 Now, we have also learned that the mass of the nucleus can be written in terms of atomic mass unit okay, so we are talking about mass energy equivalence let me write it again okay. So we have till now we have been measuring the mass of atom or mass of the nucleus in terms of units U or you can write down as AMU okay. So we do not measure mass of nucleus or atom in terms of kgs it is units which we are using which is one-twelfth of C12 atoms mass okay. So if we are measuring the mass in terms of units can we find out how much energy is released if one unit is converted into energy okay. How much is one unit? One unit is roughly equal to 1.67 into 10 raise to power minus 27 kgs okay. So quickly find out how much energy will be released if one unit gets converted into energy. Now we are dealing with subatomic particles so you have to find energy in electron volts how many electron volts just divide that energy in with charge of an electron you will get in electron volts. So we are trying to find energy that is corresponding to one unit mass if there is one unit of mass that get converted into energy how much energy will be there we have to use is equal to mc square this is 1.67 into 10 raise to power minus 27 into c square that is 9 into 10 raise to power 17 somebody is saying anything yeah sir is it 9 into 10 correct so I am getting the same okay this if I divide it with charge of an electron which is 1.6 into 10 raise to power minus 19 I will get the energy in terms of electron volt fine. So just remember this value this is very useful energy will be equal to 931.5 mega electron volt fine so this is the energy corresponding to one unit okay. So if mass defect is 0.5 units then how will you find energy corresponding to that if mass defect is 0.5 units then energy will be what half of this so this energy is corresponding to one unit fine. So if there is a mass defect of whatever number of units you just have to multiply that with 931.5 MeV and you will get the answer okay. So let us do a numerical on this write down guys you have to find out energy equivalent of this oxygen O168 this oxygen you need to find its binding energy okay. What is given is this the mass of oxygen is given as 15 point you will be dealing with a lot of weird numbers so be patient about it you know because these are easy marks the concepts are very very simple those who get frustrated get irritated will lose the marks and there lie your chance to score over them. So this is mass of O16 one atom of oxygen mass of the neutron is 1.00 866 units and mass of proton is 1.00727 units although the electron are immaterial but then still you know the value of that let me give here you know the mass of electron is very very less compared to the mass of neutrons and protons this okay. Now you need to find binding energy of 16 oxygen 8 do it first step just like you have to find suppose you know bond energy in chemistry you do what you write a reaction right similarly here also you first write down the reaction anybody got mass defect first tell me the value of mass defect the reaction which we are talking about here so it's 0.1369 yes are you using calculator hmm naman yes i'm not good that's quite accurate so eight neutrons and eight protons they give you a nucleus of eight neutrons and eight protons so this is oxygen nucleus okay so the mass defect is what see this is mass of atom given it is it's not mass of nucleus that is given they're getting it so in order to find the binding energy first i mean actually speaking you need to first subtract the mass of all the electrons that are there although that is very less but in order to maintain the correctness this is the mass of oxygen nucleus okay the mass defect is when you subtract this from eight times mass of neutron which is 00866 plus eight into 1.00727 okay so this will come out to be equal to 0.13691 are you there you got it also the final answer 127.615 mev yes correct so the energy is what for one units 931.5 mev is the energy so for 0.13691 it will be this multiplied by 931.5 mev okay this will be around 127.5 mega electron volt this is the binding energy of the nucleus are you getting it okay so can you find out they just very similar numerical we are doing now i mean this chapter you need to be comfortable with calculation so i'm giving you another similar numerical you need to find binding energy of the nitrogen the mass of nitrogen atom is given as 14.00307 units okay mass of neutron proton and electron it is already given so what is the binding energy of the nitrogen atom or a nitrogen nucleus so is it 0.975 into 8 electron volts so you're saying 9 mev yes yeah just close to that others what is the mass defect you are getting 0.108 all of you getting this mass defect yes that is seven times mass of proton plus seven times mass of neutron minus mass of 147 nitrogen i'm ignoring the mass of electron here okay fine so once you get delta m just multiply it with 931.5 mev you'll get the answer fine let us move to the next concept now we are we have already got introduced to binding energy so we will discuss more about it so binding energy if it is more it it implies that the nucleus is more stable more energy is released okay but does it actually means or does it actually give you a good indication of how stable the nucleus is because it's like this you know that if I tell you that one basket has 50 apples other basket has two oranges and if I ask you which one is heavier like apple is heavy or oranges heavy then you can't just find the weight of the entire basket and compare isn't it you have to take one unit of apple and one unit of orange and compare their masses fine so similarly here also if my intention here is to find out how stable is one proton or one neutron inside the nucleus then the total binding energy is not the correct measure of it fine I need to find out a binding energy which is corresponding to each proton or neutron here it's like this you know that suppose there is a molecule that has multiple bonds okay so if I want to find out the bond energy of one bond then I should not take the bond energy of all the bonds isn't it so I need to find out the number of similar bonds divided by sorry I need to find out the total energy that is released when these bonds are getting formed divided by total number of these similar bonds fine then only I can compare how strong the bond is similarly here also when the nucleus is getting formed I can say that some sort of nuclear bond is getting created fine so rather than measuring in an absolute manner like the total amount of energy released what I should do I should find out how much energy is released per nucleon what is nucleon nucleon means right down nucleons are the particles inside the nucleus they are particles inside the nucleus okay so what are the particles inside the nucleus neutron and proton and neutron and proton when you add it up you will get mass number okay so when you divide binding energy with mass number you will get binding energy per nucleon so binding energy per nucleon is a very good measure to compare the relative stability between the two nucleons are all of you clear about it now okay so if I take different kinds of nucleases and try to plot their binding energy and if I have to compare it I need to plot what I need to plot binding energy per nucleon okay with atomic number okay so let us say this is my x axis and let me draw my y axis y axis looks not so straight so this is my y axis which is nothing but binding energy per nucleon so BE divided by A okay so this is measured in MEVs this is MEV and here on the x axis we have mass number let us say okay so here the mass number will be zero and then you will get mass number 25 then this is 50 then you will get 100 150 you get 200 and then you will get 250 okay so like this you have the scale on the y axis for the mass number on the y axis you have two MEV you have here four MEV then six and then you have eight and ten okay so when you plot the graph although my graph is not to the scale it is better plotted in your notebook sorry in your textbook so I'll just draw roughly this is the plot that is observed okay and and it is observed that it will have a peak it will have a peak at atomic number A sorry mass number A which is 56 okay what is mass number 56 which element it is any idea no idea this is iron okay so iron nucleus is considered to be the most stable fine although the iron will take part in chemical reactions but in chemical reaction nucleus doesn't take part in in it okay it is the outermost shell electron that takes part in the reaction the nucleus remains as it is intact okay so this is the stability of the nucleus that is plotted this is you can say around here it is a uranium at 239 238 let us say okay this is hydrogen whose mass number is very less than here you can see the lithium now you can say here you know that when you go from hydrogen to helium you can see there's a huge jump this much jump is there binding energy per nucleon okay but when you have sufficient amount of nucleons let us say from atomic number let us say 30 till atomic sorry I'm again and again saying atomic number it is actually the mass number sorry for that so if mass number is greater than 30 and less than 170 you can see that the binding energy per nucleon is remaining almost the same there is not much of a variation there okay so these are the few of the direct observations you can make from the graph all right so we'll first write down the observation which are there then we will also try to derive some conclusions based on the observation we'll try to see what can be infer from the observation okay write down few of the observations binding energy the first observation is binding energy is approximately of constant value for mass number between 30 and 170 okay