### 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.