Answer:
In percentage, the sample of C-4 remains = 0.7015 %
Explanation:
The Half life Carbon 14 = 5730 year
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{5730}\ hour^{-1}[/tex]
The rate constant, k = 0.000120968 year⁻¹
Time = 41000 years
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
So,
[tex]\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}[/tex]
[tex]\frac {[A_t]}{[A_0]}=0.007015[/tex]
In percentage, the sample of C-4 remains = 0.7015 %
