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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 A 0 e ( ln ( 0.5 ) 5730 ) 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:

         A = A 0 e k t The continuous growth formula .   0.5 A 0 = A 0 e k 5730 Substitute the half-life for  t  and  0.5 A 0  for  f ( t ) .        0.5 = e 5730 k Divide by  A 0 . ln ( 0.5 ) = 5730 k Take the natural log of both sides .           k = ln ( 0.5 ) 5730 Divide by the coefficient of  k .          A = A 0 e ( ln ( 0.5 ) 5730 ) t Substitute for  r  in the continuous growth formula .

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

t = ln ( A A 0 ) 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 A 0 e 0.000121 t 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 = A A 0 e 0.000121 t . We solve this equation for t , to get

t = ln ( r ) 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, k .
  2. Substitute for k in the equation t = ln ( r ) 0.000121 and solve for the age, 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 20 % = 0.20 for k in the equation and solve for t :

t = ln ( r ) 0.000121 Use the general form of the equation . = ln ( 0.20 ) 0.000121 Substitute for  r . 13301 Round to the nearest year .

The bone fragment is about 13,301 years old.

Got questions? Get instant answers now!
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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

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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    .

Questions & Answers

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Kelvin Reply
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
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Practice Key Terms 6

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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