Question

A sample of wood from the remains of a ship was found to contain 40.0% of C-14 as of ordinary wood found now. If the half-life period of C 14 is 5770 years, find the number of years elapsed. (Round your answer to the nearest whole number)

Answer 1

Answer:

7628 y

Explanation:

Carbon-14 is radioactive and it follows the first-order kinetics for a radioactive decay. The first-order kinetics may be described by the following integrated rate law:

$ln(\frac{[A]_t}{[A]_o})=-kt$

Here:

$[A]_t$ is the mass, moles, molarity or percentage of the material left at some time of interest t;

$[A]_o$ is the mass, moles, molarity or percentage of the material initially, we know that initially we expect to have 100 % of carbon-14 before it starts to decay;

$k = \frac{ln(2)}{T_{\frac{1}{2}}}$ is the rate constant;

$t$ is time.

The equation becomes:

$ln(\frac{[A]_t}{[A]_o})=-\frac{ln(2)}{T_{\frac{1}{2}}}t$

Given:

$\frac{[A]_t}{[A]_o} = \frac{40.0 %}{100.0 %}$

$T_{\frac{1}{2}} = 5770 y$

Solve for time:

$t = -\frac{ln(\frac{[A]_t}{[A]_o})\cdot T_{\frac{1}{2}}}{ln(2)}$

In this case:

$t = -\frac{ln(\frac{40.0\%}{100.0\%})\cdot 5770 y}{ln(2)}$

$t = 7628 y$