Question

Pleaseeee someone help me . What is the half life of Caron 14 of this problem ???

Answer 1

5730 years

Step-by-step 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

`t_{(1)/(2)} \ = \ \displaystyle(\ln 2)/(\lambda) \ \ \ \ \ \ (1)`,

where λ is the reaction's rate constant modeled by the differential equation describing the kinetics of a first-order reaction

`N \ = \ N_(0)e^(-\lambda t) \ \ \ \ \ (2)`,

where N is the remaining amount of substance after time t and
`N_(0)` 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,

`\lambda \ = \ \displaystyle\frac{\ln 2}{t_{(1)/(2)}}`.

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),

`\-\hspace{0.56cm}12.5 \ = \ 100e^(-\lambda * 17190) \\ \\ \-\hspace{0.5cm} \displaystyle(12.5)/(100) \ = \ e^(-17190\lambda) \\ \\ \displaystyle(\ln\displaystyle((1)/(8)))/(-17190) \ = \ \lambda \\ \\ \-\hspace{1.05cm} \lambda \ = \ \displaystyle(3\ln2)/(17910) \ \ \ \ \ \ \ \ \ \ (\ln \displaystyle(1)/(x) \ = \ -\ln x\ ; \ \ln x^(n) \ = \ n\ln x )`

Equate both λ,

`\displaystyle\frac{\ln 2}{t_{(1)/(2)}} \ = \ \displaystyle(3\ln2)/(17190) \\ \\ \-\hspace{0.26cm} t_{(1)/(2)} \ = \ \ln 2 \ * \ \displaystyle(17190)/(3\ln2) \\ \\ \-\hspace{0.26cm} t_{(1)/(2)} \ = \ \displaystyle(17190)/(3) \\ \\ \-\hspace{0.26cm} t_{(1)/(2)} \ = \ 5730`

*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

`100\% \ \overset{t_{(1)/(2)}}\longrightarrow \ 50\% \ \overset{t_{(1)/(2)}}\longrightarrow \ 25\% \overset{t_{(1)/(2)}}\longrightarrow \ 12.5\%`.

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
`\displaystyle(n)/(3)`.