A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in 'head-on' to a particular lead nucleus and stops 6.50x10^-14 m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64x10^-27 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

Video transcript

Hey everyone. So this problem is working with potential energy. It's giving us a lot of information, but if we break it down, it's not actually as tricky as it first seems to be. So we have positrons, it gives us some information about them. They are admitted and sent, it doesn't really matter how they admitted that they are sent towards neutral copper foil. And they give us some information about copper. Um During that collision, the positrons elastically scatter and they give us the distance from the center of proper uh at which a positron stops as this value. So they're asking us for three things First, they're asking us for the potential energy of the positron when it stops. They're asking us for the initial kinetic energy of that positron and the initial speed of that positron. So the first thing that we need to do, like I said, we'll just take it one step at a time. For part one, we need to recall the potential energy between two positive charges is given as U equals K. Q one, Q two. The charges over are so we can recall that K is Fulham's constant. So that's 8.99 times 10 to the nine. And that's in newton meters squared per Coolum squared. We can recall that Q one. Um That's the charge of the positron. Actually they give it to us in this problem and that's 1.6 times 10 to the 19 problems. Q two is the charge of copper the copper atom and that is just gonna be the charge of the electron times the times E. So that's 29 times 1. times 10 to the -19 Williams. And then r is the stopping distance. And they actually give us that as well. It's 1.33 times 10 to the minus m. So from there we actually do have everything that we need to solve for potential energy. We're gonna plug and chug that into our calculator. Just write out all the terms here. So you have the columns constant, we have the charge of the positron, we have the charge of the copper atom. And then we're gonna divide all of that by our distance. And when we plug that into our calculator, we did 5.2 times 10 to the minus 17 jewels. Now, one last step here for part one, they did ask us for this in terms of electron volts. So we just need to recall the conversion factor between jules and electron volts is one point Six times 10 to the negative 19 jewels per electron volt. And that gives us 314V or the potential energy. So let's take a look at our answers here. We can actually eliminate everything except for part for answer B But let's go through the next two parts of the problem to make sure that we are on the right track. All right. So the answer to Part one we know is 314 electron volts is the potential energy. So the problem tells us that we have an elastic elastic scatter um during the collision. And so that means that due to the conservation of energy, the potential energy is all transformed into kinetic energy. So our kinetic energy is the same as our potential energy. And that's 314 electron volts. So that's really all we need to solve part two. Let's go back up to B. We see, yep, we're still kind of on track here. Both I both Part one and two are 314 bolts. And so the last step is if we recall, the kinetic energy Is given as 1/2 um B squared. They gave us mass in the problem Of the Positron. It's 9.11 times 10 to the - kg. And so we can just solve for this velocity for the speed term rearrange the problem. They're rearrange the equation there, plug in what we know, Oh except no, we're not going to plug in 314 electron volts. Why not because of the units that we're using? We need to go back to the value in jewels. So that was 5.02 times 10 to the -17 jewels all over the mass. And we get a speed of 1.05 times 10 to the seven L/s. And that is the answer for part three, let's go back up to answer B and, yep, that tracks. So the answer for this problem is the that's all I have for this problem. We'll see you in the next video.