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Question 3Hox many electrons are in an element wnose elecron configuration 5 152 252 2p" 3s2 3pt 4s2?You may use this Periodc table D15,CaZ04Question4ptsOxygen...

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Question 3Hox many electrons are in an element wnose elecron configuration 5 152 252 2p" 3s2 3pt 4s2?You may use this Periodc table D15,CaZ04Question4ptsOxygen-15 unstable and decays via beta decay: emitting = positron What daughter nucleus produced this reaction? %0

Question 3 Hox many electrons are in an element wnose elecron configuration 5 152 252 2p" 3s2 3pt 4s2? You may use this Periodc table D 15,Ca Z04 Question 4pts Oxygen-15 unstable and decays via beta decay: emitting = positron What daughter nucleus produced this reaction? %0



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Questions 1–3 (A) Alpha decay (B) Beta decay (C) Gamma ray (D) Fission (E) Fusion Select the term from above that identifies the nuclear reactions described in questions 1 to 3. $_{92}^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow_{38}^{90} \mathrm{Sr}$ $+\frac{143}{54} \mathrm{Xe}+3 \mathrm{n}+$ Energy

So we're told in this problem that we have a uranium nucleus and submitting an alfa particles the two protons and two neutrons. And the energy kinetic energy is 4.20 my girl electron volts. And then we want to know what the daughter nucleus is And, um, the mass of the daughter Adam. Um And then we can ignore the reek wild, the daughter nucleus of the kinetic energy of that. So, um, if we're starting out with 2 38 uranium geraniums, number is 92. If we lose of two neutrons into protons, that means we're losing a total of four nuclear eons. So this number is gonna now be, um, to 30. Oof! That looks like a five glitch to out. Oh, nice. Okay, so now we're gonna have 2 34 and we're gonna lose two proton. So this is going to go down to 90. And then if I looked this up on a periodic table, I see the 90 correspondences story. Oh, so that's t age. And then kind of by conservation of energy to get the final mass of the final The daughter, you can say the original mass so call us. Emma View is equal to the mass of the daughter. So I'll just call it, um th plus the mass of the alpha particle and plus the kinetic energy of the alpha particle. So, a mass, And then we should find this. We we want to find this. So, um, we can say that a moth or IAM is just the original mass minus the massive the alpha particle minus the kinetic energy of Alfa particle. And then, if go plug in numbers. So since I give us that the kinetic energy is, um, I'll put this over here kind of kinetic energy of the Alfa is 4.20 My girl electron wants, um, the 2.36 thanks. So we really don't have to use more than 36 eggs. I'll swear, actually, I guess that's more about um, that's that would be rules for multiplying. Um, but I'm still gonna just stick to three sink things. Hopefully, um, that is enough. So anyway, if I subtract thes, if not, you can always just look up more sick things. I don't think this is a critical part of the problem. So anyway, so here's the 2 38 I think the full thing was to 38.4 Um, I think it's supposed to be the actual the number past the decimal place. When you add and subtract that you have to match. I don't quite remember. Um, so sorry about that. Um, so anyway, so that's the mass of uranium. This is the mass of the alpha particle. It's like 4.0, too, I think. And then the cat ecology is 4.20 And so when I put that into a calculator, I obtained that. It was he was 2 38 minus four minus for Oh, I just realized I forgot to convert this unit. So let's go back and convert this to atomic blast units. Um, so I'm just gonna, like, do this conversion over here. So, um, 4.20 M E v. Using the conversion of, um 9 31.5 m e v proceeds squared, um, is one atomic mass unit. Then I obtained that, um, that's actually you want to subtract point for five from here, Okay. And then doing that math. So this this is the mass. The kinetic energy. An atomic classy odds. Um, so then when I plug all of that into a calculator, I got 2 34 Atomic Naciones. So that's the answer for B. And just to make this not ugly, I'm gonna box that and put a little B And then this is the answer for a

We have a discussion. The k off the oxygen 15 via electron capturing process to give us the nitrogen. 15. You click well, really to first by a process that's going on for a single particle over. Did you? Yes. All right. So basically was happening over here is that we see that the oxygen has lost and a proton right proton was lost, but the total nuclear number is unchanged Me set. I suggest that the neutron was added in its place. So it is likely that the proton is being converted into a neutron, right? So we can simply isolate up de proton over here being converted into a neutron like we don't changing the entire you know, changing t mean process off this arctic a reaction, right? So this isn t what is happening on a particle level for the second part beyond right D d k process in terms off neutral atoms. So in this case, all right, the initial equation there. Well, actually, looking eyes in terms off new clients, right? So this new place are charged because they're protons, is it? There's a charge plus eight you naturally has charge of plus seven. You know they make them into a neutral atoms. We can electrons on both sides, so we're not altering the process in any way. We're just adding bystander electrons, bumble sex. They're not affecting the direction whatsoever. Get that seven year actuals on both sites. If you do that, we see that No, the new kites can fall neutral atoms. Right. So this is simply a neutral nitrogen. 15 metal. Same Over here we have one total of electrons. Blast e oxygen. Ukraine, which is just e oxygen 15 at up from here, we can calculate what istea que value off this reaction. My just ticking de changing miss changing mess. In this case, you just in the mess off the initial rectums, which is t oxygen, oxygen, atom. But its way mess off the nitrogen atom. We're gonna assume you do funerals. Mess, Lis. Bye bye. See? Script? Yes. This guy is not 31.5 Emmy people. You you get a cue very to be to myself. Five tv in this Q value if you're to ignore the recoil off the daughter knew that s t nitrogen. Everyone ignore the required the natural choice, Mr. All the energy will go into the kinetic energy off of Ah neutrino. Therefore, the energy of the neutrino. It's just e que very off destruction, which is 2.75 MTV.

Everybody. So this one is asking us um okay let me find it. So from the energy point of view what should occur and what is the energy emitted? Okay if what is the energy admitted? If possible decays in part A. Okay. So pretty much we have when we get back here we have a and were given a c brilliant nucleus. Believe that's brilliant. We like actually write that nicely um co decay into. So I'm just gonna say it's going to decay into seven with them. Okay in the nucleus stable nucleus by capturing electron through right data DK. And so for electron capture with Q equals mm Burton minus M. D. C square. And was going to be M the brilliant minus M. Olympia C squared. And now it's gonna keep plugging and chugging. You get the numbers from your book or you can google it doesn't matter which one U minus 7.1605 You and now we have 9 31 M. E. Me. You. Okay? And that is going to equal to 9.25 times 10 negative four um times 931.5 M. E. And now we have 0.86 M E. V. And so for beta decay that's for the electron capture. Now when you have to do for beta decay, here we go. Who now we have Q equals M. P. And actually we're gonna write this lot better. Okay, brilliant. Okay minus lithium minus two M E C square. Okay, so now we know these two guys. So that is actually 0.86 M E v minus two times 0.511 And that is because M E C squared is equal to 0.511 M E B. Your book should also tell you this. And um so we end up with negative 0.16 M E P. And so only one electron is captured as possible. So it's going to be electron capture only. Is your answer for A and be is uh, and electron capture process. No neutrino will be admitted. Okay, Thank you.

So in this problem, the atomic number that is also the number off protons increases from 83 to 84 now. For this to happen, a neutron must eject a big a particle that is an electron and become a proton. So if the system nugent, it must eject a beta particle, which is an electron in order to become drop down. Now the overall mass off the atom remains virtually unchanged because the months off the ejected electron is very small. So for this reason since this is nearly equal to zero, the mass of the electron is negligible, therefore ah, and the massive neutral is nearly gold to the massive proton. Therefore, the overall months off the atom women's virtually unchanged before and after the reaction on hence of the atomic mass remains so unchanged on duh. So the light answers Oh, are the right reaction Corresponding to this problem is B because we're getting a beat. A ticket in this problem. Thank you


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