Archaeologists used pieces of burned wood, or charcoal, found at the site to date prehistoric paintings and drawings on walls and ceilings of a cave in Lascaux, France. Use the information in Example 3 of Section 3.1 to determine the appropriate age of a piece of burned wood, if it was found that 88% of the C-14 found in living trees of the same type had decayed. (Round your answer to the nearest hundred years.) years

From the question we are told that 88% of of the C-14 found in living trees of the same type had decayed , this means that the proportion of C-14 that is remaining is mathematically evaluated as

[tex] 1 - 0.88 = 0.12[/tex]

Hence

[tex] N = 0.12N_o[/tex]

Here N represents the remaining C-14 while [tex]N_o[/tex] is the original amount of C-14

Generally the half life of C-14 is [tex]h = 5730 \ years[/tex]

Generally from the formula of radioactive decay

[tex]N = N_o * 2^{-\frac{t}{h} }[/tex]

=> [tex]0.12N_o = N_o * 2^{-\frac{t}{5730} }[/tex]

=> [tex]0.12 = 2^{-\frac{t}{5730} }[/tex]

taking the natural log of both sides

=> [tex] ln [0.12] = ln[ 2^{-\frac{t}{5730} }][/tex]

=> [tex] ln [0.12] = -\frac{t}{5730}ln(2) [/tex]

=> [tex] t = 17527.5 \ years [/tex]

Cricetus Cricetus · Answer 2 · 2022-03-16T14:12:22+01:00

The appropriate age of burned wood will be "17527.5 years".

Radioactive decay

According to the question,

The proportion of C-14 = 1 - 0.88

= 0.12

then,

N = 0.12 N₀

Now,

The half-life of C-14 will be:

h = 5730 years

By using Radioactive decay formula, we get

→ N = N₀ × [tex](2)^{-\frac{t}{h} }[/tex]

By substituting the values,

0.12N₀ = N₀ × [tex](2)^{-\frac{t}{5730} }[/tex]

0.12 = [tex]2^{-\frac{t}{5730} }[/tex]

By taking "log" both sides,

ln[0.12] = ln[[tex]2^{-\frac{t}{5730} }[/tex]]

ln[0.12] = -[tex]\frac{t}{5730 \ ln(2)}[/tex]

hence,

The age will be:

t = 17527.5 years

Thus the above response is correct.

Find out more information about radioactive decay here:

https://brainly.com/question/1236735