### Video Transcript

The temperature in the core region
of a star is modeled as being 3.5 times 10 to the seventh degrees Celsius. A singly charged ion at rest could
be accelerated by a potential difference that would give the ion kinetic energy
equal to the average kinetic energy of ions found in the star’s core region. What potential difference is
required to produce that energy?

To picture the situation described
here, consider a star where we move further and further interior till we reach the
core of that star. That core has a temperature we’ll
call capital 𝑇. And we’ll use the temperature given
to us in the problem statement: 3.5 times 10 to the seventh degrees Celsius, very
hot. To get a better view of this
core. Let’s zoom in on it. Looking up close at this core of
the star, which is at a temperature 𝑇, we can imagine ions within this core which
move according to the average temperature of the core. The hotter the core temperature is,
the higher the average kinetic energy of these ions. So that’s one scenario: the very
hot core of a star.

Our other scenario is to imagine a
singly charged ion which starts out at rest. Singly charged, all that means is
that the ion has an extra proton or an extra electron. This extra charged particle gives
it an overall or a net charge so that it becomes an ion. Given this singly charged ion, we
imagine that it’s exposed to a potential difference; we can call this Δ𝑉. And we know that in the presence of
such a potential difference, a charged particle will begin to accelerate. Starting out from rest, this singly
charged ion speeds up faster and faster and faster. Eventually, the kinetic energy of
this speeded up ion will be equal to the average kinetic energy of the ions in the
core of the star. What we want to know is what is the
potential difference — that is, what is Δ𝑉, that would be required to produce this
energy equality? That is, what potential difference
Δ𝑉 would we need to speed up this ion enough so that its kinetic energy will be the
same as the kinetic energy on average of the ions in the core of the star?

Since our question is focused on
energy, let’s start to consider the energy of the ions in these two different
scenarios. First, the energy of our singly
charged ion which begins at rest. Since it starts out at rest, we
know that there is no kinetic energy initially for this particle. But it does have electrical
potential energy. The electrical potential energy of
this particle is equal to the potential difference that the particle experiences
times the charge of the particle 𝑞. Speaking of the charge 𝑞, since
this ion we’re considering is singly charged, that means it has a charge equal in
magnitude to the charge of a single proton. So we can write that for our
scenario: the energy of this singly charged ion is equal to Δ𝑉, the quantity we
want to solve for, multiplied by the charge of a single proton. And while we’re at it, we can
recall that charge is equal to 1.6 times 10 to the negative 19th coulombs.

Now let’s look at the other
scenario, which is the core of our star. In this case, we’re considering the
average kinetic energy of the ions in this core. What is that average kinetic
energy? Well we can look up that
mathematical relationship, which in fact is in terms of the temperature of the core
of the star capital 𝑇. The average kinetic energy of each
of the ions in this star’s core is equal to three-halves 𝑘 multiplied by the
temperature of the core in Kelvin 𝑇. We have to be careful here
though. This 𝑘, which we’re used to seeing
as Coulomb’s constant, is Boltzmann’s constant in this case. In fact, even though we would
normally see the equation written this way, let’s give this 𝑘 a B subscript to make
sure we differentiate it from Coulomb’s constant. And when we look up the value of
this constant, Boltzmann’s constant, we find it’s equal to 1.38 times 10 to the
negative 23rd joules per Kelvin.

This is what we’ll do next: we’re
going to equate the average kinetic energy of the ions in this core with the
electrical potential energy of this singly charged ion. In other words, we’ll write 𝑈 is
equal to KE sub avg or Δ𝑉 times 𝑞 sub 𝑝, the charge of a proton, is equal to
three-halves Boltzmann’s constant 𝑘 sub B times the temperature of the star’s core
in Kelvin. After all this, we can recall that
it’s Δ𝑉 we want to solve for: that’s the potential difference we would need to
apply to our singly charged ion to make this equality true. To solve for Δ𝑉, let’s divide both
sides of our equation by 𝑞 sub 𝑝, the charge of a proton, so that cancels on our
left-hand side. After all that then, we see that
Δ𝑉 is equal to three 𝑘 B times 𝑇 over two 𝑞 sub 𝑝.

In this equation, 𝑞 sub 𝑝 is the
charge of a proton, which we know; 𝑘 sub B is Boltzmann’s constant, which we also
know; and 𝑇 is the temperature of the core of the star. Now this is where we have to watch
out because that temperature is given to us in degrees Celsius, but we’ll need to
convert it into Kelvin. And we can see the reason why as we
look at the units in our Boltzmann’s constant. The units are joules per Kelvin,
and that means that we will need to multiply it by another temperature in Kelvin to
cancel that unit out. So how do we do that? How do we take our temperature from
degrees Celsius to Kelvin? Let’s clear a bit of space to see
how. The way to convert from one
temperature scale to another is to remember how they relate to one another
mathematically. Given a temperature in degrees
Celsius like we are in our case, if we take that temperature and add 273.15 to it,
that will give us the equivalent temperature in Kelvin.

So that’s what we’ll do; when we
plug-in for our temperature 𝑇, we’ll use the value given to us but we’ll add 273.15
to it. That’s how we’ll accomplish this
conversion. Let’s go ahead and do that now. When we do plug in for these
values, we’re careful to write out Boltzmann’s constant with its units of joules per
Kelvin and to add to our number in degrees Celsius 273.15. Now it’s in units of Kelvin. Downstairs, we have two times our
charge of a proton 1.6 times 10 to the negative 19th coulombs. With all this in place, we’re ready
to calculate Δ𝑉. And to two significant figures, we
find it’s 4.5 kilovolts. That’s the potential difference
we’d need to make the average kinetic energies of these ions equal.