Radioactive DecayDate: 10/26/2004 at 19:48:21 From: Alicia Subject: Radioactive decay The half-life of a radioactive element is 131 days, but your sample will not be useful to you after 90% of the radioactive nuclei originally present have disintegrated. For about how many days can you use the sample? Date: 10/27/2004 at 11:30:43 From: Doctor Ian Subject: Re: Radioactive decay Hi Alicia, In radioactive decay, you have atoms of one elements changing into atoms of another element because some of its neutrons turn into protons, changing the atomic number. (For example, if we start with a Carbon-14 atom, with 6 protons and 8 neutrons, and one of its neutrons becomes a proton, we end up with a Nitrogen-14 atom, with 7 protons and 7 neutrons.) The rate at which this happens isn't constant. Rather, in some given period, half of the elements will decay; then in the same period, half of the remaining elements will decay; and so on. Suppose we have an imaginary element Bobium, and it decays into the element Stanium with a half-life of two days. Then if we start with 128 grams of Bobium, after two days we'll have 64 grams of Bobium, and 64 grams of Stanium. After another two days, we'll have 32 grams of Bobium, and 96 grams of Stanium. And so on: Days Bobium Stanium ---- ------ ------- 0 128 0 2 64 64 4 32 96 6 16 112 8 8 120 10 4 124 12 2 126 14 1 127 16 1/2 127 1/2 18 1/4 127 3/4 20 1/8 127 7/8 Does this make sense? Now, in this case, suppose I need have at least 10% Bobium in order to use it. I can compute the percentage of Bobium: Days Bobium Stanium Percent Bobium ---- ------ ------- -------------- 0 128 0 100 2 64 64 50 4 32 96 25 6 16 112 12.5 8 8 120 6.25 10 4 124 12 2 126 14 1 127 16 1/2 127 1/2 18 1/4 127 3/4 20 1/8 127 7/8 So sometime between 6 and 8 days, my sample will become unusable. Could I figure this out without making a table? Well, note that after n days, the amount of Bobium I have left is 128 * (1/2)^(n/2) To check, n (days) 128 * (1/2)^(n/2) -------- ----------------- 0 128 * (1/2)^(0/2) = 128 * 1 = 128 2 128 * (1/2)^(2/2) = 128 * (1/2) = 64 4 128 * (1/2)^(4/2) = 128 * (1/4) = 32 So it looks like this captures the data in the table. What percentage of my sample will still be Bobium after n days? It will be amount of Bobium ---------------- * 100 amount of sample So that's just 128 * (1/2)^(n/2) ----------------- 128 and I need for that be more than 10%: 128 * (1/2)^(n/2) ----------------- > 0.10 128 Note that the initial amount (128 grams) isn't really doing anything here: I multiply by it, and then divide by it. Which just means that the length of time the sample will be useful doesn't depend on the size of the sample! Which is interesting, but makes sense when you think about what's going on. (If I cut something in half, and then in half again, I have 25% of what I started with, regardless of whether I started with one ounce or a million tons.) Anyway, the initial amount cancels out, and I'm left with (1/2)^(n/2) > 0.10 And now I can find the largest value of n for which this is true by trying different values (which is what I did in the table); or I can use logarithms--which is probably what you're studying now, if you've been assigned a question like this. :^D Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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