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Numerade Educator



Problem 11 Medium Difficulty

Scientists can determine the age of ancient objects by the method of radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon. $ ^{14}C $ with a halt-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates $ ^{14}C $ through food chains. When a plant or animal dies, it slops replacing its carbon and the amount of $ ^{14}C $ begins to decrease through radioactive decay. Therefore the level of radioactivity must also decay exponentially.
A discovery revealed a parchment fragment that had about $ 74\% $ as much $ ^{14}C $ radioactivity as does plant material on the earth today. Estimate the age of the parchment.


2489 years or 2500 years.


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Diego V.

April 14, 2019

Fix your handwriting please.

Video Transcript

So we have a carbon 14 dating problem and we're going to use our radioactive decay. A model exponential model m equals m knotty to the K t. Ah, Initially, we're going to find to use the half life information to find the value of K. And then once we know that, will be able to answer the question. So for the half life, we're going to use em, not as the initial amount. And then we can use half of them, not as the final amount. Put that into the equation and we put the half life in for the time. Now we can solve for K. So those are the numbers plugged into our model. So let's divide both sides of the equation by am not, and we get 1/2 which I'll just write us. 0.5 equals e to the 5730 k. Now we take the natural log of both sides and we divide both sides by 5730 on our value of K is natural. Log off 0.5 over 5730. So that goes into the model. And now our model is M equals and not I'm not e to the natural log of 0.5 over 5730 times t. So we're going to use that model as we move forward. So now we know that the sample of parchment had 74% of the normal amount of carbon 14 and a living plant, and so 74% would be 740.74 times m not is the amount that's in there, so we can substitute that in for em and we'll solve for the time. 0.74 AM not equals m not times e to the natural log of 0.5 over 5730 times teeth. We divide both sides by I am not, and we have 0.74 equals e to the natural log of 0.5 over 5730 times t. Now we take the natural log on both sides. So in the natural log of 0.74 equals the natural log of 0.5 over 5730 times T On our goal is to isolate T, so let's multiply both sides of the equation. By the reciprocal of this fraction and we end up with T equals 5730 times a natural log of 57300.74 over the natural log of 0.5. We put that into a calculator and we end up with approximately 2489. So that tells us that the parchment was, according to this method, 2489 years old.