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Uses of radioactive isotopes

These materials have a variety of uses and a selection of these are listed below.
(a) dating geological specimens, using uranium, rubidium or bismuth;
(b) dating archaeological specimens, using carbon 14
(c) paper or plastic thickness measurement using beta radiation
(d) treatment of tumours;
(e) sterilisation of foodstuffs;
(f) nuclear pacemakers for the heart;
(g) liquid flow measurement;
(h) tracing sewage or silt in the sea or rivers;
(i) checking blood circulation and blood volume;
(j) atomic lights using krypton 85;
(k) checking the silver content of coins;
(I) radiographs of castings and teeth;
(m) testing for leaks in pipes;
(n) tracing phosphate fertilisers using phosphorus 32
(o) smoke alarms
(p) sterilisation of insects for pest control.

Some isotopes used in medical physics

(a) Iodine-131 with a half-life of 8.0 days and activity of 8 mC may be taken as liquid or in a capsule.
(b)Technetium-99 with a half-life of 6 hours gives gamma-rays of 140 keV energy.
(c) Iodine-123 is suitable for medical studies since it gives no beta- radiation.
(d) Cobalt-60 sources of up to 10 000 curies have been used; such a source gives 200 R per minute at 1 m. Treatment is typically 3 grays (<2 m) a day for 20 days. (1 gray = 100 rads.)

Radioactive dating

(a) Geological dating
The dating of rocks is carried out using the decay of uranium 238 (half life 4.5x109 years), rubidium 87 (half life 4.7x109 years) or bismuth 40 (half life 1.3x109 years). The very long half-lives of these isotopes make them particularly suitable for finding the age of rocks.
For example if you consider the uranium series that the final stable isotope is lead-206, and if we assume that there was no lead in the rock when it was formed the ratio of the number of atoms of lead 206 (NPb) to the number of atoms of uranium 238 (NU) will give us the age of the sample.

Example problems
Consider a rock sample in which the ratio of lead-206 to uranium-238 is 0.7.
Now the initial number of atoms (No) is the sum of the numbers of lead(NPb) and uranium (NU) atoms present when the sample was analysed.
Therefore NU = (NU + NPb)e-lt where l is the decay constant for uranium-238 and t is the time that has elapsed since the rock was formed.
Therefore we can show that NU = (NU + NPb)e-lt and so 1 = (1 + NPb/ NU)e-lt
1 = (1 + 0.7)e-lt
1.7 = e-lt and so lt = 0.53
But l = 5 x 10-18 s-1 and therefore t = 0.53/5x10-18 = 1.06x1017 s = 3.36x109 years

(b) Archaeological dating using carbon 14
The half-life of carbon-14 (5570 years) is just the right sort of length for use in dating archaeological specimens with ages up to a few thousand years.

Carbon-14 is continually being formed in the upper atmosphere. Cosmic rays can produce neutrons, and the following reaction may then occur:

The carbon 14 is then absorbed by plants; these in turn are eaten by animals which may then be eaten by other animals. As soon as the animal dies the intake of radioactive carbon-14 stops and the proportion in the body starts to decrease. Therefore if the proportion of carbon 14 to carbon 12 is known at the start, the age of the specimen can be found once the amount of carbon 14 remaining in it has been measured. It has been found that the activity of carbon 14 in living materials is about 19 counts per minute per gram of specimen.

This method of dating can be used with success to determine not only the ages of animal remains but also those of wood, paper, cloth and other organic material.

One difficulty with this method is that it has to be assumed that the cosmic ray intensity has remained constant, and in fact this has been found not to be the case. By comparison with the tree rings in the extremely old bristle-cone pines, however, a corrected carbon date can be found for objects over about 1500 years old. The trees are themselves dated by the carbon-14 method using dead parts in the bark. A comparison between the carbon date and that due the tree rings is shown in the diagram.


Example problems
A piece of bone from an archaeological site is found to give a count rate of 15 counts per minute. A similar sample of fresh bone gives a count rate of 19 counts per minute.
Calculate the age of the specimen.

The activity A of a sample is proportional to the number of radioactive atoms within it.
Therefore A = Aoe-lt giving 15 = 19e-lt
Therefore: 0.236 = lt
t = 0.236/l = [0.236 x 5570]/0.693
t = 1897 years.
© Keith Gibbs 2010