Half Life Simulation Using Coin Flips

Half life simulation using coin flips Notes on General Chemistry In lecture we illustrated how half life decay works using coin flips to mark each half life. Here are the details of what we did and how they relate to calculations using half life decay data. We began with everyone in the class standing. The first half life consisted of the results of a coin flip: Students whose flipped coin was head down sat down. Those students whose flipped coin was head up remained standing. Students recorded the results of the half life using their CPS response pads to select "Still standing" or "Sitting down." The second and subsequent half lives consisted of the results of a coin flip by only those students who were still standing after the previous half life: Students whose flipped coin was head down sat down, and students whose flipped coin was head up remained standing. Only the students who participated in the coin flip (that is, only the students who were standing before the coin flip) recorded the results of the half life using their CPS response pads to select "Still standing" or "Sitting down." Half lives elapsed until only several students remained standing. Simulated results Here are example results for an initial "population" N0 of 256 students, in tabular form and plotted as points together with the idealized population decay curve. coin flip 0 1 2 3 4 5 6 7 still standing 256 131 76 29 19 12 5 1 Example results for an initial "population" N0 of 256 students. still standing 250 200 150 100 50 coin 1 2 3 4 5 6 7 flip Example results for an initial "population" N0 of 256 students plotted as points together with the idealized population decay curve. The results at each half life differ from the idealized curve, because with such a small sample, it is quite unlikely that exactly half of the coin flips will be head up or head down. Idealized decay curve The idealized decay curve assumes that exactly half of the population decays during each half life. This means that after n half lives the remaining population is Nn = H1 ê 2Ln N0 . Rearranging this expression and then taking the natural logarithm of both sides, we get lnHNn ê N0 L = ln@H1 ê 2Ln D = -n lnH2L, since lnH1 ê 2L = lnH1L - lnH2L = -lnH2L. Taking the antilogarithm of both sides, Nn ê N0 = ‰-n lnH2L , we can express the population after n half lives as Nn = N0 ‰-n lnH2L . The right hand side of this equation is the idealized decay curve. It is important because it allows the calculation of the population not just for integer values of n but fractional half lives too. Counting class attendance by counting half lives In lecture we found that after n = 6 coin flips N6 = 3 students remained. If we "round" this to N6 = 2, then the initial number of students—the class attendance—evaluates to N0 = N6 26 = 27 = 128. If we "round" this to N6 = 4, then the initial number of students instead evaluates to N0 = N6 26 = 28 = 256. In fact, there were 141 students. This shows that to within a power of 2 we were able to use "half life decay" to "count" the initial population size. For such a small sample, this is definitely the hard way to count. But if we had a very large sample, say the population of a city, it would be much quicker than counting each person individually. Estimating how many half lives have elapsed Sometimes we want to know how much time is required for a substance with a known half life to decay to a certain fraction of its original amount. A way to answer this kind of question is to write the fraction remaining as Nn ê N0 = H1 ê 2Ln in terms of the number n of half lives that have elapsed. By taking the logarithm of both sides, it is then easy to solve for n. For example, how much time is required for a sample with a half life of 21 minutes to decay to 40% of its original amount. We have logHNn ê N0 L = logH0.40L = log@H1 ê 2Ln D = -n logH2L Since logH0.40L = logH4.0 µ 10-1 L = logH4.0L - 1 = 0.60 - 1 = -0.40 and since logH2L = 0.30, we have -0.40 = -n 0.30 and so n = 4 ê 3. This means 4 ê 3 µ 21 = 28 minutes us required.