Question

9. Chemical analysis shows that a bone fragment contains 10 g of C. Its activity is 50

counts/min. How old is the bone fragment?

Answer 1

$N=\frac{mN_A}M=\frac{10\cdot 6.02\cdot 10^{23}}{12}=5\cdot 10^{23},$

$N_0=N\eta =5\cdot 10^{23} \cdot1.3\cdot10^{-12} = 6.5\cdot10^{11},$

$A_0=N_0\lambda = \frac{6.5\cdot10^{11 }\cdot 0.69 }{5730 \cdot 365 \cdot 24 \cdot 3600 }= 2.53~ s^{-1},$

$A=A_0e^{-\lambda t},\implies$

$\lambda t=\ln \frac{A_0}A=\ln \frac{2.53\cdot 60}{50}=1.11,$

$t=\frac {1.11}{\lambda}=\frac{1.11\cdot 5370}{0.69}=9180~\text{years}.$