Question

The half life for the decay of carbon-14 is 5.73 x 10 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 19. Bq. The activity in a similar-sized sample of fresh wood is measured to be 20. Bq. Calculate the age of the artifact. Round your answer to 2 significant digits. years X 5 ?

Answer 1

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

• For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
• $A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
• Here, $A_{t}=19Bq$, $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$

Plug-in all the given values in the above equation-

$19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}$

or, $t=4.2\times 10^{2}$

So, age of the atifact is $4.2\times 10^{2}$ years