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[xperimcntRadioactive DecavHalecQuestonLocHn190lorinutS0Cu rd onealKa ecar...

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[xperimcntRadioactive DecavHalecQuestonLocHn190lorinutS0Cu rd onealKa ecar

[xperimcnt Radioactive Decav Halec Queston LocHn 190lorinut S0Cu rd oneal Ka eca r



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Which of the following nuclides would you expect to be radioactive: ${28}^{2} \mathrm{Ni},{ }_{29}^{58} \mathrm{Cu},{ }_{47}^{108} \mathrm{Ag}$, tungsten- 184, polonium206 ? Justify your choices.

Okay, So for this question, uh, the first thing that we should do is determine the number of protons and a number of neutrons each of these atoms have. So for phosphorus, we have 15 proton because atomic number is 15. And for neutrons, it's the atomic mass. Neutrons is atomic mass. My honesty. Number off protons, which is 16 right now for ST we have proton is equal to dirty. Neutron is equal to 66 minus 30 which is 36 for chlorine. We have proton is 17. Neutron is 18 for gold. We have proton is 47 evolution? We have 61 now. That accident we should do is we should determine the the end is the ratio of the neutron to proton ratio. Right? So for phosphorus, we have 16/15 by 16. Over 15 is 1.6 of six. Repeating for zinc. We have 36/30 right? 36/30 is 1.2. For chlorine. We have 18/17 18/17 is 1.5 nine. And for a G, it's 61/47. 61/47 is equal to 1.2 98 But so, looking at all of these, we see that the neutron to proton ratio is highest for gold. So this one is the

So this question welcome. Definitely intimidating. It will take probably a little bit of time. It's really just about balancing out reactions and making sure we have the same number of atomic number of mass on each side. And so, for this first question, we start by. Let's figure out what the atomic number of authorities So you don't need appear on tape of this question. And so you looked at off. The atomic number of thorium is 90 now. An Alfa particle is like this has a mass number of four, an atomic number of too. So if we're basically if we're losing him four from our mass number, we're going to get to 30 to minus four. That's to 28. And for losing to from our atomic number, that's gonna be 88. So now we have to look up in our table. What is 88? Well, uh, 88 is radium. No Russek. Well, what is a beta particle? Beta particle is negative. One for the atomic number zero for the mass number. And so our mass is going to stay the same. But if we have 88 were you need something that minus one is immediate. That's 89. You see? 89 minus one from here gives us 88. So look up. What is 89? That's acting iam a C, and we go through one more beta emission. And that should give us that while you have that minus 10 You know, we should get 90 over here. That lines up with the auditorium is perfect. That's what we should do that. So we do this first. Now, Uranium, look up. What its atomic number is that's 92. The awful park. A little admit is gonna be like this mass number of four atomic number of two. So if we reduce the mass number by 4 to 35 minus four, that's 2 31 Reduce the atomic number by two. 90 to minus two. That's 90. So now you look that up in a table. That story. Um Hey, Max, the mission. We have this beta mission here. A beta particle is like this. The mass is going to stay the same because the mass number of the beta park was zero. So we have to 31 still. Now we're gonna need the atomic number to be 91 because we have 91 minus one should give us our atomic number of 90. So look up. When 91 is that's ah, protect idiom We live in difficult one to pounce on. I'm unsure if I'm doing track plea entirely. Um, the next Alfa particle for the next particle that we have to Mitt is now a particle. And so now this is the sort of confirmation stuff to make sure this is correct. The mass number should go down by force A to 31 minus four. That's 2 27 Perfect. And then we have 91 minus two. Should give us 89. Well, let's see is 89 act idiom. Yes, it is. So we've done this problem correctly Last last one. So now we have a little bit of working backwards where we're told we have protected me in here, which, if you look up the table, that's ah, atomic number of 91 are alpha particle is still the same outfit particle. So now I have to work backwards. What mass should have something be for if it lost four to become to 33. So in other words, to 33 plus four. That's 2 37 That was the massive. Whatever he had before the atomic number went down by two and got its to 91. 60 91 plus two is 93. So let's look up what we should have in the period table. That's Ah, Neptune iam So empty. That's what we're starting. Our next mission is a beta mission. So looks like that mass number stays the same. So it's still 2 33 sharing it down here. Um, this 91 is going to become 92 because we need to have slim at 92 minus one from the beta mission should give us 91 wherever we started with. And so you look that up with What's that in the table with? That's uranium? Um, a quick out, you realize like they're like beta particles. Sort of the opposite result of what you would expect. Generally you think, oh, decay would give us like a smaller atomic number. Not always beta particles to the office. Now, last one. Here we have an Alfa particle being admitted. 2 33 minus four. That's 2 29 90 to minus two is 90. You look it up in the table. That story, um, that's our final answer

In this problem, we have been given the half life of an isotope, a radioactive isotope and the initial quantity And we need to determine the amount of 3000 years and after 10,000 years. Now, if why is the amount remaining after time t the amount of the isotope remaining ingram's after 30 years. Then, since the initial quantities 10, well, y is equal to 10 times half to the power T divided by the half five, which is 1599 So this will be the required model. So if we want to determine the amount left after 1000 years, we will have Y is equal to 10 times half the power 1000, whereby 1599 that is we such a duty is equal to 1000 and that all of that is approximately equal to 6.48 So the amount Of the isotope remaining after 1000 years is approximately 6.48 grants. Now if you want to find the amount after 10,000 years, we need to substitute Easy go to 10,000. We hear Y is equal to 10 times half to the power 10,000, divided by 1599. And the value of that is approximately 0.13, hence the amount of the radioactive isotope remaining after 10,000 years, approximately equal to 0.13 grands.

Mhm. Okay, for about a of this question we have, we have to write the equation for the products with beat up positive emission, we have 28. Forceful 15 undergoes better positive emission which decrease the atomic number by one. So the product formation will be 28 14 silicon plus zero one better. The number of neutrons equals so number of neutrons will be equal to mass number. So mass number minus atomic number which is 28 minus 14 equals 14. So we have a number of protons equals the number of sorry. We have a number of protons equal atomic number which is 14. And so the ratio of neutron to proton, which is end by Z equal to one. Buy a beater. Positive emission Moving on to option B. We have 45 potassium 19 again. Better positive emission decrees Your atomic number By one it was 45 18 Organ to 01 better. So number of neutrons equals 45. Minus 18 equals 27. A number of protons will be equal to atomic number, which is 18 so end by Z ratio does not come equal to one by Evita Mission. We want to option C. We have zinc 73 0 in 30 again. Better emission mass remains the same copper and there will be a decrease in atomic number which is 29. The 01 vita. So number of neutrons equals 73. Minus 29 equals 44. Number of protons equals atomic number, which is equal to 29. So the ratio of neutron ratio of neutron too proton does not come does not become equal to one. So only option A, which is 28 b 15 can be expected, can be expected to decay by Dita positive emission. We know that a radioactive decay to obtain a more stable I'm or more stable product. The better positive particle is represented by Byetta, signed with mass number zero and atomic number one. The radioactive species with the neutron to proton ratio, which are closer to one, are set to be most stable. The species that come close to the ratio of one by beat a positive emission will be expected to decade by beat a positive emission only


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