Question

The half life for the decay of carbon-14 is 5.73 x 10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 53.Bq. The activity in a similar-sized sample of fresh wood is measured to be 59.Bq.
1. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer 1

Answer:

$8.93\times 10^2$ years

Explanation:

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Initial concentration [A₀] = 59 Bq

Final concentration $[A_t]$ = 53 Bq

Time = ?

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Applying values, we get that:-

$53=59\times e^{-0.00012\times t}$

$\ln \left(e^{-0.00012t}\right)=\ln \left(\frac{53}{59}\right)$

$t=-\frac{\ln \left(\frac{53}{59}\right)}{0.00012}$

t = $8.93\times 10^2$ years

$8.93\times 10^2$ years is the age of the artifact.