# Video: Describing Radioactive Rates and Products

Figure 1 shows a spinner with 12 regions. The spinner has an equal probability of landing in each region. A student has 50 of the spinner shown in Figure 1, which he uses to model the random nature of radioactive decay. He spins each spinner and removes any that land on 2. He spins the remaining spinners again. The student repeats this process a number of times. Table 1 shows his results. Give two reasons why this is a good model for the random nature of radioactive decay. The student’s results are shown in Figure 2. Use Figure 2 to determine the half-life for these spinners using this model. Show on Figure 2 how you work out your answer. The nuclear equation shows how an isotope of neptunium decays to protactinium. What is the value of 𝑚 in the equation? The nuclear equation shows the radioactive decay of an isotope of uranium. Use this equation and the information in part 3 of the question to determine the name of element X. Determine the type of radiation emitted as uranium decays into a new element, as described in part 4 of the question. Give a reason for your answer. A teacher demonstrates radioactivity to a class using a cube of radioactive material. When placing the cube down on a work surface, the teacher places the cube in a plastic tray that is on the work surface. Explain why the teacher does this.

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### Video Transcript

Figure one shows a spinner with 12 regions. The spinner has an equal probability of landing in each region. A student has 50 of the spinners shown in figure one, which he uses to model the random nature of radioactive decay. He spins each spinner and removes any that lands on two. He spins the remaining spinners again. The student repeats the process a number of times. Table one shows his results. Give two reasons why this is a good model for the random nature of radioactive decay.

When we look at table one, we see that indeed there are 50 spinners at start, at spin number zero. Then, as the spin numbers increase, more and more of the spinners land on two and are eliminated from the pool. Finally, after 10 spins, only 21 of the 50 original spinners remain. This process has been designed to model radioactive decay where an atom of one element will decay into an atom of another element.

If we consider a population of radioactive atoms, there is no way for us to tell which of these atoms in particular will decay. In the same way, given a population of spinners, there is no way we can predict ahead of time which of the spinners will land on two. That’s one reason that this model is a good one. Just as we can’t predict which of the spinners will land on two, we also can’t predict which of the atoms will decay.

There is another reason though we can say that this model is a good one. If we consider a single radioactive atom, there is no way we can predict ahead of time when that atom might decay. And similarly, if we consider a single spinner, we can’t predict on which spin the spinner will land on two and be eliminated. Given an atom then that we know will decay at some point, we can’t say when that will happen. And likewise, given a spinner that we know will be eliminated, we can’t tell on which spin number that spinner will land on two. This is a second way that our model is a good one for simulating radioactive decay. Next, let’s use this model to help us better understand the concept of half-life.

The student’s results are shown in figure two. Use figure two to determine the half-life for these spinners using this model. Show on figure two how you work out your answer.

When we talk about half-life, with respect to a population of radioactive atoms, the half-life refers to an amount of time it takes for half the atoms in this population to decay into another element. When we use spinners to model this process, that half-life would be measured not in time but in number of spins. And to solve for this half-life, we’ll use the data plotted in figure two which shows the number of spinners remaining on the vertical axis and the spin number on the horizontal axis. The half-life of our spinners will be the number of spins it takes to go from our initial population of 50 spinners down to half that value of 25 spinners.

We see that it’s at spin number zero that we start out with 50. And to find the spin number that corresponds to 25 spinners remaining, we’ll move out from that point on a horizontal line until we intersect our curve. We see that this point of intersection corresponds to a spin number of eight. This means that in going from spin zero to spin eight, we’ve cut the number of spinners in our population in half. In other words, the half-life is eight spins. That’s the number of spins it took to eliminate half of our spinners. Let’s continue now studying radioactive decay using a nuclear equation.

The following nuclear equation shows how an isotope of neptunium decays to protactinium. What is the value of 𝑚 in this equation?

We see in the equation neptunium with an atomic number of 93 decaying into protactinium and an 𝛼 particle, a helium nucleus. We want to solve for the value of 𝑚. And we see that 𝑚 is the mass number of protactinium. That is its number of protons plus its number of neutrons. To solve for 𝑚, we can use the principle that energy is conserved. And by Einstein’s equation that 𝐸 is equal to 𝑚 times 𝑐 squared, we recall that energy and mass are equivalent to one another.

Bringing both of these concepts together, that tells us that the initial mass in our nuclear equation is equal to the final mass. In other words, we can write that 237 is equal to 𝑚 plus four. That is, the atomic mass of neptunium is equal to the atomic mass of protactinium plus the atomic mass of an 𝛼 particle. We rearrange to solve for 𝑚. And we find it’s 237 minus four, or 233. That’s the mass of protactinium in atomic mass units. Now, let’s consider another nuclear equation.

The following nuclear equation shows the radioactive decay of an isotope of uranium. Use this equation and the information in part three of the question to determine the name of element X.

In this equation, we see uranium 237 decaying into an unknown element, called element X, and some radiation. We want to solve for the name of element X. And to do it, we’ll use some of the information in the previous part of this question. In particular, in part three, we saw that neptunium decayed into protactinium and an 𝛼 particle.

When we look back at the equation involving element X, we see that it has an atomic number of 93. It’s that number, the number of protons, that’s used to specifically identify an element. If we then consider the equation in part three, we see that neptunium also has an atomic number of 93. Not only that, neptunium and element X both share the same atomic mass. This tells us that not only are they the same element, they’re also the same isotope. We can write then that element X is neptunium or Np, for short. Next, let’s consider what type of radiation is emitted in this nuclear equation.

Determine the type of radiation emitted as uranium decays into a new element, as described in part four of the question. Give a reason for your answer.

We recall that part four of the question showed us a nuclear equation demonstrating the decay of uranium into an unknown element and some radiation. We now want to solve for the type of radiation that’s emitted. And to do it, we’ll look at the atomic number as well as the atomic mass number of each of these constituents. We notice that the atomic mass number of uranium, the number of its protons plus the number of its neutrons, is equal to that of the unknown element. That tells us that whatever type of radiation is emitted, it doesn’t involve protons or neutrons. Those are all accounted for.

However, we notice that the atomic number of uranium is one less than the atomic number of the unknown element. This almost seems to contradict what we just found and may make us think that the radiation must involve a negative proton. But in fact, this isn’t an issue of conservation of mass. Rather, it’s an issue of conservation of electric charge. In order to balance out this nuclear equation, the radiation emitted must have a negative charge to it.

At this point, we can consider the common types of radiation: 𝛼, 𝛽, and 𝛾 radiation. We know that 𝛼 particles are helium nuclei with two protons and two neutrons. 𝛾 radiation, on the other hand, has no charge or mass. It consists of high-energy photons. 𝛽 radiation, though, does have a charge and it’s negative. That’s because 𝛽 radiation is the emission of electrons. The symbol for 𝛽 radiation is the Greek letter capital 𝛽, with a negative one indicating its charge and a zero indicating its atomic mass.

If we insert this symbol into our nuclear equation, we see that it makes the math work out. That is, 237 equals 237 plus zero and 92 is equal to 93 minus one. We can say then that the emitted radiation is 𝛽 radiation. And the reason is that the proton number increased by one in this nuclear equation. This tells us what radiation was emitted and why. Finally, let’s consider a practical application for handling radioactive material.

A teacher demonstrates radioactivity to a class using a cube of radioactive material. When placing the cube down on a work surface, the teacher places it in a plastic tray that is on the work surface. Explain why the teacher does this?

We can picture a flat work surface used for this demonstration. And we’re told that the teacher places a plastic tray on the surface before setting down the cube of radioactive material. We want to explain why the teacher does this. And the reason has to do with the idea of radioactive contamination. We know that, in general, a radioactive material is one that’s in the process of decaying. This decay often has a by-product of high-energy radiation which is what makes radioactive materials potentially dangerous.

If our radioactive cube was in direct contact with the work surface, some of that material might remain on the surface. And that remaining material, the material left behind, would irradiate any other objects that will come in contact with the working surface. We can say then that the plastic tray keeps radioactive material from transferring to the work surface, which could lead to irradiation of other objects contacting that surface. This is why the teacher would use a barrier and what might happen if a barrier was not used.