|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
|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
= 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
|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.