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Radioactivity – decay rates and half life

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Radioactivity – decay rates and half life

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Radioactivity – decay rates and half life

presentation for April 30, 2008 by

Dr. Brian Davies, WIU Physics Dept.

- 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

- 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

- 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)

- 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.

exp(x)

+

exp(0) = 1

exp(x)<1 if x<0

x

+

exp(-lt)

exp(0) = 1

exp(-0.693) = 0.5 = ½

+

+

exp(-1) = 1/e = 0.37

lt

+

exp(-lt)

The exponential decays to ½

when the argument is -0.693

exp(-0.693) = 0.5 = ½

+

lt½

+

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

- 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.

- 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.

- 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.

- 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)

exp(0) = 1

+

exp(-0.693) = 0.5 = ½

exp(-x)

+

+

exp(-1) = 1/e = 0.37

x

- 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.