Video: Solving Nuclear Equations Involving Alpha Decay

The following nuclear equation shows how an isotope of curium decays to plutonium via alpha decay: what are the values of 𝑝 and π‘ž in the equation?

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Video Transcript

The following nuclear equation shows how an isotope of curium decays to plutonium via alpha decay. What are the values of 𝑝 and π‘ž in the equation?

Taking a look at this equation, we see that curium, symbolized Cm, is decaying into plutonium, Pu, plus an alpha particle, symbolized with the Greek letter 𝛼. This decay event means that the curium nuclei has become unstable. And it’s split up into two pieces, the plutonium plus the alpha particle. We see that, for curium as well as for plutonium, both the atomic number as well as the mass number is given. But for the alpha particle, instead of those numbers, we have a π‘ž and a 𝑝, respectively. It’s those values we want to solve for. And as we’ll see, there are two ways of going about doing this.

The first way is to recall what an alpha particle is, what it consists of. An alpha particle, the particle emitted in an alpha decay event, consists of two protons as well as two neutrons. That means if we were to represent an alpha particle as though it was its own atomic element, we would write its atomic number, the number of protons it has, two, and then its mass number, the sum of protons as well as neutrons in the particle, four. These numbers are true for any alpha particle. They always have two protons and two neutrons.

The alpha particle involved in this equation is the same way. It also has two protons and two neutrons. This will indicate to us that π‘ž is equal to two and that 𝑝 is equal to four. This is one way to find the answer to this question. But as we’ll see, there’s another way, even if we didn’t recall this about an alpha particle.

This second approach involves looking at these values, 𝑝 and π‘ž, in terms of the equation that they’re part of. This type of nuclear decay equation, where one kind of element decays into another element plus an alpha particle, involves what we could call the conservation of atomic number β€” that’s this number here to the lower left β€” as well as the conservation of mass number β€” that’s this number here to the upper left. In other words, the atomic number and the mass number on the left of this equation equal the sum of the atomic numbers and mass numbers on the right of the equation. This means there are two separate equations we can write down to help us solve for 𝑝 and π‘ž.

First, we can write down the atomic number equation. Since atomic number is conserved across this equation, it means that 96, the atomic number of curium, is equal to 94, the atomic number of plutonium, plus the atomic number of the alpha particle. That’s π‘ž. And then, likewise, for the mass numbers of these constituents, the mass number of curium, 247, is equal to the mass number of plutonium, 243, plus the mass number of the alpha particle, 𝑝. We can use these two separate equations to solve for π‘ž and to solve for 𝑝. π‘ž is equal to 96 minus 94, or two. And 𝑝 is equal to 247 minus 243, or four. Whichever of these approaches we use, we end up finding that 𝑝 is equal to four and π‘ž is equal to two.

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