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 ar

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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 ar

chaeological 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)

Chemistry

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lions [1K] · Answer 1 · 2026-03-04T18:24:59+03:00

Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in $15.3 \frac{atoms}{min-g}$

The dissolution rate of the sample is given by $9.16 \frac{atoms} {min-gram}$

The C-14 proportion within the sample can be determined as per $= \frac {9.16}{15.3} \\\\ = 0.5987$

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

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

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

Thus, the age of the sample is represented as = $n \times\ half-life$

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

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 ar

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