Questions: Off the coast of Florida has been found fossil shark teeth that have belonged to tens of thousands of years extinct ancestor of the white shark. How old is a shark's tooth, when the concentration of the radioactive isotope carbon-14 has decreased to 20.7 percent of the original concentration of the radioactive isotope carbon-14 from the living organism? The half-life of the radioactive isotope carbon14 is about 5730 years. Answer tolerancy limits have been set ± 1 a.

Solution Steps

We need to determine the age of a shark's tooth based on the decay of carbon-14. The concentration of carbon-14 has decreased to 20.7% of its original value. The half-life of carbon-14 is 5730 years. We will use the formula for exponential decay to find the age.

The formula for exponential decay is given by:

\[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

where:

• \(N(t)\) is the remaining quantity of the substance after time \(t\),
• \(N_0\) is the initial quantity of the substance,
• \(T_{1/2}\) is the half-life of the substance,
• \(t\) is the time elapsed.

Given that \(N(t) = 0.207 \times N_0\) and \(T_{1/2} = 5730\) years, we substitute these values into the decay formula:

\[ 0.207 = \left(\frac{1}{2}\right)^{\frac{t}{5730}} \]

To solve for \(t\), take the natural logarithm of both sides:

\[ \ln(0.207) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right) \]

Using the property of logarithms, this becomes:

\[ \ln(0.207) = \frac{t}{5730} \cdot \ln\left(\frac{1}{2}\right) \]

Solving for \(t\), we have:

\[ t = \frac{\ln(0.207)}{\ln\left(\frac{1}{2}\right)} \times 5730 \]

Calculate the value of \(t\):

\[ t = \frac{\ln(0.207)}{\ln(0.5)} \times 5730 \approx \frac{-1.574}{-0.6931} \times 5730 \approx 13,380 \text{ years} \]

Final Answer