A 4.78-MeV alpha particle from a 226Ra decay makes a head-on coll...

Question

A 4.78-MeV alpha particle from a 226Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Accepted Answer

Hey, everyone. So this problem is working with potential energy and kinetic energy and really the conservation of energy when we're working with electric charges. So let's see what they're giving us and what they're asking of us and we'll get into it. So we know we have alpha particles and they have a kinetic energy of six mega electron volts. They are shot at Bold foil. There is a collision head on and the particles are elastically scattered. So we know it's an elastic collision and the um charged particles are scattered but the gold atom remains at rest. OK. So it's a two part question. The first part is asking us the minimum distance um that can be reached between the two particles and then at that distance, what the force is acting on the alpha particle. All right. So first, let's recall um our conservation of energy equation. So that's just the kinetic energy and the potential energy at one point equals the kinetic energy and the potential energy at a second point. So we know at point a um where we have our alpha particles being fired at the Gold foil, the kinetic energy of the alpha particle is six mega electron volts. And um there is no potential energy there because it's so far away from the gold foil. For the um second point, this point B is when the elastic collision happens. And we know that if it's an elastic collision, there will be an infant fantastical point in time where the particle is at rest. And so the kinetic energy is zero and it's only potential energy at that point. So the next thing we need to do is recall our potential energy equation where U is given by K times Q one Q two over D. So this is good because D is the distance that we are trying to find in um part one. And so what we can write here is that the kinetic energy of the alpha particle is equal to the potential energy of the, of the same particle at point B at the, at a different point in time and space. And so that is KQ one Q two over D, we're gonna rearrange this equation for D and then look at these Q, these charge terms. So Q is the, the charge term is just given by the atomic number of the particle times elementary charge. So if we can recall that the elementary charge is um 1.6 times 10 to the minus 19 Kums, I'll write out here in a second, we actually have everything we need to solve for this distance. So um K let's recall is planks constant. No cool is the Coolum constant 8.99 times 10 to the ninth Newton meters squared ho squared. And like I said, charge one charge of the alpha particles. So we know that the atomic number of an alpha particle is two. And we'll recall that elementary charge 1.6 times 10 to the minus 19 jules for the um for Q two, the charge of the gold particle, they give us the atomic number as 79 up here. And so that's the same thing 79 times 1.6 times 10 to the minus 19 Jews and all of that over our kinetic energy of the alpha particle at point A and that was six mega electron volts. So I'll rewrite that as six times 10 to the six electron volts. And then we need to, for our units to work out, we need to convert electron volts to jewels. So let's just recall that that conversion factor is 1.6 times 10 to the minus 19 joules per electron volt. So we do all of that, plug it into our calculator and we get a distance of 3.79 times 10 to the minus 14 m. OK. So we go back up to our potential answers and we can see now that we can actually eliminate everything except for C but let's still do part two to make sure that we, you know, fully understand the problem when we're on the right track. So part two is asking us for the force acting on the alpha part goal at this point in time. So for the force we need to recall Klum's law where the force acting on the particle is given by KQ one Q two over D squared. And we actually have everything we need. The charges don't change again because they're just a function of the atomic number. We solve four distance in the first part. And so we are just going to um kind of plug and chug from here. So we're called Kums constant K and the two charges two for the alpha particle and 79 for the gold particle all over our distance squared plug that in and we get 25.3 newtons. The answer to part two, go back up and look and yes, that aligns with what is given to us in answer choice C so C is the answer for this problem. We will see you in the next video.

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