Understanding Nuclear Reactions and Binding Energy

The energy of nuclear reactions and nuclear binding energy are going to be the topics of this lesson. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. In addition to high school and college science prep, we also do DAT, OAT, and MCAT prep as well. I'll leave a link in the description below for where you can find those courses. Now, this lesson is part of my new General Chemistry playlist. I'm releasing several lessons a week, at least for a couple more weeks for the rest of this school year. And if you want to be notified every time I post a new lesson or when I start posting my next playlist, then subscribe to the channel and click the bell notification. So the energy of nuclear reaction is where we're going to start out here, and we're getting into a little bit of math, and I'm going to warn you here, the math here is not exactly fun. I'm going to give you that heads up. But the subject matter is pretty fascinating, so hopefully that'll kind of balance it out here. But it turns out Einstein's the one who kind of figured this out. But in terms of where all this energy is coming from in a nuclear reaction is that mass is actually being converted into energy. And Einstein's famous equation, E equals mc squared, or in this case we'll use delta mc squared, That's how you calculate this kind of mass-energy conversion. You plug the mass in right here, and it spits out how much energy it gets converted into. And C here stands for the speed of light, 3.0 times 10 to the 8th meters per second. That is a huge number. And so when you take a little bit of mass, but you multiply it by the speed of light squared, you get a lot of energy. And that's why it only takes a little bit of mass converted into energy to generate a ton of energy in these nuclear reactions. And so that's why, like, you know, in a nuclear reactor, we don't have to have a lot of fissible material here. to get a lot of energy harvest out of it. All right, so delta M here is called the mass defect, and if you take the difference in mass between your products and your reactants, that difference, because your products are always gonna weigh less than your reactants, is called the mass defect. So we'll find out this mass defect has another definition here in another context with nuclear binding energy towards the end, the second half of this lesson. All right, so if we take this difference in the masses, we'll get that mass defect, and then we can substitute it right into this formula. Now you might be like, well, Chad, we have a problem. If you look at those mass numbers, 235 plus 1, 236. 142 plus 91 plus 3, 236. There's no loss in mass, Chad. And again, we stated this in an earlier lesson in this chapter, but the mass numbers are going to balance. But if you take the exact masses to several decimal places, and I'll put a table up here of the exact masses for all these particles. If you take that to several decimal places, you'll find out, no, they don't actually balance perfectly. They're going to be off by just a little bit. And again, that difference in mass between the products and reactants, that's our mass defect. So if we take a look on that table here, let's put these up on the board. We got uranium here is listed at 235.04393. Neutron is 1.00866 AMUs. We got that over here as well. times three, we got barium right here, and barium's given as 141.91643. And then finally, krypton-91 is given as 90.92345. And now that we've got these exact masses, we can calculate the difference again between the reactions of products to get this mass defect and then substitute it right into here after converting it to kilograms as we'll find out let's take this difference here so we got 235.04393 plus 1.00866 i think i might need my hands for this all right then we're going to subtract off and i'll put this all in parentheses 141.91643 plus 90.92345 plus three, and keep in mind we got three neutrons, so three times 1.00, whoop, that's two points, 00866, and close my parentheses, and we're gonna get 0.18673. And technically it's customary to actually do products minus reactants that this comes out negative But the truth is we're just going to plug in the absolute value into this formula anyway, so but that's our mass defect Unfortunately, this is in atomic mass units And so in this case, which nowadays is also symbolized with just the lowercase funky u there italics u It's just an atomic mass units and again for this to come out in joules right here Your delta m has to be in kilograms, and your speed of light has to be in meters per second, because a joule is a kilogram meter squared per second squared. So we've got to convert this to kilograms. And that conversion here is that 1 amu is 1.67 times 10 to the negative... 27 kilograms and this is usually something that is provided. So if you kind of look at where this number comes from, so there's a connection between amu's and grams and the connection is Avogadro's number because it turns out there's 6.02 times 10 to the 23rd amu's is equal to one gram. And so if you take a look here well then you know that's equal to one gram. Well, then how much, how many grams is one AMU? Well, just take the inverse here. So one divided by 6.02 times 10 to the 23rd, it's going to get you 1.66. And like I said, it rounds up, I should have made the 6.022, but 1.67 times 10, that'll give 24 grams. And if you make that kilograms, 1.67 times 10 to the negative 27 kilograms, that's where that comes from. All right. So, and we'll get that there and let's do some plugging and chugging. So we got .18673 times 1.67 times 10 to the negative 27. And we're gonna get 3.11, let's just keep all these decimals, 3.118391 times 10 to the negative 28 kilograms. And now we can plug that back into equals MC squared here. Alright, so times 3e to the 8th squared, and we're gonna get 2.8. And like I said, we'll keep some sig figs here. 2.8065519. Cool. So 2.8... 065519 times 10 to the minus 11 joules. And you might be like, uh, I thought you said this gives off a lot of energy. Well, again, this is the amount of energy for just one uranium nucleus. So notice, what if you had a whole mole of uranium nuclei? Well, if you had a whole mole of these, that would weigh 235 grams, not AMUs. And so for a whole mole, how much energy would that be? Well, for a whole mole of this uranium, again, a 235 gram sample, less than a kilogram of it, a quarter, less than a quarter of a kilogram. How much energy would that be? Well, this is for one uranium. nucleus, for an entire mole, you got to multiply by Avogadro's number. So if we multiply this by 6.02 times 10 to the 23rd, again, not what the question is asking for, but if it did, so it's 1.69 times 10 to the 13th joules. It is a huge amount of energy for less than a quarter of a kilogram of uranium. This is where you see that energy payoff. So for one atom, you'd be like, that'd look like a tiny number, but for an entire mole of these nuclei, so it is a huge amount of energy when you look at it on that basis. It is ginormous. And again, you can run a nuclear power plant for a year off an amazingly small amount of fissile material. It's pretty phenomenal how much energy you can really get out of this. All right, so now we're going to talk about the nuclear binding energy, and it's the energy associated with what holds the nucleus together. Back in the first lesson in this chapter, we talked about that strong nuclear force that actually holds the protons and neutrons together in that nucleus, and they call it that strong nuclear force because it is stronger than the electrostatic force that repels the protons away from each other. And so it turns out it only operates at very short distances. At longer distances, it kind of evaporates almost to nothing, essentially. But at very short distances, it is excessively strong or exceedingly strong, especially compared to the electrostatic force. So it's at small distances. And that's why it works within a nucleus. But at bigger distances, it's almost imperceptible. Now, this nuclear binding energy oftentimes involves a calculation. And we're going to once again involve E equals mc squared, or in this case, E equals delta mc squared. And once again, we're going to... dealing with the concept of the mass defect, but a little bit different definition here. So in a nuclear reaction, the mass defect was that difference between the exact mass of the products and the reactants. Well, here, it's the difference in mass between what a nucleus does weigh and what a nucleus should weigh. That's going to sound a little bit strange. Well, it turns out we know exactly, to like several decimal places, what the mass of a proton is and what the mass of a neutron is. So, and it turns out we can... weigh the mass of a nucleus pretty exactly to several decimal places as well. And so, but notice, we know what, you know, a nucleus is made of. So many protons, so many neutrons, and we know how much those weigh. And so put them all together, we find out that the mass of the total nucleus, though... always weighs less than it should. You take it apart and you weigh out the protons and neutrons and you get a mass. Put them all together and it's not the same mass anymore. It's always smaller by a little bit than what you expect. And it turns out because a little bit of that mass of those protons and neutrons is converted into the energy that's holding that nucleus together, that energy associated with that strong nuclear force. And so the difference in mass between how much the individual protons and neutrons weigh versus how much the nucleus as a whole weighs. that's the mass defect. That's what we have to, again, convert into kilograms and then plug into our lovely equals MC squared equation here to figure out how much energy that is. And that is the nuclear binding energy. So now it turns out though, that, you know, just comparing overall total nuclear binding energy is not the most helpful thing in the world. So, because, you know, if you looked at like a, an iron nucleus versus a uranium nucleus, so you know, The energy holding these two together is not the same at all. Uranium has way more energy holding its nucleus together than iron does. But that's total energy. But also, the uranium nucleus is way bigger. It has to hold way more protons and neutrons together in there. So I would hope it would be bigger. So it's not really a fair comparison. And so what they usually do is we take a look at, say, iron 56 compared to iron, I'm sorry, compared to uranium 235. And they say, well, comparing the total energies is not really a good comparison. We should really look at it and scale it. And so what they do is they took a look at it in units of joules per nucleon. They take the total nuclear binding energy for iron, but divide it by the fact that it's holding 56 total protons and neutrons, call those nucleons collectively, together. And that way, kind of per nucleon, we get an idea of how much energy per nucleon is being used here. Same thing here with uranium-235. So in this case, we take the total nuclear binding energy we calculate using Einstein's equation. And then divide it by 235 nucleons, 235 total protons and neutrons, and that kind of scales it. So, and again, to make this a little, make a little more sense, let's just say I hired you and your best friend to babysit for me. So, and I said to babysit for me. So, however, I hired you at $50 an hour, and I hired your best friend at $20 an hour. Who got the better deal? Well, I said I hired you for me. Well, you're not babysitting me in either case. So, but, for $50 an hour versus $20 an hour, you'd say you got the better deal. $50 compared to $20. It's a no-brainer. Except it's not true because your best friend, I hired them to watch My Two Sons. And there's two of them. And so if it's $20 an hour, you'd be like, well, that's $10 an hour per kid. Okay, great. But for you at $50 an hour, I hired you to watch every single child at my church, all 50 of them. Not a fun night. And you're only getting $50 an hour. So at 50 bucks an hour with 50 children that you're watching, that's a dollar an hour per kid. Your friend got the way better deal. He or she is going to have a way better night than you are. And getting a per kid basis is definitely getting paid more. And that's kind of the comparison here. Instead of looking at the total between the two, we got to really look at it at a per nucleon basis to kind of get an idea. And what you find out is that iron 56 has the highest nuclear binding energy on a per nucleon basis. and is the most stable nucleus, whereas uranium's is actually quite a bit lower per nucleon, and is a rather unstable and radioactive nucleus. And so this idea that the nuclear binding energy, when looked at on a per-nucleon basis, it's a pretty good gauge of how radioactive something is likely to be. If it's in the ballpark of iron-56's number, that's probably pretty stable and not radioactive. If it's anywhere as low as uranium-235's number, much more likely to be radioactive. And so it's a great measure in that case. And again, the total nuclear binding energy is not a good measure of anything, but on a per-nucleon basis, we can use it to kind of gauge how stable versus radioactive that nucleus might be. And that's what we're going to kind of do here is we're going to actually calculate the nuclear binding energy per-nucleon for both iron-56 and uranium-235. All right, so we take a look at this. Iron-56 has got 26 protons and 30 neutrons. And let me make that a little bit bigger here. So 26 protons, 30 neutrons. And again, we know exactly the mass of protons and neutrons. I'm going to put them on a table there. I'll put that table up on the screen, but we can figure out the mass of both. So here we're going to take 26 times that mass of a proton, 1.00728 AMUs. And then we'll take 30 here for the neutrons, 30 times the mass of an individual neutron is 1.00866 AMUs. and we're just gonna total that up. And then we're gonna do the same thing down here for uranium. So here we got 92 protons, and we'll take 92 times 1.00728. AMUs, and then 143 neutrons, we'll take 143 times 1.00866 AMUs. And so we're going to weigh out those constituent nucleons and see like how much should this weigh if you put them all together, or how much would we expect it to weigh anyways if we put it all together. So we'll start with iron from this point on. So we've got 26 times, let's clear that out, 26 times 1.00728. and that's going to get us 26.18928 and then I'm going to add to that 30 times 1.00866 and we're going to get 56.44908 keep as many decimals as you can so combined this is 56.44908 amu's cool so that's the total mass of the protons and neutrons combined And we're going to compare this to the exact mass of the nucleus once you put it all together, and I've provided that here in this table, and that's 55.93494. See if I got that right. 55.93494. There we go, AMUs. And we can see that the nucleus weighs less than we'd expect it to based on just adding up the masses of all those protons and neutrons. And that difference between these two, again, is going to be that mass defect. And technically, we should actually probably take the nucleus minus the top and get a negative number. So, but we will take this difference and we're just going to plug in the absolute value anyways. But technically, this should come out negative because you should be taking the smaller minus the larger. So, but let's do this. So, we're going to take 55.93494 minus that last answer. And we're going to get negative 0.51414. And again, it's negative, and that's proper for a mass defect, but we're not actually going to plug in the negative. We're just going to plug in the absolute value here for delta M. But again, just like last time with nuclear reactions here, we're going to have to convert that to AMUs. I'm sorry, from AMUs to kilograms first. And so we'll multiply this by 1.67 times 10 to the negative 27. kilograms per amu and then we'll be able to plug that and multiply it by the speed of light squared to get that total energy so in this case we'll take that 0.51414 times 1.67 times 10 to the negative 27 and so in this case we're going to get e equals that mass defect of And I'm going to leave that negative out now. So 8.586138 times 10 to the negative 28 kilograms times the speed of light squared. 3.0 times 10 to the eighth meters per second squared. By the way, one of the most common mistakes students make is they forget to square the speed of light. So don't make that one. All right, so we'll take this last answer times 3e to the 8th squared, and we're going to get 7.727.73, let's just say. times 10 to the negative 11 joules. But again, that's for the entire iron 56 nucleus. That's to hold all 56 protons and neutrons together, all 56 nucleons together. What would that be in joules per nucleon? Well, we'll just take that and divide it by the 56 nucleons that we have. And so divided by 56, and now we get 1.38 times 10 to the negative 12. And again, that's joules per nucleon. Cool, and it turns out that's the highest nuclear binding energy per nucleon of any nucleus in existence. That's the highest. All right, so if we do this for uranium now, on the other hand, so I'm going to kind of do this mostly in my calculator here. So do 92 times that number. So 92 times... 1.00728 plus 143 times 1.00866 and I get 307.33598 and then right off the table I provided you we will subtract that. No that's not right then. Let's try this one more time. So 92 times 1.00728 there's my problem one point 00728, I was not pushing the zero. All right, plus 143 times 1.00866. And now we get 236.90814. And we'll subtract off the exact mass provided in the table there. 235.04393. And again, those exact masses, exact nuclear masses would have to be provided for you. And we're going to get a mass defect here, delta M of 1.86421 AMUs. But I'm gonna multiply that by 1.67e to the negative 27. So my conversion here to get it to kilograms, that's gonna get me 3.1132307 times 10 to the negative 27 kilograms. I'll multiply that by the speed of light squared, 3e to the eighth squared. And now we're gonna get a total nuclear binding energy of 2.8 times 10 to the negative 10. So notice that 2.8 times 10 to the negative 10, it's bigger than iron's total. But again, once I take that and divide it by the total, number of nucleons here. So let's write this out once. So 2.80 times 10 to the negative 10 joules. But again, that's spread out now over 235 nucleons. So we'll divide by 235. And now we're going to get 1.19 times 10 to the negative 12. And so now we see that, yeah, iron-56 has a higher nuclear binding energy per nucleon than does uranium. And notice it doesn't seem like a big amount, but it's enough to be not radioactive versus fairly radioactive. But that's nuclear binding energy per nucleon. Like I said, the math, here's a little bit of a pain in the butt. I tried to diagram it out here on your study guide and put it into four steps. Calculate that mass defect, actual mass, predictive mass, convert to kilograms by multiplying by our conversion factor there. And then plug it into E equals mc squared, so multiply by the speed of light squared, and then divide by your mass number, which is the total number of nucleons. So it's a process, again, that will be emphasized to differing degrees. Some of you guys will be totally on the hook for this. So in a few of you, your professor might just decide, end of the semester, we don't have enough time to cover that, it's hard math, and they leave it out. But the vast majority are probably on the hook for calculating this. And again, one more time, it's most common presented in joules per nucleon. and not just the totals themselves. So however those could show up as well, but most common, joules per nucleon. Now, if you found this lesson helpful, a like and a comment let me know are pretty much the best things you can do to support the channel. If you're looking for nuclear chemistry practice, if you're looking for general chemistry practice, if you're studying for finals and looking for final exam rapid reviews and practice final exams, then take a look at my general chemistry master course. I'll leave a link in the description below. Free trial is available. Happy studying.

Now, this lesson is part of my new General Chemistry playlist. I'm releasing several lessons a week, at least for a couple more weeks for the rest of this school year. And if you want to be notified every time I post a new lesson or when I start posting my next playlist, then subscribe to the channel and click the bell notification. So the energy of nuclear reaction is where we're going to start out here, and we're getting into a little bit of math, and I'm going to warn you here, the math here is not exactly fun.

I'm going to give you that heads up. But the subject matter is pretty fascinating, so hopefully that'll kind of balance it out here. But it turns out Einstein's the one who kind of figured this out. But in terms of where all this energy is coming from in a nuclear reaction is that mass is actually being converted into energy. And Einstein's famous equation, E equals mc squared, or in this case we'll use delta mc squared, That's how you calculate this kind of mass-energy conversion.

You plug the mass in right here, and it spits out how much energy it gets converted into. And C here stands for the speed of light, 3.0 times 10 to the 8th meters per second. That is a huge number.

And so when you take a little bit of mass, but you multiply it by the speed of light squared, you get a lot of energy. And that's why it only takes a little bit of mass converted into energy to generate a ton of energy in these nuclear reactions. And so that's why, like, you know, in a nuclear reactor, we don't have to have a lot of fissible material here.

to get a lot of energy harvest out of it. All right, so delta M here is called the mass defect, and if you take the difference in mass between your products and your reactants, that difference, because your products are always gonna weigh less than your reactants, is called the mass defect. So we'll find out this mass defect has another definition here in another context with nuclear binding energy towards the end, the second half of this lesson.

All right, so if we take this difference in the masses, we'll get that mass defect, and then we can substitute it right into this formula. Now you might be like, well, Chad, we have a problem. If you look at those mass numbers, 235 plus 1, 236. 142 plus 91 plus 3, 236. There's no loss in mass, Chad. And again, we stated this in an earlier lesson in this chapter, but the mass numbers are going to balance.

But if you take the exact masses to several decimal places, and I'll put a table up here of the exact masses for all these particles. If you take that to several decimal places, you'll find out, no, they don't actually balance perfectly. They're going to be off by just a little bit. And again, that difference in mass between the products and reactants, that's our mass defect.

So if we take a look on that table here, let's put these up on the board. We got uranium here is listed at 235.04393. Neutron is 1.00866 AMUs. We got that over here as well. times three, we got barium right here, and barium's given as 141.91643.

And then finally, krypton-91 is given as 90.92345. And now that we've got these exact masses, we can calculate the difference again between the reactions of products to get this mass defect and then substitute it right into here after converting it to kilograms as we'll find out let's take this difference here so we got 235.04393 plus 1.00866 i think i might need my hands for this all right then we're going to subtract off and i'll put this all in parentheses 141.91643 plus 90.92345 plus three, and keep in mind we got three neutrons, so three times 1.00, whoop, that's two points, 00866, and close my parentheses, and we're gonna get 0.18673. And technically it's customary to actually do products minus reactants that this comes out negative But the truth is we're just going to plug in the absolute value into this formula anyway, so but that's our mass defect Unfortunately, this is in atomic mass units And so in this case, which nowadays is also symbolized with just the lowercase funky u there italics u It's just an atomic mass units and again for this to come out in joules right here Your delta m has to be in kilograms, and your speed of light has to be in meters per second, because a joule is a kilogram meter squared per second squared. So we've got to convert this to kilograms. And that conversion here is that 1 amu is 1.67 times 10 to the negative...

27 kilograms and this is usually something that is provided. So if you kind of look at where this number comes from, so there's a connection between amu's and grams and the connection is Avogadro's number because it turns out there's 6.02 times 10 to the 23rd amu's is equal to one gram. And so if you take a look here well then you know that's equal to one gram.

Well, then how much, how many grams is one AMU? Well, just take the inverse here. So one divided by 6.02 times 10 to the 23rd, it's going to get you 1.66. And like I said, it rounds up, I should have made the 6.022, but 1.67 times 10, that'll give 24 grams. And if you make that kilograms, 1.67 times 10 to the negative 27 kilograms, that's where that comes from.

All right. So, and we'll get that there and let's do some plugging and chugging. So we got .18673 times 1.67 times 10 to the negative 27. And we're gonna get 3.11, let's just keep all these decimals, 3.118391 times 10 to the negative 28 kilograms.

And now we can plug that back into equals MC squared here. Alright, so times 3e to the 8th squared, and we're gonna get 2.8. And like I said, we'll keep some sig figs here.

2.8065519. Cool. So 2.8... 065519 times 10 to the minus 11 joules. And you might be like, uh, I thought you said this gives off a lot of energy.

Well, again, this is the amount of energy for just one uranium nucleus. So notice, what if you had a whole mole of uranium nuclei? Well, if you had a whole mole of these, that would weigh 235 grams, not AMUs. And so for a whole mole, how much energy would that be?

Well, for a whole mole of this uranium, again, a 235 gram sample, less than a kilogram of it, a quarter, less than a quarter of a kilogram. How much energy would that be? Well, this is for one uranium. nucleus, for an entire mole, you got to multiply by Avogadro's number. So if we multiply this by 6.02 times 10 to the 23rd, again, not what the question is asking for, but if it did, so it's 1.69 times 10 to the 13th joules.

It is a huge amount of energy for less than a quarter of a kilogram of uranium. This is where you see that energy payoff. So for one atom, you'd be like, that'd look like a tiny number, but for an entire mole of these nuclei, so it is a huge amount of energy when you look at it on that basis. It is ginormous. And again, you can run a nuclear power plant for a year off an amazingly small amount of fissile material.

It's pretty phenomenal how much energy you can really get out of this. All right, so now we're going to talk about the nuclear binding energy, and it's the energy associated with what holds the nucleus together. Back in the first lesson in this chapter, we talked about that strong nuclear force that actually holds the protons and neutrons together in that nucleus, and they call it that strong nuclear force because it is stronger than the electrostatic force that repels the protons away from each other. And so it turns out it only operates at very short distances.

At longer distances, it kind of evaporates almost to nothing, essentially. But at very short distances, it is excessively strong or exceedingly strong, especially compared to the electrostatic force. So it's at small distances. And that's why it works within a nucleus.

But at bigger distances, it's almost imperceptible. Now, this nuclear binding energy oftentimes involves a calculation. And we're going to once again involve E equals mc squared, or in this case, E equals delta mc squared.

And once again, we're going to... dealing with the concept of the mass defect, but a little bit different definition here. So in a nuclear reaction, the mass defect was that difference between the exact mass of the products and the reactants.

Well, here, it's the difference in mass between what a nucleus does weigh and what a nucleus should weigh. That's going to sound a little bit strange. Well, it turns out we know exactly, to like several decimal places, what the mass of a proton is and what the mass of a neutron is. So, and it turns out we can...

weigh the mass of a nucleus pretty exactly to several decimal places as well. And so, but notice, we know what, you know, a nucleus is made of. So many protons, so many neutrons, and we know how much those weigh. And so put them all together, we find out that the mass of the total nucleus, though...

always weighs less than it should. You take it apart and you weigh out the protons and neutrons and you get a mass. Put them all together and it's not the same mass anymore. It's always smaller by a little bit than what you expect. And it turns out because a little bit of that mass of those protons and neutrons is converted into the energy that's holding that nucleus together, that energy associated with that strong nuclear force.

And so the difference in mass between how much the individual protons and neutrons weigh versus how much the nucleus as a whole weighs. that's the mass defect. That's what we have to, again, convert into kilograms and then plug into our lovely equals MC squared equation here to figure out how much energy that is. And that is the nuclear binding energy. So now it turns out though, that, you know, just comparing overall total nuclear binding energy is not the most helpful thing in the world.

So, because, you know, if you looked at like a, an iron nucleus versus a uranium nucleus, so you know, The energy holding these two together is not the same at all. Uranium has way more energy holding its nucleus together than iron does. But that's total energy. But also, the uranium nucleus is way bigger. It has to hold way more protons and neutrons together in there.

So I would hope it would be bigger. So it's not really a fair comparison. And so what they usually do is we take a look at, say, iron 56 compared to iron, I'm sorry, compared to uranium 235. And they say, well, comparing the total energies is not really a good comparison.

We should really look at it and scale it. And so what they do is they took a look at it in units of joules per nucleon. They take the total nuclear binding energy for iron, but divide it by the fact that it's holding 56 total protons and neutrons, call those nucleons collectively, together.

And that way, kind of per nucleon, we get an idea of how much energy per nucleon is being used here. Same thing here with uranium-235. So in this case, we take the total nuclear binding energy we calculate using Einstein's equation.

And then divide it by 235 nucleons, 235 total protons and neutrons, and that kind of scales it. So, and again, to make this a little, make a little more sense, let's just say I hired you and your best friend to babysit for me. So, and I said to babysit for me. So, however, I hired you at $50 an hour, and I hired your best friend at $20 an hour. Who got the better deal?

Well, I said I hired you for me. Well, you're not babysitting me in either case. So, but, for $50 an hour versus $20 an hour, you'd say you got the better deal.

$50 compared to $20. It's a no-brainer. Except it's not true because your best friend, I hired them to watch My Two Sons. And there's two of them.

And so if it's $20 an hour, you'd be like, well, that's $10 an hour per kid. Okay, great. But for you at $50 an hour, I hired you to watch every single child at my church, all 50 of them.

Not a fun night. And you're only getting $50 an hour. So at 50 bucks an hour with 50 children that you're watching, that's a dollar an hour per kid. Your friend got the way better deal. He or she is going to have a way better night than you are.

And getting a per kid basis is definitely getting paid more. And that's kind of the comparison here. Instead of looking at the total between the two, we got to really look at it at a per nucleon basis to kind of get an idea. And what you find out is that iron 56 has the highest nuclear binding energy on a per nucleon basis.

and is the most stable nucleus, whereas uranium's is actually quite a bit lower per nucleon, and is a rather unstable and radioactive nucleus. And so this idea that the nuclear binding energy, when looked at on a per-nucleon basis, it's a pretty good gauge of how radioactive something is likely to be. If it's in the ballpark of iron-56's number, that's probably pretty stable and not radioactive. If it's anywhere as low as uranium-235's number, much more likely to be radioactive. And so it's a great measure in that case.

And again, the total nuclear binding energy is not a good measure of anything, but on a per-nucleon basis, we can use it to kind of gauge how stable versus radioactive that nucleus might be. And that's what we're going to kind of do here is we're going to actually calculate the nuclear binding energy per-nucleon for both iron-56 and uranium-235. All right, so we take a look at this.

Iron-56 has got 26 protons and 30 neutrons. And let me make that a little bit bigger here. So 26 protons, 30 neutrons. And again, we know exactly the mass of protons and neutrons.

I'm going to put them on a table there. I'll put that table up on the screen, but we can figure out the mass of both. So here we're going to take 26 times that mass of a proton, 1.00728 AMUs.

And then we'll take 30 here for the neutrons, 30 times the mass of an individual neutron is 1.00866 AMUs. and we're just gonna total that up. And then we're gonna do the same thing down here for uranium. So here we got 92 protons, and we'll take 92 times 1.00728. AMUs, and then 143 neutrons, we'll take 143 times 1.00866 AMUs.

And so we're going to weigh out those constituent nucleons and see like how much should this weigh if you put them all together, or how much would we expect it to weigh anyways if we put it all together. So we'll start with iron from this point on. So we've got 26 times, let's clear that out, 26 times 1.00728. and that's going to get us 26.18928 and then I'm going to add to that 30 times 1.00866 and we're going to get 56.44908 keep as many decimals as you can so combined this is 56.44908 amu's cool so that's the total mass of the protons and neutrons combined And we're going to compare this to the exact mass of the nucleus once you put it all together, and I've provided that here in this table, and that's 55.93494.

See if I got that right. 55.93494. There we go, AMUs.

And we can see that the nucleus weighs less than we'd expect it to based on just adding up the masses of all those protons and neutrons. And that difference between these two, again, is going to be that mass defect. And technically, we should actually probably take the nucleus minus the top and get a negative number. So, but we will take this difference and we're just going to plug in the absolute value anyways.

But technically, this should come out negative because you should be taking the smaller minus the larger. So, but let's do this. So, we're going to take 55.93494 minus that last answer. And we're going to get negative 0.51414.

And again, it's negative, and that's proper for a mass defect, but we're not actually going to plug in the negative. We're just going to plug in the absolute value here for delta M. But again, just like last time with nuclear reactions here, we're going to have to convert that to AMUs.

I'm sorry, from AMUs to kilograms first. And so we'll multiply this by 1.67 times 10 to the negative 27. kilograms per amu and then we'll be able to plug that and multiply it by the speed of light squared to get that total energy so in this case we'll take that 0.51414 times 1.67 times 10 to the negative 27 and so in this case we're going to get e equals that mass defect of And I'm going to leave that negative out now. So 8.586138 times 10 to the negative 28 kilograms times the speed of light squared. 3.0 times 10 to the eighth meters per second squared. By the way, one of the most common mistakes students make is they forget to square the speed of light.

So don't make that one. All right, so we'll take this last answer times 3e to the 8th squared, and we're going to get 7.727.73, let's just say. times 10 to the negative 11 joules.

But again, that's for the entire iron 56 nucleus. That's to hold all 56 protons and neutrons together, all 56 nucleons together. What would that be in joules per nucleon?

Well, we'll just take that and divide it by the 56 nucleons that we have. And so divided by 56, and now we get 1.38 times 10 to the negative 12. And again, that's joules per nucleon. Cool, and it turns out that's the highest nuclear binding energy per nucleon of any nucleus in existence.

That's the highest. All right, so if we do this for uranium now, on the other hand, so I'm going to kind of do this mostly in my calculator here. So do 92 times that number.

So 92 times... 1.00728 plus 143 times 1.00866 and I get 307.33598 and then right off the table I provided you we will subtract that. No that's not right then. Let's try this one more time.

So 92 times 1.00728 there's my problem one point 00728, I was not pushing the zero. All right, plus 143 times 1.00866. And now we get 236.90814.

And we'll subtract off the exact mass provided in the table there. 235.04393. And again, those exact masses, exact nuclear masses would have to be provided for you.

And we're going to get a mass defect here, delta M of 1.86421 AMUs. But I'm gonna multiply that by 1.67e to the negative 27. So my conversion here to get it to kilograms, that's gonna get me 3.1132307 times 10 to the negative 27 kilograms. I'll multiply that by the speed of light squared, 3e to the eighth squared.

And now we're gonna get a total nuclear binding energy of 2.8 times 10 to the negative 10. So notice that 2.8 times 10 to the negative 10, it's bigger than iron's total. But again, once I take that and divide it by the total, number of nucleons here. So let's write this out once. So 2.80 times 10 to the negative 10 joules. But again, that's spread out now over 235 nucleons.

So we'll divide by 235. And now we're going to get 1.19 times 10 to the negative 12. And so now we see that, yeah, iron-56 has a higher nuclear binding energy per nucleon than does uranium. And notice it doesn't seem like a big amount, but it's enough to be not radioactive versus fairly radioactive. But that's nuclear binding energy per nucleon.

Like I said, the math, here's a little bit of a pain in the butt. I tried to diagram it out here on your study guide and put it into four steps. Calculate that mass defect, actual mass, predictive mass, convert to kilograms by multiplying by our conversion factor there.

And then plug it into E equals mc squared, so multiply by the speed of light squared, and then divide by your mass number, which is the total number of nucleons. So it's a process, again, that will be emphasized to differing degrees. Some of you guys will be totally on the hook for this.

So in a few of you, your professor might just decide, end of the semester, we don't have enough time to cover that, it's hard math, and they leave it out. But the vast majority are probably on the hook for calculating this. And again, one more time, it's most common presented in joules per nucleon. and not just the totals themselves.

So however those could show up as well, but most common, joules per nucleon. Now, if you found this lesson helpful, a like and a comment let me know are pretty much the best things you can do to support the channel. If you're looking for nuclear chemistry practice, if you're looking for general chemistry practice, if you're studying for finals and looking for final exam rapid reviews and practice final exams, then take a look at my general chemistry master course. I'll leave a link in the description below.

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Transcript for:Understanding Nuclear Reactions and Binding Energy

Transcript for:
Understanding Nuclear Reactions and Binding Energy