Question 792092
Wood deposits recovered from an archaeological site contain 36% of the C-14 they originally contained.
 How long ago did the tree from which the wood was obtained die?
 (Round your answer to the nearest whole number.)
:
The radio-active decay formula: A = Ao*2^(-t/h), Where:
A = resulting amt after t time
Ao = initial amt (t-0)
t = time in yrs
h = half-life of substance (carbon 14 half-life is 5730 yrs)
:
Let initial amt = 1, then resulting amt = .36
:
{{{2^(-t/5730)}}} = .36
Using nat logs
{{{-t/5730}}}*ln(2) = ln(.36)
{{{-t/5730}}} = {{{ln(.36)/ln(2)}}}
{{{-t/5730}}} = -1.47393
t = -5730 * -1.47393
t = 8,446 yrs ago