 So here is another question. This question is from radioactivity chapter. So let us see what this question is about. Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme. Now what is this decay scheme? There are two reactions that are given here. Nitrogen whose mass number is 12, it decays into C12 but the C12 is in excited state right now. So there is a star mark on C12 plus a positron and a neutrino. And then this carbon in its excited state, it goes into the ground state. So this nucleus is in excited state. This is not electron. So nucleus which is in excited state goes to the ground state and it is accompanied with emission of gamma particle. So this is what the scenario is and let us see how we can visualize it and try to solve it. So ideally what will happen here is that this mass, mass of C12 positron and let us say this neutrino, some of these three masses will be less than nitrogen 12. So whatever is the difference in mass, that mass will get converted into energy. So this is the energy of C12 positron and this neutrino. Now we want to find the maximum kinetic energy of beta particle here. This energy that is emitted will be in the form of kinetic energy of these three particles. Now neutrino is like almost massless. So we can just ignore whatever energy it carries and C12, we can say that it is very heavy. So velocity is very less. So maximum amount of energy will be carried away by this positron. And if we say that entire energy is carried away by this positron, then I can say that that is the maximum possible kinetic energy this positron will have. So how much energy will be released in this particular reaction? The energy that will be released will be equal to mass of nitrogen minus mass of C12. Now this C12 mass in its excited state will be different from carbon 12 mass in its ground state. This into 931 MeV. So when I am finding mass defect, I am ignoring mass of positron and neutrino because they are very, very less. Now mass of 12 nitrogen is given but mass of C12 in its excited state is something which is not known. Now do you know what is mass of C12 in its ground state? It will be exactly equal to 12 AMU. That is how the atomic mass unit is defined. It is one-twelfth of C12 atoms mass. So C12 mass will be 12 AMU. Now if you look at the second equation, you can see that 4.43 mega electric volt is the energy of this gamma particle that is emitted in this particular reaction. So there must be some amount of mass of this carbon in excited state that got converted into the energy of the gamma radiation. So we can write this down in equation like this. C12 mass in its excited state minus mass of C12 in its ground state this multiplied by 931. Why I am multiplying again and again by 931? Because one atomic mass unit is equivalent to 931 MeV of the energy. So this is equal to 4.43 MeV. So mass of C12 in its excited state is given as 4.43 divided by 931 plus mass of carbon in its ground state which is 12. So now we have mass of carbon here. So this we can say that equation number one. So if we substitute everything in the equation, let us call this equation number two. What we get here is that mass of 12 nitrogen which is given 12.018613 minus mass of carbon in its excited state that is taken from equation number one. So this minus 12. Actually here you can in this equation number two, you can also take into account the mass of positron to be more precise. So like that you can write here mass of positron. Mass of positron is basically equals to the mass of electron. It is mass wise equivalent to the electron but charge wise it is opposite. So this multiplied by 931 MeV, this will be the answer. Mass of positron in terms of AMU will be mostly given to you in a particular question. So once you substitute it here, you will get the answer. So this is how you solve this particular question. So I hope you have learned something today and in case you have any doubts, please feel free to get in touch with us. Okay friends, so here is another question from Nuclear Chapter. So let us see what this question is all about. So the masses of C11, this is a carbon element but the mass is 11. So this must be an isotope of carbon. C11 and B11, the masses are given in terms of AMU. We have to find maximum energy a positron can have in beta plus decay of C11 to B11. Now first of all, we should be writing the equation. So what should be the equation here? Carbon atomic number should be 6 and mass number for this isotope is 11. This should disintegrate and give boron. Now you have to be very careful here because the boron atomic number is 4. So it will emit 2 positrons. It will not emit a single positron. It will emit 2 positrons. This is where most of the students get it wrong. When they read the question, they are in a hurry to find the answer. They just consider that only one positron is emitted and then they get a wrong answer. So along with 2 positron, a neutrino will be emitted. Now the amount of energy that is released in this particular equation or in this particular reaction will be what mass of C11 minus mass of boron minus 2 times mass of a positron. Now mass of a positron is same as mass of electron. So you can use mass of electron over here in terms of AMU and once you substitute it, you will get the mass defect and that once you multiply with 931, you will get amount of energy released in this particular reaction. Now this energy that is releasing here will be distributed among these three elements boron, positron and neutrino. Now we need to find maximum energy a positron can have. So if we imagine a scenario in which all the energy is taken away by the positron only, then that is the maximum possible energy positron will have. So whatever answer you will be getting here, delta E will be the maximum energy a positron can have in this particular reaction. So that is it for this particular question. I hope you have learned something today. I am here with another question. So this question is from chapter nuclei. So let us straight away jump into the question. We need to calculate the energy that can be obtained from 1 kg of water through the fusion reaction. So this is the fusion reaction that is given. Assume that 1.5 into 10 is power minus 2 percent of natural water is heavy water. So it is given that 1.5 into 10 is power minus 2 percentage of water is D2O. And all deuterium is used for the fusion. Now whenever such question comes, we should not look at the compound as such. We should just look at the nuclei because this is a nuclear reaction that is happening. So even though this is D2O, but then you should just look at how many deuterium nucleases are here. There are 2 deuterium nucleases out here that can undergo a nuclear reaction. So what is this reaction? 2H1 plus 2H1, this gives tritium 3H1 plus proton. Now the masses of tritium and deuterium and proton, these masses will be given in a particular question. So I will assume that masses are given. So in this one reaction, how much energy is liberated? You can find out using mass defect. So mass of these 2 deuterium should be less than, sorry, should be more than the mass of these. So mass of the product is less than the mass of reactant. So whatever is a mass defect, which is what? Mass of tritium plus mass of proton minus 2 times mass of deuterium. So this is the mass difference between reactants and product. This if you find out using atomic mass units for a particular atom, then this into 931 MeV will be the energy liberated in one reaction. So this is the energy in one reaction. And one reaction needs how many deuterium? One reaction requires 2 molecules or sorry, 2 nucleases of deuterium. Fine, it requires 2 deuterium, fine. Now I need to find out in 1 kg of water, you know, how much energy will be liberated? So first I need to find out how many deuterium nucleases are there in 1 kg water. Now water, I can take molecular mass or the, you know, molar mass to be 18. Molar mass of the water is let us say 18 grams. So in 1 kg, how many moles of water molecules will be there? That will be 100 divided by 1000, sorry, 1 kg has 1000 grams, 1000 divided by 18 moles. These many moles are there. So how many molecules of water are there? You can just multiply number of moles which is 1000 divided by 18 with Avogadro number. So this into Avogadro number. These are number of water molecules in 1 kg of water. Now how many deuterium molecules will be there? We know that the percentage of deuterium or the heavy water is given which is 1.5 into transfer minus 2. So if these are the number of molecules of the natural water, I can just find out 1.5 into transfer minus 2 percentage of this, that will be the number of heavy water molecules. So let me quickly do that. So 1.8, 1.5 into 10 ratio power minus 2 percentage. So this divided by 100 into number of natural water molecules. So these are the number of molecules of heavy water. And one molecule has how many nucleus of deuterium? It has 2 nucleases of deuterium, 2. And we need only 2 nucleases for one reaction. So number of heavy water molecule will be equal to number of nuclear reactions. So these many nuclear reactions will happen. These many. So if these many nuclear reactions are happening, so how much energy will be liberated? This multiplied by whatever energy you have found out for one reaction. This is the entire energy. Now this energy you will be getting in MEV. You can convert this into joules by multiplying it with 10 ratio power 6 and with charge of electron. So like this you can find out the energy in terms of joules that will be liberated by 1 kg of water using this particular nuclear fusion reaction. Thank you.