Respuesta :
Answer : The age will be, 14205.59 year
Explanation :
First we have to determine the [tex]^{14}C[/tex] decay constant (k).
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{5700yr}[/tex]
[tex]k=1.216\times 10^{-4}yr^{-1}[/tex]
Now we have to determine the number of [tex]^{12}C[/tex] nuclei.
Number of [tex]^{12}C[/tex] nuclei = [tex]\frac{\text{Mass of }^{12}C}{\text{Molar mass of }^{12}C}\times 6.02\times 10^{23}mole^{-1}[/tex]
Number of [tex]^{12}C[/tex] nuclei = [tex]\frac{200g}{12g/mole}\times 6.02\times 10^{23}mole^{-1}=1.003\times 10^{25}[/tex]
Now we have to determine the number of [tex]^{14}C[/tex] nuclei.
Ratio of [tex]^{14}C[/tex] to [tex]^{12}C[/tex] = [tex]1.10\times 10^{-12}[/tex]
[tex]\frac{\text{Number of }^{14}C\text{ nuclei}}{\text{Number of }^{12}C\text{ nuclei}}=1.10\times 10^{-12}[/tex]
[tex]\frac{\text{Number of }^{14}C\text{ nuclei}}{1.003\times 10^{25}}=1.10\times 10^{-12}[/tex]
[tex]\text{Number of }^{14}C\text{ nuclei}=1.1003\times 10^{13}[/tex]
Now we have to determine the initial activity.
[tex]\text{Initial activity}A_o=k\times \text{Number of }^{14}C\text{ nuclei}[/tex]
[tex]\text{Initial activity}A_o=(1.216\times 10^{-4})\times (1.1003\times 10^{13})=1.338\times 10^9\text{ decay per year}=42.1948\text{ decay per second}[/tex]
conversion used : [tex]1year=3.171\times 10^{7}second[/tex]
Now we have to determine the current activity.
[tex]\text{Current activity}A=7.5\text{ decay per second}[/tex]
Now we have to determine its age.
[tex]A=A_o\times e^{(-kt)}[/tex]
[tex]7.5=(42.1948)\times e^{[(-1.216\times 10^{-4})t]}[/tex]
[tex]t=14205.59year[/tex]
Therefore, the age will be, 14205.59 year
