 In Rutherford's alpha scattering experiment, an alpha particle with a kinetic energy of 7 mega electron volts approaches a gold nucleus. It comes momentarily to rest and reverses its direction. The question is to figure out the distance of closest approach. As always pause the video and give this a try. Alright so over here we have an alpha particle which is moving because it has some kinetic energy that is 7 mega electron volts and it is approaching a gold nucleus. Now it comes momentarily to rest and then reverses its direction. So if we were to draw its path it would look somewhat like this. This alpha particle has a velocity so it's moving to the right, it's moving to the right, it slows down because of repulsive forces between positive charges, comes to rest and then reverses its direction and starts going in the opposite direction. And we need to figure out the distance of closest approach. That is when the alpha particle is momentarily at the state of rest. Then what is the distance? The distance between the two centers, the centers of the gold nucleus and the center of the nucleus of the alpha particle which has two neutrons and two protons. That distance would be this. This is the distance between the centers of these two nuclei. And this is the distance that we need to figure out. Now we can ask ourselves the alpha particle had some kinetic energy to begin with. Let's say that was the initial state and let's call this a final state when the alpha particle is at a state of rest. So where did that kinetic energy go? From the conservation principle we can say that energy would be conserved in this system. We can think of the entire system as containing the alpha particle and the gold nucleus. We can say that the initial, the total initial mechanical energy that should equal the total final mechanical energy. And total initial energy is the sum of kinetic energy and potential energy. And final is the sum of the final kinetic energy and the final potential energy. We know that there is some initial kinetic energy and we can assume that the initial potential energy is, this is zero. And we can say that there was no interaction between the alpha particle and the gold nucleus to begin with because they were far away. Finally when the particle is at a state of rest, it has no kinetic energy because velocity is zero, it is at rest. So we see that all the kinetic energy that the alpha particle had has been changed to the potential energy of the system. And in this case, the system is both the alpha particle and the gold nucleus. So initial kinetic energy, what is that? That is seven mega electron volts. So because it has a mega, we can write seven into 10 to the power six. This is electron volts. And one electron volt that is equal to 1.6 into 10 to the power minus 19 joules. So we can write that over here. This is equal to the potential energy of the system. And the potential energy between any two charges that was given by this relation. It's 1 by 4 pi epsilon naught into q1, q2 divided by the distance between the centers, r. In this case, that would be d, the distance that we are interested in, distance of closest approach. And 1 by 4 pi epsilon naught, this had a value. This was, this was nine into 10 to the power nine. So I'm just going to write nine into 10 to the power nine in place of this huge constant. So this is nine into 10 to the power nine into q1. So okay, let's look at the charge of the alpha particle. It has two protons. And we know that the magnitude of the charge on proton and electron is the same. The sign is different, but magnitude is the same. And it has two protons. So the total charge on this nucleus that is two into e into the total charge of the gold nucleus that can be reduced from the atomic number of gold. Atomic number is 79. So that means there are 79 protons in the gold nucleus. So the total charge would be 79 into e. This is divided by, this is divided by d. Now we know everything in this equation. And the charge of the electron e that is 1.6 into 10 to the power minus 19 coulombs. So we can place this value at both of these places. And I would encourage you to pause the video, carry out the calculation, and see what value of d you get. All right. The value of d that you should get that is 32.5 into 10 to the power minus 15 meters. At the scale of a nucleus, the units that we often use is femtometers. So one FM, one femtometer that is given by 10 to the power minus 15 meters. So we can express this, we can write this as 32.5 femtometers. Now this distance of closest approach, this number, it tells us something. It tells us that the size of the target nucleus, in this case gold, it has to be of the order 10 to the power minus 15 meters. It cannot be more than that. It cannot be 10 to the power minus 14 or minus 13. If the radius of the nucleus would have been of the order of minus 14 or minus 13, then the distance of closest approach would have been greater. It would have been again either 10 to the power minus 14 or 10 to the power minus 13. And you can check that from the diagram. If we increase the size of this nucleus, if we make it larger, the alpha particle will not come this close to the larger nucleus. It will go back much before. So even your d, even your d will increase, even your d will increase. So the distance of closest approach, it gives us some idea about the size of the nucleus, but it does not accurately tell us what exactly is the radius of the target nucleus. There are other experiments to figure out the radius.