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likoan [24]
3 years ago
9

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 ?
Chemistry
1 answer:
Elena-2011 [213]3 years ago
6 0

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

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