### 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.