Answer:
17047.54 years or 17048 y
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of the carbon-14 =5670 years
t= time elapsed
Ao= activity of living C-14
A= activity if the sample under study
But A= 1/8 Ao
Hence;
0.693/5670= 2.303/t log(Ao/1/8Ao)
1.22×10^-4 = 2.303/t log 8
1.22×10^-4 = 2.0798/t
t= 2.0798/1.22×10^-4
t= 17047.54 years
