Does radioactive dating with isotopes of uranium and thorium provide

19-Aug-2016 00:41 by 5 Comments

Does radioactive dating with isotopes of uranium and thorium provide

Imagine we have an undiscovered element, Parentium, that has a radioactive isotope, Parentium-123, which decays to stable Daughterium-123.This is the only way Parentium-123 decays, and there is no other source of Daughterium-123.

If there were such a pair of isotopes, radiometric dating would be very simple.

We could be sure that a mineral containing parentium originally had no daughterium.

If the mineral contained 1 part per million Parentium-123 and 3 parts per million Daughterium-123, we could be sure all the Daughterium-123 was originally Parentium-123.

In other words there was originally 4 parts per million Parentium-123 and 0 parts per million Daughterium-123.

Since there is now only 1/4 of the original amount of Parentium-123, we know that two half-lives of Parentium-123 have elapsed.

But there are some questions that come to mind: Calculus students typically meet this problem somewhere in the second semester.

It is one of the simplest examples of a differential equation.What radioactive materials actually do is decay according to a law: Decays/Time = K * Number of atoms K is a constant called the decay constant.Let t stand for time and N(t) stand for the number of atoms at time t .In calculus terms, we write: d N(t)/dt = -K * N(t) or d N(t)/N(t) = -K dt The minus sign means that each decay decreases the total number of atoms.Integrating both sides, we get: ln N(t) = -Kt + C C is the constant of integration that we can often ignore, but not here.When t = 0, ln N(0) = C Taking exponentials of both sides, we get N(t) = N(0)exp(-Kt) If t = one half life, then N(t)/N(0) = 1/2 = exp(-Kt), and: ln(1/2) = -ln2 = -Kt, so t = ln2 / K So what we do in practice is determine the decay constant and calculate half life from it.

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