The most stable nucleus in terms of binding energy per nucleon...

Question

The most stable nucleus in terms of binding energy per nucleon is ⁵⁶Fe. If the atomic mass of ⁵⁶Fe is 55.9349 amu, calculate the binding energy per nucleon for ⁵⁶Fe, in joules. The mass of a hydrogen atom is 1.0078 amu, and the mass of a neutron is 1.0087 amu. (1 J = 1 kg･m²/s², 1 amu = 1.66 × 10^-27 kg)

Accepted Answer

Alright folks, welcome back. So here we have a naturally occurring scandia um it's going to consist of a single isotope which is candy in 45. So its atomic weight is 45. So the atomic mass of scanning 45 is exactly 44.95591. To your aim, you units atomic mass units. We need to calculate the binding energy per nucleons for this isotope. So the binding energy is actually going to be the change in the energy. Um and this is going to be equation, this is going to be the Einstein equation equals M C squared. Okay, so this is the equation we're going to be using now. This is the binding energy that we want to solve for dr t ah see we have because that is speed of light but um we actually don't have we don't have the masks. So we're going to go ahead and take some steps in order to find the mass. Alright, so first we need to find the mass defect. So the change in the mass um of the the isotope. So, if we take a look at candy um 45, right? Atomic Masses. It has an atomic number of 21. And what this consists of is 21 protons. Right, because its atomic number is 21 is going to have that many. A number of protons. And then how many neutrons do we have? We have 24 neutrons? They're neutral. Why do we have 24 neutrons? Because mass here is 45. Um And mass consists of a number of protons. And um neutrons. Right, so this is the number of protons and neutrons we have in one Scan diem nucleons. So we actually need to take a look at the masses of the protons and the neutrons. So since here we have 21 protons And one proton Ways. 1.007-8 Atomic mass units. We're going to go ahead and multiply the mass of one proton by 25. So here we're going to be solving for Just the atomic mass unit of this um isotope here. Okay, so we're going to go ahead and say we have 21 protons and there are masses are 1.007- Atomic mass units. Atomic mass unit. Okay, and then plus the number of neutrons we have 24 of them and their weight per one is 1. 867 atomic mass units. Okay, so what this will actually give us is 45 point oh my goodness, Come on. 45. 6096. So we're going to try to use as many significant figures as possible here. So this is atomic mass units of this isotope. Now we're going to go ahead and find the mass defect which is like basically the chain. The difference in the weight that we got here and the weight of this isotope that's given to us. Okay, so change in mass. This is the mass defect nasty effect. Sorry, my friend was having a hard time writing right now. Do you effect? Okay, so the change in mass here is going to be we're going to take the 45 that we just got 45.3606 Atomic Mass Units. Okay, hold on a second. Alright, so we're going to take the mask that we got from this calculation minus the the mass that is given to us, which is 44 . 912 atomic mass unit. Okay, so the mass defect here is going to be 0. 05 048 atomic mass units. So that is the mass defect that we just found. Now of course we can't take this mass here and just plug it into here. We actually need that mass to be in kilograms. So we need to convert. So this was our step one right here. This was step two. And now we're gonna go ahead and do step three, basically just converting the mask so that we can plug it into the Einstein equation. Alright, so this change right here, the mass defect . 048 Atomic mass units. And we have a conversion factor to go from atomic mass units two kg. So we are going to put amy on the bottom and then kilograms on top. So we know that the conversion factor is one a.m. U unit is going to be 1.6 Times 10 to the -27 kg. All right, so amy units cancel out There we go. And that will give us an answer of 6. 6227. And Times 10 to the negative kilograms. Now this is the mass per one nucleus or nucleons. Alright, so per one nucleus. And that's what we need. Alright, so now we're going to go ahead and plug this into the um to the Einstein equation. So we're gonna go ahead and do step four now which will be basically our last step. So step one was to find the atomic mass unit of this isotope using the number of protons, the number of neutrons that actually has then to find the mass defect, which is just the difference and the masses that we found between the mass that was given to us. Right. And then step three was to find or basically to convert that atomic mass unit into kg. And of course this was always per one nucleus. And then step three, we're going to go ahead and use this equation. So the change in energy or the binding energy equals to M C squared. Alright, so realize that the unit of jewels. Remember, it can also have units of kilograms times mass squared over seconds squared. Ok, so you're going to see why we need that. Alright, so the change in energy or the binding energy is going to equal to the mass that we found for one nucleus. So it's 6.72 7 to 6. 2 to 7. It's a lot of numbers. And so it is kilograms per one nucleus. Alright, so this is our mask that we found times the speed of light which is 3.00 Times 10 to the 8th and it is meters per second. And of course it's c squared. So this whole thing is squared. So when we find the answer to this, it's going to be actually in jewels per nucleus. Why? In jewels? Because here we have kilograms we have meters squared and we have seconds squared. Right? So whenever we multiply these two numbers together, these are the units that we get. So instead of writing those units, we're just going to plug in the jewels, Right? Because these are equivalent. So Delta E here is going to be 6. 36 times 10 To the -11. And it's jules still per one nucleus. Right, Alright, So these are not the correct units that we want. We want it in mega electron volts. So we're just gonna go ahead and convert. So this is going to be our last conversion. Last calculation before we get to our final answer. So jules will go on the bottom and then mega electro volts on top. So in one mega electron volt There are 1.602 Times to the -13 jewels. Okay, so jules here, we can cancel out. And we're going to have mega electron volts per one nucleus. And that's exactly what the question was asking. So let's go ahead and divide. Make sure you put all of these numbers that are in scientific notations um into parentheses when you put them into the calculator And our final answer here will be 377 0.88 mega electron volts. Okay, you don't have to write per nucleus. It's just implied here. That's exactly the number that we found. Okay, so this is going to be the final answer. 3 77.88. Alright folks, so as you can see, we had quite a bit of calculations here, but we knew that the binding energy was these delta H from the I'm sorry delta P from the Einstein equation. We have to find the mass on the mass is the mass defect. So we had to calculate basically the difference between the isotope um masses, isotopic masses here, Step three was just converting from a mule units two kg. And then finally stepped forward. Able to use our equation but of course that um answer was in jewels per nucleus and they wanted a different units. So again, we were doing a conversion. So not too difficult. It's just a lot of steps, I hope you're able to follow along. So um let's go ahead and say that a here is going to be our correct answer, right? It's the closest one that we got um, to to any of these. Alright folks, we'll see you in the next problem.

Struggling with General Chemistry?

Watch next

Nuclear Binding Energy practice set