Question

Archeologists have several methods of determining the age ofrecovered artifacts. One method is radioactive datingAll matter is made of atoms. Atoms, in turn, are made of protons,neutrons, and electrons. An "element" is defined as an atom withgiven number of protons, Carbon, for example, has exactly protons,Carbon atoms can, however, have different numbers of neutrons,These are known as "sotopes" of carbon, Carbon 12 has 6 neutrons,carbon-13 has 7 neutrons, and carbon-14 has 8 neutrons, Allcarbon-based life forms contain these different isotopes of carbon.Carbon-12 and carbon-13 account for over 99% of all the carbon inliving things, Carbon 14. however, accounts for approximately 1 partEssentialExper trillion or 0.0000000001% of the total cacbon in living things. More importantly, carbon 14is unstable and has a half-life of approximately 5700 years. This means that, within the span of5700 years, one-half of any amount of carbon will "decay" into another atom. In other words, ifyou had 10 g of carbon-14 today, only 5 g would remain after 5700 years.But, as long as an organism is living, it keeps taking in and releasing carbon-14, so the level ofit in the organism, as small as it is, remains constant. Once an organism dies, however, it nowe know how much carbon-14 an organism had when it was alive, as well as how long it takeslonger ingests carbon-14, so the level of carbon-14 in it drops due to radioactive decay. Becausefor that amount to become half of what it was, you can determine the age of the organism byUse the information presented to create a function that will model the amount of carbon-14 ina sample as a function of its age. Create the model C(n) where C is the amount of carbon 14 inparts per quadrillion (1 part per trillion is 1000 parts per quadrillion) and n is the age of thesample in half-lives. Graph the model.One of thIn ordercalculatoThe statisa set of dcomparing these two values.AEwIt

Answer 1

We have that the exponential decay model is the following:

`y=A_0e^(kt)`

Where:

`\begin{gathered} A_0\text{ : starting value} \\ e\colon\text{ exponential function} \\ k\colon\text{ rate of decay} \\ t\colon\text{ time} \end{gathered}`

Now, to find the half life of a function describing exponential decay, we have the following equation:

`(1)/(2)A_0=A_0e^(kt)_{}`

In this case, we have that the half life is t=5700, and we want to find the rate of decay (k). Then, we have:

`\begin{gathered} (1)/(2)A_0=A_0e^(5700k) \\ \text{Dividing both sides by A}_{0\text{ }}\colon \\ \Rightarrow(1)/(2)=e^(5700k) \\ \text{Taking natural logarithm on both sides:} \\ \Rightarrow\ln ((1)/(2))=5700k \\ \Rightarrow k=(\ln((1)/(2)))/(5700) \end{gathered}`

therefore, the model C(n) is:

`C(n)=A^{}_0\cdot\exp (((\ln((1)/(2)))/(5700))\cdot n)`

and the graph looks like this for A_0 = 1000: