A piece of ancient wood shows an activity of 3.9 disintegrations per second per gram of $C^{14}$. The age of wood, if half life of $C^{14}$ is 5568 years, and activity of fresh $C^{14}$ is 15.6 disintegrations per second per gram, is (take $\log _e 2$ = 0.693).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 11,136 years
To determine the age of Ancient Wood,
$A=A_0e^{-λt}$ [Radioactive decay]
where,
A = Current Activity = 3.9
$A_0$ = Initial Activity = 15.6
The Decay Constant (λ) is -
$λ=\frac{\log 2}{T_{1/2}}=\frac{0.693}{5568}$
$≈0.001244$ per year
$∴\frac{A}{A_0}=e^{-λt}$
$\log(\frac{A}{A_0})=λt$
$t=\frac{1}{λ}\log(\frac{A}{A_0})=\frac{1}{0.001244}×\log(\frac{15.6}{3.9})$
$=11,136$ years