# Lesson Video: Rutherford Scattering Physics • 9th Grade

In this video, we will learn how to calculate the distance of closest approach between two charged objects by considering energy conservation.

15:03

### 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.