### Video Transcript

A block of an unknown material
emits charged particles. These particles are accelerated in
a vacuum across an electric potential difference of six volts. The kinetic energy of each particle
is found to be 1.92 times 10 to the negative 18th joules once it has crossed this
potential difference. What is the charge of each
particle?

Okay, let’s say that this is our
block of material. And we’re told that this block is
emitting particles. And along with that, we’re told
that there is a potential difference of six volts, which accelerates these particles
through a vacuum. If we follow one of these particles
as it moves across this six-volt potential difference, once it reaches the barrier
of that six-volt potential difference, we’re told its kinetic energy, we can call it
𝐾𝐸, is 1.92 times 10 to the negative 18th joules. Knowing all this, we first want to
answer this question. What’s the charge of each one of
these particles?

Now, notice that we don’t know what
the particles are. And we don’t know what their mass
is. Nonetheless, given what we’re
given, it’s still possible to answer this question. We’ll do it by recalling a
mathematical relationship between particle kinetic energy, potential difference, and
particle charge. According to this relationship, if
we have a particle with charge 𝑞 and that particle moves through a potential
difference 𝑉, then if we take the product of those two values, we’ll find the
kinetic energy that the particle acquires. Now, in this scenario, we’re told
that kinetic energy. And we’re also told the potential
difference 𝑉. And we want to solve for the charge
𝑞. So let’s do that. Let’s rearrange this equation so
that we have 𝑞 by itself on one side.

To do that, we can divide both
sides by the potential 𝑉, which means that cancels out on this right-hand side. And we see that particle charge 𝑞
is equal to kinetic energy divided by potential difference. Our next step is to substitute in
the given values for kinetic energy and potential. And once we do that and calculate
this fraction, we find a result of 3.2 times 10 to the negative 19th coulombs. So that’s the answer we’ll give to
this first question of what is the charge of each of the particles emitted from this
block of unknown material. Each of these particles has a
charge of 3.2 times 10 to the negative 19th coulombs. Now let’s move on to the second
part of our question.

Our second question says, using a
value of 1.6 times 10 to the negative 19th coulombs for the charge of a proton, what
is the relative charge of each particle?

And talking about each particle, we
mean the particles emitted from this block of unknown material, the particles whose
charge we’ve solved for in the earlier part. To answer this question, we need to
know that the relative charge of a proton, we’ll symbolize it with 𝑃, is positive
one. So rather than 1.6 times 10 to the
negative 19th coulombs, which is a somewhat clumsy number being used for that
charge, as a shorthand, we can say that that amount of charge, 1.6 times 10 to the
negative 19th coulombs, is equal to one charge unit. And the question is, based on the
fact that a proton has a relative charge, we could call it a plus one, we want to
know the relative charge of these unknown particles.

Here’s how we can figure that
out. We can take the charge of our
unknown particles that we solved for earlier, that charge is 𝑞. And then we’ll divide that charge
by the charge of a proton, not the plus one but by the true charge of a proton. That is, its charge in units of
coulombs. So to figure out the relative
charge of these unknown particles, we’ll substitute in the value we found for
𝑞. And then, when we perform this
division, we find that the answer is simply positive two. And that this, therefore, is the
relative charge of these unknown particles. They have a charge magnitude twice
as big as the charge of a proton and the same sign charge, positive. Knowing all this, let’s answer a
third and final question about this scenario.

It is found that the charged
particles are being emitted by radioactive decay of the material in the block. What kind of particles are being
emitted by the block?

Okay, so back to our block, we’re
told that the particles emitted by it, which right now we don’t know what they are,
are being emitted, thanks to radioactive decay of some material, in the block. Now, based on that, that narrows
our options for what kind of particles these particles being emitted by the block
are. The particles could be neutrons or
beta particles, electrons or positrons. Or, they could also be alpha
particles. The way we’ll figure out which one
of these particle types is actually being emitted will be based on the charge value
of these particles that we’ve solved for previously. We know that whatever these
particles are, they have a charge of 3.2 times 10 to the negative 19th coulombs. In other words, a relative charge
of plus two, two times the charge of a proton.

Now if these particles have an
overall charge of two times the charge of a proton, then it makes sense that we
would imagine that two protons are contained within these particles. And it’s at this point that we can
recall that an alpha particle, also known as a helium nucleus, is a particle that
has two protons as well as two neutrons in it. Now, since these neutrons, the
green dots, have no electric charge, they don’t contribute anything to the overall
charge of this particle. It still has a charge of plus
two. So this is an alpha particle, two
protons plus two neutrons. And the charge of an alpha
particle, we see, is the same as the charge we solved for, 𝑞. This charge is different from the
charge of a beta particle and, of course, different from the charge of a neutron,
which has no charge. This then is our answer to this
final question. The particles being emitted by this
block of unknown material are alpha particles.