Dinosaur fossils 

Dinosaur fossils has helped in the study of dinosaurs. Although dinosaurs lived many millions of years ago, we know that they existed because some of them turned into fossils when they died. These fossils have been found almost all over the world. 

Dinosaur fossils are  traces of ancient life that have been preserved by natural processes, from spectacular skeletons to tiny sea shells. Imprints, tracks and trails can also become fossilised, like dinosaur footprints or worm burrows. These are called trace fossils. By studying the remains of life and the traces it left behind we can learn a lot about how animals and plants lived and behaved millions of years ago. 

Question

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this case, approximately 1.25 billion years). Suppose the minimum detectable amount is 0.1% and a dinosaur fossil is dated with 40K to be 69 million years old. Is this possible? What is the maximum age of a fossil that we could date using ⁴⁰K? 

Answer

We can use the following model for finding the amount of K-40 after t years.

$$\begin{array}{c}Q\left(t\right)=Q\left(o\right)e^{-kt}\end{array}$$

Here, Q(o) is the initial quantity and k is the rate of decay.

We need to find the rate of decay. We know that after approximately 1.25 billion years half of K-40 is left. So,

$$\begin{array}{cc}& Q\left(o\right)e^{-1250000000k}=0.5Q\left(o\right)\\ & -1250000000k=\ln 0.5\\ & k\approx 5.55452\ast {10}^{-10}\end{array}$$

We know that the half-life of K-40 is 1.25 billion years. Therefore, after 1.25 billion years, 50% of the initial quantity is left. This also means that after 69 million years more than 50% will be left which is considerably more than the minimum detectable amount of 0.1%. Therefore, it is possible to date the dinosaur.

The maximum possible age of the fossil will be the time it takes for the initial quantity to be reduced to 0.1%. This time will be:

$$\begin{array}{cc}& Q\left(o\right)e^{-5.55452\ast {10}^{}t}=0.001Q\left(o\right)\\ & -5.55452\ast {10}^{-10}t=\ln 0.001\\ & t\approx 12436277624.31702\text{years}\end{array}$$

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