Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of \({ }^{14} \mathrm{C}\) by neutron capture in the upper atmosphere.
\(
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
\)
\({ }^{14} \mathrm{C}\) is absorbed by living organisms during photosynthesis. The \({ }^{14} \mathrm{C}\) content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of \({ }^{14} \mathrm{C}\) in the dead being, falls to the decay which \({ }^{14} \mathrm{C}\) undergoes.
\(
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
\)
The half life period of \({ }^{14} \mathrm{C}\) is 5770 years. The decay constant \((\lambda)\) can be calculated by using the following formula \(\lambda=\frac{0.693}{t_{1 / 2}}\).
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 \({ }^{14} \mathrm{C}\) to \({ }^{12} \mathrm{C}\) in living matter is \(1: 10^{12}\).
Question:
A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(\mathrm{C}^{14}\) 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

1. A The age of the fossil will increase at the place where explosion has taken and
   \(
   T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}
   \)
2. B The age of the fossil will decrease at the place where explosion has taken place and \(T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}\)
3. C The age of fossil will be determined to be same
4. D \(\frac{T_1}{T_2}=\frac{C_1}{C_2}\)