Radioactivity decay rates and half life

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 - 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

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)lt1 if xlt0
x
7
Graph of the exponential exp(-lt)

exp(0) 1
exp(-0.693) 0.5 ½

• exp(-lt)

exp(-1) 1/e 0.37
lt
8
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
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(-1) 1/e 0.37

• exp(-x)

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.