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A proton I5 ejected from @ very far d stance (where olectre porental energy 0) {owards an Alpha particle . The proton was complelely stopped at a distance T 3 X10 &...

Question

A proton I5 ejected from @ very far d stance (where olectre porental energy 0) {owards an Alpha particle . The proton was complelely stopped at a distance T 3 X10 "m of the Alpha particlo; How much kinetic energy must the prolon have initially to get this close %0 the alpha particle? (charge of Alpha particle +20)

A proton I5 ejected from @ very far d stance (where olectre porental energy 0) {owards an Alpha particle . The proton was complelely stopped at a distance T 3 X10 "m of the Alpha particlo; How much kinetic energy must the prolon have initially to get this close %0 the alpha particle? (charge of Alpha particle +20)



Answers

A proton (mass $1.67 \times 10^{-27} \mathrm{kg},$ charge $+e$ ) is fired directly at a lithium nucleus (mass $1.16 \times 10^{-26} \mathrm{kg}$ charge $+3 e$ ). If the proton's velocity is $5.24 \times 10^{5} \mathrm{m} / \mathrm{s}$ when it is far from the nucleus, how far apart will the two particles be when the proton is at rest, just before it turns around? Assume the nucleus is free to recoil.

This problem wants you to find the final kinetic energy of a particle that starts at rest in travels one centimeter under the influence of an electric field, the most basic formula for work. His work is equal to force times distance. This will be useful since we know all of the work that the electric field is doing on this particle is going to go into kinetic energy, so our work will be the same as kinetic energy. We have distance. It's one centimeter we don't have force Force due to an electric field on a charged particle is f equals Q times. The magnitude of the electric field. Both of these air given to E e. Being the charge of an electron is the charge of our alpha particle, and our electric field is 10,000 volts per meter. This multiplies to 3.2 times 10 to the minus 15. Nunes. Going back to this formula are kinetic. Energy is equal to the force. We just found times the distance of one centimeters this sequel to 3.2 times 10 to the minus 17 jewels. There's your answer. Good luck

16.20. So we have a proton and an Alfa particle, which has twice the charge of a proton and four times its mass. Because it's made out of two protons and a neutral two protons and two neutrons, we have them separated by four centimeters four times 10 to the negative 15 meters, and we have them initially arrested. Then we release both of them simultaneously. So first we want to, uh, explain why, if they're released simultaneously, why you can't find what they're speeds at. Infinity are using only conservation of energy. Then we want to suggest another conservation law that might be applicable to help us overcome that problem. And then we want to actually find what their speeds are at infinity. So because the Alfa particle is more massive them, then the pro Todd so roughly four proton masses is equal to an Alfa particle mass. The reason that this is only roughly equal is something to learn about later in your physics career. So they're they're speeds at any time. Well, first of all, this, uh, this be so they'll have an equal and opposite force is that they're experiencing from Newton's third Law But this means that the acceleration of the proton is not going to be equal to the acceleration of the Alfa particle, which means that there their speeds, we're going to be different at all times. So because we don't have a symmetry here, where we can argue that the speed is going to be the same for both of them at any given point in time or if we're holding one of them still let me know its speed is always zero we could, you know, then use conservation of energy. But here it's not good enough because we have one equation. You know, the conservation of energy. Hey, with two unknowns, which is the two speeds and you know this doesn't work, it can't. You can't determine to unknown things. With only one equation, you can find a relationship for them so you could express the speed of divinity of the proton and say, as a function of the speed of the Alfa particle. But that doesn't determine either one, uniquely, at least, so now the other conservation weaken Conservation, while we can use is the conservation of linear momentum, which is, you know, useful even in situations as we've seen, where energy is not necessarily concerned in a way that can be accounted for any given situations like any elastic collisions. Conservation of momentum still works, even though you would have a difficult time modeling the loss of energy that occurs from the, uh, you know, whatever you're hitting together, deforming any elastic Lee or even them like sticking together, et cetera. So now let's use that along with the conservation of energy, because we don't have anything in the last going on here. Of course, they're not even hitting each other. Um, so we can use that too. Now determine what the speeds are because we have two equations and two unknowns. So the conservation of energy is going to give us that the kinetic energy initially. All right, I'm sorry at the end. Well, at all times. But, you know, we're just concerned about the kinetic energy of the end. Eso won half em proton, proton squared, plus the mess of the Alfa Times its speed squared okay, minus zero. Because the initial the initial kinetic energy zero because their arrest plus the difference in potential energy, we have zero initial zero potential energy and infinity minus the initial, which is okay, you Proton. You off, huh? Over. The initial separation is equal to zero. So then this tells us we're gonna start a new age here. So then this tells us that, um, the sum of the M V squared is always going to be a constant, the constant being two times cool arms, constant times the product of their charges provided by their initial separation. And this evaluates to 2.3 times 10 to the negative. 13th. Jules, you can see the jewels. They're not a very appropriate unit of energy for subatomic physics here, because this is ridiculously tiny. But, you know, that's again. Something will be alert about later on. So then the conservation of momentum gives us that the momentum of the alpha particle plus the momentum of the proton is always going to be zero, because that's the initial momentum. So the the speed of the alpha particle is always going to be equal to the ratio of the masses times the speed of the proton. So now, if we substitute this into our first equation to eliminate the speed of the health of particle, we get em Alfa Times this mass ratio squared spends the protons speed squared plus the mass of the proton times its speed squared is equal to Let's just call this e for the total energy. So then simplifying a little bit, we get MP over and, uh, squared plus one. Oh, no. Sorry. We don't even both of those cans. Sorry. Uh, it's just empty over. I'm out because one factor of, um, Alfa, um, cancels their then this gets multiplied by MP. There we go. B squared equals e. And so then just solving that for the speed we get that it's 1.5 solving it for the speed of them, evaluating at 1.5 times 10 to the seventh meters per second just pretty fast. And now we can Sorry about that. Now we can find that the speed of the alpha particle. We already know that it's the ratio of these two masses times the speed of the proton, where we know the mass of the proton and the mass of the alpha particle. As we're told these things and again, the ratio is going to be roughly 1/4. But you should use the actual numbers that that are available so that you know that approximately equals thing doesn't cause a problem. And so the speed of our Alka particle at infinity will be 2.64 times 10 to the sixth meters per second, which is less, as you'd expect, because it's more massive. And so, in order for it have equal Valentim. It has to be going slower, but still still pretty fast.

Hi guys. Now we will solve question 34 here, the alpha particle is accelerated through revolt. And for this alpha particle the kinetic energy will be converted into potential energy. Now we know potential energy is he goes to one by four pi epsilon, not Q one Q two by our where Q one and Q two are the charge of alpha particle and the nucleus, respectively. And R. Is the distance of closes approach. And the kinetic energy will be Q one V. Where key one is the charge of the alpha particle and V is the vault. Okay, so we can say that key one V is because to one by four by epsilon. Not he wants you to buy art here. In this situation we can cut this to anyone from opposite sides. So from here we will get distance of closes approach equals to one by four pi epsilon. Not queue to buy. We So from this equation one we can see that the distance of closest approach is independent of the charge of alpha particle. So we can say that the distance of closest approach for proton will be Exactly equal to the ar of alpha particle because it's independent of the charge Q1 and we are using the same potential v. So the distance off closest approach of the proton will be are too therefore option air are is the correct answer. That's all. Thank you

In this question we have content scattering involving proton. So um Mhm Yeah, two parts in this question. We want to calculate the energy of the incoming photon and I can have the energy that that the proton received team. Okay, so to solve this problem. Okay first we know that we are working with proton. So first we calculate the content with playing proton. Okay so this is a church over MPC. Okay, so just putting the num relevant values. Okay. And you calculate this to be 1.32 times 10 to the negative 15. Yes. Then using the contents gathering equation he will say that I'm down mine islam that No. Is he going to come to see in this case proton one minus for science data. Okay. And then the scattering angle is 45 degrees. Okay everyone to calculate the energy of the incoming photons. So we'll calculate Bandana. Okay so lambda lambda see yeah one minus society data. So now you can put in the values. Okay, so London is given to be 6.2 times 10 to the negative 13 m. He then you I'm going to come to the thing on proton. The anger is 45 degrees. So you calculate this to be um oh sorry. So this should be minus key and then you get the London are to be 6.196 hands tend to die negative 13 m. And then the energy he is H. C. Over from that. Not. And then putting the relevant comes values. Okay, using your calculator, you get the energy to be 3.21 times 10 to the Hello, negative 13 shoes and convert to MTV. You get 2.1 Thank God, electron volts. Okay. Okay. So we can choose to live your answer in jewels. Okay. Or in mega electron words. Okay, fine. Okay. And then you put me you want to calculate that um, energy at the proton received the Canada energy. Okay. So using conservation of energy case of the K of the proton, is he going to die? You know, I minus E uh, skeptic the energy of the incoming photon and uh, uh, energy of the scattered proton. Okay. So, you know, it's 2.1 And maybe he brightness hc over lambda. Okay. Substitute the relevant quantities. Yeah. And also convert this into E. V. So I would be getting 1.25 K E V. As the answer. Okay. So this is the kinetic energy that the proton receive. So this is the answer for bobby and that's all for this question.


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