Answer: 7546.76 years
Explanation:
This can be solved by the following equation:
[tex]N_{t}=N_{o}e^{-\lambda t}[/tex] (1)
Where:
[tex]N_{t}[/tex] is the number of atoms of carbon-14 left after time [tex]t[/tex]
[tex]N_{o}=255 Bq/kg[/tex] is the defined atmospheric carbon-14 (the number of atoms of C-14 in the original sample)
[tex]\lambda[/tex] is the rate constant for carbon-14 radioactive decay
[tex]t[/tex] is the time elapsed
On the other hand, [tex]\lambda[/tex] has a relation with the half life [tex]h[/tex] of the C-14, which is [tex]5730 years[/tex]:
[tex]\lambda=\frac{ln(2)}{h}=\frac{ln(2)}{5730 years}=1.21(10)^{-4} years^{-1} [/tex] (2)
In addition, we can calculate the value of [tex]N_{t}[/tex] knowing the mass [tex]m[/tex] of the sample and the decay rate [tex]d[/tex]:
[tex]m=500 g \frac{1 kg}{1000 g}=0.5 kg[/tex]
[tex]d=3070 \frac{decays}{min} \frac{1 min}{60 s}=51.16 \frac{decays}{s}=51.16 Bq[/tex]
Then:
[tex]N_{t}=\frac{d}{m}=\frac{51.16 Bq}{0.5 kg}=102.32 Bq/kg[/tex] (3)
Now, we have to find the age of the sample [tex]t[/tex] from (1):
[tex]t=ln(\frac{N_{t}}{N_{o}})(-\frac{1}{\lambda})[/tex] (4)
Substituting (2) and (3) in (4):
[tex]t=ln(\frac{102.32 Bq/kg}{255 Bq/kg})(-\frac{1}{1.21(10)^{-4} years^{-1}})[/tex] (4)
Finally:
[tex]t=7546.76 years[/tex] This is the age of the sample
