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Radioactivity – decay rates and half life

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

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Radioactivity – decay rates and half life

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  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: 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

  3. Radioactive decay constant l • The rate law DN = - l N Dt can also be written: • DN/N = - lDt 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 = - lDt = - (1x10-6 s-1) x (60s) = - 6x10-5 • Equivalently: l = 6x10-5 min-1 and Dt = 1 min

  4. 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)

  5. 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.

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

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

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

  9. 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

  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(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.

  11. 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.

  12. 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.

  13. 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)

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

  15. 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.

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