Question

In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an archaeological site has a C-14 disintegration rate of 9.16 atoms per minute per gram of carbon. Estimate the age of this sample in years. The half-life of C-14 is 5730 years. (enter only the number of years in standard notation, not the unit years)

Answer 1

Answer:

The answer is "4,241 .17 years"

Explanation:

The disintegration rate, which shows in C-14 atoms =  $15.3 \frac{atoms}{min-g}$

Rate of sample disintegration =$9.16 \frac{atoms} {min-gram}$

The digit proportion of C-14 can be determined that is included in the sample $= \frac {9.16}{15.3} \\\\ = 0.5987$

5730 years from half-life.  

The number with half-lives (n) which are repelled must be determined:

$(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987$

$\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740$

So, the age of the sample is given by = $n \times\ half-life$

                                                 $= 0.740 \times 5730 \ years \\\\=4241.17 \ years$