Respuesta :
a) The exponential function that models the situation is [tex]n(t) = n_{o}\cdot e^{-1.2097\times 10^{-4}\cdot t}[/tex].
b) The mass of Carbon-14 after 800 years is approximately 181.552 grams.
c) Half of the 200 g-Carbon 14 will remain after approximately 5729.91 years.
a) Mass of radioactive isotopes decay exponentially in time. The decay model is described below:
[tex]n(t) = n_{o}\cdot e^{-k\cdot t}[/tex] (1)
Where:
- [tex]n_{o}[/tex] - Initial amount of Carbon-14, in grams.
- [tex]n(t)[/tex] - Current amount of Carbon-14, in grams.
- [tex]k[/tex] - Decay rate, no units.
- [tex]t[/tex] - Time, in years.
If we know that [tex]k = 1.2097\times 10^{-4}[/tex], then the exponential function that models the situation is:
[tex]n(t) = n_{o}\cdot e^{-1.2097\times 10^{-4}\cdot t}[/tex] (1)
The exponential function that models the situation is [tex]n(t) = n_{o}\cdot e^{-1.2097\times 10^{-4}\cdot t}[/tex].
b) We determine the current amount of Carbon-14 by simply replacing all known variables: ([tex]n_{o} = 200\,g[/tex] and [tex]t = 800\,yr[/tex])
[tex]n(800) = 200\cdot e^{-1.2097\times 10^{-4}\cdot 800}[/tex]
[tex]n(800) \approx 181.552\,g[/tex]
The mass of Carbon-14 after 800 years is approximately 181.552 grams.
c) If we know that [tex]n = 100\,g[/tex] and [tex]n_{o} = 200\,g[/tex], then the time passed is:
[tex]100 = 200\cdot e^{-1.2097\times 10^{-4}\cdot t}[/tex]
[tex]0.5 = e^{-1.2097\times 10^{-4}\cdot t}[/tex]
[tex]t \approx 5729.91\,yr[/tex]
Half of the 200 g-Carbon 14 will remain after approximately 5729.91 years.
To learn more on radioactive decay, we kindly invite to check this verified question: https://brainly.com/question/4124868