Answer:
The age of the wood is [tex] t = 17527.5 \ years [/tex]
Step-by-step explanation:
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]
The appropriate age of burned wood will be "17527.5 years".
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
