Problema Solution
A wooden artifact from an ancient tomb contains 60 percent of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.)
Answer provided by our tutors
Let A(0) denote the amount of C-14 present at t = 0. Let t2 denote the half-life. Then we
have A(t2) = (1/2)A(0)
Using the formula A(t) = A(0) e^(kt) we have
(1/2)A(0) = A(t2) = A(0) e^(kt2) divide by A(0)
1/2 = e^(k∗t2)
take the natural log of both sides
− ln 2 = t2 ∗ k
k = (−ln 2)/t2
Thus, since t2 = 5730 for C-14 this radioactive element satisfies the law A(t) = A(0) e^((-t ln2)/5730)
Now we need to find t is A(t) = 0.60 A(0) that is
0.60A(0) = A(0) e^((-t ln2)/5800)
e^(-t ln2)/5730) = 0.60
(-t ln2)/5730 = ln 0.60
t = - (5730 ln 0.60)/ ln2
t = 4222.82 years
the artifact was made 4222.82 years ago.