Question

While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake. Samples taken from one of the fossils have carbon‑14 activities of 40.0 counts/min. A reference sample of the same size, from a plant alive today, has a carbon‑14 activity of 160.0 counts/min.

If carbon‑14 has a half‑life of 5730 years, what is the age of the plant fossil in years?

Answer 1

11552.45 years

Step-by-step explanation:

Given that:

Half life = 5730 years

`t_(1/2)=(\ln2)/(k)`

Where, k is rate constant

So,

`k=(\ln2)/(t_(1/2))`

`k=(\ln2)/(5730)\ years^(-1)`

The rate constant, k = 0.00012 years⁻¹

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration

Given that:

The rate constant, k = 0.00012 years⁻¹

Initial concentration
`[A_0]` = 160.0 counts/min

Final concentration
`[A_t]` = 40.0 counts/min

Time = ?

Applying in the above equation, we get that:-

`40.0=160.0e^(-0.00012* t)`

`e^(-0.00012t)=(1)/(4)`

`-0.00012t=\ln \left((1)/(4)\right)`

`t=11552.45\ years`