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Problem 60 Hard Difficulty

A by-product of some fission reactors is the isotope $_{94}^{299} \mathrm{Pu}$ , which is an alpha emitter with a half-life of 24 000 years:
$$_{94}^{239} \mathrm{Pu} \quad \rightarrow \quad_{92}^{235} \mathrm{U}+_{2}^{4} \mathrm{He}$$
Consider a sample of 1.0 kg of pure $^{239}_{94} \mathrm{Pu}$ at $t=0$ . Calculate (a) the number of $^{239}_{94} \mathrm{Pu}$ nuclei present at $t=0$ and $(b)$ the initial activity of the sample. (c) How long does the sample have to be stored if a “safe” activity level is 0.10 Bq?

Answer

a. 2.5 \times 10^{24}
b. 2.3 \times 10^{12} B q
c. 1.1 \times 10^{6} y r

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Video Transcript

for number 60. We have this polonium 2 39 one kilogram of it. And we know the half life of that is 24,000 years. And we want to start by finding how many nuclei that is. I'm just gonna do a little like I was converting units. Just do one big long conversion. So, you know, I'm starting with one kilogram. I'm trying to get that the number of nuclear. If I get the number of Adam's, that will be the number of nuclear. So that's where I'm heading. You know, for this in here, one kilogram is 1000 grams. So kilogram canceled. I know the one u it is the same as 1.67 times 10 to the negative 24th grams. So Graham counsel and I'm in use, and I know it's pulling plutonium 2 39 So I looked it up in the Appendix B and I get that. The mass of one Adam is 2.239 point 05 to wants 156 I don't need all this. I do that when I get the Q valiant step for this. I probably don't need that many significant things without so many use are in one, Adam. We're also one and each Adam is one nuclear. I said this will be my number of nuclear. So times that divide by that divide. By that I get to point 50 times 10 to the 24th nuclei. So this party, poor baby. We want to know what's the activity of that? Well, and in my equation, it relates activity to number, radio, uh, number of radio times the decay constant is the activity. So I have the number of nuclear. I just need to figure out the decay constant. Well, that's the other relationship I know is that half life, this is natural luggage too, which is 0.693 over the decay constant. So I'm gonna rearrange this and plug it in over here to get this are so I know my and I found that already it's me. Number nuclei. 2.5 times 10 to 24. We're gonna multiply that by Medicaid constant. I'm just gonna switch places of these two. So 0.693 on top, and the bother is gonna be the half life. But the half life has to be in seconds for this. I'm just gonna throw my conversion right in here. So I'm starting with 24 1000. Remember, that's years. I'm going to multiply that by 3 65 Change that two days. Multiply that by 24 to change at the hours multiplied by 30. 600 to go from ours, too. Seconds. So that's on the bottom. There. Just put my conversion rate in there and I get that. Or so this will always are. So our equals 2.29 times 10. 12. And they would be measuring back row and then from your last part port. See, You know, how long does it take until this only has an activity of 0.1? So right now, the activities this and I wanted to be able to 0.1 back room. So money is my equation. I always like to use words. Number nuclei is the original number of nuclei. Times 1/2 raise the number of half life. Now I know I want activity, not nuclear. But remember, if I multiply both of these bye, that decay constant both sides of the equation. I could do that. Algebra. Now these are activity instead. So I want my new activity to be 0.1. My original activity was this. 2.29 times 10 to 12 on delivery times, 1/2 raised to the number of half Life's. Well, if I can figure out the number of half lifes, then I will know how long it is because I know how long half life is over here. So I'm going to take the log of both sides when we divide this first and then divide this. I'm dividing both sides. I'm taking this. So that said the equation dividing and I get we're point three, 67 times 10 to the negative 14. So that will equal this 1/2. We're gonna write to this 0.5 now. 0.5, raise to the end. So I'm gonna take the log of both sides. Look of this number log of this lover. Remember how logs work when you have raised the exponents? I could bring that out in front of being coefficient instead. So now since I'm running out of room here, I'm gonna say a log of this divided by log 0.5 and no equal may end so n I get 44 0.38 Remember what that is? That's not years. That's the number of half lives. So and I know 1/2 life is 24,000 years, so I'm gonna multiply that by 24,000 and I get that the time period is one point. 07 time instead of the six. And over the years, that's my desperate point. Oh, I can't put it down here, but, you know, said, um, so they were gonna number 60.

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

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Liev B.

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Marshall S.

University of Washington

Aspen F.

University of Sheffield