Question

The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample is measured to be 0.574 times the Carbon-14 activity of living plants. What is the age of the sample in years? (The half-life of the Carbon-14 isotope is 5730 years.)

Answer 1

Answer:

4589.05 year

Explanation:

The relation of activity is given by $A=A_0e^{\lambda t}$ we have given $\frac{A}{A_0}=0.574$

Half life $T=5730\ years$

We know that half life $T=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{5730}=1.2094\times 10^{-4}\ per\ year$

So $0.574=e^{-1.2094\times 10^{-4}t}$

$e^{1.2094\times 10^{-4}t}=\frac{1}{0.574}=1.742$

${1.2094\times 10^{-4}t}=ln1.742=0.555$

$t=\frac{0.555}{1.2094\times 10_{-4}}=4589.05\ year$