 Hello friends. So I am here with another question. So this question is from chapter nuclei. Alright, so let's 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. Okay. Assume that 1.5 into 10 is power minus 2% of natural water is heavy water. Okay. So it is given that 1.5 into 10 is power minus 2% of water is D2O. Okay. And all deuterium is used for the fusion. Fine. 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. Okay. So even though this is D2O, but then you should just look at how many deuterium nucleases are here. There are two deuterium nucleases out here. Alright. That can undergo a nuclear reaction. Okay. So what is this reaction? 2H1 plus 2H1. Okay. This gives tritium, 3H1 plus proton. Fine. Now the masses of tritium and deuterium and proton, these masses will be given in a particular question. Alright. 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. Fine. So mass of these two 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 the mass defect, which is what? Mass of tritium plus mass of proton minus 2 times mass of deuterium. Okay. So this is the mass difference between reactants and product. This if you find out using, you know, 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. Fine. And one reaction needs how many deuterium? One reaction requires two molecules or sorry, two nucleases of deuterium. Fine. It requires two 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. Okay. Now water, I can take molecular mass or the, you know, molar mass to be 18. Okay. Molar mass of the water is let us say 18 grams. Fine. 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. Alright. 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. Okay. So this into Avogadro number. These are number of water molecules in 1 kg of water. Alright. 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 10 to the power minus 2. Okay. So if these are the number of molecules of the natural water, I can just find out 1.5 into 10 to the power minus 2 percentage of this. That will be the number of heavy water molecules. Okay. So let me quickly do that. So 1.8, 1.5 into 10 to the power minus 2 percentage. So this divided by 100 into number of natural water molecules. Okay. So these are the number of molecules of heavy water. Fine. And 1 molecule has how many nucleus of deuterium? It has 2 nucleases of deuterium. 2. Fine. And we need only 2 nucleases for one reaction. Fine. So number of heavy water molecules will be equal to number of nuclear reactions. Fine. So these many nuclear reactions will happen. These many. Fine. 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. Fine. 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 to the power 6 and with charge of electron. Fine. 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. Okay. Thank you.