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.Ndt

dN

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.

Ndt

dN

This is the exponential law of radioactive decay. This first order differential equation can be integrated to show that

Ndt

dN

teNN 0

•e is a pure number = 2.718•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

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-livesso 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

λt0 eNN

so the equation becomes

When one half life has passed, t = t1/2 and N = N0/2

λt0 eNN

2/10

02

teNN 2/1

21 te ie

Taking natural logarithms of both sides we obtain:

ln (1/2) = -t1/2

and so t1/2 = ln(2) / t1/2 = 0.693 /

The count rate is often denoted as R and will be directly proportional to the rate of decay.

EXAMPLEUranium 238 has a half life of 4.51 x 109 years. Calculate the number of particles of uranium 238 that will exist after 2.255 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 365.25 x 24 x 60x60 = 1.423 x 1017sNotice 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 = 0.693 / so = 0.693 / t1/2 = 0.693 / 1.423 x 1017s = 4.869 x 10 -18 s-1

t = 2.255x109 years = 2.225x109 x 365.25 x 24 x 3600t = 7.1 x 1016 s

so =teN

N 0

=16-18 x7.1x104.87x10718.2 =

16-18 x7.1x104.87x10718.2

= 0.707

ie 70.7% is left BUT how many were there to start with?

so =teN

N 0

=16-18 x7.1x104.87x10718.2 =

16-18 x7.1x104.87x10718.2

= 0.707

1 mole is 238g

238g is NA

0.238 kg is NA

1kg is NA / 0.238 = 6.02x1023 / 0.238 = 2.53 x 1024 particles