The tutor continues about the phenomenon of half life and one of its well-known applications.

In my post from yesterday I introduced the topic of carbon dating and the principles that make it feasible. Today I’ll give a few more technical details to facilitate its mathematical use.

Example: A sample taken from a mammoth bone is found to contain 17% of the radioactivity of a freshly dead bone. Estimate the time of the mammoth’s death.

Solution:

First, convert the percent to a decimal: 17%=0.17

The general formula for time since death, t, is

t=5730*(log(decimal remaining)/(log0.5))

We plug our decimal remaining, 0.17, into the formula:

t=5730*(log(0.17)/log(0.5))=14648 years

By the reckoning of the carbon dating process, the mammoth died 14648 years ago.

Incidentally, the radioactive form of carbon is carbon-14, whereas the (predominant) stable form is carbon-12:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Tutoring science, and also for general interest, this topic is relevant.  The tutor introduces it with an example.

Most people realize that scientists can “date” animal or plant remains – that is, discover, approximately, when they died.  How is it done?

Carbon dating is one name for the process, and it’s done by detecting the radioactivity of the dead sample.  Its radioactivity is compared to that of a living sample (perhaps surprisingly, the living sample will be more radioactive than the dead one).  From that comparison, the time of death can be approximated.

The baseline radioactivity of living things is due to their constant intake of fresh carbon. Plants do so by photosynthesis; animals, by eating.  Carbon dioxide in the open air is constantly radiated by the sun.  The cosmic rays maintain a constant abundance, in the atmosphere, of about 1 radioactive carbon atom per 0.75 trillion.

Once a body dies, it ceases taking in fresh carbon.    Therefore, the radioactive carbon in the body is no longer replenished.  It constantly decays with a half life of 5730 years – which means that, every 5730 years, only half of the radioactive carbon remains compared to previously.

 number of half lives elapsed 0 1 2 3 4 5 % radioactive carbon remaining 100 50 25 12.5 6.25 3.125

From the table, if a specimen retains about 3% radioactivity relative to a living sample, it’s around 5 half lives old, or 5(5730)= 28650 years old.

What if the sample retains a percentage radiation not on the table, like 60% or 35%? The math for such cases will be covered next post:)

Source: Ken Sejkora

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.