Radio active decay reactions follow first order rate kinetics.
a) The half life and decay constant for radio active decay reactions are related by the equation:
[tex] t_{\frac{1}{2}} = [/tex][tex] \frac{ln 2}{k} [/tex]
[tex] t_{\frac{1}{2}} = \frac{0.693}{k} [/tex]
Where k is the decay constant
b) Finding out the decay constant for the decay of C-14 isotope:
[tex] Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}} [/tex]
[tex] k = \frac{0.693}{5230 years} [/tex]
[tex] k = 1.325 * 10^{-4} yr^{-1} [/tex]
c) Finding the age of the sample :
35 % of the radiocarbon is present currently.
The first order rate equation is,
[tex] [A] = [A_{0}]e^{-kt} [/tex]
[tex] \frac{[A]}{[A_{0}]} = e^{-kt} [/tex]
[tex] \frac{35}{100} = e^{-(1.325 *10^{-4})t} [/tex]
[tex] ln(0.35) = -(1.325 *10^{-4})(t) [/tex]
t = 7923 years
Therefore, age of the sample is 7923 years.
