Half-Life Calculations

Half-life calculations Notes on General Chemistry Using coin flips to count people Everyone in of a group of 202 people is initially standing. Each person flips a coin, and sits down if the coin lands heads up. Each person still standing flips again, and sits down if the coin lands heads up. After 5 coin flips, 7 people remain standing. How many people are in the large group? The fundamental expression Nn = H1 ê 2Ln N0 . becomes 7 = H1 ê 2L5 N0 . We rearrange this for the number of people present, then the initial number of students—the class attendance—evaluates to N0 = 7 μ 25 = 7 μ 64 = 448. This is different from the actual number of people, 202, but the result shows that, to within a factor of two, we can use "half life decay" to "count" an initial population size. Why is our result so different form the number of people present? For such a small sample, this is definitely the hard and not very accurate way to count. But if we had a very large sample, say the people at a Boston Red Sox game in Fenway Park, it would be much quicker than counting each person individually. Half-life Half-life, thalf , is defined as the amount of time required for the amount of a substance to be reduced by 50%. Half-life a useful concept if its value does not depend on how much material is present. Such a decay process is called d first order. Nuclear decay is first order. Let's say a substance undergoes first order decay with thalf = 1 hour, and that we begin with 60 units of the substance. This means têhours 0 1 2 3 units remaining 60 30 = 50% of 60 15 = 50% of 30 7.5 = 50% of 15 2 Half-life calculations Here is a plot of the amount remaining versus time. units remaining 60 48 30 15 7.5 hours ? 1 2 3 Decay of 60 units with thalf = 1 hour. What is the time, marked ?, at which 48 units remain? The curve connecting the points is the known as the first-order decay curve. We can derive an expression for this curve, and so the amount remaining at any time (not just a multiple of thalf ). First, we rearrange the fundamental relation Nn = H1 ê 2Ln N0 . and then take the natural logarithm of both sides to get lnHNn ê N0 L = ln@H1 ê 2Ln D = -n lnH2L, since lnH1 ê 2L = lnH1L - lnH2L = -lnH2L. Then, by taking the antilogarithm of both sides, Nn ê N0 = ‰-n lnH2L , we can express the population after n = t ê thalf half lives as Nn = N0 ‰-n lnH2L = N0 ‰-t lnH2Lêthalf = N0 ‰-0.69 têthalf . For numerical calculations without calculator, it is more convenient to work with base-10 logarithms. Using the same procedure as above, you will get the expression Nn = N0 10-n logH2L = N0 10-0.30 têthalf . Show that this expression is correct. Given half-life, how long to decay by a x%? Let's determine the missing time on the graph above. Given 60 units of a substance that decays with thalf = 1 hour, how much time must elapse for 48 units to remain? We can use the fundamental expression Half-life calculations Nn = H1 ê 2Ln N0 . to write 48 = H1 ê 2Ln 60, and then solve this for the number of half-lives that have elapsed. Taking logs of both sides and rearranging, we get -n logH2L = logH48 ê 60L = logH0.80L = logH23 μ 10-1 L = -1 + 3 logH2L. This means 1 - 3 logH2L 1 - 0.90 0.10 n logH2L = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 1 ê 3. logH2L 0.30 0.30 Since thalf = 1 hour, in 1 ê 3 hour = 20 minutes, 12 units will have decayed. Given a decay amount in a given time, what is half-life? If we know the amount of decay that has taken placed in a given time, we can determine the half-life. A sample decays by 90.% in 20. min. What is its half-life? As always, let's begin with the fundamental expression Nn = H1 ê 2Ln N0 . In this case we know that in 20. min, Nn 0.10 N0 H1 ê 2Ln = ÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 0.10. N0 N0 Solving for n we get -n logH2L = logH0.10.L = logH1.0 μ 10-1 L = -1, n = t ê thalf = 1 ê logH2L = 1 ê 0.30 = 20 min ê thalf , and so that thalf = 0.30 μ 20. min = 6.0 min. 3