, the different types of radiative decays, the nuclear structure, stability and the nuclear models, mainly liquid drop model and shell model. Today we will start the details of the nuclear decays like alpha, beta and gamma. In today's lecture I will discuss the alpha decays. So, as we discussed in the introductory lecture, there are three types of decays alpha, beta and gamma. There are others also like spontaneous fission and so on, but mainly we will be discussing the three types of decays like alpha, beta and gamma. So, let us see the energetics. What are the types of energies that are involved when a heavy nucleus undergoes alpha decay? So, as you know already by the group displacement laws, during alpha decay, the atomic number of a heavy nucleus is decreased by 2 whereas the mass number is decreased by 4 because the alpha, alpha particle is coming out of the nucleus. Just an example, uranium 38 undergoing alpha decay to thorium 234. To calculate the energy of the alpha particle that is limited in alpha decay, we calculate the mass difference between the parent and the dot products. So, the mass of uranium 238 minus the sum of the masses of 234 and alpha and now if you write in terms of the atomic mass units, then you can multiply by 931 MeV or you can write in terms of the mass defect. So, mass defect is nothing but delta m equal to m minus a where m is the actual mass in atomic mass units a is the mass number. So, this into c square becomes actually the mass defect and the mass tables actually give the masses in terms of mass defect. So, what you see here is the 47.308 is the mass defect that means minus a into c square. So, when you have the alpha decay since the mass number is conserved, essentially it will give you the difference in the masses of the parent and the dot products. So, 238 uranium 238 this is the mass 234 thorium and alpha particle. So, if you see the difference between this is equal to 4.27 MeV that is the Q value of this alpha decay. Q value means the heat or the energy liberated in this process. So, it is a positive Q value this much energy is immediate. This energy is now shared between the alpha particle and the dot product that is 234. So, how it is shared how to calculate the energy of alpha? The basic concept is that since the 238 uranium is stationary when it is alpha particle immediate then the momentum of this nucleus is 0. So, when the it is written into two particles the net momentum should be again 0. And so, the momentum that is the linear momentum is mass into velocity. So, m alpha p alpha equal to m 234 p 234 that means the momenta of the alpha particle and thorium 234 are same they will be in opposite direction of course. So, the net momentum will be 0. Now, mv if you can convert it to m alpha m alpha e alpha if you if you take a square of this one like for example, mp you take square and 1 by 2. So, it will become m 1 by 2 mv square. So, essentially it becomes mass into energy. So, you can convert this relationship between the momentum equal to relationship between the mass and energy. So, m alpha e alpha equal to m 234 e 234 and you can see here e alpha upon e 234 equal to m 234 upon m. That is what I was telling that the Q value is shared between the two particles in the inverse ratio of their masses. That means, e 1 by e 2 equal to m 2 by m 1. So, you have two equations energy is equal to 4.27 mv and the ratio. So, you can now calculate energy of alpha equal to Q value into the ratio of the masses 234 by 238 that is 4.198 and energy of 234 thorium the other part is 0.027 mv. So, you can see here the lighter particle takes the major share of energy because energy is shared in the inverse ratio of their masses. So, lighter particle gives major portion of energy and the heavier particle takes the smaller portion of the energy. So, that is why you will see that sometimes now we say Q value is close to alpha because the thorium 234 is very very small energy. So, alpha particles that are limited now in this alpha decay have the energies of the 4 to 8 mv. Second important property of these alpha particles is the alpha spectra. So, when the two when a heavy nucleus undergoes alpha decay and if you detect this alpha particles in the detectors then what type of spectra we will see in the alpha spectra. So, here the even even nuclei that we are discussing like 234 238 uranium and 234 thorium they have their ground state spins 0 0 they are even even nuclei and so, mostly the alpha decay between the two even even nuclei will go from ground state to ground state and so, there will be only one alpha. So, what you see here is that this is a typical alpha spectrum counts versus energy of alpha and you will get a single line in the spectrum. Actually you should have got a small very thin line in one line because there is only one energy, but the detector had its own resolution which you will discuss in the relation detection measurements and so, because of that there is a broadening in the alpha spectrum. There will be in fact, a left hand tail. So, that this happens because of the instrumentation aspect. When we have in some many cases you will find that the apart from the ground state to ground state transitions there are also transitions to the excited states. Now, radium 224 the total product of thorium 228 had the excited states they are called the rotational states 0, 2, 4, 6 have just touched upon the collective model which is how nuclei can also have their collective motions of rotation and vibration and so, apart from the ground to ground transition 72.7 percent they can there can also be transitions to higher energy states, but they are hindered with respect to the ground to ground transitions and in such cases where there are multiple alpha particles emitted the alpha spectrum will be something like this. So, you have the main peak this is the high energy peak 72 percent and then you have the lower energy this energy. So, if you say 1, 2, 3 so, this will be 1, 2 and 3 energy is decreasing this way. So, from this data also you can make out how many alphas are being emitted in a particular radioactive peak. Now, let us see how the half-lives of the heavy nuclei depend upon energy of the alpha particles. The typical values of alpha particle energy or heavy nuclei which are emitting alpha particles goes from 1.8 MeV which is 144 nidinium one of the lightest nuclei to emit alpha particles and the almost the highest energy alpha particle is emitted in 11.7 MeV for a 412 polonium which is a metastable state of polonium 212 and you can see here the corresponding half-lives. So, if you have the alpha energy very low then the half-life become very high and for 15 years and when we have the alpha energy very high half-lives become very small and we will discuss this part in more details when we will see the how to explain the decay of alpha particle for heavy nuclei in terms of the penetration of a coulomb barrier. So, the higher the energy of the alpha lower is the half-life for the alpha decay just to see an examples different examples here all the even even nuclei of thorium, radium, redon again redon 220 polonium 212 and you can see the alpha energy is increasing whereas the half-life is decreasing from here to here. So, as the alpha energy increases the half-life decreases and the normal range for the alpha decay alpha energies is 4 to 8 MeV though there are exceptions like 11.7 MeV or 1.8 MeV but by and large in the mass region of 200 to 240 50 you will see the alpha decay alpha energies are of the order of 4 to 8 MeV and the half-lives accordingly go from microseconds to 10 power 15 16 years or so. Now, you must be observing here that why there are no alpha emitters having masses below 144 or in fact there are none. So, the question I will just put here why is alpha decay not observed in low and medium A B B I this is a question which must be coming to the mind of the students. So, the reason is that as you go down and down in terms of the mass number the Q value becomes very small if you recall the binding energy curve as a function of mass number B by A. So, when we are here every mass region let us say the 150 maybe somewhere here then alpha decay is happening like this a heavy nucleus is going to a lighter slightly lighter nucleus and alpha particle limited. So, the binding energy is increasing. So, when the binding energy is going down when the going up means mass is coming down and so Q value will be positive. But if you see in this region when you when there is alpha decay the binding energy is decreasing. So, Q values are negative. So, below mass number 150 you will find the Q values become negative or very low and the Coulomb barriers are high. So, then you will find because of this region very low Q value or negative Q value and the very high Coulomb barrier the leads to the not no alpha decay between for the heavy nuclear lighter nuclei masses more than 150. In fact, already around mass 150 half lives are in the range of power 15 or 16 years. In fact, very interesting correlations were obtained by Geiger and Nuttel they called Geiger-Nuttel law in 1911 at that time did not have substituted detectors to detect the energies of alpha particles. Instead they were measuring the range of alpha particles in different matters and what they found the range actually is how much distance the alpha particle will travel in a material. So, that is what they call the range and this range is actually related to the energy of the alpha. By let us say P is power Q where E is the energy or is the range like alpha particle travel in few centimeters in air. So, what they found that the four different natural radioactivity series that way the lambda is increasing with the range. That means, as the range increases the decay constant increases or the half life decreases and this range can be correlated with the E alpha. So, as the energy of the alpha particle increases the decay constant increases that means the half life goes down. So, the same thing is shown here for different isotopes of heavy elements like polonium, thorium and fermium the half life and Q alpha are inversely correlated. As the Q alpha increases half life goes down same correlation that you see from the bigger neutral. So, the half life of the alpha emitter will increase if the energy of the alpha particle decreases that means the main lower alpha and Q alpha higher half life. Now, to alpha particles why it is a problem that why this half life I cannot immediately the alpha particles. So, as you know alpha particle is a charged particle and heavy nuclei. So, the nucleus offers a potential. So, they are all in the attractive potential of the heavy nucleus alpha particle is formed inside the nucleus to come out of the nucleus alpha particle has to cross a coulomb area. So, this is a typical example that we have the attractive potential well or the nucleons or if there is a preformed alpha particle the alpha particle is sitting inside the nucleus in a attractive potential well. And now let us see the so, suppose the energy of the alpha particle is this order it is coming out to this energy, but there is a coulomb barrier. The coulomb barrier and alpha particle alpha energy we discussed of the order of 4 to 8 MeV and 4 to 38 uranium alpha decay we found it is 4.27 MeV whereas the coulomb barrier. Now, let us try to calculate the coulomb barrier. The coulomb barrier when they are like here know it is like the heavy nucleus and the small nucleus when they are in contact it will say like Z1, Z2. So, at contactor the coulomb energy of this system can be calculated by Z1, Z2 e square by R1 plus R2. And so, you can calculate this by this formula 1.4382 this is a term depends upon that this is a pre term for this formula Z1 in a charge atomic number of alpha particle Z2 atomic number of thorium and into so, this is R0 A113 plus A213. Now, why this 1.4382 has come this comes in terms of MeV. So, what I will try to see here that if you see e square. So, only terms here are the in the charge of the electron square upon the radius in centimeter e square by centimeter. So, ESU square little static upon centimeter and 1 ESU. So, it will become 4.8 10 to the power minus 10 square upon centimeter it will become arcs. So, this is what I have done here that if you put this value it will come in arcs and now you can put this in terms of this you can write 10 to the power minus 13 centimeter this is like the radius of a nucleus and then you convert into arcs into joule. So, 10 to the power minus 7 joule for arcs multiplied by that and joule to MeV it will be 1.602 10 to the power minus 13 joule for MeV 1 MeV 1.602 10 to the power minus 13 joules. So, like that if you do this put the different units and their actors the whole thing will come to 1.48. So, instead of now every time putting this numbers you straight away put 1.4 or you can even make 1.44 into 4 atomic numbers of the alpha and the thorium upon 1.4 this is r0 a1 1 third a2 1 third. So, this can be this will come out to be 23.2 MeV. So, you can see here a alpha particle of energy 4.27 MeV it has to come out of a coulomb variable which is in the potential well where the coulomb barrier is 23.2 MeV. So, classically it is not possible for a 4.27 MeV alpha particle to escape from a potential well of 23.2 MeV, but quantum mechanically yes alpha particle can penetrate this coulomb barrier and come out with a finite probability that is what is the problem of barrier penetration. So, by quantum mechanical tunneling one can in fact calculate the decay constant for the alpha decay of heavy nuclei. So, the decay constant for the alpha decay lambda is equal to the penetration probability for alpha decay into the frequency of the alpha particle. So, you see the alpha particle is formed here alpha particle is formed and it is striking the coulomb barrier. So, there is a frequency of striking at the coulomb barrier that is f and there is a penetration probability. So, with this is the alpha here it is coming to come out of this what is the probability that this alpha particle will penetrate and it is assumed that the alpha particle is already preformed inside. So, that means inside the nucleus there are clusters of alphas 2 protons and 2 neutrons combined inside the nucleus to form a alpha particle the probability is assumed to be 1 for even even nuclei. So, it is easily possible. Now, how to calculate this one these two factors. So, the frequency of striking the barrier is you can calculate based on the de Broglie wavelength de Broglie wavelength h cross upon mu v where mu is the reduced mass alpha t is the velocity. So, this is equal to de Broglie wavelength is close to the radius of the nucleus. So, because the distance that alpha has to travel inside the potential well will be close to the radius of the nucleus. So, this is the radius of the nucleus heavy nucleus. So, de Broglie wavelength equal to r 1 velocity. So, you can see the velocity of alpha particle in the nucleus will be you can put from here cross upon mu r 1. So, this is now you can find out probability the frequency v equal to n lambda n is the frequency. So, lambda is given and v is given you can find out the n. So, n the frequency equal to frequency v equal to n lambda. So, n equal to upon lambda. So, v is the velocity with terms of h cross by mu r lambda is equal to r. So, it becomes h upon 2 mu r 1 square. So, this is the frequency with which the alpha is striking the barrier and the penetration probability. So, the barrier penetration from can be obtained from by considering the wave function of alpha inside the nucleus in this region and outside the nucleus by calculating a transmission coefficient that is nothing but the transmitted flux upon the incident flux. So, there is a flux of alpha striking the barrier the flux of alpha going out. So, there are two three areas there are different wave functions in these three areas you can calculate the transmission coefficient we will not go into details, but the product of these two quantities will be the. So, there is a integration here in fact, from r 1 to r 2 v r potential at a function of r minus the t the kinetic energy to the power half. Now, essentially it is the area under this a this graph this curve. So, the curvature the height of the fifth coulomb barrier and the the just in the thickness of this barrier it determines the lambda. So, this has been also the integral can be solved here the formula h cross upon 2 mu r square the frequency into e raised to minus 8 pi z 1 z 2 e square by h v cross inverse t by b t is the kinetic energy b is the barrier the power half minus t by b. So, power half minus into 1 minus t by b 1 by half. So, this is called the Gamow-Gerlin Condon formula and it is in fact, this formula it will found to be very successful reasonably successful in predicting the uplives or the lambda values of the nuclei. We will just see here the decay constant values lambda calculated and the lambda experimental value in second inverse. So, right from N 144 to Parmium 254 alpha indices are given here which are sedimentally known their radii are known and you can see here that in terms of the order of magnitude 2.7 minus 24 1 minus 23 1 minus 6 5.8 minus 7 minus 8 8 minus 9 8.4 minus 9 7.8 minus 19 1.2 minus 18. So, you will find that this the barrier penetration formula of Gamow-Gerlin and Condon reasonably gives the prediction about the decay constant and therefore, the half-lives which are very close to the experimental determined half-lives. So, now this is very fine for even even nuclei the formula just now we derived or just now we saw the final form of it gives values close to the experimental values that is possible for the even even nuclei. See very close to that, but so why it is because the success of it lies because alpha particle is carrying no angular momentum. So, l value is 0 because alpha particle spin is 0 and the two states like 0 to 0 plus to 0 plus of union to 32 to 28. So, 0 to 0 transitions and with alpha having 0 spin easily you can display it without that formula just now we saw. So, S wave alpha particles there is no problem there is no hindrance. Once we come to odd A or ordered nuclei then what was found that the calculated lambda values are much more than the experimental values because there is a hindrance. What is the hindrance? One is the centrifugal barrier if the alpha particle has to carry certain angular momentum then there is a additional barrier apart from the coulomb barrier there is a centrifugal barrier that is l l plus 1 h cross square upon 2 mu r square is like a rotational energy of a system. So, that hinders the alpha decay. So, alpha decay half-lives are much longer or lambda values are shorter. So, because of this for example, here apart from the ground to ground transition there are decays to excited states to 2 plus 4 plus. So, the probability to populate 2 plus 4 plus state will be much lower than 0 plus state because now alpha has to carry 2 units of angular momentum here 4 units of angular momentum and wherever alpha has to carry angular momentum then there is a centrifugal barrier and therefore even for even even nuclei the hindrance factors go from right up to and for 4 because of the requirement of carrying angular momentum. Whereas for odd A nuclei and ordered nuclei even to the ground state because the now the spins of the heavy nucleus and the the daughter and parent are different like here i by 2 minus i by 2 plus and the excited states are also there. So, the in in the case of order mass or ordered nuclei not only that the there is a centrifugal barrier even the pre-formation probability. Now, the probability that the other particle will be formed inside the nucleus will be much less because for alpha to form a pair may have to break. So, there is an unpaired nucleon. So, because of that the nucleus is left in an excited state. So, invariably you will find that when there is an alpha decay from a odd mass nucleus or ordered nucleus population of excited state is more than the ground states like Amritschium 241 probability is more to this excited state and which undergoes gamma decay of 60 kV to the ground state of Natunium 237. So, what we saw that for the even in a nuclei the penetration formula can reasonably explain the half-lives of these many several nuclei, but for odd mass and ordered nuclei there are entrances due to the pre-formation of alpha and the centrifugal barrier. Therefore, the predicted half-lives are much longer than the calculated. The experimental half-lives are much longer than the calculated ones. Now, lastly we see the systematics of alpha decay because they are very useful to explain we need to know what is the reason. So, among the isotopes of an element the same atomic number, but varying mass number the Q value decreases with the increasing mass number. You can see here for a same element like polonium as the mass number is increasing the Q value of the alpha decay or the energy of alpha is decreasing and this trend is seen for all elements. So, you can see here also uranium 232 to 238 alpha energy is decreasing and this can be explained even by the liquid drop mass formula. Later on we will see the facility in the fission there you will see also the facility parameter the same happens with the fission half-life. In isobars same mass number and varying atomic number 2 alpha increases with increasing z you can see here 238 uranium 238 polonium isobars, but with increasing z the alpha energy is increasing and now cell effects the nuclear cell effects influence the half-lives. So, if the daughter product after alpha decay whatever daughter is formed having a magic number of protons or neutrons then because daughter has got a lower half-life when we say magic number means the nuclei are having more stability so masses are low. So, there the half-life the alpha energies long high and therefore the half-lives are short. Typical example polonium 212 z equal to 84 goes to lead 208. In fact, it is a doubly magic number so the masses mass of the lead 208 is much smaller and therefore q alpha is very high correspondingly the half-life is very short. And other aspect is if the parent is having magic number so the parent mass is low like 1 and equal to 126 though here z equal to 82, but the neutron cell effects more dominant here because it is a higher value. And so you will find that the again the half-lives are long because the q alpha is short. So, if the parent is a magic number then the parent mass is low the q alpha is low and hence the half-life is high. So, just see polonium 210 to polonium 210 138 days polonium 212 microseconds. So, this is the systematics the cell effects influence the half-lives of alpha decay and also the among the isobarth and isotopes of the half-life change we have just now seen. So, this is all about alpha decay. I hope I could give you a flavor of the alpha decay why alpha decay is not seen in the lower mass seen in the heavy nuclei all that and the systematics we can explain using the simple concepts. So, I will stop here in the next lecture and take the video. Thank you very much.