The bones of a prehistoric man found in the desert of new Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

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The bones of a prehistoric man found in the desert of new Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

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Answer 1 · 2017-09-25T08:52:02

$\approx 8678.5$ $~~8678.5$ years ago

Explanation:

Quick note: The half life of C-14 is more commonly 5730 years, the value I will be using.

To find the age of an object with a radioactive element still present, we use this formula:
$t = \frac{t_{\frac{1}{2}} ln (\frac{N_{t}}{N_{0}})}{- ln 2}$ $t=(t_(1/2)ln(N_t/N_0))/-ln2$ , where $t$ $t$ is the age of the object, $t_{\frac{1}{2}}$ $t_(1/2)$ is the half life of the element, $N_{0}$ $N_0$ is the initial quantity of the element (usually 100), $N_{t}$ $N_t$ is the remaining quantity of the element after time, and $ln$ $ln$ is the natural logarithm (base of $e$ $e$ ).
http://mathcentral.uregina.ca/beyond/articles/ExpDecay/Carbon14.html

As you can see, we have every value except for $t$ $t$ . Plug our known variables into the equation, and we get
$t = \frac{5730 ln (\frac{35}{100})}{- 0.693}$ $t=(5730ln(35/100))/-0.693$
$t = \frac{\frac{4011 log}{2}}{- 0.693}$ $t=((4011log)/2)/-0.693$
$t \approx 8678.5$ $t~~8678.5$

Therefore the bones of the prehistoric man are roughly 8678.5 years old.

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The bones of a prehistoric man found in the desert of new Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

1 Answer

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