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slavikrds
3 months ago
15

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
1 answer:
lions [2.9K]3 months ago
6 0

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\\\\

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Clarification:

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