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).

11,136 years

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