Radiometric dating half life examples


This is the only way Parentium-123 decays, and there is no other source of Daughterium-123.Furthermore, Parentium and Daughterium are so different in chemical properties that they don't otherwise occur together.

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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.

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.

If the decay constant is very small, even tiny amounts of contamination by other radioactive materials can be very significant.