Charcoal used to make prehistoric drawings on cave walls in France was analyzed. The results were 4 mg carbon-14 (parent isotope) and 60 mg nitrogen (daughter isotope). The half-life of carbon-14 is 5,730 years. How old are the cave drawings?

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Charcoal used to make prehistoric drawings on cave walls in France was analyzed. The results were 4 mg carbon-14 (parent isotope) and 60 mg nitrogen (daughter isotope). The half-life of carbon-14 is 5,730 years. How old are the cave drawings?

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Answer 1 · 2017-09-15T05:51:46

The cave drawings are 22920 years old.

Explanation:

$A = A_{o} \cdot {(\frac{1}{2})}^{\frac{t}{h}}$ $A = A_o * (1/2)^(t/h)$ , where $A$ $A$ is the final amount of radioactive anything, $A_{o}$ $A_o$ is the initial amount, $t$ $t$ is the elapsed time, and $h$ $h$ is the half-life of the radioactive material.

At the time of the drawing, there were 0 g of nitrogen 14 and 64 g of Carbon 14.

There are now 4 g.

Plugging everything in, we have:

$4 = 64 \cdot {(\frac{1}{2})}^{\frac{t}{5730}} \to \frac{1}{16} = {(\frac{1}{2})}^{\frac{t}{5730}} \to$ $4=64*(1/2)^(t/5730) -> 1/16 = (1/2)^(t/5730) ->$
${log}_{\frac{1}{2}} (\frac{1}{16}) = {log}_{\frac{1}{2}} {(\frac{1}{2})}^{\frac{t}{5730}} \to 4 = \frac{t}{5730} \to 22920 = t$ $log_(1/2)(1/16)=log_(1/2) (1/2)^(t/5730) -> 4=t/5730 -> 22920 = t$ .

The cave drawings are 22920 years old.

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Charcoal used to make prehistoric drawings on cave walls in France was analyzed. The results were 4 mg carbon-14 (parent isotope) and 60 mg nitrogen (daughter isotope). The half-life of carbon-14 is 5,730 years. How old are the cave drawings?

1 Answer

Explanation:

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