Question

$^{14} C$ is absorbed by living organisms during photosynthesis. The ${ }^{14} C$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14}{C}$ in the dead being, falls due to the decay which ${ }^{14} C$ undergoes
${ }_{6}^{14} C \rightarrow{ }_{7}^{14} N+\beta^{-}$
The half-life period of ${ }^{14} C$ is $5770$ years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{\frac{1}{2}}}$
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than $30,000$ years. The proportion of ${ }^{14} C$ to $^{12} C$ in living matter is $1: 10^{12}$.
A nuclear explosion has taken place leading to increase in the concentration of $C^{14}$ in nearby areas. $C^{14}$ concentration is $C_{1}$ in nearby areas and $C_{2}$ in areas far away. If the age of the fossil is determined to be $T_{1}$ and $T_{2}$ at the places respectively then

• A. The age of the fossil will increase at the place where explosion has taken place and $T_{1}-T_{2}=\frac{1}{\lambda}$ $\ln \frac{C_{1}}{C_{2}}$.
• B. The age of the fossil will decrease at the place where explosion has taken place and $T_{1}-T_{2}=\frac{1}{\lambda}$ $\ln \frac{C_{1}}{C_{2}}$
• C. The age of fossil will be determined to be the same.
• D. $\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}$

Answer 1

Answer: (A)

Radioactive decay of all elements follows first-order kinetics.
So, in accordance with first-order kinetics, the decay constant is
$\lambda=\frac{1}{T_{1}-T_{2}} \ln \frac{C_{1}}{C_{2}}$
$ \Rightarrow T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}$
The rate of decay is proportional to the concentration, it would be faster near the explosion as concentration of $C$ is more.
Hence, the age of fossil will increase.