Question

The half life of 14C is 5730 years. 11,460 years after an organism dies, what percent of the initial 14C will still be present in the organism's bones?

Answer 1

we are going to write the radioactive decay equation as:
\[N(t)=N_{0}*e ^{-k*t}\]
Where
\[k=\frac{ \ln(2) }{ \tau }\]
Here is:
τ: the half life
N(t): the number of atom cores.
N0: the number of atom cores to the t=0
t: the time

We start by calculating k:
\[k=\frac{ \ln(2) }{ 5730 }=1.210*10^{-4}\]
Now as we do not know N0 from the question we can remove it by using percentage called x to the time t so we say that:
\[N_{i}=x*n(y)\]
Doing so we can rewrite the equation:
\[x*N _{0}=N _{0}*e ^{-k*t}\]
We see that N0 goes out with each other:
\[x=e ^{k*t}\]
We insert the information about t and k and calculation x in the following equation:
\[x=e ^{-1.210*10^{-4}*11460}=0.25\]
In other words there are 25% left of the 14C left.

Answer 2

Correction in equation 4:
\[N _{i}=x*N_{0}\]

Answer 3

Sorry I forgot all the units.. but this math thingy is not that unit friendly.