Answer:
5730 years
Explanation:
In a chemical reaction, including radioactive decay, the half-life of a species is the time taken for the substance to decrease to exactly one-half its initial value.
For a first-order reaction, the half-life of the reactant is
[tex]t_{\frac{1}{2}} \ = \ \displaystyle\frac{\ln 2}{\lambda} \ \ \ \ \ \ (1)[/tex],
where λ is the reaction's rate constant modeled by the differential equation describing the kinetics of a first-order reaction
[tex]N \ = \ N_{0}e^{-\lambda t} \ \ \ \ \ (2)[/tex],
where N is the remaining amount of substance after time t and [tex]N_{0}[/tex] is the initial amount of substance before the chemical reaction proceeds.
Since the value of λ is constant, rearrange equation (1) with λ as the subject,
[tex]\lambda \ = \ \displaystyle\frac{\ln 2}{t_{\frac{1}{2}}}[/tex].
Given that only 12.5% of carbon-14 remained after 17190 years assuming that initially there was 100% of carbon -14, substitute in the values into equation (2) and solve for λ (make λ the subject of the equation),
[tex]\-\hspace{0.56cm}12.5 \ = \ 100e^{-\lambda \times 17190} \\ \\ \-\hspace{0.5cm} \displaystyle\frac{12.5}{100} \ = \ e^{-17190\lambda} \\ \\ \displaystyle\frac{\ln\displaystyle(\frac{1}{8})}{-17190} \ = \ \lambda \\ \\ \-\hspace{1.05cm} \lambda \ = \ \displaystyle\frac{3\ln2}{17910} \ \ \ \ \ \ \ \ \ \ (\ln \displaystyle\frac{1}{x} \ = \ -\ln x\ ; \ \ln x^{n} \ = \ n\ln x )[/tex]
Equate both λ,
[tex]\displaystyle\frac{\ln 2}{t_{\frac{1}{2}}} \ = \ \displaystyle\frac{3\ln2}{17190} \\ \\ \-\hspace{0.26cm} t_{\frac{1}{2}} \ = \ \ln 2 \ \times \ \displaystyle\frac{17190}{3\ln2} \\ \\ \-\hspace{0.26cm} t_{\frac{1}{2}} \ = \ \displaystyle\frac{17190}{3} \\ \\ \-\hspace{0.26cm} t_{\frac{1}{2}} \ = \ 5730[/tex]
*Note that the above calculations demonstrated that carbon-14 went through 3 half-lives to yield a remaining 12.5% of its original amount, since
[tex]100\% \ \overset{t_{\frac{1}{2}}}\longrightarrow \ 50\% \ \overset{t_{\frac{1}{2}}}\longrightarrow \ 25\% \overset{t_{\frac{1}{2}}}\longrightarrow \ 12.5\%[/tex].
Therefore, if n is the time taken for carbon-14 to half its initial amount trice to remain 12.5% of its initial amount. Then, the half-life of carbon-14 is [tex]\displaystyle\frac{n}{3}[/tex].
