1. Radioactivity – decay rates and half life presentation for April 30, 2008 by Dr. Brian Davies, WIU Physics Dept.

2. 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:  N = - N  t where N is the number of nuclei in the sample is the probability that each nucleus will decay in one interval of time (for example, 1 s)  t is the interval of time (same unit of time, 1s)  N is the change in the number of nuclei in  t

3. Radioactive decay constant The rate law  N = - N  t can also be written:  N/N = -  t As an example, Suppose that the probability that each nucleus will decay in 1 s is = 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  t = 60 s to get:  N/N = -  t = - (1x10 -6 s -1 ) x (60s) = - 6x10 -5 Equivalently: = 6x10 -5 min -1 and  t = 1 min

4. Rate of radioactive decay Now write the rate law  N = - N  t as:  N/  t = -  N ( -  N/  t is the rate of decay) Stated in words, the number of nuclei that decay per unit time is equal to  N, the decay constant times the number of nuclei present at the beginning of a (relatively short)  t. Example: if = 1x10 -6 s -1 and N = 5x10 9. then  N/  t = -  N = -1x10 -6 s -1  5x10 9 = - 5x10 3 s -1 If each of these decays cause radiation, we would have an activity of 5000 Bq. (decays per second)

5. Decrease of parent population  N = - N  t represents a decrease in the population of the parent nuclei in the sample, so the population is a function of time N(t).  N/  t = -  N can be written as a derivative: dN/dt = -  N and this can be solved to find: N(t) = N o exp(- t) = N o e - t Recall that e 0 = 1; we see that N o is the population at time t = 0 and so the population decreases exponentially with increasing time t.

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

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

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

9. Half-life of the exponential exp(- t) 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(- t ½ ) = exp(-0.693) = 0.5 = ½ The quantity t ½ is called the half-life and is related to the decay constant by: t ½ = 0.693 or t ½ = 0.693/ In our previous example, = 1x10 -6 s -1 The half-life t ½ = 0.693/ = 6.93x10 5 s = 8 d

10. 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(t ½ ) = N o  exp(- t ½ ) = ½ N o In our example, the half-life is t ½ = 8 d, so half the nuclei decay during this 8 d interval. In each subsequent interval equal to t ½, half of the remaining nuclei will decay, and so on.

11. Half-life of activity of radioactive nuclei Because the activity (the rate of decay) is proportional to the population of radioactive parent nuclei:  N/  t = -   N(t) but N(t) = N o e - t the activity has the same dependence on time as the population N(t) ( an exponential decrease ):  N/  t = (  N/  t) o  e - t In our example, the half-life is t ½ = 8 d, so the activity is reduced by ½ during this 8 d interval.

12. Multiple half-lives of radioactive decay The population N(t) decays exponentially, and so does the activity  N/  t. After n half-lives t ½, the population N(t) = N(n. t ½ ) is reduced to N o /(2 n ). In our example, the half-life is t ½ = 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/(2 10 ) = 1/1024 = 0.001 times (approximately) their starting values. After 20 half-lives, there is about 10 -6 times N o.

13. Plotting radioactive decay (semi-log graphs)  N/  t = (  N/  t) o  e - t or N(t) = N o  e - t 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( N o  e - t ) = ln(N o ) + ln(e - t ) ln(N) = - t + ln(N o ) which has the form of a straight line if y = ln(N) x = t and m = - y = m. x + b with b = ln(N o )

14. Semi-log graph of the exponential exp(-x) exp(-x) x + exp(-1) = 1/e = 0.37 exp(-0.693) = 0.5 = ½ + exp(0) = 1 +

15. Plotting radioactive decay (semi-log graphs) Starting with N(t) = N o  e - t, we want to plot this on semi-log paper based on the common logarithm log 10. We previously had: ln (N) = ln (N o  e - t ) = ln (N o ) + ln (e - t ) = - t + ln (N o ) This unfortunately uses the natural logarithm, not common logarithm. We can use this result: log 10 (e x ) = (0.4343)x Repeat the calculation above for the common log 10 log(N) = log(N o ) + log(e - t ) = -(0.4343) t + log(N o ) 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).