Lesson Video: Beta Decay | Nagwa Lesson Video: Beta Decay | Nagwa

# Lesson Video: Beta Decay Physics

In this video, we will learn how to solve nuclear equations involving beta decay.

12:39

### Video Transcript

In this video, we’re talking about beta decay. Beta decay is a radioactive process where an atomic nucleus emits a beta particle. And as we’ll see, this emission has an effect on the nucleus that’s decaying. As we get started, we can consider that when an atomic nucleus becomes unstable, that nucleus will often emit or give off something in order to try to become more stable.

One type of particle a nucleus can emit to do this is called a beta particle. And this particle has certain qualities to it. It has mass, although its mass is very small. And it has a relative electric charge of negative one, the charge of an electron. In fact, a beta particle is an electron. So beta decay is the process of an unstable radioactive nucleus emitting a beta particle, that is, an electron. And as we mentioned, the goal of this emission, as with any radioactive emission, is to create a more stable atomic nucleus. The way that beta decay accomplishes this is quite interesting.

Let’s consider a nuclear equation involving this process. Say that we have an isotope of carbon-14. In other words, we have a nucleus with six protons and 14 minus six or eight neutrons. Now, if carbon-14 undergoes beta decay, that means it emits a beta particle, an electron. And left over from that emission is the resulting nucleus. Let’s consider the electron, the beta particle, first. The Greek letter 𝛽 looks like this. If we then add an atomic number of negative one and a mass number of zero, then this is how we represent a beta particle in a nuclear equation.

Now, we can probably see why the mass number of a beta particle is zero. We know that that number indicates the number of protons plus neutrons in the object being considered. And an electron has none of them, so that makes sense. But what about this atomic number of negative one? Does that indicate that a beta particle has negative one protons? Not exactly, but it does mean that the relative charge on a beta particle, an electron, is negative one. So even though a beta particle has no protons, it still has a nonzero atomic number of negative one because that’s the relative charge of this particle. So that’s the beta particle symbol.

But if we look at it as part of this equation, we can see that the equation as it stands is incomplete. That’s because when it comes to nuclear equations, the total atomic number and total mass number on one side must equal the totals on the other side. And as things stand currently, we don’t have matches for atomic number or mass number. When carbon-14 undergoes beta decay, it emits an electron, and it becomes nitrogen-14. Now, there’s something very interesting going on here because the nucleus that emitted the beta particle, carbon-14, has the same mass number after that emission. But it has a greater atomic number by one. That is, the atomic number went from six to seven.

This is telling us that if we add up all the protons and neutrons here for carbon-14 and add up all the protons and neutrons for nitrogen-14, we’ll get the same total number. But that in this transition, from carbon to nitrogen, one of those particles, specifically one of the neutrons, turned into a proton. That’s what explains why this atomic number is one higher than the original value. If we were to draw this out using pictures, we could let blue circles represent protons with a positive charge and green circles represent neutrons. Using that convention, this represents carbon-14. There are six blue circles, protons, and eight green ones, neutrons.

When carbon-14 undergoes beta decay, it emits an electron. And what’s left over is a nucleus that has seven neutrons and seven protons. This means, as strange as it may seem, that one of the neutrons from carbon-14 has actually changed into a proton in this process. It’s gone from being one type of subatomic particle to another. As we look at the nuclear equation describing this process, we can see why this is necessary. On the left-hand side, with carbon-14, we have a total atomic number of six. Then, since the atomic number of a beta particle is negative one, that must mean we add seven to it to balance out the atomic number on the left- and right-hand sides of this equation. But then, that extra proton ⁠— we could call it ⁠in nitrogen — had to come from somewhere because in terms of mass number, we have 14 equaling zero plus 14. And now, we can start to see why, in order for this nuclear equation to balance, a neutron on the reactant side needed to transform into a proton on the product side.

So if we compare the original nucleus with the one that’s left over after the emission, here’s what we can say. The mass number of the nucleus remains the same, while the atomic number of the nucleus increases by one. This means that the relative charge of the nucleus goes up by one. These relationships of the mass number staying the same while the atomic number and relative charge increasing by one apply to more than just this specific beta-decay equation. In fact, they apply to all beta-decay processes.

Let’s look at another beta-decay example to see how this works. Uranium-235 is a common isotope used in nuclear fission facilities. This isotope can undergo beta decay. And when it does, it gives off a beta particle, an electron. And the nucleus that remains behind is neptunium-235. Notice that, just like before, the mass number of the nucleus stays the same, while the atomic number, and therefore the relative charge, increases by one. In these two examples, we’ve seen that one way to represent a radioactive decay process is through a nuclear equation. It turns out though that this is not the only way. Decay transitions can also be indicated by a nuclear decay chart. These charts show the number of neutrons in a nucleus versus the number of protons.

If we wanted to indicate this particular beta-decay process on our nuclear decay chart, we could begin doing that by populating the tick marks with the appropriate number of neutrons and protons according to this nuclear decay process. We find the number of neutrons by subtracting the atomic number, the number of protons, from the mass number, the number of protons plus the number of neutrons. 235 minus 92 is 143. So that defines the top of our range of number of neutrons. And then, as for the number of protons, that’s simply equal to the atomic number. And we can see on this chart, we have a range from 91 to 95.

Then, following this particular nuclear decay process, we start out by locating our original isotope on our nuclear decay chart. That means an isotope with an atomic number of 92 and a mass number of 235, which indicates a number of neutrons, as we saw, of 143. If we draw a vertical line up from 92 protons and a horizontal line out from 143 neutrons, we see where they intersect. And that is where our uranium-235 isotope is located. So that’s our start point. But then, according to this equation, we know that this isotope undergoes beta decay. It emits a beta particle. It ends up as neptunium-235. So the nucleus has now transitioned to having 93 protons, but the same number of protons plus neutrons.

Now, if we’ve gained one proton, going from 92 to 93, and we’ve kept the same mass number, the sum of protons plus neutrons, then that must mean that we’ve lost one neutron in the process. We must go from 143 to 142. So then, drawing lines out from these two points, we see where they intersect. And this is where our nucleus ends up after the decay occurs. Our decay pathway then is from this original point to this final point. This then is how we would use a nuclear decay chart to indicate the decay process shown in this nuclear equation. We go from uranium to neptunium keeping the mass number the same, indicating the loss of a beta particle.

Knowing all this about beta decay, let’s get some practice through an example exercise.

Calcium-40 is created through the beta decay of potassium, as shown in the nuclear equation here. What are the values of 𝑝 and 𝑞 in the equation?

Okay, taking a look at this equation, we see potassium is decaying into calcium and a beta particle, an electron. All of the numbers are filled in in this equation except for two, potassium’s atomic number, labelled 𝑞, and its mass number, labelled 𝑝. It’s those two values that we want to solve for. We can start to do this by recognizing the fact that because this is a nuclear equation, that means the total atomic number and total mass number on one side of the equation must equal the total atomic and mass numbers on the other side. That has to be true in order for this to be an equation, and we’re told that it is.

If we consider first the atomic numbers involved in this equation, we know that 𝑞, whatever that value is, must be equal to the sum of 20 and negative one. We can even write that as an equation down at the bottom of our screen. And then, if we consider next the mass numbers, represented for potassium by the letter 𝑝, we know that 𝑝 must be equal to the sum of the mass numbers of calcium and our beta particle. That’s 40 plus zero, so we’ll write that out as an equation. So based on the fact that this is a nuclear equation, we now have an equation to solve for 𝑞 and one to solve for 𝑝. 𝑞 is equal to 20 plus negative one, which is equal to 19. And 𝑝 is equal to 40 plus zero, which is equal simply to 40. And with that, we’ve solved for the values we wanted to find. 𝑞 is equal to 19 and 𝑝 is equal to 40 in this equation.

Let’s look now at a second example exercise.

When an atomic nucleus emits a beta particle, how much does the atomic number of the remaining nucleus change by?

Okay, so in this example, we have an atomic nucleus. And this nucleus decays and, in doing so, emits a beta particle, which is an electron. In the process of giving off the electron, the nucleus goes through a change. To see what that change is, let’s represent this emission process using a nuclear equation. We’ll call this element that we start off with element X. And we’ll say that it has an atomic number of 𝑧 and a mass number of 𝐴. We knew that this nucleus emits a beta particle, represented by the Greek letter 𝛽, with an atomic number of negative one and a mass number of zero. In addition to this, there is some leftover nucleus after this emission. We’ll just refer to it using a star symbol. And like any nucleus, this one has a particular atomic number and a mass number.

In this example, we’re specifically wondering how the atomic number of the remaining nucleus ⁠— that’s this one over here ⁠— changes compared to the original nucleus over on the left-hand side. All that to say, we want to figure out what goes here, what is the atomic number of our resulting nucleus, and how that compares to the original atomic number what we’ve called 𝑧. To make this comparison, we can recall that this is a nuclear equation. That means, for one thing, that the total atomic number on the left-hand side is equal to the total atomic number on the right side. In other words, 𝑧, the atomic number of our original nucleus, is equal to negative one, the atomic number of our beta particle, plus whatever goes in the blank, the atomic number of our resulting nucleus.

When it’s written out this way, we can see that the only thing that can go in this blank that makes this equation true is 𝑧 plus one. When we have 𝑧 plus one as the atomic number of our resulting nucleus, then the plus one and the negative one here cancel one another out. And then, we have an equation saying 𝑧 is equal to 𝑧, which is true. So if we start with an atomic number of 𝑧 and then that nucleus goes through beta decay, we wind up with a nucleus with an atomic number of 𝑧 plus one. This means that, in general, when a nucleus emits a beta particle, the atomic number of the remaining nucleus goes up by one. And that’s our answer. The change in the remaining nucleus’s atomic number is plus one.

Let’s summarize now what we’ve learned about beta decay. In this lesson, we saw that beta decay is a radioactive process in which an atomic nucleus emits a beta particle. We saw that, for a nucleus undergoing beta decay, the mass number of that nucleus remains the same through the process, while its atomic number and relative charge both increase by one. Additionally, we saw that beta particles are electrons. And they’re represented in nuclear equations this way, using the Greek letter 𝛽, and have an atomic number of negative one and a mass number of zero. Finally, we saw that beta decay can be represented using a nuclear equation or a nuclear decay chart.

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