Question

The half-life for decay of radioactive ${}^{14}C$ is $5730$ years. An archaeological artifact containing wood has only $80\%$ of the ${}^{14}C$ activity as found in living trees. The age of the artifact would be:
[Given: $\log 1.25=0.0969$]

• A. 1845 years
• B. 184.5 years
• C. 19 years
• D. 190 years

Answer 1

Answer: (A)

Radioactive decay follows first order kinetics.
Let $[A{]}_0=100$
$\therefore [A]=80$
Decayconstant $(k)=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}$
$t=\frac{2.303}{k}\log \frac{[Ao]}{[A]}$
$t=\frac{2.303}{\frac{0.693}{5730}}\log \frac{100}{80}$
$t=\frac{2.303\times 5730}{0.693}\times \log 1.25$
$t=\frac{2.303\times 5730}{0.693}\times 0.0969=1845yrs$