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The half-life of $^{131} \mathrm{I}$ is 8.04 days. ( a ) Convert the half-life to seconds. (b) Calculate the decay constant for this isotope. (c) Convert 0.500 $\mu \mathrm{Ci}$ to the SI unit the becquerel. (d) Find the number of $^{131} \mathrm{I}$ nuclei necessary to produce a sample with an activity of 0.500$\mu \mathrm{Ci} .(\mathrm{e})$ Suppose the activity of a certain $^{131} \mathrm{I}$ is 6.40 $\mathrm{mCi}$ at a given time. Find the number of half-lives the sample goes through in 40.2 $\mathrm{d}$ and the activity at the end of that period.

a. 6.95 \times 10^{5} s

b. 9.97 \times 10^{-7} s^{-1}

c. 1.9 \times 10^{4} B q

d. 1.9 \times 10^{10} \text { nuclei }

e. 0.200 \mathrm{mCi}

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for number 19 we have iodine 1 31 which has 1/2 life of eight corner for days and important. They were asked to convert, but the half life in two seconds. So I'm just gonna do a little factor label method here. So eight point, both for days. I know that one day is the same as 24 hours. So days canceled. And no one hour is the same as 3600 seconds. So our skins, I just need multiple I and I get that it is 6.95 come down here. Six points 95 times 10 to the fifth, second in part B m to find that it decay constant. Well, my relationship here is that the half life is the natural log, too, over the decay rate. So I'm just gonna rearrange my equation here. So my decay rate is gonna be the natural log of two over the half Life natural log of two is 0.693 My half life in seconds is 6.95 times time of food. So I get my decay rate is 9.97 claims 10 to the negative seven and that is basically per second. So second, the negative one. Quite See, it's just another conversion. I'm converting micro curies to becquerels the tree two girls. So I'm gonna start with what I'm given. 0.5 Micro Curious. I was just going to do a step to give her my prefix. I know that, um, One micro curie is the same as 10 to the negative six. Curious. So Michael Curry canceled, and I want to go from curious about girls. Um, I know there's 3.7 times 10 with Tim back where Els is the same as one Curie. So curious. Cancel and I'm left in. Big girls come down here. So that is 18,000 500. That girl's and Port de We want to know how many nuclear we need to have this activity. The 0.5 micro Curie, which is the 18,000 back row, Um, in the relationship for that, it was just with number of nuclear is equal to the activity over the decay constant. And we found the decay constant here. We're looking for an We just want to make sure we have the activity in back row. So that's the 18,500 divided by the decay constant 9.97 times 10 to the negative seven. And that gives us that we need 1.86 times. Tend to attempt nuclei, Ellen Party. They want to know if you started with an activity of 6.5 Micro Curie, what would be the activity be after 40.2 days? So I like to use this relationship. I think it's the easiest to plug in the calculator home that the number of nuclear that you start with 1/2 raised to the number of half life's gives you the number of nuclear you end with now this question and asked me for nuclear. But look, if I just multiply both sides of the equation here by, well, Dick Avery Now, this is Look, my equation here. Now, this is our So now you can think of this as the new activity. The original activity 1/2 of raised to the her may have left. I have well, 42 days appear 42 days divided by we're just going to keep it in days. Remember, my half life was eight point of four days, so divided by eight point of four days. I get that. That will be five. Half life's exactly five. So, red flag in my equation here, I'm looking for the activity. If the original activity was 6.5, actually in the problem is 6.4, 6.4 1/2 raised to the 5/2 lifes. And here it didn't matter that I kept this in. In Micro Curie, it just means my anchovy. In micro curies, you tell an equation like this wasn't just multiplying. That works. And I get that my new activity is 0.2. What grew curie?

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