 High science 30s, we're going to take a look at this nuclear energy and radiation handout where we're doing some of the calculations to figure out how much energy is released in a nuclear reaction. To do this, you're going to need three things. You're going to need this handout, which you can find on my website, ldindustries.ca. You're going to need a data booklet, specifically page eight, and I've got a digital copy of that on the website as well, and you're going to need a calculator. We're going to work out how much energy is released in the first calculation in kilojoules. We'll do three calculations and the answers for the others I'll post on my website so you can see them to check your work. We're going to do lanthium-146, which is going to fission to form barium-141, a proton, and five neutrons. So to start off with here, we're going to just write out what the formula looks like, and we're just translating it from the words that we see here into chemical symbols and nuclear isotope notation, which sounds a lot harder than it is. It's basically just reading off the data sheet. So let's find lanthium. There it is in the data sheet, left-hand side, kind of part of the way down, and it is 146 on the top, the top number, and 57 on the bottom, and then a big LA. Now the 146 here is called the mass number, and this is the protons and the neutrons added together. The bottom number here is called the atomic number, and this is just the number of protons that are in the element. Now the other thing we're going to do over here is there's a number next to that compound. It is 145 decimal, 9258. That's kind of an annoying number to write down. That's the mass of this compound in funny units. It's 10 to the negative 3 kilograms per mole. All right, so I'm going to write that number down as well. Now this compound, it says it's going to fission. So that means it's going to break apart into barium 141. That's right at the top of the table on page 8. So I'm just going to basically copy that out. And while I'm there, I'm also going to write down the mass of it. Be careful when you do this. You want to try to get those masses exactly right, so take your time. All right, we're also going to get a proton. Now the proton is on your chart, but it's sort of in a funny spot. It's up at the top table, right at the top of the page on the right hand side. And it also has a mass not very big compared to some of these other isotopes we're dealing with. Same units though, 10 to the negative 3 kilograms per mole. And then the other thing we're going to get is 5 neutrons. Now here's how we write 5 neutrons. We put down the 5 first, and then we put down the symbol for a single neutron like that. And I'll put down the mass of a neutron, which is really close to the mass of a proton. All right. Now that we've got all of the data book that's stuff out of the way, it's going to look a little bit like Hess's law. We're going to look at any compounds where there is more than one of them. And we're going to multiply the mass by how many there was. And the only one we've got to worry about here for this is that neutron at the end. Because there's 5 neutrons, I have to multiply the mass of the neutron by 5. So into my calculator, that's going to give me a mass of 5.0433 times 10 to the negative 3 kilograms per mole. All right. So we've got our masses all worked out. And any multiplication that needed to be done has been done. Now what I'm going to do is I'm going to do some totaling up. I'm going to total up all the mass on the left-hand side of the equation. And I'm going to total up all the mass on the right-hand side of the equation. Now the left-hand side of the equation is easy because there is only one nuclear isotope there. So nothing really to do. I'm just rewriting it out to make it look a little more complete. On the right-hand side, I do have some numbers that I can total up here. So typing those through my calculator gives me a total of 146.96499 times 10 to the negative 3. And technically at this step we're at kilojoules, not kilojoules per mole, because I took the number of kilojoules per mole I originally had and I multiplied it by how many moles of each compound I had. That's what I did when I did that multiplying by five. And technically there's sort of a one in front of each of these other compounds. So as we multiply the masses by one we cancel out the moles. Now we're nearly done. All we have to do now is one more calculation. And the formula for that calculation is on page H as well. It's at the bottom of the page and it's the very famous E equals MC squared calculation. Here's what it actually looks like though. It's got two extra little things in it from what you're probably used to hearing. It's got these little triangles, okay? So the triangle is the Greek symbol delta. It means change in. How much did the mass change and how much did the energy change? So use this formula. I'm going to put in the masses that I had, but I need to make sure first that I include not just what the mass on the left-hand side was or what the mass on the right-hand side was, but how much the mass changed by. So here's how I'm going to do that. I'm going to substitute in that equation and in place of delta M, I'm going to subtract the two masses. I'm going to figure out what the difference is or the change in mass is, okay? So there's my first mass minus my second mass. Really important that as you do this you put in those units of times 10 to the negative three kilograms. If you miss that then your answers will all be a thousand times too small. Last thing we're going to do is multiply this by c. So c is the speed of light, three times 10 to the eight meters per second and that is on your data sheet as well and we've got to square that. All right, here comes the moment of truth. We are going to type all of this through our calculator now. So I get from my energy released 9.35 two seven one times 10 to the 13 kilojoules. So there's our first nuclear energy calculation and all the other ones work essentially the same way. The only difference is you're dealing with different amounts of each of the different compounds or different starting isotopes. Let's take a look at the second page where we're writing a nuclear decay reaction and then we'll do one more problem maybe we'll do four and five. So here's a review of just writing a nuclear reaction. We're going to do uranium 235. So uranium 235 is decays alpha. So what does that look like? Well uranium 235 I've got to find on my table on page eight. So I can see it's actually at the bottom right hand side. This is what it looks like. The u for uranium 235 is the mass number, the number of protons and neutrons and 92 is the atomic number just how many protons are present. Now it says that decays alpha so it's going to make an alpha particle and the alpha particle is up in the upper left hand corner of page eight. It looks like this. You can either write it as a helium nucleus he the four and two or which I think is a little bit more fun because I feel like I'm kind of really being a physicist when you do this. I'm going to write it using the Greek symbol alpha. Either one is fine. So there's our alpha particle but there's more than just an alpha particle produced here as well. There's some other compound that other unknown here will have some sort of chemical symbol and it'll have some sort of unknown mass number and it'll have some sort of unknown atomic number. So how are we going to figure out what these are? Well to do this we use what's called the law of conservation of mass and the law of conservation of nucleons. It's a pretty simple idea actually. It's just this idea that we have to have like a balance between the numbers that we've written down here. I had 92 as my original mass number that means I have 92 protons to start off with in this reaction. Later on after the alpha decay occurs I have only two protons. So like where did all of the other protons go? Well there's 90 protons left over and those go into the we call that the daughter isotope. So I go 92 minus two and I get the 90 that would be present for my unknown. I'm going to do the same thing now on the top. So here's my number of protons and neutrons together. I started off with 235 of them. Now I've got four left over that went into the alpha particle. So the mass number left over here has to be 230, oops, get 231. So I just went 235 minus 4. Now we're nearly done all I have to do now is look in my data sheet to see if I can find something that has a atomic number of 90 and if it has a mass number close to or exactly 230 that's going to be nice as well. So really I don't need to look too far. If I go right above uranium 235 I see thorium and thorium has a number of protons which is 90 so that's how I know it's going to be thorium and it's got a mass number of 230 in the data sheet which is really nice and close to the one I'm looking for. So that gives me a hint too that I've picked the right element. So there's the uranium 235 alpha decay. Last one we're going to tackle here is we've got a nitrogen 15 beta decay and we're going to figure out how much energy is released. So we're going to kind of put both of those ideas we just did together into one big step. So first of all let's write out what nitrogen 15 looks like as a beta decay. So I'm going to go find nitrogen 15 on my data sheet. It looks like this and with a 15 is the mass number and a 7 as the atomic number and I'm going to write down the mass of that as I go 15.00011 times 10 to the negative 3 kilojoules per mole. Now when you make a beta decay you form a beta negative particle like so. That beta negative particle is also on your data sheet. It can have a little E for its symbol or it can have the beta symbol. Now we're going to figure out what else do we get produced here. So we're going to do that sort of unknown detective work trick again. So I had 15 for my number of protons and neutrons to start off with and then afterwards I had zero in the beta particle. So that means I must also have 15 here in my unknown. Now the next thing I'm going to do is I'm going to think about the bottom numbers. I had seven to start off with and there's a negative one in the beta particle. So what I need to do is think what number should I put in this black box so that if I added negative one to it I'd get seven so that the totals on either side of the equation are the same. I need to put an eight in. That is going to give me a nice balance of numbers of protons on either side of the equation. That also tells me which element I'm looking at now or which isotope which is oxygen. And there is a mass for oxygen here on the data sheet as well which I can write down. It is going to be 15.00307 times 10 to the negative three kilojoules per mole. Now one thing you're going to notice is that's really close to the original mass. Hasn't changed by a whole lot. Actually this one because the masses are so small we definitely need to include the mass of this beta negative particle because the differences here are pretty tiny. All right so now let's do our totaling up step. So pretty easy to total the left hand side. It's not going to change and I'm going to go and add together the two numbers on the right hand side as well. If you're doing this properly what you should find is that your numbers are very close to one another and that's a good thing. The differences in mass we're talking about here are very small. Now there are no coefficients that we had to multiply by. There is no numbers of moles in there so I don't have to do that step in this particular problem and that means I can cancel the moles. And now I'm going to go through and put these into the equation. E equals mc squared. So it's E equals delta mc squared. The change in mass is 15.003619 times 10 to the negative 3 kilograms minus 15.0011 times 10 to the negative 3 kilograms. And I'm going to multiply that all by the speed of light which is 3 times 10 to the 8 meters per second square. So let's see how much energy is released in this one. I totally messed up the original calculation but here's an edit. You're going to get an energy of 3.1581 times 10 to the 11 joules. So I hope this helps. Check out the website for the key on this and email me if you have any questions.