Americium-241 is used in smoke detectors. It has
a first-order ra...

Question

Americium-241 is used in smoke detectors. It has
a first-order rate constant for radioactive decay of
k = 1.6 * 10-3 yr-1. By contrast, iodine-125, which is
used to test for thyroid functioning, has a rate constant
for radioactive decay of k = 0.011 day-1. (c) How much of a 1.00-mg sample of each
isotope remains after three half-lives?

Accepted Answer

Hey everyone today, we're being asked to find how much of a 1.25 mg sample of each isotope of Berkeley um To 48. And Seaborg e um to 60 remains after four half lives for each of the respective elements. So in the question itself were given a few key details that I'd like to mention were given the rate constant or decay rates for both Berkeley um and see board game respectively. As well as the mass of the sample in question. So with that in mind let's go ahead and take a look at the half life or the amount remaining after four half lives for Berkeley. Um First so we'll do that. Let's do this in blue will just be a little more fun. 2 48 B. K. So we can recall that for a half life Which is 1st. Which is a first order reaction. We can find the half life by taking the natural log of two, Dividing by the decay rate. Which in this case for Berkeley um is 0. per hour. This would be Lawn two divided by 0.29 0.0292. My bad 292 negative one. Which gives us that the half life is 23.73-9 hours. So almost one whole day. And we know that we're being asked for after four lives or four half lives. So multiply this by four and we get 94. hours now to find the amount of or the mass of the sample remaining. After four half lives. We need to use a formula. Where is the lawn in this case of the Berkeley um will be equal to forgot. A subscript will be equal to the rate constant. The negative rate constant times the total number of half lives. We'll just write that as H. For now plus lawn of the mass with em of the sample. So inserting these values, we get negative 0.0 to 9 to 0 to 9 to Times 94. hours plus the natural log of The mass of our sample which is 1.25 mg. Simple find this. We get that the lawn or the natural log of Berkeley ums four total half lives is equal to negative 2.5486. Which therefore means that the mass of constant or the mass of Berkeley um left after four. Half lives Solving for Berkeley um is 0.07817 mg. So that's our first answer. We can do a similar process for cyborg and I'm just gonna scroll. So we have a little more space and let's do this in red. So we follow a very similar process. Again. The half life is equal to the natural log of two divided by the rate constant which in this case excuse me, is 0.17 30 milliseconds. 173 milliseconds. Which means that our or sorry milliseconds. The negative one, which means that our total half life is 4.58 milliseconds. We know that again. We're dealing with 1.25 mg and four half lives would therefore be 16.231 milliseconds using the same formula as above except adapting it, adapting it. For sea board game, this is S. B. Or C. S. G. It'll be the negative decay rate 1.73 milliseconds. One Multiplied by the total number of half lives. 16.023 or the time it will take for four half lives to pass. In this case which is milliseconds plus the natural log of 1. milligrams. Which gives us a value of negative 2.5486. And therefore The mass of Seaborg left after 4/2 lives will be 0.07817 mg. I hope this helps. And I look forward to seeing you all in the next one

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