Radiological Dating and Half-Lives
The determination of the age of a geological deposit or an archeological artifact can be found through the use of radionuclides in the sample. This technique is called radiological dating. It takes advantage of the known half-lives of the radionuclides, and the premise that these half-lives have been constant throughout the entire period in question. This premise is strongly supported by the finding that half-lives are insensitive to all external forces such as heat, pressure, magnetic, or electrical stresses.
In geological dating, a pair of isotopes is sought that are related as a "parent" and "daughter" in a radioactive disintegration series such as U-238 and Pb-206. A sample whose age is desired has the concentration of U-238 and Pb-206 determined. The ratio of these concentrations together with the t½ of U-238 can then be used to calculate the age of the rock. For dating organic remains we restrict ourselves to C-14 dating. A new method using an accelerator mass spectrometer counts each particle of a sample and separates all the isotopes. It is far more effective and efficient and gives better results. The older method uses C-14, a beta emitter, that has been made from cosmic radiation in atmospheric nitrogen.
           10n + 147N ---> 157N ---> 146C + 11p
The C-14 migrates to the lower atmosphere where it gets used by all organic life. As long as an animal is alive the C-14:C-12 ratio is constant. At death, the ingestion of food ceases and the ratio of C-14 to C-12 begins to change.

Calculation of half-life
The determination of the half-life of an isotope is critical in understanding just how radioactive an isotope is. Let's say we have two isotopes. One has a half-life measured in years, the other in seconds. Which one is more radioactive. The one in seconds of course. If we start with equal amounts, then in seconds ½ of the second isotope is gone. In the next half-life ½ of what is left is gone, and so on.


The half-life equation is: 
where        Ao is the initial amount of substance at time zero
                  At  is the amount of substance left
                  t½  is the half-life
                  t     is the amount of elapsed time in the same units as the half-life.
A more useful rearrangement of this formula can be used to find the amount of time elapsed.


If we find an artifact that as a C-14 activity of 708 Bq/g and if the normal C-14 is 918 Bq/g, what is the age of the artifact?
t½ for C-14 = 5730 y

therefore t = 5730 y X ln 918 Bq/g
                     0.693          708 Bq/g

= 8268.40 y X ln 1.30

= 8268.40 y X 0.262364

= 2169.33 years

= 2.2 x 103 years

This is of course the more difficult way to do it. A more elegant way is to see if the half-life divides evenly into the time elapsed.
A sample of C-14 decays through a time period of 34380 years. If you started with 1000 g how much of it is left?

How many half lives? 34380 y = 6 half-lives
                                   5730 y

Therefore Amount left = Amount at start X ½half-lives
                                  = 1000 g X ½6
                                  = 1000 g X 1
                                 = 15.625 grams are left after 34380 years.
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