Question

$C^{14}$ in a fossil of 'X' is presently 1/5 times of that in present atmosphere. Given that due to nuclear explosion after death of 'X' there was 25% increase in $C^{14}$ content in atmosphere. Calculate correct age of fossil (in years).
Given $T_{1/2}$ for $C^{14}$ = 6000 yrs.
Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

Answer 1

Its a 1st Order Reaction. Consider 100 is present in Atmosphere. Then 20 is present in $C^{14}$. Due to nuclear explosion , the $C^{14}$ content is 25.

$t_{1/2}=\frac{0.693}{k}$

$k = 1.155 \times {10}^{-4}\space {yrs}^{-1]

$t=\frac{2.303}{1.155\times {10}^{-4}}log\frac{25}{20}$

$t=1932yrs$

Sum of Digits till the single digit answer is 6.