Question

The activity of ${}^{14}C$ in a living animal has an equilibrium value of $15.3dmi{n}^{-1}{g}^{-1}$. The skeleton of a prehistoric animal was found to have ${}^{14}C$ activity as $2.42dmi{n}^{-1}{g}^{-1}$. $t_{1/2}$ of ${}^{14}C=5720year$. The animal lived for:

• A. 15224.24 year
• B. 54624.24 year
• C. 55325.14 year
• D. None of these

Answer 1

The correct option is A 15224.24 year

$t=\frac{2.303\times t_{1/2}}{0.693}\times log\frac{N_0}{N}=\frac{2.303\times 5720}{0.693}\times log\frac{15.3}{2.42}=15224.24year$
Thus, the animal lived for 15224.24 years.