# 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 |
4.5x10^{9} years |

Thorium 234 |
24 days |

Protactinium 234 |
1.2 minutes |

Uranium 234 |
2.5x10^{5} years |

Thorium 230 |
8x10^{4} 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 dN

_{1}/dt = -

l_{1}N

_{1} = dN

_{2}/dt = -

l_{2}N

_{2} = dN

_{3}/dt = -

l_{3}N

_{3} = dN

_{4}/dt = -

l_{4}N

_{4} = 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:

l_{1}N_{1} = l_{2}N_{2} etc. or lN = a constant
Because

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

lwe have for a decay chain:

Decay chains: N_{1}/N_{2} = T_{1}/T_{2}
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 = T_{Pb}/T_{Bi} = 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)N

_{A} T

_{1}/T

_{2} = N

_{1}/N

_{2} = (m

_{1}/M

_{1})/(m

_{2}/M

_{2}) = (m

_{1}/m

_{2})x(M

_{2}/M

_{1})

where m is the mass of the isotope in the sample, M its molecular mass and N

_{A} is Avogadro’s number (6.02x10

^{23} particles per mole).

## Proof of A = A_{o}/2^{n}

This can be proved as follows.

Start with the
standard radioactive decay law and take logs to the base e:

A = A

_{o}e

^{-lt}
ln A = ln A

_{o} -

lt = ln A

_{o} – ln(2t/T) where T is the half life.

Therefore:

ln A =
ln[A

_{o}/2

^{n}) where n = t/T and so A =
A

_{o}/2

^{n}