Carbon-14 has a half-life of 5730 years. In a plant fossil, you find that the 14C has decayed to 1/4.00 of the original amount. How long ago was this plant alive?
We use the radioactive decay formula to solve this problem. It is expressed as:
An = Aoe^-kt
where An is the amount left in time t, Ao is the initial amount, k is the constant.
We first find the value of k from the half-life data. We do as follows:
An = Aoe^-kt
0.5 = e^-k(5730)
k = 1.21x10^-4
An = Aoe^-kt
1/4 = e^-1.21x10^-4t
t = 11456.98 years
Answer:
The plant was alive for 1155 years
Explanation:
Any radioactive decay process follows the exponential law which is mathematically expressed as:
\(N(t)= N(0)e^{-kt}------(1)\)
where N(0) = initial amount of the radioactive material
N(t) = amount at time t
k = decay constant
The decay constant is in turn related to the half-life t1/2
\(k = \frac{0.693}{t1/2}----(2)\)
It is given that t1/2 of 14C = 5730yrs. Therefore:
\(k = \frac{0.693}{5730}=1.21*10^{-4}yr-1\)
In addition it is mentioned that:
N(t) = 1/4*N(0)
substituting this in equation (1) gives:
\(1/4*N(0)= N(0)e^{-1.21*10^{-4}t}\)
t = 1155 years