Your measurements indicate that a fossilized skull you unearthed has a carbon-14/carbon-12 ratio about 1 ⁄ 16 that of the skulls of present-day animals. What is the approximate age of the fossilized skull?

All organisms have a constant concentration of C-14 accumulated in them.Once an organism dies, C-14 accumulation stops. As a result, no new C-14 gets added.

One of the well-known processes of determining a fossil’s age is radiometric dating. It is also known as radiocarbon dating. In this process, the radioactive isotope C-14 is involved. The rate of decay from C-14 to C-12 in a fossil can help measure the fossil’s age.

Half-life is the time required for half of a radioactive substance to disintegrate or decay. It is the rate of decay denoted by\({t_{\frac{1}{2}}}\).

Carbon-14 atoms get decayed to nitrogen atoms.The half-life of carbon-14 is 5730 years. It implies that a dead organism requires 5730 years for the decay of half of its carbon-14 atoms.

The approximate age of a fossil can be determined using the carbon-14 to carbon-12 ratio. This method works for fossils up to 75000 years old.

Given:

\(\frac{{^{14}C}}{{^{12}C}} = \frac{1}{{16}}\)

We know:

\(\frac{{^{14}C}}{{^{12}C}} = \frac{N}{{{N_o}}}\), where\(N\)is the amount of radioisotope left after a time ‘t.’\({N_o}\)is the original or initial amount of radioisotope during the initial time.

\(\frac{{^{14}C}}{{^{12}C}} = \frac{N}{{{N_o}}} = \frac{1}{{16}}\)

\( \Rightarrow {\left( {\frac{1}{2}} \right)^4} = {\left( {\frac{1}{2}} \right)^n}\)

\( \Rightarrow n = 4\)

After n half-lives, the age of a fossil (t) is given by:

\(t = nT\), where ‘T’ is the half-life of C-14.

Substituting the value of\(n = 4\)and the half-life of C-14 as\(T = 5730\), we have:

\(\begin{aligned}{c}t &= 4 \times 5730\\ \Rightarrow t &= 22920years\end{aligned}\)

Therefore, the approximate age of the fossilized skull is 22,920 years.