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

7628 y

Step-by-step 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(([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_{(1)/(2)}}` is the rate constant;

`t` is time.

The equation becomes:

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

Given:

`T_{(1)/(2)} = 5770 y`

Solve for time:

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

In this case:

`t = -(ln((40.0\%)/(100.0\%))\cdot 5770 y)/(ln(2))`

`t = 7628 y`