Question

Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atmosphere, but when an organism dies the carbon-14 decays at a half-life of 5,730 years. If an archeologist measured the amount of carbon-14 in an organism and it was 25% of the total amount of atmospheric C-14, what is the age of the organism? 1432.5 years 5,730 years 11,460 years 17,190 years

Answer 1

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

$\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} }$ (1)

Where:

$n_{o}$ - Initial amount of carbon-14.

$n(t)$ - Current amount of carbon-14.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

Then, we clear the time within the formula:

$t = -\tau \cdot \ln \frac{n(t)}{n_{o}}$ (2)

In addition, time constant can be calculated by means of half-life of carbon-14 ($t_{1/2}$), measured in years:

$\tau = \frac{t_{1/2}}{\ln 2}$

If we know that $\frac{n(t)}{n_{o}} = 0.25$ and $t_{1/2} = 5730\,yr$, then the age of the organism is:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.25$

$t \approx 11460.001\,yr$

The age of the organism is approximately 11460 years.

Answer 2

Answer:

8

Explanation: