Question

question 16 (4 points)
a sample from an archeological artifact contained 7.97 x 10^(-11) g of c - 14. a similar sample from a present day object contained 2.55 x 10^(-9) g of c - 14.
this represents 1 half lives.
how old is the artifact if the half - life of c - 14 is 5,730 years? 2
a. 0 b. 1. c. 2 d. 3 e. 4 f. 5 g. 6
h. 7 i. 5,730 y j. 11,460 y k. 17,190 y l. 22,920 y
m. 28,650 y n. 34,380 y o. 40,110 y p. not old at all
q. same age as your professor

Explanation:

Step1: Calculate the ratio of C - 14 amounts

Let $N$ be the amount of C - 14 in the archaeological sample and $N_0$ be the amount in the present - day sample. $N = 7.97\times10^{-11}\text{ g}$ and $N_0=2.55\times10^{-9}\text{ g}$. The ratio $\frac{N}{N_0}=\frac{7.97\times 10^{-11}}{2.55\times 10^{-9}}=\frac{7.97}{2.55\times10^{2}}\approx0.031$.

Step2: Use the half - life formula $N = N_0(\frac{1}{2})^n$

We want to find the number of half - lives $n$. Rearranging the formula $n=\log_{\frac{1}{2}}\frac{N}{N_0}$. Since $\frac{N}{N_0}\approx0.031$, and $(\frac{1}{2})^5=\frac{1}{32}\approx0.03125$, so $n = 5$.

Step3: Calculate the age of the artifact

The half - life of C - 14 is $T_{1/2}=5730$ years. The age $t$ of the artifact is given by $t=nT_{1/2}$. Substituting $n = 5$ and $T_{1/2}=5730$ years, we get $t=5\times5730 = 28650$ years.

Answer:

1. F. 5
2. M. 28,650 y