# 6.7 Exponential and logarithmic models  (Page 3/16)

 Page 3 / 16

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after $t$ years is

$A\approx {A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}$

where

• $A$ is the amount of carbon-14 remaining
• ${A}_{0}$ is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

To find the age of an object, we solve this equation for $t:$

$t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}$

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let $r$ be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation $A\approx {A}_{0}{e}^{-0.000121t}$ we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is $r=\frac{A}{{A}_{0}}\approx {e}^{-0.000121t}.$ We solve this equation for $t,$ to get

$t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}$

Given the percentage of carbon-14 in an object, determine its age.

1. Express the given percentage of carbon-14 as an equivalent decimal, $\text{\hspace{0.17em}}k.$
2. Substitute for k in the equation $\text{\hspace{0.17em}}t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}\text{\hspace{0.17em}}$ and solve for the age, $\text{\hspace{0.17em}}t.$

## Finding the age of a bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute $\text{\hspace{0.17em}}20%=0.20\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ in the equation and solve for $\text{\hspace{0.17em}}t:$

The bone fragment is about 13,301 years old.

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

less than 230 years, 229.3157 to be exact

## Calculating doubling time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time    .

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