### Video Transcript

In this video, our topic is
Rutherford scattering. We’re going to learn what this
scattering means, how it’s different from people at the Rutherford’s family reunion
scattering from one another. And we’ll also explore different
pathways that various scattered particles follow.

To get right into it, we can define
Rutherford scattering as elastic scattering between charged particles. So say that we have two particles
both with a net positive charge and that one of these particles is approaching the
other. Because both of these particles are
charged, we know that they will experience a coulomb force between them. And because these charges have the
same sign — in this case, they’re both positive — we know that force will tend to
push them apart from one another.

Now, this whole idea of Rutherford
scattering arose at a time when scientists were experimenting with bombarding atomic
nuclei with alpha particles. We can recall that an alpha
particle is a helium nucleus with two protons and two neutrons. And the idea was that this particle
could be projected at a high speed towards the nucleus of an atom. If we adopt this framework, we can
say that this is an incoming alpha particle and that this is a relatively much more
massive atomic nucleus. And we know that if a much smaller
mass approaches a much larger one, then the larger mass, thanks to its greater
inertia, will tend to move very little. In fact, it’s not unreasonable to
model the larger mass as being stationary, not moving at all throughout the
interaction.

So if we say that our much larger
atomic nucleus is fixed in place, then as our alpha particle, which is moving toward
that nucleus, gets closer and closer, its speed will decrease as the repulsive force
on it increases. And eventually, just for a moment,
that incoming alpha particle will be stopped. It has used up, we could say, all
of its initial kinetic energy getting closer and closer to this larger nucleus. When the alpha particle does come
to a stop, we say that it’s arrived at its distance of closest approach. This is the minimum distance
between the alpha particle and the nucleus it was getting closer to.

For an alpha particle moving on a
straight line directly towards a larger positive charge, the distance of closest
approach is determined by two things, first by the initial kinetic energy of the
alpha particle. And we can recall that, in general,
kinetic energy is equal to one-half an object’s mass times its speed squared. The other quantity this distance
depends on is the electric potential energy between these two charged particles. That potential energy is equal to
the product of the two charges divided by four times 𝜋 times 𝜀 naught, the
permittivity of free space, all multiplied by the distance between the two
charges.

Now the electric potential energy
between two like charges — and in this case, we have two positive charges — will
tend to keep the charges away from one another. And we see, consistent with this,
that the coulomb force between these two positive charges is repulsive. Not only that, but that repulsion
gets stronger and we can see our electric potential energy between the two charges
increases as 𝑟, the distance between the charges, gets smaller and smaller. It’s this increasing repulsion that
decreases the kinetic energy, that is, the speed, of our incoming alpha
particle.

When our alpha particle gets to the
point of being stopped completely — having a speed of zero — it’s at that point
exactly that we can say that the initial kinetic energy of the alpha particle is
equal to the electric potential energy it experiences. And so we can write this. When that initial kinetic energy is
equal to the electric potential energy between our two charged particles, then 𝑟,
the distance between them, is equal to what we’ll call 𝑑 sub 𝑐. That’s the distance of closest
approach. So then if our two particles are
that distance, 𝑑 sub 𝑐, apart, we can set the initial kinetic energy of our alpha
particle equal to the electric potential energy of our particles at this
distance. This means that one-half 𝑚𝑣
squared, where 𝑚 and 𝑣 refer to the mass of the alpha particle and 𝑣 is its
initial speed as it approaches the larger nucleus, is equal to the product of the
charge of the alpha particle and the charge of the larger nucleus all divided by
four 𝜋𝜀 naught times 𝑑 sub 𝑐, the distance of closest approach. We can then rearrange this
expression so that 𝑑 sub 𝑐 is the subject of the equation.

Getting back to our two charged
particles, which are now at that distance of closest approach, we can say what will
happen next. Remembering that our larger
nucleus, we’re saying, is fixed in place, the mutual force of repulsion on these two
charges will tend to push the alpha particle back where it came from. So the alpha particle, after coming
in directly at the larger nucleus, will stop some distance away, that’s called the
distance of closest approach, and then will start moving straight back in the
direction it came from. Note that this progression of
motion will only happen if the alpha particle was initially moving directly at the
center of our larger nucleus.

If we think about it though, there
are lots of other possible approaches besides this one. To see this, let’s clear our screen
and then consider a situation where we have separate alpha particles on different
paths that individually approach a relatively large atomic nucleus. As we consider the paths that these
various alpha particles will follow, we already know how this one will travel. It will move right towards the
larger fixed nucleus, moving more and more slowly as it goes, until eventually it
will stop at a point we’ve called the distance of closest approach. After it comes to rest, it will
begin to move backward away from this positive, mutually repulsive charge. But we can see that each of these
other alpha particles, which start out moving at the same speed as this one, won’t
have that same trajectory.

As each one moves left to right, it
will experience the mutual electrical repulsion from this positive charge. But the farther away our alpha
particles get from this positive charge, the weaker that interaction will be. Let’s say that this alpha particle
follows the dotted line we’ve drawn in. We can see that there’s been an
angular deflection in the direction this particle follows. We can say that the size of this
angle, we can call it alpha, is a measure of just how much this alpha particle is
scattered through its interaction with the larger nucleus. The larger alpha is, the more a
particle is scattered. And if we consider the motion of
this alpha particle once more, we could say that this is an extreme limiting case
where alpha is 180 degrees. So the bigger alpha is, the more
scattering has occurred. And the smaller alpha is, the
less.

If we then consider our next alpha
particle, this one here, in this case we could say that the angle of deflection is
this angle right here. And just by eye, we can tell that
this is smaller than the angle we’ve marked out as alpha. In other words, this alpha
particle, which is more distant from our positive charge, was deflected less. And this trend will continue as we
consider this and then this incoming particle. So we can make this summary
statement for alpha particles approaching larger positive charges at some fixed
speed. The more the direction of an
incoming alpha particle points away from another positive charge, the less its
angular deflection, that is, its scattering, will be.

And this makes sense because the
farther away from our central positive charge our alpha particle passes, the weaker
the force of interaction between those two charges will be. In this example, we’ve considered a
series of alpha particles which all are moving in different directions relative to a
central positive charge, but they all had the same initial speed. Having varied the direction then
but kept the speed constant, what if we do the opposite?

What if we have an alpha particle
pass by our central positive charge always moving in the same direction but now with
different speeds: in one case, a speed indicated by this blue arrow, in another, by
this orange one, and in a third, by this pink one? The greatest of these speeds is
indicated by the longest vector. That’s the pink one. And if we picture the path our
alpha particle would follow when it starts out with that speed, we can see that
there definitely is deflection, scattering. But that scattering is minimized
because our particle just doesn’t spend very much time near to our central positive
charge. This is thanks to the fact that
it’s moving relatively quickly.

But what if, instead, our alpha
particle had a speed indicated by this orange vector? This relatively slower-moving
particle would have more time to be strongly influenced by the coulomb repulsion
between it and the central positive charge. Therefore, its path would deviate
even more than the pink path. And our slowest moving charge, the
one with the blue vector, would demonstrate this most strongly. We could summarize all this this
way by saying that the more slowly an alpha particle approaches a positive charge,
the more it will be scattered, that is, deflected. And we said this comes down to the
fact that a slower-moving particle will have more time to strongly interact with
that central positive charge.

Keeping all this in mind, let’s
look now at an example exercise.

An alpha particle is fired toward a
fully ionized gold nucleus. The figure shows five possible
paths along which the alpha particle could initially move. For which path would the alpha
particle undergo the least deflection due to the gold nucleus?

Looking at our figure, we see here
the gold nucleus. And we’re told that this gold atom
is fully ionized. That is, all of its electrons are
stripped off. So there are no negative charges
here, only the positively charged nucleus. And then we have these five
possible pathways, path A, path B, C, D, and E, for an alpha particle to follow as
it approaches the nucleus. We want to know along which of
these five paths will the alpha particle be deflected least due to its interaction
with the gold nucleus.

To get a sense of what this
interaction will be like, let’s recall that an alpha particle is a helium
nucleus. That is, it consists of two
protons, we’ve drawn them here in blue, and two neutrons, we have them in green. So overall, an alpha particle has a
positive electric charge and so, we see, does our gold nucleus. Because these two objects have like
electrical charges, that means they’ll repel one another, push one another away.

What we’re going to assume though
is that because our gold nucleus is so much more massive than our much smaller alpha
particle, relative to the moving alpha particle, we’ll say that our gold nucleus is
stationary, fixed in place. So as our alpha particle travels,
potentially, all these different paths, it will move, but the gold nucleus will
not.

Let’s start looking at our path
options by considering path C. This path, if we follow it forward, leads directly to
the center of the gold nucleus. But as we think about the
interaction between an incoming alpha particle and that nucleus, we know that
physically the alpha particle will not reach the nucleus. The reason is that the closer the
alpha particle gets, the more strongly it will be repelled by an electrostatic
force, the coulomb force. So no matter how fast the alpha
particle is moving initially, that repulsion between it and the gold nucleus will
slow it down until eventually it comes to a stop. It will only be stopped for an
instant though. After that, it will start to move
back in the direction it came.

If we think about this motion in
terms of particle deflection, we can see that this is an extreme case. It wouldn’t be possible for the
alpha particle to be deflected more than it is here, where it goes in one way and
comes out 180 degrees opposite that. Path C, then, is an example of the
most deflection possible. And therefore, it won’t be our
answer for the path that undergoes the least deflection.

Now we saw that path C’s direction
was on axis with the center of the gold nucleus. But what if we move off of that
axis slightly? That is, what if we follow, say,
path B? In that case, our alpha particle
won’t come to a complete stop, because it’s not moving directly toward the nucleus,
but will instead be scattered or deflected. We see though that this deflection
is much less than that experienced by an alpha particle following path C. This is
because an alpha particle following path B is passing by the nucleus at a greater
distance. In fact, it’s generally true that
the more distance there is between our particle’s path and the gold nucleus, the
less that particle will be deflected.

We could show that this way. If we draw a line that passes right
through the center of the gold nucleus, since all five of our paths involve motion
that’s parallel to this blue line, then if we draw a line that’s perpendicular to
this horizontal blue one, the farther away from the blue line a particle’s path is,
the less that particle will be deflected as it follows that path. And that’s exactly what we want to
find. It’s the path that leads to the
least deflection of the alpha particle.

To figure out which of our five
paths this is, let’s extend this vertical line we’ve drawn a bit. And now we can pretend that it’s
the vertical axis of a graph. So this point right here will be
our origin. And then using a ruler, we’ll make
evenly spaced markings along this. Relative to these tick marks, these
markings, which we’ll leave without numbers or any units, let’s figure out where our
four remaining path candidates, paths A, B, D, and E, lie.

Starting at the top, we can see
that here is path A. That’s one, two, three, a little bit more than three tick marks
up away from our origin. Then right here, there’s where path
B crosses this vertical line. We can see that that’s less than
one tick mark. And then path D crosses it here,
which is slightly less than two tick marks away from our origin. In this case, whether we’re above
or below the origin doesn’t matter, only that total distance. And then path E, we see, intersects
this line here, at about two and a half tick marks from our origin. So of all five paths, path A is
farthest away from the origin. And that means an alpha particle
traveling this pathway will be farthest away from the gold nucleus and, therefore,
be deflected by that nucleus the least.

And so that’s our answer to this
question. An alpha particle following along
path A would undergo the least deflection of any of these path options.

Let’s summarize now what we’ve
learned about Rutherford scattering. In this lesson, we saw that
Rutherford scattering is elastic scattering of charged particles due to electrical
interaction. We consider the case of a
relatively smaller positive charge, an alpha particle, moving directly toward a
larger positively charged object. And we saw that as this goes on,
the alpha particle moves more and more slowly until eventually it comes to a
stop. When this happens, the two positive
charges are separated by a minimum distance called the distance of closest
approach. We called it 𝑑 sub 𝑐. This distance depends on the
initial kinetic energy of the alpha particle and the electrical potential energy
between this particle and the larger positive charge.

Along with this, we also considered
cases where incoming positive charges, alpha particles, were moving in directions
other than directly toward the larger positive charge and also at different
speeds. We saw that particles moving closer
to the central positive charge or at slower speeds would both tend to be deflected
more, while particles approaching more quickly or at a greater distance from that
central positive charge are deflected or scattered less. This is a summary of Rutherford
scattering.