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Radioactive decay chains

An unstable nucleus may emit an alpha or beta particle to form another element, or a gamma ray and become less excited nucleus. This daughter product as it is called may also be radioactive and so a whole chain of decay products can be formed.
One of the most well known of these is the decay chain of uranium 238 (shown below).


Isotope Half life
Uranium 238 4.5x109 years
Thorium 234 24 days
Protactinium 234 1.2 minutes
Uranium 234 2.5x105 years
Thorium 230 8x104 years
Radium 226 1620 years
Radon 222 3.8 days
Polonium 218 3.1 minutes
Lead 214 27 minutes
Bismuth 214 20 minutes
Polonium 214 1.6x10-4 s
Lead 210 19 years
Bismuth 210 5.0 days
Polonium 210 138 days
Lead 206 STABLE


Since all the nuclei are in dynamic equilibrium in a decay series - that is there can be no build up of any particular element the rate of decay of all the components in the series must be the same - in other words dN/dt for every component is the same.


Therefore dN1/dt = -λ1N1 = dN2/dt = -λ2N2 = dN3/dt = -λ3N3 = dN4/dt = -λ4N4 = etc. so:-
One consequence of this is that the elements with the long half lives will be present in larger quantities than those with short half lives because:

λ1N1 = λ2N2 etc.      or      λN = a constant


Because λN is a constant and the half life (T) of a radioactive isotope in the decay chain is proportional to 1/λ

we have for a decay chain:

Decay chains:     N1/N2 = T1/T2


The number of nuclei of a particular radioactive isotope in a decay chain will be proportional to the half-life of that isotope when equilibrium conditions have been reached.

Example problem
Find the ratio of the anmount of lead 214 to bismuth 214 in the above series.

Ratio of numbers of nuclei = TPb/TBi = 27/20 = 1.35 times as much Lead 214 as there will be Bismuth 214.

Notice that we have to make an adjustment if we want to deal with the relative masses of components in the series.
Since N = (m/M)L T1/T2 = N1/N2 = (m1/M1)/(m2/M2) = (m1/m2)x(M2/M1)

Proof of A = Ao/2n

This can be proved as follows.
Start with the standard radioactive decay law and take logs to the base e:

A = Aoe-λt
ln A = ln Ao -λt = ln Ao ln(2t/T) where T is the half life.

Therefore:

ln A = ln[Ao/2n) where n = t/T and so A = Ao/2n
 
 
 
© Keith Gibbs 2013