Video: Solving Nuclear Equations Involving Nuclear Fission

The nuclear equation shows the fission of uranium-235 into xenon and strontium. We can see that uranium-235 is decaying into xenon-140 and strontium-94, as well as some neutrons. How many neutrons are produced in this reaction?

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Video Transcript

The following nuclear equation shows the fission of uranium-235 into xenon and strontium. We can see that uranium-235 is decaying into xenon-140 and strontium-94, as well as some neutrons. How many neutrons are produced in this reaction?

So basically, having seen this nuclear reaction, we need to try and work out how many neutrons there are over here. And the way to do this is to remember that a correct nuclear equation will be based on the principles of conservation of mass number. Now, what the conservation of mass number tells us is that the total mass number before a reaction is equal to the total mass number after the reaction. And this of course applies to the reaction that we’re considering here. But then what do we even mean by mass number?

Well, we can recall that the mass number of an entity is equal to the total number of protons and neutrons in that entity. So for example, if we just think of the uranium nucleus that we started with, then the mass number of this uranium nucleus is the total number of protons and neutrons in that nucleus. Now, the mass number for a nucleus in its chemical symbol is traditionally given in the top left-hand corner of that nucleus chemical symbol.

So for this uranium nucleus, we see that the mass number is 235. In other words then, we see that the total number of protons and neutrons in this uranium nucleus is 235. And actually, because we only have the uranium nucleus before the reaction happens, the total mass number before the reaction must be 235 because there’s nothing else on the left-hand side of the equation. And so using a conservation of mass number, we can say that 235, the total number of protons and neutrons before the reaction or on the left-hand side, is equal to the total number of protons and neutrons on the right which must be equal to 140, which is the mass number of xenon, plus 94, which is the mass number of strontium, plus the number of neutrons that are released.

And so in the equation that we’re creating, we’ll say that 235 is equal to 140 plus 94 plus 𝑥 which is what we call the number of neutrons produced. Now, it’s important to note that the nuclear equation is telling us that this particular xenon nucleus has 140 protons and neutrons in total. This particular strontium nucleus has 94 protons and neutrons in total. And only neutrons are produced apart from this. In other words, no protons are produced or no other nuclei are produced. And so the conservation of mass number must apply to everything on the right-hand side that’s been given to us.

And since neutrons do form part of mass number, remember mass number is total number of protons and neutrons, we can therefore use this to calculate the number of neutrons, which we’re calling 𝑥. So all we need to do to calculate 𝑥 then is to rearrange this equation at the bottom. We simply need to subtract 140 and 94 from both sides of the equation because, this way, the 140 and minus 140 cancel. And same is true for 94 minus 94. So on the right, we’re just left with 𝑥. And on the left, we’ve got negative 140 minus 94 plus 235. So evaluating the expression on the left, we find that one is equal to 𝑥. In other words then, the total number of neutrons produced in this reaction is one.

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