Question

A sample of preserved wood was found to have the specific activity of 4.11 disintegration $mi{n}^{-1}{g}^{-1}$, while a fresh sample was found to have a specific activity of 15.3 disintegrations $mi{n}^{-1}{g}^{-1}$. Age of the preserved wood is :
$(Given,t_{1/2}{}^{14}C=5570year)$

• A. $[1.056\times {10}^{4}year]$
• B. $[6.655\times {10}^{4}year]$
• C. $[2.3540\times {10}^{4}year]$
• D. None of these

Answer 1

The correct option is A $[1.056\times {10}^{4}year]$

Age of preserved wood $\left(t\right)$
$=\frac{2.303\times t_{0.5}}{0.693}\log \frac{Activityof{}^{14}Cinfreshsampleofwood}{Activityof{}^{14}Cinthesampleofpreservedwood}$
$=\frac{2.303\times 5570}{0.693}\times \log \frac{15.3}{4.11}=18510.40\times \log 3.72=1.056\times {10}^{4}years$