Respuesta :
Answer:
Age of the atifact is [tex]4.2\times 10^{2}[/tex] years
Explanation:
- For first -order radioactive decay- [tex]A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
- [tex]A_{t}[/tex] represents activity of radioactive nuclide after t time, [tex]A_{0}[/tex] represents initial activity of radioactive nuclide and [tex]t_{\frac{1}{2}}[/tex] represents half-life
- Here, [tex]A_{t}=19Bq[/tex], [tex]A_{0}=20Bq[/tex] and [tex]t_{\frac{1}{2}}=5.73\times 10^{3}years[/tex]
Plug-in all the given values in the above equation-
[tex]19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}[/tex]
or, [tex]t=4.2\times 10^{2}[/tex]
So, age of the atifact is [tex]4.2\times 10^{2}[/tex] years
