1. The Random Nature of Radioactive Decay
It is impossible to predict when an individual atom will decay.
We do know that if we have twice as many, the rate of decay (dN/dt) will double. ie
The minus sign indicates the decay.
The constant of proportionality is called the decay constant  (what are its units?).
(s-1) It is fixed for any particular nuclear decay. The rate of decay is measured in bequerel (Bq). 1 Bq = 1 decay per second.

2. The half life is the average time taken for half the nuclei to decay
This is the exponential law of radioactive decay. This first order differential equation can be integrated to show that
e is a pure number =
N is the number of particles left after time t
N0 is the original number of particles.
Radioactive Half-Life
For a particular decay process, it will always take the same amount of time for the number of nuclei to halve. It could be from 100 to 50 or from 106 to 5x105.
The half life is the average time taken for half the nuclei to decay

3. There is no known way of changing the half-life of a nuclide.
eg. The half life of a substance is 2 years. How much will be left after 8 years?
soln. 8 years = 4 half-lives
so the number of particles left will be 1/2 x 1/2 x 1/2 x1/2
= 1/16th
Note that if you do not have a whole number of half-lives, you must use the previous equation

4. t1/2 = 0.693 /  When one half life has passed, t = t1/2 and N = N0/2
so the equation becomes ie
Taking natural logarithms of both sides we obtain:
ln (1/2) = -t1/2
and so t1/2 = ln(2) / 
t1/2 = / 
The count rate is often denoted as R and will be directly proportional to the rate of decay.

5. EXAMPLE
Uranium 238 has a half life of 4.51 x 109 years. Calculate the number of particles of uranium 238 that will exist after x 109 years from a sample of 1 kg. (NA = 6.02x1023 mol-1)
Solution:
We need the half life in seconds
T1/2 = 4.51x109 x x 24 x 60x60 = x 1017s
Notice that it is not a whole number of half lives so we have to use BUT we don’t know  - examiners frequently do this!
Fortunately we know that t1/2 = / 
so  = / t1/2 =
0.693 / x 1017s
 = x s-1
t = 2.255x109 years = 2.225x109 x x 24 x 3600
t = 7.1 x 1016 s =
so = =
= 0.707

6. The End ie 70.7% is left BUT how many were there to start with? so = =
= 0.707
1 mole is 238g
238g is NA
0.238 kg is NA
1kg is NA / = 6.02x1023 / = 2.53 x 1024 particles
The End