Coal mine Carbon

Question

While exploring a coal mine, scientists found plant fossils which had been preserved by an earthquake in the ceiling of the mine. Samples taken from one of the fossils had carbon‑14 activities of  A reference sample of the same size from a plant alive today has a carbon-14 activity of  If carbon‑14 has a half-life of 5730 years, what is the age of the plant fossil in years?

Answer 1

In copying your question, it appears that the relevant numbers didn't copy correctly. Even so, I can give you a path to the solution.

Any exponentially decaying quantity can be described as

s(t)=s₀exp(-ln(2)*t/t_1/2)

where s(t) is the amount of the sample remaining after time t has elapsed, s₀ is the initial amount of the sample, and t_1/2 is the half-life, the time it takes half of the sample to decay.

Suppose we know how much of the sample remains remains after t seconds, s_f. We can find the time elapsed by solving this expression for t

s_f=s₀exp(-ln(2)*t/t_1/2)

s_f/s₀=exp(-ln(2)*t/t_1/2)

ln(s_f/s₀)=-ln(2)*t/t_1/2

t_1/2ln(s_f/s₀)=-ln(2)*t

t=-t_1/2 ln(s_f/s₀)/ln(2)

t=t_1/2 ln(s₀/s_f)/ln(2)

So, whatever your initial and final samples were, substitute those for s₀ and s_f, along with the half-life and you will find how long the sample took to decay to this point. The units will be the same as the half-life, so since that is listed in years, so too will the time elapsed.