dN is proportional to:-

N - the number of radioactive nuclei present at that moment

dt - the time over which the measurement is made

the element, represented by a constant (l) called the

So:

the minus sign is there because the number of radioactive nuclei decreases as time increases.

The quantity dN/dt is the rate of decay of the source or the activity of the source and is the number of disintegrations per second.

This is measured in units called Becquerels (Bq) where 1 Bq = 1 disintegration per second.

A larger and more traditional unit is the Curie (Ci) 1 Ci = 3.7 x10^{10} Bq.

A larger and more traditional unit is the Curie (Ci) 1 Ci = 3.7 x10

The disintegration constant or decay constant (l) can be defined as the probability of a nucleon
decaying in the next second.

The number of nuclei in a sample can be related to the mass of the source (m) using the molar mass (M) and Avogadro's number (L) by the formula:-

m = MN/L

and we can use it to find out the mass of a given source if we know its activity.

A school has a radium 226 source with an activity of 5 mCi (5x10

What is the mass of the source?

Avogadro constant (L) = 6.02x10

(see later for an explanation of this term)

Using the formula dN/dt = -lN = -l(m/M)L so 5x10

Therefore m = (5x10

Returning to the formula

dN/dt = -lN and rearranging gives:

dN/N = -ldt which when integrated between the limits N = N

If we plot ln(N) against t we have a straight line graph with gradient -l) and an intercept on the ln(N) axis of ln(N

Returning to the equation and taking antilogs of both sides gives:

A graph of N against t would give an exponential decay graph, and if background radiation were ignored the line would tend towards N = 0 as time goes by. Since N is directly proportional to the activity (A) and the mass (m) of the sample we have three alternative forms of this formula.

It can be expressed as:

When N = N

Therefore when t = T N = N

Taking the inverse gives 2 = e

Decay constant (l) = 0.693/T

The decay constant (l) of a particular isotope (radon 220) is 1.33x10

How long will it take for the activity of a sample of this isotope to decay to one eighth of its original value?

Half life = 0.693/l = 0.693/1.33x10-2 = 52 s.

Number of half lives required to reduce the activity to one eighth = 3

Therefore time needed = 3 x 52 = 156 s = 2m 36s

Isotope |
Half life |
Decay constant (s^{-1}) |

Uranium 238 | 4.5x10^{9} years |
5.0x10^{-18} |

Plutonium 239 | 2.4x10^{4} years |
9.2x10^{-13} |

Carbon 14 | 5570 years | 3.9x10^{-12} |

Radium 226 | 1622 years | 1.35x10^{-11} |

Free neutron 239 | 15 minutes | 1.1x10^{-3} |

Radon 220 | 52 seconds | 1.33x10^{-2} |

Lithium 8 | 0.84 seconds | 0.825 |

Bismuth 214 | 1.6x10^{-4} seconds |
4.33x10^{3} |

Helium 5 | 6x10^{-20} seconds |
1.2x10^{19} |

A = A

ln A = ln

Therefore: ln A = ln(A

1. A sample of material is found to contain 2 g of the isotope gold 199.

How much will remain 10 days later?

The half life of gold 199 is 3.15 days and so the decay constant is therefore 0.693/3.15x86400 = 2.55x10

Using the formula:- m = m

2. A sample of carbon 14 has an activity of 2.5 Bq when corrected for background radiation.

If the initial count rate of a sample of the same mass was 3.7 Bq how old is the sample?

(Half life of carbon 14 = 5570 years).

Using the formula A = A

Taking logs gives: ln(2.5/3.7) = -(0.693/5570)t this gives t = 3151 years.