Radioactivity – decay rates and half life

Probability of radioactive decay Radioactive decay obeys an exponential decay law because the probability of decay does not depend on time: a certain fraction of nuclei in a sample (all of the same type) will decay in any given interval of time. The rate law is: DN = - l N Dt where N is the number of nuclei in the sample l is the probability that each nucleus will decay in one interval of time (for example, 1 s) Dt is the interval of time (same unit of time, 1s) DN is the change in the number of nuclei in Dt

Radioactive decay constant l The rate law DN = - l N Dt can also be written: DN/N = - l Dt As an example, Suppose that the probability that each nucleus will decay in 1 s is l = 1x10-6 s-1 In other words, one in a million nuclei will decay each second. To find the fraction that decay in one minute, we multiply by Dt = 60 s to get: DN/N = - l Dt = - (1x10-6 s-1) x (60s) = - 6x10-5 Equivalently: l = 6x10-5 min-1 and Dt = 1 min

Rate of radioactive decay Now write the rate law DN = - l N Dt as: DN/Dt = - l.N ( -DN/Dt is the rate of decay) Stated in words, the number of nuclei that decay per unit time is equal to l.N , the decay constant times the number of nuclei present at the beginning of a (relatively short) Dt. Example: if l = 1x10-6 s-1 and N = 5x109. then DN/Dt = - l.N = -1x10-6 s-1 . 5x109 = - 5x103 s-1 If each of these decays cause radiation, we would have an activity of 5000 Bq. (decays per second)

Decrease of parent population DN = - l N Dt represents a decrease in the population of the parent nuclei in the sample, so the population is a function of time N(t). DN/ Dt = - l.N can be written as a derivative: dN/dt = - l.N and this can be solved to find: N(t) = No exp(-lt) = No e -lt Recall that e0 = 1; we see that No is the population at time t = 0 and so the population decreases exponentially with increasing time t.

Graph of the exponential exp(x) + exp(0) = 1 exp(x)<1 if x<0 x

Graph of the exponential exp(-lt) + exp(0) = 1 exp(-0.693) = 0.5 = ½ + exp(-lt) + exp(-1) = 1/e = 0.37 lt

Half-life of the exponential exp(-lt) + The exponential decays to ½ when the argument is -0.693 exp(-0.693) = 0.5 = ½ + exp(-lt) lt½ + lt

Half-life of the exponential exp(-lt) Because the exponential decays to ½ when the argument is -0.693, we can find the time it takes for half the nuclei to decay by setting exp(-lt½) = exp(-0.693) = 0.5 = ½ The quantity t½ is called the half-life and is related to the decay constant by: lt½ = 0.693 or t½ = 0.693/l In our previous example, l = 1x10-6 s-1 The half-life t½ = 0.693/l = 6.93x105 s = 8 d

Half-life of number of radioactive nuclei Because the exponential decays to ½ after an interval equal to the half-life this means that the population of radioactive parents is reduced to ½ after one half-life: N(t½) = No . exp(-lt½) = ½ No In our example, the half-life is t½ = 8 d, so half the nuclei decay during this 8 d interval. In each subsequent interval equal to t½ , half of the remaining nuclei will decay, and so on.

Half-life of activity of radioactive nuclei Because the activity (the rate of decay) is proportional to the population of radioactive parent nuclei: DN/Dt = - l . N(t) but N(t) = No e -lt the activity has the same dependence on time as the population N(t) (an exponential decrease): DN/Dt = (DN/ Dt)o . e -lt In our example, the half-life is t½ = 8 d, so the activity is reduced by ½ during this 8 d interval.

Multiple half-lives of radioactive decay The population N(t) decays exponentially, and so does the activity DN/Dt . After n half-lives t½, the population N(t) = N(n.t½) is reduced to No/(2n). In our example, the half-life is t½ = 8 d, so the activity is reduced by ½ during this 8 d interval, and the population is also reduced by ½. After 10 half-lives, the population and activity are reduced to 1/(210) = 1/1024 = 0.001 times (approximately) their starting values. After 20 half-lives, there is about 10-6 times No.

Plotting radioactive decay (semi-log graphs) DN/Dt = (DN/ Dt)o . e -lt or N(t) = No . e -lt can be plotted on semi-log paper in the same way as the exponential decrease of intensity due to absorbers in X-ray physics. ln(N) = ln( No . e -lt ) = ln(No) + ln(e -lt) ln(N) = -lt + ln(No) which has the form of a straight line if y = ln(N) x = t and m = - l y = m . x + b with b = ln(No)

Semi-log graph of the exponential exp(-x) + exp(-0.693) = 0.5 = ½ + + exp(-1) = 1/e = 0.37 exp(-x) x

Plotting radioactive decay (semi-log graphs) Starting with N(t) = No . e -lt , we want to plot this on semi-log paper based on the common logarithm log10. We previously had: ln (N) = ln (No.e -lt ) = ln (No) + ln (e -lt) = -lt + ln (No) This unfortunately uses the natural logarithm, not common logarithm. We can use this result: log10(ex) = (0.4343)x Repeat the calculation above for the common log10 log(N) = log(No) + log(e -lt) = -(0.4343)lt + log(No) If we plot this on semi-log paper, we get a straight line for y = log(N) as a function of t, with slope -(0.4343)l.